Bihar Board Class 10 Science Solutions Chapter 12 Electricity

Bihar Board Class 10 Science Chapter 12 Electricity InText Questions and Answers

In-text Questions (Page 200)

Question 1.
What does an electric circuit mean?
Answer:
A continuous conducting path consisting of wires and other resistances (like electric bulb, etc.) and a switch, between the two terminals of a cell or battery, along which an electric current flows, is called a circuit.

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Question 2.
Define the unit of current.
Answer:
The SI unit of electric current is ampere (which is denoted by the letter A). When 1 coulomb of charge flows through any cross-section of a conductor in 1 second, the current flowing through it is said to be 1 ampere.

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
The charge on an electron, e = 1.6 × 10^-19 c. No. of electrons constituting 1 coulomb is given by

In-text Questions (Page 202)

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
An electrical cell (or a battery).

Question 2.
What is mean by saying that the potential difference between two points is 1 V?
Answer:
The potential difference between two points is said to be 1 volt (1 V) if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
W = QV = 1C × 6V = 6J

In-text Questions (Page 209)

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The electrical resistance of a conductor (or a wire) depends on the following factors:

• length of the conductor,
• area of cross-section of the conductor (or thickness of the conductor),
• nature of the material of the conductor, and
• temperature of the conductor.

Question 2.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
The current will flow more easily through a thick wire than through a thin wire of the same material when connected to the same source (like a battery). This is due to the fact that the resistance of a wire is inversely proportional to the square of its diameter. A thick wire has greater diameter and hence lesser resistance making the current to flow through it more easily. On the other hand, a thin wire has smaller diameter and hence greater resistance to the flow of current through it.

Question 3.
Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
The current through the component will also decrease to half of its former value.

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The coils (or heating elements) of toasters and electric irons are made of an alloy rather than a pure metal because:

• the resistivity of an alloy is much higher than that of a pure metal, and
• an alloy does not undergo oxidation (or burn) easily even at high temperature, when it is red hot.

Question 5.
Use the data in Table on page 207 of this book to answer the following:
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer:
(a) The electrical resistivity of iron is 10.0 × 10^-8 Ω m whereas that of mercury is 94.0 × 10^-8 Ω m. Since the resistivity of iron is less than that of mercury, therefore, iron is a better conductor than mercury.
(b) Silver metal has the lowest electrical resistivity (1.60 × 10^-8 Ω m), therefore, silver metal is the best conductor of electricity.

In-text Questions (Page 213)

Question 1.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω, resistor, and a plug key, all connected in series.
Answer:
A schematic diagram of a circuit consisting of a battery of three 2V cells, a 5Ω resistor, an 8Ω resistor and a 12Ω resistor and a plug key, K; all connected in series, is shown in figure given below.

Question 2.
Redraw the circuit of Question 1 putting an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the reading in the ammeter and the voltmeter?
Answer:
The circuit diagram of question 1 with an ammeter and voltmeter V across the 12 Ω resistor is shown in figure given below.

Since three resistance are in series.
Total resistance of circuit, R = 5Ω + 8Ω + 12Ω = 25Ω

Potential difference across 12 Ω resistor i.e. V = current × resistance = 0.24 × 12 = 2.88V
Reading of ammeter is 0.24 A and reading of voltmeter is 2.88 V.

In-text Questions (Page 216)

Question 1.
Judge the equivalent resistance when the following are connected in parallel:
(a) 1 Ω and 10⁶ Ω
(b) 1 Ω, 10³ Ω and 10⁶ Ω
Answer:
(a) The equivalent resistance of two resistances 1 Ω and 10⁶ Ω connected in parallel will be less than 1 Ω. This is because when a number of resistances are connected in parallel, then their equivalent resistance is less than the smallest individual resistance (which is 1 Ω in this case).
(b) The equivalent resistance of three resistances 1 Ω,10³ Ω and 10⁶ Ω connected in parallel will be less than 1 Ω.

Question 2.
An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
The combined resistance R of three resistors (or electrical devices) R₁, R₂ and R₃, connected in parallel is given by the formula:
\(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)
Here, Resistance of electric lamp, R₁ = 100 Ω
Resistance of toaster, R₂ = 50 Ω
and, Resistance of water filter, R₃ = 500 Ω
Putting these values of R₁, R₂ and R₃ in the above formula, we get:

Thus, the resistance of electric iron which takes the same current as drawn by all the three appliances will be 31.25 Ω.
Now, Potential difference, V = 220 V
Current, I = ? (To be calculated)
and, Resistance, R = 31.25 Ω (Calculated above)

Thus, the current passing through the electric iron is 7.04 amperes.

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
The advantages of connecting electrical devices in parallel are the following:
(i) In parallel circuits, if one electrical appliance stops working due to some defect, even then all other appliances keep working normally. On the other hand, in a series circuit, if one electrical appliance stops working due to some defect, then all the other appliances also stop working (because the whole circuit is broken).

(ii) In parallel circuits, each electrical appliance has its own switch due to which it can be turned on or off independently, without affecting the other appliances. In a series circuit, all the electrical appliances have only one switch due to which they cannot be turned on or off independently.

(iii) In parallel circuits, each electrical appliance gets the same voltage as that of the battery due to which all the appliances work properly. In a series circuit, the appliances do not get the same voltage as that of the battery because the voltage is shared by all the appliances.

(iv) In the parallel connection of electrical appliances, the overall resistance of the circuit is reduced due to which the current from the battery is high and hence each electrical appliance can draw the required amount of current. In the series connection of electrical appliances, the overall resistance of the circuit cannot draw sufficient current for their proper working.

Question 4.
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, and (b) 1 Ω ?
(a) In order to obtain a total resistance 4 Ω from three resistors of 2 Ω, 3 Ω and 6 Ω:
(i) First connect the two resistors of 3 Ω and 6 Ω in parallel to get a total resistance of 2 Ω. This is because in parallel combination:

(ii) Then the parallel combination of 3 Ω and 6 Ω resistors (which is equivalent to 2 Ω resistance) is connected in series with the remaining 2 Ω resistor to get a total resistance of 4 Ω. This is because in series combination:
R = R₁ + R₂
R = 2 + 2
R = 4Ω
The arrangement of three resistors of 2 Ω, 3 Ω and 6 Ω which gives a total resistance of 4 Ω can now be represented as follows:

Thus, we can obtain a total resistance of 4 Ω by connecting a parallel combination of 3 Ω and 6 Ω resistors in series with 2 Ω resistor.

(b) In order to obtain a total resistance of 1 Ω from the three resistors of 2 Ω, 3 Ω and 6 Ω, all the three resistors should be connected in parallel. This is because in parallel combination:

Question 5.
What is (a) the highest, and (b) the lowest, total resistance that can be secured by the combination of four coils of resistance 4 Ω, 8 Ω,12 Ω and 24 Ω ?
Answer:
(a) The highest resistance can be secured (or obtained) by connecting all the four coils in series. In this case:
R = R₁ + R₂ + R₃ + R₄
R = 4 + 8 + 12 + 24
R = 48 Ω
Thus, the highest resistance which can be secured is 48 ohms.

(b) The lowest resistance can be secured by connecting all the four coils in parallel. In this case:

Thus, the lowest resistance which can be secured is 2 ohms.

In-text Questions (Page 218)

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The heating element of an electric heater is made of an alloy (such as nichrome) which has high resistance whereas the cord is made of copper metal which has very, very low resistance. The heating element glows because it becomes red-hot due to the large amount of heat produced on passing current (because of its high resistance). On the other hand, the connecting cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing the same current (due to its extremely low resistance).

Question 2.
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Answer:
First of all we will calculate the current (I) by using the given values of charge (Q) and time (t). We know that:
Current, I = \(\frac{\mathrm{Q}}{t}=\frac{9600 \mathrm{C}}{1 \mathrm{~h}}=\frac{96000 \mathrm{C}}{60 \times 60 \mathrm{~s}}\)
or, I = 26.67 A …………(1)
We will now calculate the resistance using Ohm’s Law
Resistance, R = \(\frac{V}{I}=\frac{50 \mathrm{~V}}{26.67 \mathrm{~A}}\)
or R = 1.87 Ω …….(2)
and, Time, t = 1h = 60 × 60s = 3600 s ……(3)
Heat generated, H = I² × R × t = (26.67)² × 1.87 × 3600 J = 4788397 J = 4.79 × 10⁶ J or 4.8 × 10⁶ J
Thus, the heat generated is 4.8 × 10⁶ joules.

Question 3.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
Here, Current, I = 5 A
Resistance, R = 20 Ω
and, Time, t = 30 s
Now, Heat produced, H = I² × R × t ⇒ H = (5)² × 20 × 30
H = 25 × 20 × 30 ⇒ H = 15000 J
Thus, the heat developed is 15000 joules.

In-text Questions (Page 220)

Question 1.
What determines the rate at which energy is delivered by a current?
Answer:
Electric power of the appliance.

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Here, Voltage, V = 220 V and, Current, I = 5 A
Now, Power, P = V × I
So, P = 220 × 5 W
∴ P = 1100 W
Thus, the power of the motor is 1100 watts.
Now, Power, P = 1100 W (Calculated above)
Time, t = 2h = 2 × 60 × 60 s = 7200 s
So, Energy, E = P × t
E = 1100 × 7200 J
E = 7.92 × 10⁶ J
Thus, the energy consumed is 7.92 × 10⁶ joules.

Bihar Board Class 10 Science Chapter 12 Electricity Textbook Questions and Answers

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is:
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer:
After cutting the wire, resistance of each part = \(\frac{\mathrm{R}}{5}\)
Equivalent resistance when these parts connected in parallel.

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer:
Electric power
P = VI = (IR) I = I²R = \(\left(\frac{\mathrm{V}^{2}}{\mathrm{R}^{2}}\right) \mathrm{R}=\frac{\mathrm{V}^{2}}{\mathrm{R}}\)
So, IR² does not represent electric power in circuit.

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:

Here, operating voltage = 110 V
P = \(\frac{(110)^{2}}{484}\) = 25 W
(d) is correct answer.

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be
(a) 1: 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
Since both wires are made up of same material and all other parameters are identical,
so, resistance of both the wires is same. Let it be R. When connected in series, their equivalent resistance is given by
R_eq = R₁ + R₂ = R + R = 2R
When connected in parallel, their equivalent resistance is given by

(c) is the correct answer.

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is always connected in parallel in the circuit across the points between which the potential difference is to be determined.

Question 6.
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^-8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:
Given, diameter of wire, D = 0.5 mm = 0.5 × 10^-3 m
Resistivity of copper, ρ = 1.6 × 10^-8 Ω m
Resistance = 10 Ω


As R ∝ \(\frac{1}{r^{2}} \propto \frac{1}{D^{2}}\), when D doubled, resistance becomes \(\frac {1}{4}\) times.

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:

Taking two points A and B
A(0.5 A, 1.6 V) and B (1.0 A, 3.4 V)
Slope = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}\)
Here, ΔV = 1.8 V, ΔI = 0.5 A
R = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}=\frac{1.8 \mathrm{~V}}{0.5 \mathrm{~A}}\) = 3.6 Ω

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Here V = 12 V, I = 2.5 mA = 2.5 × 10^-3 A
Resistance = \(\frac{V}{I}=\frac{12 V}{2.5 \times 10^{-3} \mathrm{~A}}\) = 4800 Ω = 4.8 kΩ

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer:
Equivalent resistance = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω
= 13.4 Ω
Current through circuit, I = \(\frac{\mathrm{V}}{\mathrm{R}_{e q}}=\frac{9 \mathrm{~V}}{13.4 \Omega}\) = 0.67 A
In series, the current flowing through all resistor is same. Thus, current flowing through 12 Ω resistor = 0.67 A.

Question 10.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Given, I = 5 A, V = 220 V
Resistance of the circuit, R = \(\frac{\mathrm{V}}{\mathrm{I}}=\frac{220 \mathrm{~V}}{5 \mathrm{~A}}\) = 44 Ω
Resistance of each resistor, r = 176 Ω
In case of parallel combination of identical resistors

Number of resistors required = 4

Question 11.
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω (ii) 4 Ω.
Answer:
1. For getting equivalent resistance R_eq = 9Ω two resistors will be connected in parallel and the third resistors will be connected in series with this parallel combination.

2. For getting equivalent resistance R_eq = 4 Ω two resistors should be connected in series. The combination of which will lead to equivalent of 12 Ω afterwards connecting this 12 Ω resistor with 6 Ω resistor in parallel, the overall equivalent resistance will be 4 Ω.

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Ans. Given, potential difference = 220 V
Resistance of each coil R_A = R_B = 24 Ω
(i) When each coil A or B is connected separately, current through each coil i.e.,

(ii) When coil A and B connected in parallel equivalent resistance in the circuit

(iii) When coil A and B are connected in series equivalent resistance

Question 14.
Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:

In a parallel circuit, potential (V) across each resistor is the same and is equal to the potential applied across the circuit.
Power dissipation through 2 Ω resistor

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:

Question 16.
Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?
Answer:
Energy used by 250 W TV set in 1 hour = (250 W) (1 h) = 250 Wh
Energy consumed by 1200 W toaster in 10 min = \(\left(\frac{1}{6} h\right)\) = 1200 W × \(\frac {1}{6}\) h = 200 Wh
Energy consumed by 1200 W toaster in 10 minutes is less than energy consumed by 250 W TV set in 1 hour.
Hence energy consumed by the TV set is more.

Question 17.
An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:
Given, I = 15A, R = 8Ω, t = 2h
Rate at which heat is developed i.e., electric power
P = I²R = (15)² × 8 = 1800 W = 1800 J/S.

Question 18.
Explain the following:
(а) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
(a) Tungsten has high melting point (3380°C). It does not melt and becomes very hot emitting light.
(b) The resistivity of an alloy is generally higher than that of pure metals and alloys do not oxidise readily at high temperatures.
(c) In series arrangement, if any one of the appliances fails or is switched off, all the other appliances stop working.
(d) Resistance of a wire is inversely proportional to its area of cross-sectional area (A)
R ∝ \(\frac{1}{\mathrm{~A}}\)
(e) Copper and aluminium posses low resistivity and hence low energy losses.

Bihar Board Class 10 Science Chapter 12 Electricity Textbook Activities

Activity 12.1 (NCERT Textbook Page 203-204)

Discussion and Conclusion:
The graph of V and I is a straight line passing through origin and slope of graph is equal to resistance

Activity 12.2 (NCERT Textbook Page 205)

Discussion and Conclusion:
The readings of ammeter are different for different components. This is due to the reason that resistances of all the components are different.

Activity 12.3 (NCERT Textbook Page 206)

Discussion and Conclusion:

• When the length of a wire is doubled, the resistance of the wire becomes twice as R ∝ l and current becomes half as I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
• When the wire is replaced by a thicker nichrome wire, the resistance of wire decreases as R ∝ \(\frac{1}{\mathrm{~A}}\) and current increases as I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
• When the nichrome wire is replaced by a copper wire of same length and cross-section, current decreases as resistivity of nichrome is greater as R ∝ p.
• Resistance of a wire depends on the resistivity of the material, length of wire and area of cross-section. R = ρ\(\frac{l}{\mathrm{~A}}\)

Activity 12.4 (NCERT Textbook Page 210)

Discussion and Conclusion:

• Here total resistance, R = 1Ω + 2Ω + 3Ω = 6Ω, Reading of ammeter is found to be around 6V/6Ω = 1A.
• When the position of the ammeter is changed, the reading remains the same.

Activity 12.5 (NCERT Textbook Page 211)

Discussion and Conclusion:

• Potential difference across the series combination is equal to potential difference across the two terminals of the battery.
• Relation between potential drop across battery and resistors V = V₁ + V₂ + V₃.

Activity 12.6 (NCERT Textbook Page 213-214)

Discussion and Conclusion:

• Potential difference across each resistor is same.
• Net current is equal to the sum of current through each resistor.

Bihar Board 12th Pali Grammar Important Questions

विभत्ति प्रकरण

कारक को पालि में विभति कहा जाता है। हिन्दी की तरह ही पालि में भी विभक्ति (कारक) का प्रयोग शब्दों के साथ किया जाता है। परन्तु पालि में कभी-कभी विशेष ढंग से इसका प्रयोग किया जाता है।

पठमा विभत्ति – किसी वाक्य के कर्ता के साथ ‘पठमा’ विभत्ति का प्रयोग होता है। जैसे-बालको, पठति, कञा लिखति।

दुतिया विभत्ति – किसी वाक्य के कर्ता के साथ कर्तृवाच्य के कर्म में ‘दुतिया’ का प्रयोग होता है। जैसे- सिस्सा पोत्थकं पठन्ति, दासी चुल्लिं लिम्पति आदि।

लगातार धि, अभि, पति, अनु, लिवना, अञत्र आदि के योग में दुतिया’ विभत्ति होती ‘ है। जैसे-

लगातार – सिस्सा दिवसं पोत्थकं पठन्ति। कोसं रुक्खो।
धि – धि अलसं पुत्त।
अभि – बालको नगरं अभिधावति।।
पति – सिस्सो पसन्नो आचरियं पति।
अनु – भगवन्तं अनुगच्छति आनन्दो।
विना – आचरियं बिना सिस्सा न पठन्ति।
अल्लत्र – तथागतं अज्ञत्र को अञ्जो ‘लोकनायको’
अन्तरा – अन्तराचं राजगह, अन्त राच नालकं अम्बंलट्टिका होति।

ततिया विभत्ति – ‘ततिया विभत्ति का प्रयोग भाववाच्य तथा कर्मवाच्य के कर्ता, करण कारक क्रिया विशेषण, साथ तुलना लक्षण, हेतु बिना, अञत्र, पृथक आदि के अर्थ में होता है। जैसे-

भाववाच्य – भूपेन अत्र भूयते।
कर्मवास्य – आचरियेन पोत्थकं दीयते।
करण कारक – सिस्सो मसिना लिखति।
क्रिया विशेषण – पञ्चकेन गजो किणाति।
साथ – आचरियेन सदिसो सिस्सो।
लक्षण – नयनेन कोणों मनुस्सो।
हेतु – याचकों इध अन्नेन वसति।
बिना – खीरेन बिना सिसु रोदति।
अञत्र – तथागतेन अञत्र को अञ्जो लोकनायको।
पृथक – पृथुजेव नगरेन भिक्खु अरचं अधिवसति।

चतुत्थी विभत्ति – सम्प्रदान उसके लिए रुचि, क्रोध, ईर्ष्या के अर्थ में चतुत्थी विभत्ति का उपयोग किया जाता है, जैसे-

सम्प्रदान – भगवा भिक्खून धम्म देसेति।
उसके लिए – वहुजन सुखाय बुद्धो धम्म देसेति।
रुचि – बालकांन अनज्झायो रूच्चति।
क्रोध – आचरियो सिस्सांनं कुप्पति।
ईर्ष्या – सिस्सा, सिस्मानं इस्सन्ति।

पञ्चमी विभत्ति – अपादान, भय, सा, ग्रहण, ऋण, पृथु, पति, बिना अत्र आदि के योग में पञ्चमी विभत्ति लगाई जाती है।

अपादान – पण्णानि रुक्खस्मा, पतन्ति।
रक्षा – बलवा फरिसो चोरस्मां जामं रक्खति।
ग्रहण – माणवको आचरियस्मा सिक्खं गणहाति।
ऋण – सो सतस्मा बद्धो।
पृथु – पृथुगेव गामस्मा इसयो ‘पब्बत स्स गुहायं अधिवसन्ति।’
पति – अजा तसत्तस्मा पति सुनीधो।
बिना – हातस्मा बिना दीपो न झायति।
अञत्र – हाथागतस्मा अज्ञत्र को अज्ञो लोकनायको ?

छटठी विभनि – सम्बन्ध, कृदन्त एवं हेतु के अर्थ में तथा अनेक जाति, गण अथवा क्रिया में से किसी एक को चुना जाय तो, छट्ठी विभत्ति होती है। जैसे-

सम्बन्ध – आमस्स जना बुद्धस्स धर्म सुणोन्ति।
कृदन्त – विनयस्य आतारो भिक्खथो आगच्छत्ति।
हेतु – याचको उदरस्स हेतु इहा वसति।
जाति – अनुस्सानं खत्तियो सेद्धो।
गुण – सीहो, पसून बल सम्पन्नो भवति।
क्रिया – दानानं धम्मदानं सेट्ठ।

सत्तमी विभत्ति – क्रिया के आधार (अधिकरण) में, निमित्त के अर्थ में, एक काम के बाद दूसरे काम के होने पर अधिक होने के अर्थ में ‘उप’ शब्द के योग में स्वामी होने के अर्थ में ‘अधि’ शब्द के योग में तथा अनेक जाति, गुण या क्रिया में से किसी एक को चुनना हो तो ‘सत्तमी’ विभक्ति का प्रयोग किया जाता है। जैसे-

क्रिया के आधार में-वातयानानि आकासे डयन्ति।
निमित्त के अर्थ में-अजनम्हि मिगं हति।
एक काम के होने पर दूसरे काम के होने पर एवं वुत्ते भिक्खुणियों मंकु अहेंसु।
अधिक होने के अर्थ में ‘उप’ शब्द के योग में- उप – भिक्खुणीस भिक्खवो संघ सन्ति।
स्वामी होने के अर्थ में ‘अधि’ शब्द के योग में- मगधेसु अजातसत्रु वेदेहिपुतो।
जातिनिर्धारण में – मनुस्सेसु खत्तियौ सहो।
गुण निर्धारण में – कण्हा गावीसु सम्पन्न खीरतमा।
क्रिया निर्धारण में – दानेसु धम्मदानं सेट्ठ।

आलपन विभत्ति – ‘आलपन’ विभत्ति आमन्त्रण करने के अर्थ में प्रयुक्त होती है। जैसे-आयाम-नन्द ! देमें, भिक्खवे, अन्ता पब्बजितेन न सेवितब्बा। कभी-कभी पठमा विभत्ति का भी प्रयोग आलपन में हो जाता है। जैसे-रे धुत्ता ! जे अम्बपालि ! आदि।

संधि

दो अक्षरों (स्वर व्यंजन या निग्गइति) के परस्पर सम्मिलन को संधि कहते हैं। इनके इस सम्मिलन से प्रायः शब्दों के आकार और ध्वनि में अन्तर आ जाता है।
जैसे – सु + आगतं = स्वागतं आदि।

संधि के तीन भेद किए गए हैं-

1. सर (स्वर) संधि
2. व्यंजन संधि, और
3. निग्गहीत सन्ध

1. स्वर सन्धि

दो स्वरों के मिलन स्वरूप जो विकार उत्पन्न होता है, उसे स्वर सन्धि कहते हैं। जैसेते + अज्ज = त्यज्जा।
स्वर संधि की व्याख्या निम्न प्रकार से की जा सकती है-

(क) स्वर के बाद यदि स्वर आवे तो कभी-कभी पूर्व स्वर लुप्त हो जाता है, जैसे-
तत्र + इमे = तत्रिमे, पुरिस + उत्तमी = पुरिसुत्तमो, पञान + इन्द्रिय = पद्धिन्द्रिय, नोहि + एतं = नोहेतं, भिक्खूनी + ओवादो = भिक्खुनोवादो, भोणी + इन्दो = भोगिन्दो, समेतु + आयस्या = समेतामस्या , अभिभू + आयतनं = अभियातेनं, ने + अलि = मथ, असन्तो + एत्थ. = असन्तेथ ऐसी + ओवुसो = एसावुसो, महा + इच्छो = महिच्छो आदि।

(ख) यदि स्वर के बाद स्वर आता है, तो कभी-कभी पर स्वर का लोप हो जाता है-
छाया + इव = छायाव, सा + एव = साव, ते + अहं = तेहं, आकासेन + इव = आकासेव, चत्तारो + इमे = चत्तारो मे, ततो + एव = ततोव, यतो + उढकं = यतोढकं, वसलो + इति । = वसलोति, सो + अपि = सोपि।

(ग) स्वरोपरान्त स्वर के पर्दापण होने पर कभी-कभी सन्धि की अवस्था में दोनों स्वर बने रहते हैं। जैसे-
कञान + इव = काइव, छाया + इव = छायाइव, लता + इव लताइव आदि।

(घ) कभी – कभी लुप्त स्वर अर्थात अकारान्त शब्द के बाद आने वाले ‘इ’ का ‘ए’ तथा ‘उ’ का ओ हो जाता है। जैसे – तस्स + इदं = तस्सेदं, वात + ईरितं = वातरितं, वाम + ऊरु – वामोरु, सीत + उदकं = सीतोदक।

(ङ) यदि ‘इ’ तथा ‘उ’ के बाद कोई स्वर आता है तो कभी-कभी ‘इ’ का ‘य’ एवं ‘उ’ का ‘व’ हो जाता है। जैसे- इति + अस्स = इत्यस्स, वि + आकतो = वयाक्तो, बदु + आवाधो = वहादो। सुन + आगतं = स्वागतं आदि।

(च) पुनः यदि ‘ए’ या ‘आ’ के उपरान्त कोई स्वर आता है, हो क्रमशः ‘ए’ का य तथा ‘ओ’ का व बन जाता है। जैसे- ते + अज्ज = त्यज्ज, मेन + अयं = म्याय, पब्बते + अहं = पब्बल्याहं, सो + अहं = स्वाहं, आदि।

(छ) यदि ‘गो’ शब्द के बाद कोई स्वर आवे तो ‘गो’ का गव आदेश हो जाता है। जैसे- गो + अलं = गवास्सं, गो + एलकं = गवेलकं आदि।

अपवाद यस्मा + इह = यस्माति, मनसान + अञा = मनसादञ्जा, चिरंन + आयति = चिरन्नायति, लहुनएस्सति = लहुमेससति, न + इमस्स = नयिमस्स, सब्भि + एव = सब्मेिरेव, . – तथा + एव = तथाखि, यथा + एव = यथखि, छ + अभिंञा = छ अभिञा आदि।

2. व्यञ्जन संधि

व्यञ्जन के बाद स्वर या व्यञ्जन हो, तो उनके मिलन से पैदा होने वाले विकार को व्यञ्जन – सन्धि की संज्ञा प्रदान की जाती है। जैसे- प + बजं = पब्बजं इसके नियम निम्नांकित हैं-

(क) यदि किसी ‘ह्रस्व स्वर’ के बाद कोई व्यञ्जन आवे तो वह ‘स्वर’ दीर्घ और दीर्घ स्वर के बाद व्यञ्जन आवे तो ‘स्वर’ ‘ह्रस्व’ हो जाता है। यथा + खन्ति + परम = खन्ती परमं, जायति + सोको जायतीसोकी, मुनि + चर = मुनीचेर माल + भारी = मालभारी।

(ख) स्वर के बाद यदि व्यंजन हो, तो कभी-कभी उस व्यञ्जन का द्वित्व हो जाता है। जैसे- इध + पमादो = इधप्पमादो, दु + कतं = दुक्कतं (दुक्कटं, प + गहो = प्पगहो, प + बजं = पब्बजं, वि + गहो = विग्गहो आदि।

(ग) यदि किसी वर्ण के पूरे किसी वर्ण का ‘द्वितीय’ या ‘चतुर्थ’ वर्ण आवे तो सन्धि के समय ‘द्वितीय’ पूर्व उसी वर्ग का प्रथम एवं ‘चतुर्थ’ वर्ग पूर्व उसी वर्ण का ‘तृतीय’ वर्ग का आगम होता है। यथा-अ + खन्ति = अक्खन्ति सरोवत + ज्ञानफलटो = एसोवतन्झानफलो, ,नि + ठान = निट्ठानं, सेत + छतं = सेतच्छतं, यस + थेरो = थसत्थेरे, अनफुटं = अप्फुटं आदि।

(घ) ‘ए’ तथा ‘ओ’ के बाद यदि कोई वर्ण आता है तो ‘ए’ तथा ‘ओ’, ‘अ’ में परिवर्तित हो जाते हैं। यथा- अकरम्हसे + ते = अकरम्हसते, अब्मो + अखायति = अग्मक्वायति, एसो + अत्तथो = एसत्थो, एसी + धम्मो = एसधम्मो, याचके + आगते = याचकमागते, सो + सीलवा = ससीलवा आदि।

3. निग्गहीत – सन्धि

निग्गहीत के बाद आने और जुड़ने वाले स्वर या व्यञ्जन के कारण होने वाले परिवर्तन को ‘निग्गहीत संधि’ के नाम से पुकारा जाता है। इसके प्रमुख नियम इस प्रकार हैं-

(क) दो वर्गों के मिलन स्वरूप कभी-कभी ‘निग्गहीत’ (अं) का आगम होता है। यथाअव सिरो = अवसिरो, चक्खु + उदपादि = चक्खुंउदपादि (चक्खुमुदपादि) त + खणे, = तंखेणे, त + सभावो = तंसभावो पुरिम + जाति = पुरिमं जाति, याव + विध = यावविध आदि।

(ख) कभी-कभी दो वर्गों में सन्धि होते समय पूर्व शब्द के अन्तिम वर्ण पर रहने वाले ‘निग्गहीत’ का लोप हो जाता है। जैसे- एवं + अहं = एवाह, कथं + अहं = कयांह, गन्तुं + कामां = गन्तुकामो, वुद्धानं + सासनं = वुद्धानसासन, सं + स्तो = सास्तो, सं + रम्भो = सारम्भो, सं + रागो = सारागो. आदि।

(ग) कभी-कभी निग्गहीत के उपरान्त आनेवाला ‘स्वर’ लुप्त हो जाता है। यथा-अभिनन्द + इति = अभिनन्दति, अलं + इदानि = अलन्दानि, इदं + अपि, इदम्पे, किं + इदानि = किन्दानि, किं + इति = कित्ति, त्वं + असि = वसि, वीजं + इव = वीजंत आदि।

(घ) निग्गहीत से परे यदि कोई वर्गीय वर्ण रहता है, तो वह निग्गहीत उसी वर्ग के अन्तिम वर्ग में परिणत हो जाता है। जैसे- तं + करोति = तङ्गरोति, तं + चरति = चञ्चरति, तं + कानं = तण्ठान, तं + धनं = तन्धन। तं + पाति = तम्पाति आदि।

(ङ) कहीं-कहीं निग्गहीत के बाद आने वाला स्वर लुप्त हो जाता है और लुप्त स्वर के बाद व्याख्या संयुक्त व्यञ्जन असंयुक्त हो जाता है। जैसे- एवं + अस्स = एवंस, पुफ्फं + अस्सा = पुफ्फंसा आदि।

(च) निग्गहीत के बाद यदि ‘य’, एव या हि रहे तो कभी-कभी निग्गहीत का न हो जाता है। यथा-यं + एल = यज्ञदेव, तं + एव = तजेव, तं + हि = तज्हि आदि।

(छ) यदि ‘स’ शब्द के बाद ‘य’ अक्षर आवे तो ‘सं’ का निग्गहीत ‘अ’ बन जाता है। यथा- सं + यतो = सञतो,, सं + यमो = सञमो।

(ज) स्वर से परे यदि ‘निग्गहीत’ हो तो कहीं-कहीं निग्गहीत का ‘म’, ‘य’ या ‘द’ आदेश ‘हो जाता है, जैसे-तं + अहं = तमहं, तं + इदं = तायिदं, तं + अलं = तदल आदि।

अपवाद = छ + अग्गं = छलग्गं छ + आयतन = छलायतन, तं + इमिना = तदमिना, साकि + आगामी = सकदागामी, एक + इध + अहं = एकमिदाह, संविधाय + अवहारो = संविदावहारो, वारिनो + वाहको = वलाहको, जीवन + मूतो जीमूतो, छव + सयनं = सुसानं आदि।

समास

दो या दो से अधिक विभिन्न अर्थक पदों का एकार्थक एक पद हो जाने को समास कहते हैं। जैसे- जितानि इद्रियानि येन सो-जितिन्द्रयो आदि इसके छः भेद हैं-
1. अव्ययी भाव
2. बहुब्रीहि
3. तत्पुरुष
4. कर्मधारम
5. द्विगु और
6. द्वन्द

1. अव्ययीभाव-अव्यय (उपसर्ग) के साथ जब किसी अन्य शब्द का योग होता है, तब उसे ‘अव्ययीभाव समास की संज्ञा दी जाती है। यथा-
अव्यय – समास – विग्रह

अधि – अधित्थि – इत्थीसु कथा पवत्त।
स – सब्रह्मं – सम्पन्न ब्रह्मा
सु – सुभिक्खं – समिद्धि भिक्खानं
उप – उपकुम्भं – कुम्भस्स समीपं
सह – सतिणं – सहतिणेन
नि – निम्नक्खिकं – आभावो मक्खिकानं
अनु – अनुरुप – रुपस्स योग्गं
अनु – अनुरथं – रथस्स पच्छा
सह – सचक्क – सहचक्केन

2. बहुब्रीही – जब सभी पद अपने अर्थ को मूल, एक होकर अन्य अर्थ प्रकट करने लगते हैं तब उसे ‘हम’ बहुब्रीहि समास कहते हैं। जैसे-
समारन – विग्रह – अर्थ

अपुत्तो – न सन्ति पुत्तायस्य, ससो – पुत्र रहित
कुमारभरिया – कुमारी भरिया यस्स, सो – युवक
जितिन्द्रियो – जितानि इन्द्रियाग्नियस्स, सो – बुद्ध
बहुतापसो – वहवो तापसा यस्यिं, सो – मठ (आश्रम)
बहुधनो – बहुनि धनानियस्स, सो – धनवान
पतितेपण्णो – पतितं पण्णं यस्स, तं – ठूठ (वृक्ष)
महद्धना – महन्तानि धनानि यस्स, सो – तथागत
लम्बकणो – लम्बा कण्णा यस्स, सो – धनी
अजिरपाणि – वजिरं पाणिम्हि यस्स, तो – इन्द्र
विजितमारो विजितं मारं यस्स, सो – बुद्ध

3. तत्पुरुष – इस समास में अन्तिम पद प्रधान हो जाता है। इसमें दुतिया से लेकर सत्तमी तक की विभक्तियों का योग होता है। यथा-
समास – विग्रह – अर्थ
गांमगतो – गामंगतो – गाँव को गया हुआ
कुम्भकारो – कुम्भयंकार को – कुम्हार
राजहतो – राहतो – राजा द्वारा मारा गया
असिच्धिनो – असिना छिन्नो – तलवार से कटा हुआ
दधिभोजनं – दधिना उपसित्तं भोजन – दही से परिपूर्ण भोजन
संघगतं – संधस्स भत्तं – संघ के लिए भोजन
गामनिग्गतो – गामस्मा निग्गतो – गाँव से निकला हुआ
चोरभय – चोरस्मा भयं – चोर से भयभीतं
नदितीरं – नदियातीरं – नदी तट
कारूपं – कज्ञायरुप – कन्या का रूप
वनचरो – वने चरितं – वनचर
धम्मरतो – धम्मेरतो – धर्म में लीन।

4. कर्मधारय – जब विशेषण पद का विशेष्य पद से समास होता है, तब उसे ‘कर्मधारय समास कहा जाता है। जैसे- मुनिसीहो – मुनी च सो सीहो चाति। नीलुप्पल – नीलञ्च तं उप्पलं। सीलधनं – सीलमेव धुनं। पुथुज्जनो = अरियेहि पुथुणेवायें जनोति। निक्कोसम्बि निग्गतो कोसम्बिया आदि।

5. द्विगु – जब किसी संख्यावाचक पद का मिलन अन्य पद से होता है, तब हम उसे ‘द्विगु’ समास कहते हैं। जैसे- साहं (छाहँ) – छन्नं अहानं। सलायतन – धन्नं आयतमान। तिलोक – तिन्नं लोकान। तिरतनं – तयो रतनानां चतुसच्चं – चतुन्नं सच्चान। तिभवा तयो भवा आदि।

6. छन्द-विभिन्न पदों के संयुक्त होने के उपरांत भी जब सभी पद समान रूप से प्रधान होते हैं, तब वह ‘द्वन्द’ समास कहा जाता है। जैसे- चक्खुसोतं – चक्खुंच सोतं च! नामरुपं = नामं च रुपं च। जरामरणं – जरा च मरणं च। युगनंगलं = युगं न नंगल च। असिचम्म = अखि च चम्मं च। मातोपितरो = माता च पिता च। गीतवादितं = गीत च वादितं च। दासिदासं = चासिंच च दासं च। समणब्राह्मणा = समणो च ब्राह्मणोच आदि।

उपसर्ग

ऐसे शब्द, जिनका अपना स्वतंत्र स्तित्व नहीं होता है, पर अन्य शब्दों, विशेषकर धातुओं के साथ जुड़कर प्रायः उनके अर्थ में विशेषता ला देते हैं, उपसर्ग कहलाते हैं। उप का अर्थ पूर्व या पहले होता है, अतः ये शब्दों के पूर्व ही अपना स्थान ग्रहण करते हैं। उपसर्ग की संख्या बीस है।

अशुद्ध वाक्यों का शुद्धिकरण

अशुद्ध वाक्य – शुद्ध वाक्य
1. रामो चारेन भायति – रामो चोरस्सा भायति।
2. आचरियं सदिसो सिस्सं अत्थि – आचरियेन सदिसो सिसो अत्थिा।
3. सां अजिनाय मिजं छचति – सो अजिनस्मिं मिंग हञ्जति।
4. जनक सदा पुत्रे गच्छति – जनकेन सह पुत्रो गच्छति।
5. आगतस्स आचरियस्स सिस्सा उट्ठदन्ति – आगते आचरिये सिस्सा उट्ठहन्ति।
6. भूपो दलिदं धनं ददाति – भूपो दलिट्ठय धनं ददाति।
7. सुरियस्स आलोके कस्सकस्स दारका कीलति – सुरियस्स आलोके
– कस्सकस्स दारका कीलन्ति
8. दारको गामेन गच्छति – दारको गामं गच्छति।
9. कोसोन कुटिला नदी – कोसं कुटिला नदी।
10. सिस्सा आचारयेन पठति – सिस्सा आचरियेन पठन्ति।
11. सो मासेन पोत्थकं पठन्ति – सो मासं पोत्थकं पठति।
12. दिवसेन गेहो सुजे तिट्ठन्ति – द्विवसं गहो सुङ्गे तिट्ठति।
13. बालको दण्डस्या सत्सं पहरति – बालको दण्डेन सप्पं पहरति।
14. जलस्स बिना रूक्खो सुक्खन्ति – जलं बिना रूक्खो सुक्खति।
15. बालको चन्दो दिस्सन्ति – बालकेन चन्दो दिस्सति।
16. धम्मस्मा यसो वड्ढति – धम्मेन चसो वट्ठति।
17. लोकहितस्मिं बुद्धो धम्मे देसेति – लोकहिताय बुद्धो धम्म देसेति।
18. मनुस्सेन खतियो सेट्ठो – मनुस्सानं खत्तियो सेट्ठो।
19. दानस्स धम्मदानं सेट्ठ – दानानं धम्मंदानं सेट्ठ।
20. आकासेन सकुणा विचरति – आकासे सकुणा विचरन्ति।
21. थोरस आगतस्स समणेरा उड्ढहति – थेरे आगते समणेरा उट्ठहन्ति।
22. माणवको अनज्झयो रूचन्ति – माणवकाय अनज्झायो रूचति।
23. जनको गेहेन आगच्छन्ति – जनको गेहस्सा आगच्छति।

शब्द – रूप
अकारान्त पुल्लिंग शब्द ‘बालक’

टिप्पणी – गज, जनक (पिता), नर, वानर, बुद्ध, पुत्र आदि का रूप बालक शब्द के समान होगा।

आकारांत स्त्रीलिंग शब्द ‘लता’

विभक्ति – एकवचन – अनेकवचन

पठमा – लता – लतायो
दुतिया लतं – लतायो
ततिया – लताय – लताहि
चतुत्थी – लताय – लतानं
पञ्चमी – लताय – लताहि
छट्ठी – लताय – लतानं
सत्तमी – लतायं – लतासु
आलपन – लते – लतायो

टिप्पणी – अज्जिना कल्ला, गाया, तण्हा, पञ, महिला, माला, विज्जा आदि का रूप लता के समान होगा।

इकारांत पुल्लिंग शब्द ‘मुनि’

विभक्ति – एकवचन – अनेकवचन

पठमा – मुनि – मुनयो
दुतिया – मुनि – मुनयो
ततिया – मुनिना – मुनीहि
चतुत्थी – मुनिस्स – मुनीनं
पञ्चमी – मुनिस्स – मुनीहि
छट्ठी – मुनिस्स – मुनीन
सत्तमी – मुनिस्मिं – मुनीसु
आलपन – मुनि – मुनयो

टिप्पणी – निम्नलिखित शब्दों के रूप मुनि के समान चलेंगे-अग्नि (आग), अरि (शत्रु), असि (तलवार), कपि (बन्दर), कवि, गेहपति (गृहस्थ), ढीपि (बाघ), निधि, पागि (हाथ), भूपति, रवि, सेनापति, हरि आदि।

इकरांत स्त्रीलिंग शब्द “पालि’ (पंक्ति)

विभक्ति – एकवचन – अनेकवचन

पठमा – पालि – पालियो
दुतिया – पालि – पालियो
ततिया – पालिया – पालिहि
चतुत्थी – पालिया – पालिनं
पञ्चमी – पालिया – पालिहि
छठी – पालिया – पालिनं
सत्तमी – पालियं – पालिसु
आलपन – पालि – पालियो

टिप्पणी – इद्धि (त्रिद्धि), कित्ति (कीर्ति), खन्खि (सहनशीलता), नन्दि (तृष्णा), बुद्धि, बोधि (ज्ञान), भूमि, भट्ठी (लाठी), रत्ति, बुद्धि, बुट्ठ (वृष्टि), सति (स्मृति) आदि पालि के समान चलेगा।

ईकारान्त स्त्रीलिंग शब्द ‘नारी’

विभक्ति – एकवचनं – अनेकवचन

पठमा – नारी – नारियो
दुतिया – नारि – नारियो
ततिया – नारिया – नारीहि
चतुत्थी – नारिया – नारीनं
पञ्चमी – नारिया – नारीहि
छट्ठी – नारिया – नारीनं
सत्तमी – नारियं – नारीसु
आलपन – नारि – नारियो

टिप्पणी – इन शब्दों का रूप भी नारी के समान होगा- अर्णा (बकरी), इत्थी (स्त्री), कदली (केला), कुमारी, कोमढी (चाँदनी), गब्भिनी (गर्भवति), तरुणी (युवती), दासी, देवी, नदी, पठवी, ब्राह्मणी, मही (पृथ्वी), यक्खी (यक्ष स्त्री), रजनी (रात), वानरी, वापी (कुंआ), वारुणी, सीही (सिंहनी), सुकरी, हंसी आदि।

उकारान्त पुल्लिंग शब्द “सिसु’

विभक्ति – एकवचन – अनेकवचन

पठमा – सिसु – सिसवो
दुतियां – सिसु – सिसवो
ततिया – सिसुना – सिसहि
चतुत्थी – सिसस्स – सिसूनं
पञ्चमी – सिसुस्मा – सि सिसूहि
छट्ठी – सिसूस्स – सिसून
सत्तमी – सिसुस्मिं – सिसूसू
आलपन – सिसु – सिसवी

टिप्पणी – अंसु (किरण), उच्छ (ईख), कारु (वश्वकमी), केत (पता का), कोत्थ (गीदर), जन्तु (पशु), पसू (पशु), बन्धे (भाई), भानु (सूर्य), मच्चु (मृत्यु), मेरु (पर्वत), रेणु (पराग), वैलु (बाँस), सत्रु (शत्रु), साधु, सेतु (पुल), हेतु (कारण) आदि शब्दों का रूप सिसु के समान चलेगा।

सर्वनाम शब्द ‘अम्ह’ (मैं)

विभक्ति – एकवचन – अनेकवचन

पठमा – अहं – अम्हे
दुतिया – ममं – अम्हे
ततिया – मया – अम्हेहि
चतुत्थी – मम – अम्हाकं
पञ्चमी – माम – अम्हेहि
छट्ठी- मन – अम्हाक
सत्तमी – ममि – अम्हेसु

टिप्पणी -1. सर्वनाम को सम्बोधित नहीं किया जा सकता, अतः आलपन विभक्ति का प्रयोग नहीं किया जाता है।
2. ‘अहं’ शब्द का रूप तीनों लिंगों में एक समान होता है।

सर्वनाम शब्द ‘तुम्ह’ (तू)

विभक्ति – एकवचन – अनेकवचन

पठमा – तवं – तुम्हे
दुतिया – तवं – तुम्हें
ततिया – त्वया – तुम्हेहि
चतुत्थी – तव – तुम्हाकं
पञ्चमी – त्वया – तुम्हेहि
छट्ठी – तव – तुम्हाकं
सत्तमी – त्वमि – तुम्हेसु

टिप्पणी – 1. सर्वनाम संबोधन कारक का प्रयोग नहीं किया जाता है, अतः आलपन में ‘तुम्ह’ का कोई रूप नहीं होगा।
2. ‘तुम्ह’ शब्द का रूप तीनों लिंगों के लिए एक ही होता है।

सर्वनाम पुल्लिंग शब्द ‘त (वह)

विभक्ति – एकवचन – अनेकवचन
पठमा – सो – ते
दुतिया – तं – ते
ततिया – तेन – तेहि
चतुत्थी – तस्स – तेसं
पञ्चमी – तस्मा – तेहिं
छट्ठी – तस्मिं – तेसु
सतनो- तस्स – तेसं

टिप्पणी – सर्वनाम शब्द का प्रयोग आलपन में नहीं होता है। इसलिए आलपन में कोई रूप नहीं होगा।

सर्वनाम स्त्रीलिंग शब्द ‘ता’ (वह)

विभक्ति – एकवचन – अनेकवचन
पठमा – सा – तायो
दुतिया – तं – तायो
ततिया – ताय – ताहि
चतुत्थी – ताय – तासं
पञ्चमी – तायं – ताहि
छट्ठी – ताय – तासं
सत्तमी – तायं – तांसु

टिप्पणी – सर्वनाम में आलपन का प्रयोग नहीं किया जाता है।

सर्वनाम नपुंसक लिंग ‘त’ (वह)

विभक्ति – एकवचन – अनेकवचन

पठमा – तं – तानि
दुतिया – तं – तानि
ततिया – तेन – तेहि
चतुत्थी – तस्म – तेसें
पञ्चमी – तस्मा – तेहि
छट्ठी – तस्स – तेसें
सत्तमी – तस्मिं – तेसु
टिप्पणी- सर्वनाम में आलपन का प्रयोग नहीं किया जाता है।

धातु-रूप

‘अस’ (होना) धातु (वर्तमान काल)

पुरिस (पुरुष) – एकवचन – अनेकवचन
पठम (अन्य) – (सो) अस्थि (वह है) – (ते) सन्ति (वे हैं)
मज्झिम (मध्यम) – (त्वं) असि (तुम है) – (तुम्हे) अत्य (तुमलोग हो)
उत्तम (उत्तम) – (अहं) अस्मि (मैं हूँ) – (अम्हे) अस्म (हमलोग हैं)

‘कर’ धातु (वर्तमान काल)

पुरिस – एकवचन – अनेकवचन
पठम – करोति – करोन्ति
मज्झिम – करोसि – करोथ
उत्तम – करोमि – करोम

‘चुर’ धातु (वर्तमान काल) एकवचन

पुरिस – एकवचन – अनेकवचन
पठम – चोरेति, चोरयति – चोरन्ति, चोरयन्ति
मज्झिम – चोरेसि, चोरयसि – चोरेथ, चोरयथ
उत्तम – चोरेनि, चोरयानि – चोरम, चोरयाम

पठ’ धातु (वर्तमान काल)

पुरिस – एकवचन – अनेकवचन
पठम – पठति – पठन्ति
मज्झिम – पठसि – पठथ
उत्तम – पठामि – पठाम

“लिख’ धातु (वर्तमान काल)

पुरिस – एकवचन – अनेकवचन
पठम – लिखति – लिखन्ति
मज्झिम – लिखति – लिखय
उत्तम – लिखासि – लिखाम

‘सु’ धातु (वर्तमान काल)

पुरिस – एकवचन – अनेकवचन
पठम – सुणोति – सुणोन्ति
मज्झिम – सुणोस – सुणोथ
उत्तम – सुणोम – सुणोम

अनागत काल
‘कर’ धातु ( अनागत काल)

पुरिस – एकवचन – अनेकवचन
पठम – करिस्सति – करिस्सन्ति
मज्झिम – करिस्ससि – करिस्सथ
उत्तम – करिस्सामि – करिस्साम

‘चुर’ धातु (अनागत काल)

पुरिस – एकवचन – अनेकवचन
पठम – चोरेस्सति, चोरयिस्सति – चोरेस्सन्ति, चोरयिस्सन्ति
मज्झिम – चोरेस्ससि, चोरयिस्ससि – चोरेस्मथ, चोरयिस्सथ’
उत्तम – चोरेस्सामि, चोरयिस्सामि – चोरेस्साम, चोरयिस्माम

“पठ’ धातु (अनागत काल)

पुरिस – एकवचन – अनेकवचन
पठम – पठिस्सति – पठिस्सन्ति
मज्झिम – पठस्ससि – पठस्सिथ
उत्तम – पठिस्साम – पठिस्साम

“लिख’ धातु (अनागत काल)

पुरिस – एकवचन – अनेकवचन
पठम – लिखिस्सति – लिखिस्सन्ति
मज्झिम – लिखिस्ससि – लिखिस्सथ
उत्तम – लिखिस्सानि – लिखिस्साम

‘सु’ धातु (अनागत काल)

पुरिस – एकवचन – अनेकवचन
पठम – सुणिस्सति – सुणिस्सन्ति
मज्झिम – सुणिस्ससि – सुणिस्सथ
उत्तम – सुणिस्सामि – सुणिस्साम

अतीत काल
‘कठ’ धातु (अतीत काल)

पुरिस – एकवचन – अनेकवचन
पठम – करि – कसिंसु
मज्झिम – करि – करित्थ
उत्तम – करि – करिम्हा

‘चुर’ धातु (अतीत काल)

पुरिस – एकवचन – अनेकवचन
पठम – चोरयि – चोरमिंस
मज्झिम – चोयमि – चोरमित्थे
उत्तम – चोरयिं – चोरयिम्हा

‘पठ’ धातु (अतीत काल)

पुरिस – एकवचन – अनेकवचन
पठम – पठि – पठिंसु
मज्झिम – पठि – पठित्थ
उत्तम – पठिं – पठिम्हा

‘लिख’ धातु (अतीत काल)

पुरिस – एकवचन – अनेकवचन
पठम – लिखि – लिखिंसु
मज्झिम – लिखि – लिखित्थ
उत्तम – लिखिं – लिखिम्हा

‘सु’ धातु (अतीत काल)

पुरिस – एकवचन – अनेकवचन
पठम – सुणि – सुणिंसु
मज्झिम – सुणि – सुणित्थ
उत्तम – सुणिं – सुणिम्हा

तत्सम शब्द से पालि

पालि से हिन्दी अनुवाद

1. इदं खो पन, भिखवे, दुक्खं अरियसच्यं। जाति पि दुक्खा, व्याधि दि दुक्खो, मरणं पि दुक्खं, अप्पियेहि सम्पयोगो दुक्खो, पियेहि विप्पयोगे दुक्खो। यम्पिच्छं न लभति तं पि दुक्खं। संखितेन, पञ्चुपादानक्खन्धा दुक्खा। इदं खो पन भिक्खवे, दुक्ख-समुदयं अरियसच्चं-स तण्हा पोनोब्भविका नन्दिरागसहगता तत्रतत्राभिनन्दिनी, सेय्यथीदं काम तण्हा भवतण्हा, विभतण्हा।

हिन्दी – भिक्षुओं ? दुःख आर्य सत्य है-जन्म भी दुःख है जरा भी दुःख है व्याधि भी दुःख है, मरण भी दुःख, अप्रिय का वियोग भी दु:ख है प्रिय का वियोग भी दुःख है, इच्छित वस्तुओं का न मिलन भी दुःख है, संक्षेप में पाँच उपादान स्कंध भी दुःख है, यह दुःख समुदाय आर्य सत्य है, यह जो तृष्णा है फिर जन्म लेने की, प्रसन्न होने की, राग सहित जहाँ-तहाँ प्रसन्न होने वाली जैसाकि-काम तृष्णा, भव तृष्णा, विभव तृष्णा।

2. यावकीवं चमे भिक्खवे ‘इमेसु चतुसु अरियसच्चेसु एवं तिपरिवर्ड्स द्वादसाकां यथा. भूतं त्राणदम्सनं न सुविसुद्ध अहोसि, नवे तावारं भिक्खवे, सदेवके लोके समार की सब्रह्मके ससमणब्रह्मणिया पंजाय सदेवमनुस्साय अनुत्तरं सम्मासम्बोधि अभिसम्बुद्धो नि पञ्चालि। यतो च खो में भिक्खवे इमेसु चतुसु अरियसच्चेसु एवं तिपखिट्ट द्वादसाकारं यथाभूतं आणदस्सनं सुविसुद्धं अहेसि, अथाहं भिक्खवे, सदैवके लोके संसारके सब्रह्मके सस्समण ब्राह्मणिया पजाय सदेवनुस्साय अनुत्तर सम्मासम्बोधि अभिसम्बुद्धो ति प्रज्यत्रालि त्राणं च पन में दस्सनं उदपादि अकुव्सा में विमुक्ति, अयमत्मा जाति, नत्थि दानि पुनष्भवा ति 12 दमत्रीया भगवा अत्तमन पञवाग्गिया भिक्खु भगवतो भाषितं अभिन्नयु।

हिन्दी – भिक्षुओं जबतक मैं इन चार आर्य सत्यों को इस प्रकार विहार कर बारह प्रकार के यथार्थ ज्ञानदर्शन से शुद्ध-विशुद्ध नहीं हुआ तब तक मैंने देवता सहित, मार सहित, ब्रह्मा सहित, धमण-ब्राह्मण सहित लोक में प्रजा के लिए, देवता और मनुष्य के लिए अनन्त्र सम्यक् सम्बोधि से अभिव्यक्त हुआ, प्रज्ञपित नहीं किया। भिक्षुओं, जब मैं इन चार आर्य सत्यों को तिहरा कर बारह प्रकार के यथार्थ प्रज्ञा दर्शन से शुद्ध विशुद्ध हुआ तब मैं देवता, मार, प्रज्ञा, श्रमण, ब्राह्मण सहित संसार में प्रजा देवता देवता और मनुष्य के लिए अनुत्तर सम्यक सम्बोधि से पूर्ण हुआ, प्रज्ञापित किया। मुझमें ज्ञान और दर्शन उत्पन्न हुआ। “मुझे अचल से विमुक्ति मिली, यह मेरा अन्तिम जन्म है, अब मेरा पुनर्जन्म नहीं होगा।” प्रसन्न मन से पंचवर्गीय भिक्षुओं ने भगवान . से कथन का अभिवादन किया।

3. “इदम वोच भगवा इदं वत्वा सुगतो अथापा सदवोच सत्या-”
मातापित-दिसा पुब्बया, आचरिया दक्खिणा दिशा।
पुत्तदारा दिशा पच्छा मित्तमच्चा च उत्तरा ॥
दासकम्मकारा हेट्ठा, उद्धं समणब्राह्मण ।
एतादिसा नमस्सेय्य, अलमतो कुले गिही ॥

हिन्दी – भगवान ने ऐसा कहा। इतना बोलकर सगत शास्ता ने इसके उपरान्त भी ऐसा कहा-
माता-पिता पूर्व दिशा है, गुरू दक्षिण दिशा। पुत्र-स्त्री पश्चिम दिशा एवं मित्रमात्य उत्तर दिशा। नौकर चाकर पताल दिशा तथा श्रमण-ब्राह्मण आकाश दिशा है। गृहस्थ को इस प्रकार से दिशाओं का नमस्कार अपने कुल में करना चाहिए।

4. अथ.खो भगवा पुब्बण्हासमयं निवासेत्वा पत्तचीवरमादाय राजगह पिण्डाय पाविसि। अद्दसा खो भगवा सिगालकं गहपतिपुतं कालस्सेव उटाय, राजगहा निक्खमित्वा, अलवत्थं अल्लकेसं पतलिकं पुथुदिशा नमस्सन्तं पुररंथमं दिसं, दक्खिणं दिसं, पच्छिम दिशं, उत्तरं दिसं, हेट्ठिम दिसं, उपरिमं दिसं ति।

हिन्दी – तब भगवान बुद्ध पूर्वाह्न समय में वस्त्र धारण कर पात्र और चीवर लेकर भिक्षाटन के लिए राजगृह में प्रविष्ट हुए। भगवान-बुद्ध ने गृहपति-पुत्र सिंगाल को प्रातःकाल ही उठकर राजगृह से निकलकर भीगे वस्त्र, भींगे केश, हाथ जोड़े विभिन्न दिशाओं-पूर्व दिशा, दक्षिण दिशा, पश्चिम दिशा, उत्तर दिशा, आकाना दिशा और पाताल दिशा को नमस्कार करते देखा।

5. एवं वुत्ते सिंगालको गहपलिपुन्नों भगवन्तं एतदवोच-अभिक्कन्तं भन्ते, अभिकन्तं भन्ते, सेय्यथापि भन्ते, निकुन्जितं वा उक्कुज्जेय पटिच्छन्न वा विवरेय्य मुल्हस्स वा रेग्गं आचिक्खेय्य, अन्धकारे वा तेलपज्जातं धामिरय्य-चक्खुमन्तो. रूपानि दक्खन्तोति, एवमेव भगवता अनेक परियायेन धम्मो पकासितो। एसाहं भन्ते भगवन्तं सरणं गच्छामि, धमञ्च भिक्खुसंघञ्च उपासकं मं भगवा धारेतु अञ्जगो पाणुपेतं सरणं अंतति।

हिन्दी – ऐसा कहने जाने पर गृहपति-पुत्र सिंगाल ने भगवान से ऐसा कहा-आश्चर्य है भन्ते, अद्भूत है भन्ते, जैसा महाराज, अंधे की सीधा कर दे, ढंके खोल दे, मूर्ख को मार्ग बतला दे, अन्धकार में दीपक दिखा दे, जिससे कि आँख वाला रूप को देख ले, उसी प्रकार से भगवान बुद्ध ने अनेक उदाहरणों के द्वारा धर्म को प्रकाशित किया। अतः भन्ते मैं भगवान की शरण जाता हूँ, (साथ ही) धर्म और भिक्षु संघ का भी आज से आजीवन भगवान मुझे .. अपनी शरण में आया उपासक स्वीकार करें।

6. अथ खो भगवा उट्ठायासना पक्कामि ? अथ खो राहुल कुमारो भगवन्त विद्वितो पिट्ठितों अनबन्धिदायज में समण देहि, दायन्जं में समण, देही ति। अथ खो भगवा आयस्मन्तं सारिपुत्तं आमन्तेसि- तेन हि त्वं सारिपुत्रं राहुल कुमारं पब्बाजोही, ति। कथाहं भन्ते, राहुल कुमारं पब्बोजेसी, ति ? अथ खो भगवा एतस्मिं निदाथे एतस्मिं पकरणे धम्मिं कथं कत्वा भिक्खु आमन्तेसि अनुजामि भिक्खवे, तीहि सरणगमनेहि सामणीदूपब्बन्ज। अथ खो आयस्मा सारिपुत्रों राहुल कुमारं पब्बाजेसि।

हिन्दी – तब भगवान आसन से उठकर चल दिये। राहुल कुमार भी भगवान के पीछे-पीछे चलने लगा। “श्रवण ? मुझे दायज्ज दें, श्रमण ? दयाज्ज दें।” तब भगवान आयुष्मान् सारिपुत्र से कहा-“तो सारिपुत्र.? राहुल कुमार को प्रब्रजित करो।” भन्ते किस प्रकार राहुल कुमार को प्रब्रजित करूं ? इसी मौके पर इस प्रकार मैं धार्मिक कथा कहकर भगवान ने भिक्षुओं को सम्बोधित किया ‘भिक्षुओं से? तीन कारण गमण से अमणेर प्रब्रज्जा का अनुज्ञा देता है। तब आयुष्मान सारिपुत्र ने राहुल कुमार को प्रव्रजित किया।

7. “अथ खो भगवा राजगहे यथाभियन्तं विहरित्वा येन कपिलवत्थु तेन चारिक पक्कामि। अनुपुब्बेन चारिकं चारमानो येन कपिलवत्थु तदवसरि। तत्र सुदं भगवा सक्सेसु विहरति कपिलवत्युस्मिं निग्रहेधारामे। अथ खो भगवा पुवण्हसमयं निवासेत्वा पत्रचीवरं आदाय येन सुद्धोदनस्स सक्कस्स निवे सावं तेनुपसंकमि उपसमित्वा पञत्ते आसने निसीदि।

हिन्दी – तब भगवान ने राजगृह में इच्छानुसार विहार कर जहाँ कपिलवस्तु थी, वहाँ के चारिका के लिए प्रस्थान किया। क्रमशः भ्रमण करते हुए जहाँ कलिपवस्तु थी, वहाँ पहुँचे। वहाँ भगवान बुद्ध कपिलवस्तु अवस्थित निग्रोधाराम में शाक्यों के बीच विहार करते थे। तब भगवान दो पहर के पूर्व वस्त्र पहन, पात्र चीवर लेकर जहाँ शाक्य शुद्धोधन का निवास स्थान था वहाँ पहुँचे, पहुंचकर विछे हुए आसन पर बैठ गये।

8. अथाखो ते लिच्छवी येन भगवा तेनुपसंकमिसु। अद्वसां खो भगवा ते लिच्छवी दुरतोव आगच्छते। दिस्बान भिक्खू आमन्तेसियेहि भिक्खवे भिक्खूहि देवा तावतिंसा अदिट्ठपुब्बा आलोकेथ भिक्खवे, लिच्छविपरिसं, उपसंहरथ, भिक्खवे लिच्छविपरिसं तावतिसंपरिसं ति। अथ खो ते लिच्छवी यावतिका यानस्स भूमि, यानेने गन्त्वा याना पच्चारोहित्वा पत्रिकाव येन भगवा तेनुषसंकमिसु, उपसंकमित्वा भगवन्तं अभिवादेत्वा एकमन्तं निसीदिस।

हिन्दी – तब वे लिच्छवी जहाँ भगवान थे, वहाँ पहुँचे। तब भगवान ने दूर से ही उन लिच्छवियों को आते देखा। देखकर भिक्षुओं को आमन्त्रित किया। भिक्षुओं, देखो, तायस्त्रिंश लोक के देवों का अभूतपूर्व दृश्य। लिच्छवियों की परिषद का अवलोकन करो, भिक्षुओं लिच्छवी परिषद को समझो। तब वे लिच्छवी रथ से जाने योग्य भूमि तक रथ से जाकर, फिर रथ से । उतरकर पैदल ही जहाँ भगवान थे, वहाँ पहुँचे। पहुँचकर भगवान का अभिवादन कर एक ओर बैठ गये।

9. “अथवा खो वे लिच्छवी अम्बपालि गणिकं एतदवोच किस्स जे अम्बपालि दहरानं-दहरानं लिच्छवीन…….. अक्खेन अक्खं पटिवट्टेसी ति ? तथापि पन मया अय्यपुत्ता स्वातनाय बुद्धप्रमुखो भिक्खूसंघो निमन्तितो, ति। देहि जे अम्बपालि, अम्हाकं एक भत्तं सतसहस्सेन, ति सचे में अय्यपुत्ता वेज्ञाति साहारं दन्जेय्याथ, नेव दज्जा अहं तं भत्तं, ति अथ खे ते लिच्छवी अंगुली फोटेसु-जितम्हा वत भी अम्बकाया पराजितम्हा वत भो अम्बकाया, ति।

हिन्दी – तब लिच्छवीयों ने अम्बपालि गणिका से ऐसा कहा-क्यों री, अम्बपालि युवक” लिच्छवीयों के घरे से धूरा युग से युग, चक्के से चक्का और आँख से आँख लड़ती है ? आर्य पुत्रों क्योंकि बुद्ध प्रमुख संघ कल के भोजन के लिए मेरे द्वारा आमन्त्रित है। हे अम्बपालि यह भोजन हमें (सौ) हजार (कार्यपण) से (में) दे दो। आर्यपुत्री यदि वैशाली जनपद भी दिया जाय, तो मैं इस भार को नहीं दूंगी। तब उन लिच्छवियों ने (दाँते तले) अँगुलियाँ दबायी। अरे अम्बपालि ने हमें जित लिया, अम्बपालि ने हमलोगों को पराजित कर दिया।

10. तेन खो पन समयेन तस्सा रलिया अच्चयेन चातुद्वीपकमोमेघो पावसि अथ खो भगवा भिक्खू आमन्तेसि यथा भिक्खवे, जेतवने वस्सति, एक चतुसु दीपेसु वस्सति। ओवस्सपिथ भिक्खवे कायं। अयं पच्छिम को चातुद्वीपको महामेघो ति। एवं भन्ते ति खो ते भिक्खू भगवतो पटिस्सुत्वा निक्खित ‘चीवरा कायं ओवस्सपिन्ति।

हिन्दी – उस समय उस रात्रि के अवसान होने पर चातुद्धीपिक महामेघ वरसने लगा। तब भगवान ने भिक्षुओं को सम्बोधित किया-भिक्षुओं, जैसी वर्षा जेतवन में हो रही है, उसी प्रकार चारों द्वीपों में वर्षा हो रही है। भिक्षुओं काया शीतल करो। यह अन्तिम चातुद्वीपिक महामेघ है। ठीक है भन्ते वे भिक्षु भगवान को प्रत्युत्तर देकर वस्त्र फेककर शरीर शीतल करने लगे।

11. अथ खो भगवा भिक्खु आमन्तेसि-संनहथ भिक्खवे पत्तचीवरं, कालो भत्तस्साति। एवं भन्ते, ति खो ते भिक्खू भगवतो पक्वस्सोसुं। अथ खो भगवा पुब्बण्हसमयं निवासेत्वा पत्रचीवर-मादाय, सेय्यथापि नाम बलवा पुरिसो समिञ्जितं वा वाहं पसारेय्य, पसारितं वा वाहं समिओय्य एवमेव जेतवने अन्तरहितो विशाखाय। मिगारमातुसा कोठेके पातहोसिा। निसीद्धि भगवा पञत्ते आसने सद्धि भिक्खू संघेन।

हिन्दी – तब भगवान बुद्ध ने भिक्षुओं को सम्बोधित किया-भिक्षुओं, पात्र-चीवर तैयार कर लो। भोजन का समय हो गया। अच्छा भन्ते, कह भिक्षुओं ने भगवान को उत्तर दिया। तब भगवान पूर्वाह्न में, वस्त्र धारण कर पात्र-चीवर लेकर जैसे कोई बलवान पुरुष अपनी मोड़ी हुई बाँह को पसार ले, पसारी हुई बाहों को समेट ले वैसे ही जेतवन से अतध्यान होकर विशाखा मिगार माता के कोठे पर प्रकट हुए। भिक्षु संघ के साथ विछे आसन पर बैठ गये।

12. जयं वेरं पसवति, दुक्खं सेति पराजितो ।
उपसन्तो सुखं सेति, हित्वा जय पराजयं ॥
हिन्दी – विजय वैर (शत्रुता) को जन्म देती है और पराजित (मनुष्य) दुःख की नींद सोता है। (अतः) उपशान्त (मानव) जय और पराजय से परे रहकर सुख की नींद सोता है।

13. साधु दस्सनमरियानं, सन्निवासो सदा सुखी ।
अदस्सनेन बालान, निच्चमेव सुखी सिया ॥
हिन्दी – आर्य का दर्शन उत्तम और संगति सदा सुखदायी होते हैं। मूखों के दर्शन से दूर रहने से मनुष्य सदा सुखी रहता है।

14. नत्थि रागसमो अक्गि, नत्थि दोससमो कलि ।
नल्यि खन्धसभा दुक्खा, नत्थि सन्ति परं सुखं ॥
हिन्दी – आसक्रि के समान आग नहीं है, द्वेष के समान कोई मल नहीं है, पञ्चस्कन्ध से बढ़कर कोई दुःख नहीं है शान्ति से बढ़कर (संसार में) कोई सुख नहीं है।

15. यथा माता-पित भाता, अजे. वापि च आतक।
गावो नी परमा मित्ता, यासु जायन्ति ओसध ॥
हिन्दी – जिस प्रकार माता-पिता, भाई या दूसरे सम्बन्धी, उसी प्रकार गाये हमारी परम मित्त है, जिनसे (अनेक) औषधियाँ उपलब्ध होती हैं।

16. अन्नदा बलदा चेता, वण्णदा सुखदा तथा ।
एतमत्थवसं ञत्वा, नास्सु गावो हनिंसु ते ॥
हिन्दी – (गायें) अन्न, बल, चेतना, वर्ण और सुख देने वाली है। इस बात को जानकर वे गायों की हत्या नहीं करते थे।

17. एवं धम्मे वियापन्ने, विभिन्ना सुदृवेस्सिका ।
पुथु विभिन्न खत्तिया, पति भरिया अवमअथा ॥
हिन्दी – इस प्रकार धर्म में खलल पड़ने पर शुद्धों और वैश्यों में फूट पड़ गयी क्षत्रिय भी विभिन्न भागों में बँट गये और स्त्रियाँ पति का अपमान करने लगी।

18.’न भजे. पापके मित्रे, न भजे पुरिसाधमे ।
भजेथ मित्रे कल्याणे भजेय पुरिसुत्रमे ॥
उत्तर – न पापी मित्रों के साथ रहो और न नराधमों की संगति करो । कल्याणकारी मित्रों की संगीति करो और नर श्रेष्ठ के साथ रहो।

19. उदकं हि न्यदन्ति नेतिक, उसुकारा नमयन्ति तेजनं ।
दारूं नमयन्ति तच्छका, अत्रानं दमयन्ति पण्डिता ॥
उत्तर – नहर निर्माता (सीधे ढंग से) पानी ले जाते हैं। वाणकार वाण को ठीक (सीधा) करते हैं। बढ़ई लकड़ी को सीधा करते हैं, (परन्तु) पंडित लोग अपने आपको सीधा (दमन) करते हैं।

20. यथा हि स्वतःरद्धो गम्भीरी, विप्पसन्नो अनाविलो ।
एवं धम्मानि सुत्वान, विप्पसीदन्ति पण्डिता ।।
हिन्दी – जैसे एक गम्भीर जलाशय प्रशान्त व निर्मल रहता है, वैसे ही धार्मिक प्रवचन सुनकर पण्डित लोग शान्त हो जाते हैं।

हिन्दी से पालि अनुवाद

हिन्दी वाक्य – पालि अनुवाद
1. बुद्ध का धर्म लोक में प्रकाशित होता है- बुद्धस्स धम्मो लोकस्मि आलोकत्ति।
2. पाप से संसार में मनुष्य को दुःख होता है- पापने लोके मनुष्यं दुक्खो हीति।
3. मैं बुद्ध को नमस्कार करता हूँ- अहं वुद्धं नमामि।
4. बुद्ध के धर्म से संसार में सुख होता है- बुद्धस्स धम्मेन संसार सुखो होति।
5. वह बुद्ध की शरण में जाता है- जो बुद्धं सरणं गच्छति।
6. देवता नगर से निकलते हैं- सुरो नगरस्या निक्खमति।
7. ऋषि लोग जंगल में रहते हैं- इसयो अरजएिंग अधिवसन्ति।
8. प्राचीन काल में राजगृह एक बड़ा नगर था- प्राचीनकाले राजगहो एको महानगरे आसि।
9. जहाँ मगध की राजधानी थी- तत्र मगधस्स राजधानी अस्ति।
10. भगवान बुद्ध के समय वहाँ बिम्बिसार नामक राजा रहता था- भगवान बुद्धस्स समये तत्र बिम्बिसारे नाम भूपो वासि।

11. आज यहाँ एक वन है जिसका नाम वेणुवन है- अज्ज इमस्मिं एको वनो अत्थि तस्स नाम बुलुलनो।
12. इस वन में बुद्ध अपने शिष्यों के साथ आते थे- इसस्मिं वने बुद्धो उत्तना सिस्सेहि स आगहिंसा।
13. राजगृह और नालन्दा के बीच अम्बलिका वन है- अन्तराच राजगहं अन्तरा च राजगह अम्बवालिकावने अस्थित।
14. राजगृह के पर्वत नगर की रक्षा करते हैं- राजगहस्स पब्बता नगरस्स रक्खनित।
15. वैभार पर्वत पर सप्तपर्णी गुफा है- विभारपब्बतो सत्रपण्णी गुहो अत्थि।
16. पूर्वदिशा में रत्नागिरि पर्वत है- पुरत्थिमे दिसे रत्नागिरि पब्बतो अत्थिा
17. इस पर्वत पर एक शान्ति स्तुप है- इमस्मि पब्बते, एको शान्ति स्तूपो अत्थि।
18. नालन्दा एक पुराना नगर है- नालन्दा एको प्राचीननगरो अत्थिा।
19. प्राचीनकाल में यहाँ बौद्धधर्म की पढ़ाई होती थी- प्राचीन काले स्तस्मिं बुद्धधम्मस्स अन्झापं आसि।
20. आज भी यहाँ एक महाविद्यालय है- अज्ज पि एतस्मि एको महाविन्जालयो अत्थि।

21. उसका नाम नवनालन्दा महाविहार है- अम्हे नवनालन्दा महाविहारो अत्थिा।
22. हमलोग नालन्दा महाविहार में पढ़ेंगे- अम्हे नवनालन्दामहाविहारस्मिं पठिस्साम।
23. तुम बुद्ध को नमस्कार करते हो- त्वं बुद्ध नमसि।
24. वह पुस्तक पढ़ता है- सो पोत्थकं पठाति।
25. मैं मित्र के साथ रहता हूँ- अहं मितेहि सह वसामि।
26. आचार्य के साथ शिष्यगण रहते हैं- आचरियेन सह विस्सा वसन्ति।
27. आजातशत्रु बुद्ध का उपासक था- आजातसतु बुद्धस्स उपासको आसि।
28. वे लोग मन्दिर में पुजा करते हैं- ते देवालयस्मि अच्चन्ति।
29. मैं खाकर पढूंगा- अहं भुज्जित्वा पठिस्समि।
30. तुमलोग भाइयों के साथ खेलोगे- तुम्हें. सहादरेहि सह कीलिम्ह।

31. हमलोग मित्र के साथ खेलेंगे- अम्हें मित्रेहि सह कीजिस्साम।
32. बालक बुद्धधर्म को जानता है- बालको बुद्धस्स धर्म जानति।
33. लड़के विद्यालय में पढ़ते हैं- बालका विज्जालयस्मि पठन्ति।
34. वह खेत में दौड़ता है- सो खेत्तस्मं धावति।
35. मैं घर में सोता हूँ- अहं गेहस्मिं सयामि।
36. राजगृह में भिक्षु रहते हैं- राजगहे भिक्खो वसन्ति।
37. भगवान बुद्ध को उपदेश देते हैं- भगवा भिक्खो वसन्ति।
38. गृहस्थ बुद्ध को दान देता है- गहपति बुद्धाय दानं ददाति।
39. नालन्दा में एक विश्वविद्यालय है- नालन्दा एको विस्सविज्जालयो आसि।
40. आज वहाँ लोग पालि पढ़ते हैं- अज्ज तत्र पुरिसा पालिं पठन्ति।

41. ‘त्रिपिटक बौद्ध धर्म के महत्वपूर्ण हैं- तिपिटका बुद्धधमस्स महतपुण्णगन्था सन्ति।
42. लोक हित के लिए बुद्ध धर्म का उपदेश करते हैं- लोकहिताय बुद्धो धम्म देसेति।
43. गुरू के आने पर शिष्य खड़े हो जाते हैं- आचरिये आगते सिस्सा उट्ठहन्ति।
44. गृहस्थ लोग बुद्ध की शरण में जाते हैं- गहपतयो बुद्धो सरणं गच्छन्ति।
45. आग वृक्ष को जलाता है- अनलो रूक्खं झापेति।
46. नेवला साँप को खाता है- नकुलो सप्पं खादत्ति।
47. तुम घर में रहते हो- त्वं गेहस्मिं वससि।
48. हमलोग जंगल में रहते हैं- अम्हे अरमि अधिवसाम।
49. मैं शील की रक्षा करूंगा- अहं शीलं रक्खिस्सामि।
50. हाथी ताल में नहाता है- गजो सरोवरो ओवस्सापेति।

51. चोर धन चुराता है- चोरा धनं चोरेन्ति।
52. हमलोग महाविद्यालयों में पढ़ते हैं- अम्हे महाविज्जालयस्मि पठाम।
53. राजा राष्ट्र को जीतता है- भूपो रटुं जिनाति।
54. वन में फूल खिलते हैं- वने पुप्फनि पुकान्ति।
55. उद्यान में फूल शोभते हैं- उदयाने पुप्फानि सोभन्ति।
56. शोक से शरीर गलता है- शोकेन शरीरं जीरति।
57. दान से शील बढ़ता है- दानेन शीलं बहुति।
58. तुमलोग आचार्य को प्रणाम करते हो- त्वं आचरियं पनमथ।
59. भारत गाँवों का देश है- भारतो गामनं देशो अत्थिा
60. बुद्ध धर्म को प्रकाशित करते हैं- बुद्धो धम्म देसेति।

61. बुद्ध के साथ शिष्य लोय रहते हैं- बुद्ध सह सिस्सा अधिवसन्ति।
62. हमलोग ध्यान करते हैं- अम्हे झानं कराम।
63. पटना विहार की राजधानी है- पटना विहारस्स राजधानी अत्थिा .
64. पटना का पुराना नाम पाटलिपुत्र था- पटना पुरानो नाम पाटलिपुत्रे आसि।
65. अशोक पाटलिपुत्र का सम्राट था- अशोकं पाटलिपुत्रस्स सम्राटो आसि।
66. जंगल में बन्दर वृक्ष पर दौड़ते हैं- अरजस्मि वानरा रूक्खहिंम धावन्ति।
67. आग की लपेट जंगल को जलाती है- अग्गीनं अच्चीनी अरखं झापेति।
68. ऋषि जंगल में ध्यान करते हैं- इसमो अरमिं झायन्ति।
69. वह खाने की इच्छा करता है- सो खापितुं इच्छति।
70. बुद्ध निर्वाण प्राप्त करते हैं- बुद्धो निब्बानं निब्बायति।

71. भगवान धर्मचक्र घुमाते हैं- भगवा धम्मचक्कं पवति।
72. ध्यान से प्रज्ञा पूर्ण होती है- झानेन पञ्च परिपुरति।
73. भगवान बुद्ध देखते हैं- भगवा बुद्धे पस्सति।
74. सारिपुत्र निर्वाण प्राप्त किया- सारिपुत्रो निब्बानं पापुणि।
75. बुद्ध धर्म के साथ विहार करेगे- बुद्धो धम्मेन सह विहारस्सिति।
76. भिक्षु लोग उपदेश देते रहेंगे- भिक्खवो उपदेशो देस्सिसन्ति।
77. मैं धर्म की शरण जाता हूँ- अहं धम्म सरणं गच्छामि।
78. तुमलोग पाटलिपुत्र गये- तुम्हे पाटलिपुत्रो यच्छित्थ।
79. हमलोग बुद्धत्व प्राप्त करेंगे – अम्हे सम्बोधि गहिस्साम।
80. तुम सूर्य लोक से आये थे- त्वं सूरिय लोकस्स आगच्छि।

81. वे लोग आनन्द के पास गये- वे आनन्दं समीपं गच्छिसु।
82. आनन्द ने बुद्ध से कहा- आनन्दी बुद्धेन कथि।
83. बुद्ध पालि भाषा में धर्म को प्रकाशित करते हैं- बुद्धो पालिभाषायं धम्म देसेति।
84. बुद्ध प्रव्रज्या के लिए घर से निकलते हैं- बुद्धो पब्बजाय गेहस्सा निक्खमति।
85. शिष्य आचार्य के साथ-साथ जाते हैं- शिस्सा आचरियेन मद्धिं गच्छन्ति।
86. जल विन पेड़ सुख रहा है -जलं बिना रूक्खो सुक्खति।
87. शील से ध्यान उत्पन्न होती है- सीलेन झानो उप्पज्जति।
88. ध्यान से प्रज्ञा उत्पन्न होती है- झनेन पञो उप्पज्जति।
89. प्रज्ञा से मुक्ति होती है- पञ्जन विभूति होति।
90. भगवान ने भिक्षुओं को देखा- भगवा भिक्खवो पस्सि।
91. आकाश में पंक्षी उड़ते हैं- आकाशे शकुणा विचरन्हिता
92. गृहाना के लड़के स्याही से लिखते हैं- गृहपतिस्स दारका मस्यिा लिखन्ति।

Bihar Board Class 10 Science Solutions Chapter 16 Management of Natural Resources

Bihar Board Class 10 Science Chapter 16 Management of Natural Resources InText Questions and Answers

In-text Questions (Page 269)

Question 1.
What changes can you make in your habits to become more environment friendly?
Answer:
The changes to become more environment friendly are:

• Reduce the use of fossil fuels.
• Repairing the water taps if leakage is seen.
• Planting trees.
• No use of plastic bags.
• Reuse the empty bottles of pickles and jams.
• Recycle the recyclable materials for use.

Question 2.
What would be the advantages of exploiting resources with short-term aims?
Answer:
Advantages for exploiting resources for short-term aims are:

• Increase profits without any accountability.
• Fulfil the needs of the present generation.

Question 3.
How would these advantages differ from the advantages of using a long-term perspective in managing our resources?
Answer:
A long-term perspective would be more advantageous as it will ensure the fulfilment of needs of the present generation and conserve the resources for the benefit of the future generations too.

This approach will also be helpful in avoiding overexploitation of resources for short term gains.

Question 4.
Why do you think there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Answer:
Equitable distribution of resources is necessary so that benefits are also enjoyed by the poor people instead of being in hands of the rich and powerful people.

The social structure, the monetary gap and the social status work against equitable distribution of resources between the poor and the rich and powerful people.

In-text Questions (Page 273)

Question 1.
Why should we conserve forest and wildlife?
Answer:

• To preserve the biodiversity as its loss disturbs the ecological stability.
• Forest trees prevent floods and soil erosion.
• Various benefits like fuelwood, fruits, medicines, etc. are obtained by people from forests.

Question 2.
Suggest some approaches towards the conservation of forests.
Answer:

• Felling of trees indiscriminately should be banned.
• Creating awareness among people for forest conservation.
• Involving local people in forest conservation as was done in Arabari forests of West Bengal.
• Banning hunting of wild animals to protect wildlife.
• Encouraging planting more trees.

In-text Questions (Page 276)

Question 1.
Find out about the traditional systems of water harvesting/management in your region.
Answer:
The traditional systems of water harvesting are ponds, tanks, lakes and other water reservoirs which were called differently based on the regions. Example: Kulhs in Himachal Pradesh, Khadins, tanks and Nadis in Rajasthan, Kattas in Karnataka, etc. Baoli were used in Delhi region to store water in olden days.

Question 2.
Compare the above system with the probable systems in hilly/mountainous areas or plains or plateau regions.
Answer:
In hilly areas like Himachal Pradesh, the local system of canal irrigation called kulhs were developed in which water flowing in the streams was diverted to man-made channels. These channels took water to different villages down the hillside.

In level terrains like plains or plateau, crescent shaped earthen embankments or straight concrete and rubble ‘check dams’ are built to store running water.

Question 3.
Find out the source of water in your region/locality. Is water from this source available to all people living in that area?
Answer:
Water drawn from the rivers or tubewells is supplied by government through pipes to the homes of people in our locality. Only the areas approved by the local municipal bodies get the supply of water in this way.

Bihar Board Class 10 Science Chapter 16 Management of Natural Resources Textbook Questions and Answers

Question 1.
What changes would you suggest in your home in order to be environment-friendly?
Answer:
The following of 3 R’s is necessary to be environment-friendly in home, which are:
Reduce: Use less and not to waste any resource. For example, Water and electricity.
Recycle: Segregate the waste that can be recycled and make required things by using them. For example, Paper and plastic bottles.
Reuse: Use the things again and again. For example, Bottles of jams and pickles, plastic carry bags.

Question 2.
Can you suggest some changes in your school which would make it environment friendly?
Answer:

• Switch off the lights and fans when not necessary.
• Plant trees in the school campus.
• Recycle waste paper.
• Segregate wastes into biodegradable and non-biodegradable and put them in separate dustbins.

Question 3.
We saw in this chapter that there are four main stakeholders when it comes to forests and wildlife. Which among these should have the authority to decide the management of forest produce? Why do you think so?
Answer:
The stakeholders in the conservation of forests are:

• Local People: Those who are dependent on forests for their survival.
• Forest Department: Government who owns the land and controls resources
• Industrialists: Those who use various forests
• Wildlife enthusiasts: Those who want to conserve nature.

I am of the opinion that the local people should have the authority to decide the management of forest produce as they never overexploit the forest resources. They utilise them as per their needs and try to replenish them too. Their efforts can help in fulfilling and balancing the needs of the other three stakeholders in the conservation of forests.

Question 4.
How can you as an individual contribute or make a difference to the management of
(a) forests and wildlife, (b) water resources and (c) coal and petroleum?
Answer:
As an individual we can contribute or make a difference to the management of
(a) Forests and wildlife: By creating awareness among people for their conservation and planting more and more trees.
(b) Water resources: By turning off tap when not in use, repair leaking taps, use treated water for washing cars.
(c) Coal and petroleum: By using renewable sources of energy like solar energy, switching over to cleaner fuels like CNG, using solar heaters, using LED lights, less use of private vehicles.

Question 5.
What can you as an individual do to reduce your consumption of the various natural resources?
Answer:
As an individual we can reduce our consumption of the various natural resources by

• Taking a bus, public transport or walking/cycling instead of a private vehicle.
• Using LED bulbs in homes instead of normal bulbs.
• Taking the stairs instead of lift.
• Wearing an extra sweater on cold days instead of burning fossil fuels for warmth.

Question 6.
List five things you have done over the last one week to
(a) Conserve our natural resources.
(b) Increase the pressure on our natural resources.
Answer:
(a) Five things done to conserve our natural resources are:

• Switched off fans when not in use.
• Changed the lights in home to LED bulbs.
• Used public transport.
• Recycled waste paper.
• Reused the empty jam bottles to store kitchen grocery.

(b) Five things done to increase pressure on our natural resources are:

• Kept the lights switched on unnecessarily.
• Washed the car with water fit for drinking.
• Used car to travel even for short distances.
• Burnt paper wastes.
• Threw away old envelopes which could have been recycled.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your life-style in a move towards a sustainable use of our resources?
Answer:
The changes which I would incorporate in my life-style to move towards a sustainable use of our resources will be:

• Conserve water and electricity.
• Recycle the waste paper, plastic polybags, etc.
• Do water harvesting to recharge ground water level.
• Minimise use of fossil fuels and switch over to solar energy.
• Use public transport for travel instead of private vehicle.

Bihar Board Class 10 Science Chapter 16 Management of Natural Resources Textbook Activities

Activity 16.1 (Textbook page 266)

Euro norms refer to the permissible emission levels from both petrol and diesel vehicles, which have been implemented in Europe. However in India, the government has adopted the Euro norms for available fuel quality and the method of testing. Euro norms have been given a name of Bharat-I, Bharat-IV, etc. Refer-https://en.wikipedia.org/wiki/European_emission_standards https://en.wikipedia.org/wiki/Carbon_credit

The ways in which you can reduce the carbon emission levels are:
(i) Use public transport (ii) Conserve energy (iii) Manage resources in a better way (iv) Conservation of nature and natural resources (v) Creating awareness among people regarding need of conservation (vi) Plant more trees (vii) Reduce the carbon emissions from industries (viii) Strictly adhere to the pollution under control norms (ix) Use devices in industries to reduce carbon emissions (x) Proper planning and switching over to sources that emit less carbon

Activity 16.2 (Textbook page 266)

Refer-http://www.findouter.com/Asia/India/Science and Social Science/Environment/Environmental Organisations

We can contribute towards the same cause by actively participating in the activities of such organisations and creating awareness among the people of our locality and society.

Activity 16.3 (Textbook page 268)

Yes, the pollution levels of water can be indicated by the levels of pH of the water body. However, certain other variables like the Coliform count, Total Dissolved Solids and Conductivity of Suspended sediment etc. can supplement the data regarding the levels of pollution of the water.

Activity 16.4 (Textbook page 268)

The changes one would observe are that

• Forests cleared to use the land for agricultural fields.
• New houses have been constructed in place of mud huts.
• The non-metalled roads have replaced the metalled roads.

The destruction of forests to make agricultural fields should be minimised.

Activity 16.5 (Textbook page 270)

The forest produce which we use are- resins, gums, timber or wood, fruits, medicines, etc.
The persons living near a forest would use:

• timber-wood
• fruits
• spices
• wood to make paper
• tendu leaves for making bidis
• products of plantation of one species of plants
• resins and gums

The persons living in forest would use them for:

• fire-wood to be used as fuel
• fruits
• collection of honey
• fodder for cattle
• making implements for agriculture and construction
• medicines
• material for thatch roof of their huts

Activity 16.6 (Textbook page 271)

• Tendu leaves are used for making bidis
• Wood obtained from trees used in paper industry, furniture industry; wood from Salix species for making cricket bats etc.

No, these industries are not sustainable in the long run as they are being overexploited for human needs. So, we need to control our consumption of the products obtained from such trees.

Activity 16.7 (Textbook page 273)

(a) A lot of trees and vegetation is needed to be cut down for building rest houses for tourists in national parks. This is not good for the ecological balance of the area. Making such rest houses in National parks increases the levels of air pollution, noise pollution, etc. and the uncontrolled littering in such areas is a common sight these days which is very detrimental for the animals living in the area.

(b) Grazing animals in the national parks if carried out in an uncontrolled manner can lead to soil erosion as the grasses which bind the soil particles will not be there and so soil would be easily washed away during rains. The water holding capacity and ability to sustain the growth of larger trees will also be affected in absence of grass cover. Even the animals residing in the forest would be deprived of their food if large scale overgrazing is carried out.

(c) Tourists throwing plastic bottles/covers and other litter in national parks plays a havoc on the nature as these substances are non-biodegradable and they take several years to get completely degraded. The toxins which are released during their degradation further pollute the soil of the national parks. Plastics can choke the alimentary canals of the animals swallowing them and lead to their death. These substances also give an unpleasant look to the area due to their littering here and there.

Activity 16.8 (Textbook page 273)

The use of water resources merely for entertainment of few people when a large chunk of people from villages are suffering from chronic water shortage is very pathetic and shameful. Such a misuse of water should be stopped and the water should be diverted to the village for the use by the people during shortage of water. We need to manage our resources in a way that the distribution is equitable and useful to a large section of the society rather than catering to the needs of a few people. A big solution can be provided to the chronic water shortages in the village if rainwater is properly channelized, wastage of water is reduced, people are sensitised towards the need for conservation of water and the water resources should be managed in the most efficient way to fulfil the goals of sustainable development.

Activity 16.9 (Textbook page 273)

The maximum rainfall is received in the North east part of India and the western coastal regions of India. Mawsynram is the place where highest rainfall has been recorded till date. The area with least amount of rainfall are the Northwestern parts of India which include the states like Rajasthan, Haryana, parts of Gujarat and Punjab, so problems of water scarcity is more prevalent in these regions. The map showing the rainfall patterns in India is shown below:

Activity 16.10 (Textbook page 277)

The consumption of coal and petroleum can be reduced in the following ways:

• More use of public transport than personal vehicles.
• Conserving electricity by switching off fans and lights, when not in use.
• Use of solar cookers for cooking food in the areas which get good amount of sunlight.
• Use alternative sources of energy like the solar energy, wind energy etc.
• Use of biogas to generate electricity and to use as domestic fuel.
• The energy efficient devices like LED’s, higher Bureau of energy efficiency rating products etc.
• Save fuel by using bicycles or walking for covering short distances.
• An extra sweater can be worn in winters to save fuel/electricity used in a heating device.
• Less use of lavish lightning and decorations in weddings and public functions.
• Keeping the engines of vehicles well tuned and serviced to save fuel.

Activity 16.11 (Textbook page 278)

Euro norms refer to the permissible emission levels from both petrol and diesel vehicles, which have been implemented in Europe (EU-European Union). The Euro norms require manufacturers to reduce the existing polluting Emission Levels from the exhausts in a more efficient manner by making certain technical changes in the engines of the vehicles. The Euro-I norms was implemented from the year 1992-93. Euro-1 norms in India are known as India 2000 since it was implemented from 1/4/2000 in India. The changes required to be made by manufacturers for a euro I compliant car focussed primarily on: (a) catalyser capacity increase (b) tri-metal coating in the catalyser (c) exhaust gas recirculation (d) carburettor retuning (e) secondary air intake. The Euro-I and Euro-II norms are given in the table below:

Bihar Board Class 10 Science Solutions

Bihar Board Class 10 Science Solutions Chapter 15 Our Environment

Bihar Board Class 10 Science Chapter 15 Our Environment InText Questions and Answers

In-text Questions (Page 257)

Question 1.
Why are some substances biodegradable and some non-biodegradable?
Answer:
The bacterial and fungal enzymes degrade the complex organic compounds present in the dead remains of plants and animals into simpler substances, so such wastes are called biodegradable. However, some wastes like glass, plastics, etc. cannot be degraded by these enzymes due to which they are persistent in the environment for many years. These are called non-biodegradable substances.

Question 2.
Give any two ways in which biodegradable substances would affect the environment.
Answer:
The biodegradable wastes can affect the environment as:

• Their huge heaps act as breeding grounds for flies and mosquitoes which lead to spread of diseases like typhoid, tuberculosis, cholera, etc.
• During decomposition of these biodegradable wastes, foul smell is produced which makes people feel uncomfortable.

Question 3.
Give any two ways in which non-biodegradable substances would affect the environment.
Answer:
The non-biodegradable substances would affect the environment as:

• Some of the harmful pesticides and chemicals lead to loss of soil fertility and creates imbalance in the food chains.
• Their increased volumes in the garbage dumps due to non-decomposition create problems for many animals. For example, plastics eaten by cows can choke them to death.

In-text Questions (Page 261)

Question 1.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:
Each step or level in the food chain where transfer of energy takes place is called a trophic level. Example: Grass → Insect → Frog → Snake
First Trophic Level : Grass is Producer;
Second Trophic Level : Insect is Primary consumer;
Third Trophic Level : Frog is Secondary consumer;
Fourth Trophic Level : Snake is Tertiary consumer

Question 2.
What is the role of decomposers in the ecosystem?
Answer:
The complex organic substances present in the dead remains of plants and animals are broken down and converted into simpler inorganic substances by the decomposers. Thus they help in recycling of nutrients and cleansing the environment.

In-text Questions (Page 264)

Question 1.
What is ozone and how does it affect any ecosystem?
Answer:
Ozone (O₃) is a molecule formed by three atoms of oxygen. The ozone present in the lower parts of the atmosphere is a deadly poison and harmful to organisms. But, the ozone present in the higher levels of the atmosphere is good ozone as it shields the organisms against the harmful ultraviolet rays of the Sun (that cause skin cancer) by absorbing them. It is formed by the action of ultraviolet radiation on oxygen molecule to convert it into free oxygen atom. These free oxygen atoms combine with molecular oxygen to form ozone.
\(\mathrm{O}_{2} \stackrel{\mathrm{UV}}{\longrightarrow} \mathrm{O}+\mathrm{O}\)
\(\mathrm{O}+\mathrm{O}_{2} \longrightarrow \underset{(\mathrm{Ozone})}{\mathrm{O}_{3}}\)

Question 2.
How can you help in reducing the problem of waste disposal? Give any two methods.
Answer:
A collective effort by all the people can be very instrumental in reducing the problem of waste disposal. This can be done by:

• Recycling and reusing the wastes before disposing them off completely in the garbage bins.
• Segregate the various wastes as bio-degradable, non-biodegradable and hazardous wastes by using separate bins for each kind of waste.

Bihar Board Class 10 Science Chapter 15 Our Environment Textbook Questions and Answers

Question 1.
Which of the following groups contain only biodegradable items?
(a) Grass, flowers and leather
(b) Grass, wood and plastic
(c) Fruit-peels, cake and lime-juice
(d) Cake, wood and grass
Answer:
(a) Grass, flowers and leather

Question 2.
Which of the following constitute a food-chain?
(a) Grass, wheat and mango
(b) Grass, goat and human
(c) Goat, cow and elephant
(d) Grass, fish and goat
Answer:
(b) Grass, goat and human

Question 3.
Which of the following are environment-friendly practices?
(a) Carrying cloth-bags to put purchases in while shopping
(b) Switching off unnecessary lights and fans
(c) Walking to school instead of getting your mother to drop you on her scooter
(d) All of the above
Answer:
(d) All of the above

Question 4.
What will happen if we kill all the organisms in one trophic level?
Answer:
If we kill all the organisms in one trophic level, then it will cause imbalance in the ecosystem as transfer of food energy to the next level will stop. An overpopulation of the organisms in the previous trophic level will also take place. This will disturb the ecological balance.

Question 5.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:
The impact of removing all the organisms in a trophic level will be same as the impact for different trophic levels because it would lead to

• Starvation and death of organisms in the next trophic level as energy transfer to the next level would stop. For example, if all producers are killed, herbivores will be affected as no food will be available for them.
• Overpopulation in the previous level as no organism will be available in the next trophic level to feed on them.

Since all the organisms are interdependent on each other in an ecosystem, it will result in an imbalance in the ecosystem if organisms of any trophic level are removed.

Question 6.
What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:
The harmful chemicals like pesticides are neither metabolised nor excreted by the organisms. They get accumulated in the body of the organisms due to which their concentration goes on increasing in the successive trophic levels. This phenomenon is called biological magnification.

The level of this magnification goes on increasing as we move up the trophic levels. So, the concentration of these chemicals is highest in the human beings as they occupy the topmost trophic level.

Question 7.
What are the problems caused by the non-biodegradable wastes that we generate?
Answer:
The following problems are caused by the non-biodegradable wastes:

• They are not decomposed and get accumulated in the environment. They contaminate the soil and water resources.
• Their increased volumes in the garbage dumps due to non-decomposition create problems for many animals. Plastics eaten by cows or other stray animals can choke them to death.
• Their accumulation makes the land unfit for other purposes, makes it infertile.
• The rag pickers are exposed to harmful chemicals and toxins present in these kinds of wastes.
• They enter the food chain and become harmful to human beings and other animals.

Question 8.
If all the waste we generate is biodegradable, will this have no impact on the environment?
Answer:
Even the biodegradable wastes have an impact on the environment in the following ways:

• A foul smell is released during their decomposition. Also, some harmful gases are released which cause air pollution.
• The heaps of such biodegradable wastes act as a breeding ground for disease causing organisms like flies and mosquitoes.

Question 9.
Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Answer:
Ozone present in stratosphere acts as a shield which absorbs the harmful ultraviolet rays of the Sun. Damage to the ozone or its depletion will lead to non absorption of UV rays which cause skin cancer, ageing, cataract, etc. Damage to ozone is most prominent in Antarctica region where it’s visible as a ozone hole.

Steps to limit ozone damage:

• The use of substances called CFCs (Chlorofluorocarbons) as refrigerants and in fire extinguishers should be reduced and these should be replaced by environment friendly substances.
• UNEP (United Nations Environment Programme) has forged an agreement to freeze CFCs production at 1986 levels by all countries.

Bihar Board Class 10 Science Chapter 15 Our Environment Textbook Activities

Activity 15.1 (Textbook page 256)

The materials like metallic containers, plastic bottles, bubble packs, broken footwear, etc. remain unchanged for longer periods of time. The materials which change their form and structure overtime are the empty cartons, kitchen wastes, torn clothes etc.

The fastest change is seen in spoilt food, vegetable peels, used tea leaves.

Activity 15.2 (Textbook page 257)

The non-biodegradable substances last from a few years to several million years. Biodegradable plastics are plastics that can be decomposed by the action of living organisms like bacteria. Under proper conditions, some biodegradable plastics can degrade to the point where microorganisms can completely metabolise them. They do not harm the environment in the way the non-biodegradable plastics harm because they are decomposed at a faster rate by the microorganisms and once decomposed they are no more a problem.

Activity 15.3 (Textbook page 258)

An aquarium becomes a self sustaining system once the above essential materials are provided in it. The water, oxygen and food are the resources which help in the life processes of organisms present in the aquarium. The aerator keeps providing fresh supply of oxygen in it. The biotic and the abiotic components interact among themselves to sustain the artificial ecosystem. We cannot leave the aquarium as such for a longer time as the growth of microorganisms like bacteria and fungi occurs which need to be removed in order to avoid infection of the fishes. Lakes and ponds are large ecosystems which do not need to be cleaned so frequently. But sometimes they need to be cleaned in order to avoid siltation or excessive growth of aquatic plants. Example is the recent cleaning of the Pushkar lake in Ajmer.

Activity 15.4 (Textbook page 259)

Yes, we have to take care not to put an aquatic animal which would eat others. Otherwise the ecosystem will not remain self-sustaining and can lead to imbalances in the processes and the whole ecosystem (aquarium).

The food chain present in the ecosystem would also consist of primary producers, primary consumers, secondary consumers, etc. Primary producers will be the phytoplanktons and the algae. Primary consumers will be zooplanktons. Secondary consumers will be small fishes followed by the larger fishes as the tertiary consumers. These groups are dependent on each other as the energy produced by the organisms at one level gets transferred from the lower level to the higher level. The aquatic organisms in the chain would be will be:
Phytoplanktons (producers) → zooplanktons (primary consumers) → small fishes (secondary consumers) → large fishes(tertiary consumers)

The producers can be considered of primary importance in an ecosystem. But, an aquarium being an artificial ecosystem, the role of human being is more in maintaining the supply of nutrients, aeration, energy, etc. to keep the ecosystem self sustaining type.

Activity 15.5 (Textbook page 261)

There is a big debate going on regarding the use of chemical pesticides as most of them like DDT once taken in by animals through the food products they consume, can neither be excreted nor metabolised. It starts accumulating in the body of the organism and as we go higher up the food chain, the concentration of harmful toxicant reaches higher levels. This phenomenon is called as biological magnification which is harmful for the health of human beings and other organisms too. Source of pesticides in the food items are the pesticides sprayed on plants or crops by humans to increase their levels of agricultural production. Yes, pesticides can enter our body from this source through other food products too as the various organisms are dependent upon each other for their survival.

The methods which can be applied are

• use of bio-control agents to protect loss of crop
• development and use pest resistant crops
• practice of organic farming

Activity 15.6 (Textbook page 262)

The chemical which has been highly detrimental to the ozone layer is the use of chlorofluorocarbons in the refrigerator, deo-sprays, etc. The harmful effect of these chemical was highlighted in 1985 when the ozone hole was detected in Antarctica. The various countries of the world met and signed the Montreal protocol in 1987 which laid down the roadmap to control the emission of ozone depleting substances.

Activity 15.7 (Textbook page 262)

The wastes generated in our homes are segregated in the various bins designated for keeping biodegradable, recyclable, non-biodegradable and hazardous wastes separately. The scavengers take away the wastes from the homes and dispose it off properly at the places meant for them so that they do not cause harm to other living organisms. The resident welfare association have designated scavengers who take the wastes away from the home. The recyclable materials are either recycled or sent to units which recycle the wastes for various other processes. The non-biodegradable and hazardous substances are disposed off by sending them to the landfill sites or by incineration.

Activity 15.8 (Textbook page 263)

Approximately 2-3 kg of waste is generated at home in a day. Approximately half of it is biodegradable. The waste generated per day in a classroom is approximately 1 kg. Almost all the waste generated in the classroom is biodegradable. The biodegradable waste can be put in compost pits and converted to compost or the recyclable ones can be recycled.

Activity 15.9 (Textbook page 263)

The sewage is treated in the sewage treatment plants. The plans are being made to ensure that no untreated sewage is discarded in the local water bodies. The local industries send their wastes to the sewage treatment plants where the harmful chemical and constituents are removed from it, before discharging it into the water bodies. Though some mechanisms are there to prevent soil and water from getting polluted by such wastes, other advanced systems are being implemented to control such pollution.

Activity 15.10 (Textbook page 263)

The electronic items thrown as e-wastes contain many harmful and hazardous substances like lead, cadmium, chromium, brominated flame retardants or polychlorinated biphenyls (PCBs).
Refer- http://ewasteguide.info/hazardous-substances
http://listverse.com/2013/01/27/10-ways-recycling-hurts-the-environment/http://www.livestrong.com article/181103-the-effect-of-recycling-plastic-water-bottles-on-the-environment/

Bihar Board Class 10 Science Solutions Chapter 14 Sources of Energy

Bihar Board Class 10 Science Chapter 14 Sources of Energy InText Questions and Answers

In-text Questions (Page 243)

Question 1.
What is a good source of energy?
Answer:
A good source of energy should have the following qualities:

• does a large amount of work per unit mass.
• cheap and easily available.
• easy to store and transport.
• safe to handle and use.
• does not cause environmental pollution.

Question 2.
What is a good fuel
Answer:
A good fuel should have the following qualities:

• high calorific value (give more heat per unit mass).
• bums without giving out any smoke or harmful gases.
• proper ignition temperature.
• cheap and easily available.
• convenient to store.
• bums smoothly.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
I will use LPG (Liquefied Petroleum Gas) for heating my food; because it is easily available in my kitchen, gives no smoke on burning and takes less time to do the job.

In-text Questions (Page 248)

Question 1.
What are the disadvantages of fossil fuels?
Answer:
Disadvantages of fossil fuels:

• They are non-renewable.
• They produce air pollutants on burning.

Question 2.
Why are we looking at alternate sources of energy?
Answer:
Because of growing population, the energy demand is rising. Fossil fuels are going to be exhausted in the near future and burning them is causing pollution. Hence, we need to find an alternate source of energy which is renewable and environment friendly.

Question 3.
How has the traditional use of wind and water energy been modified for our convenience?
Answer:
To use energy of flowing water, large dams are built to store huge amount of water at a height. The stored water is allowed to fall from a height which runs the turbine at the bottom of the dam to generate electricity. Similarly, the wind energy is used to generate electricity. For this purpose, the rotatory motion of windmill is used to turn the turbine of the electric generator.

In-text Questions (Page 253)

Question 1.
What kind of mirror; concave or convex or plain would be best suited for used in a solar cooker? Why?
Answer:
A solar cooker uses heat of the sunlight to cook food. A concave mirror is used in order to reflect and focus sunlight at a particular area. The mirror focuses all the incident sunlight at a point. The temperature at the point increases, thereby cooking and heating the food placed in that particular area.

Question 2.
What are the limitations of the energy that can be obtained from the oceans?
Answer:
Limitations of the energy obtained from the oceans are

• There are very few locations where dams can be built to utilise tidal energy.
• Cost of installation of power houses is extremely high and efficiency of plants is comparatively small.
• Power plants built in oceans or at sea-shores will need high continuous maintenance as chances of corrosion are extremely high.

Question 3.
What is geothermal energy?
Answer:
The energy which can be harnessed from the heat inside the earth is called geothermal energy.

Question 4.
What are the advantages of nuclear energy?
Answer:
Advantages of nuclear energy:

• A small amount of radioactive material can generate a huge amount of energy.
• It does not produce air pollution.
• A nuclear power plant is more efficient than other power plants.

In-text Questions (Page 253)

Question 1.
Can any source of energy be pollution free? Why or why not?
Answer:
No sources of energy can be pollution free. This is because operation of a device like in a solar cell and some other sources of energy may be pollution free but their installation may cause pollution.

Question 2.
Hydrogen has been used as a rocket fuel. Would you consider it a cleaner fuel than CNG? Why or why not?
Answer:
Hydrogen is a much cleaner energy source than CNG as it produces water vapour on burning. CNG (compressed natural gas) is derived from biomass and hence burning the CNG causes air pollution.

In-text Questions (Page 254)

Question 1.
Name two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:
Biomass and solar energy because these sources can be easily replenished.

Question 2.
Give the names of two energy sources that you would consider to be exhaustible. Give reasons for your choices.
Answer:
Coal and petroleum are two exhaustible sources of energy. These fuels were formed over millions of years ago and there are only limited reserves. If we continue to use them as at present rate, then these reserves will be exhausted very soon.

Bihar Board Class 10 Science Chapter 14 Sources of Energy Textbook Questions and Answers

Question 1.
A solar water heater cannot be used to get hot water on:
(а) sunny day
(b) a cloudy day
(c) hot day
(d) a windy day
Answer:
(b) a cloudy day

Question 2.
Which of the following is not an example of a biomass energy source?
(a) wood
(b) gobar-gas
(c) nuclear energy
(d) coal
Answer:
(c) nuclear energy

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the Sun’s energy?
(a) geothermal energy
(b) wind energy
(c) nuclear energy
(d) biomass
Answer:
(c) nuclear energy

Question 4.
Compare and contrast fossil fuels and the sun as direct sources of energy.
Answer:
Fossil fuels:

1. The reserves of fossil fuels are limited i.e., exhaustible.
2. Burning of fossil fuels cause pollution.
3. Fossil fuels can provide energy at any required time.

Solar energy:

1. Solar energy is available in abundance i.e., inexhaustible.
2. Solar energy is pollution free.
3. Solar energy is available during daytime only and when there in no cloud.

Question 5.
Compare and contrast biomass and Hydroelectricity as direct sources of energy.
Answer:
Biomass:

1. The energy from biomass can be obtained by using a challah or a gobar gas plant.
2. Biomass is pollution free when only biogas is used.

Hydroelectricity:

1. Hydroelectricity requires construction of dam.
2. Hydroelectricity is totally pollution free.

Question 6.
What are the limitations of extracting energy from: (a) the wind (b) waves (c) tides?
Answer:
(a) Limitation of wind energy

• Wind speed is not uniform always.
• Big amount of investment is needed.
• Need a large area for plant installation.

(b) Limitation of wave energy

• Power output is variable in nature.
• Wave energy generation is expensive.

(c) Limitation of tidal energy

• Power output is variable.
• Power generation is intermittent.

Question 7.
On what basis would you classify energy sources as
(a) renewable and non-renewable?
(b) exhaustible and inexhaustible?
Are the options given in (a) and (b) the same?
Answer:
Yes, options are same.
Renewable and inexhaustible sources of energy are similar to each other while non-renewable and exhaustible sources are similar.

Renewable sources of energy are those which are inexhaustible i.e., which can be replaced as we use them and can be used to produce energy again and again, Examples: (a) solar energy (b) wind energy

Non-renewable sources of energy are those which are exhaustible and cannot be replaced once they have been used.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
Qualities of an ideal source of energy:

• Has high calorific value
• Produces less smoke
• Produces less residue after burning
• Easy availability
• Inexpensive
• Easy to store and transport

Question 9.
What are the advantages and disadvantages of using a solar cooker? Are there places where solar cookers would have limited utility?
Answer:
Advantages of solar cookers:

• Ecofriendly
• Renewable
• Used in rural areas
• Retains all the nutrients in food due to slow cooking

Disadvantages of solar cooker:

• Silicon cells are expensive.
• Solar radiations are not uniform over the earth’s surface.
• Cannot be used at night or on cloudy days.
• Cannot be used to make chapattis for frying as these require a temperature of 140°C or more. (Maximum temperature of 100°C only can be achieved in a solar cooker)

Question 10.
What are the environmental consequences of the increasing demand for energy? What steps would you suggest to reduce energy consumption?
Answer:
Environmental consequences:

1. Air pollution is increasing due to burning of fossil fuels.
2. Construction of dam destroys large ecosystem.

Steps to reduce energy consumption:

1. Use energy efficient devices.
2. Extract maximum possible energy from fossil fuels.

Bihar Board Class 10 Science Chapter 14 Sources of Energy Textbook Activities

Activity 14.3 (Textbook Page 244-245)

Discussion and conclusion:
The tennis balls acts like a turbine and starts rotating. It transfer its rotational energy to the dynamo, which converts it rotational energy to electricity. The electric current from the dynamo passes through the bulb and lights it up.

Activity 14.5 (Textbook Page 249)

Discussion and conclusion:
The conical flask which is painted black is hotter than the one painted white. Black colour absorbs more heat.

Bihar Board Class 10 Science Solutions Chapter 13 Magnetic Effects of Electric Current

Bihar Board Class 10 Science Chapter 13 Magnetic Effects of Electric Current InText Questions and Answers

In-text Questions (Page 224)

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
Magnetic compass needle and bar magnet both have magnetic field. When they are brought near each other, these magnetic fields interact with each other showing deflection in the needle.

In-text Questions (Page 228)

Question 1.
Draw magnetic field lines around a bar magnet.
Answer:

Question 2.
List the properties of magnetic lines of force.
Answer:
Properties of magnetic lines of force are listed below:

• Outside the magnet, the magnetic field lines are directed from N-pole of magnet towards S-pole. However, inside a magnet the field lines are directed from S-pole to N-pole. Thus magnetic field lines from a closed loop.
• The magnetic field line at any point, points in the direction of magnetic field at that point.
• The relative strength of magnetic fields is given by degree of closeness of the field lines. The magnetic field is strong in the region where the field lines are crowded.
• No two magnetic field lines can ever intersect with each other.

Question 3.
Why don’t two magnetic lines of force intersect each other?
Answer:
Magnetic field-lines do not cross each other. If they did, it would mean that at the point of intersection, the compass needle would point towards two directions, which is not possible.

In-text Questions (Page 229-230)

Question 1.
Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
The magnetic field lines have been shown in figure given below. As per right-hand rule, we find that inside the loop, the magnetic field lines are directed perpendicular to the plane of paper in the inward direction. Outside the loop, magnetic field lines are directed out of the plane of paper.

Magnetic field lines of the field produced by a current carrying circular loop

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
The uniform magnetic field is represented by parallel, equidistant lines of equal length as shown in figure.

Question 3.
Choose the correct option. The magnetic field inside a long straight solenoid-carrying current (a) is zero. (b) decreases as we move Uniform Magnetic Field Lines towards its end. (c) increases as we move towards its end. (d) is the same at all points.
Answer:
(d) is the same at all points.

In-text Questions (Page 231-232)

Question 1.
Which of the following property of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.) (a) Mass (6) Speed (c) Velocity (d) Momentum.
Answer:
(c) Velocity and (d) Momentum.

Question 2.
In Activity 13.7, how do we think the displacement of rod AB will be affected if (i) current in rod AB is increased; (ii) a stronger horse-shoe magnet is used; and (iii) length of the rod AB is increased?
Answer:
(i) If current in rod AB is increased, the displacement will also increase, (ii) If we use a stronger horse-shoe magnet then the displacement of rod AB will increase, (iii) If length of the rod is increased, force acting on it will increase and, hence, displacement of the rod will increase.

Question 3.
A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is (a) towards south (b) towards east (c) downward (d) upward.
Answer:
(d) upward in accordance with Fleming’s left-hand rule, the direction of magnetic field is vertically upward.

In-text Questions (Page 233)

Question 1.
State Fleming’s left-hand rule.
Answer:
According to Fleming’s left-hand rule, stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular. If the first finger points in the direction of magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.

Question 2.
What is the principle of an electric motor?
Answer:
The working principle of an electric motor is based on the magnetic effect of current. A current-carrying loop experiences a force and rotates when placed in a magnetic field. The direction of rotation of the loop is given by the Fleming’s left-hand rule.

Question 3.
What is the role of the split ring in an electric motor?
The split ring in the electric motor acts as a commutator. The commutator reverses the direction of current flowing through the coil after each half rotation of the coil. Due to this reversal of the current, the coil continues to rotate in the same direction.

In-text Questions (Page 236)

Question 1.
Explain different ways to induce current in a coil.
Answer:
Different ways to induce current in a coil are as given below:

• If a magnetic field is changed around a coil then an induced current is set up in the coil. It can be done by taking a bar magnet and bringing it closer to the coil or taking it away from the coil.
• If a coil is moved in a magnetic field, then again an induced current is set up in the coil.
• If a coil is rotated in a uniform magnetic field, it may also produce an induced current in the coil.

In-text Questions (Page 237)

Question 1.
State the principle of an electric generator.
Answer:
An electric generator is based on the principle of electromagnetic induction. When a rectangular coil is rotated in a uniform magnetic field, an induced voltage is generated between the ends of the coil.

Question 2.
Name some sources of direct current.
Answer:
Some sources of direct current are a cell, a battery and a D.C. generator.

Question 3.
Which sources produce alternating current?
Answer:
A.C. generator and invertors (used in house for emergency power supply) produces alternating current.

Question 4.
Choose the correct option.
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(a) two revolutions
(b) one revolution
(c) half revolution
(d) one-fourth revolution
Answer:
(c) When a rectangular coil of copper wire is rotated in a magnetic field, the direction of the induced current changes once in each half revolution.

In-text Questions (Page Page 238)

Question 1.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Two safety measures are:

1. Use of earth wire and proper earthing.
2. Use of fuse (now-a-days fuse wire is replaced by MCB).

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.
Answer:
Power rating of electric oven P = 2 kW = 2000 W Supply voltage V = 220 V
So, the current drawn by the electric oven I = P/V = 2000 W/220 V = 9A

As the current rating of domestic electric circuit is only 5 A and the oven draws a current 9 A, which is more than the current rating, hence the circuit will be damaged due to overheating/overloading.

Question 3.
What precaution should be taken to avoid the overloading of domestic electric circuits?
Answer:
The precautions that should be taken to avoid the overloading of domestic circuits are as follows:

• Too many appliances should not be connected to a single socket.
• Too many appliances should not be used at the same time.
• Faulty appliances should not be connected in the circuit.
• Fuse should be connected in the circuit.

Bihar Board Class 10 Science Chapter 13 Magnetic Effects of Electric Current Textbook Questions and Answers

Question 1.
Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centred on the wire.
Answer:
(d) The field consists of concentric circles centred on the wire.

Question 2.
The phenomenon of electromagnetic induction is
(а) the process of charging a body.
(b) the process of generating magnetic field due to a current passing through a coil.
(c) producing induced current in a coil due to relative motion between a magnet and the coil.
(d) the process of rotating a coil of an electric motor.
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil.

Question 3.
The device used for producing electric current is called a
(a) generator
(b) galvanometer
(c) ammeter
(d) motor.
Answer:
(a) generator

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electromagnet while a DC generator has permanent magnet
(b) DC generator will generate a higher voltage
(c) AC generator will generate a higher voltage
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(d) AC generator has slip rings while the DC generator has a commutator.

Question 5.
At the time of short circuit, the current in the circuit
(a) reduces substantially
(b) does not change
(c) increases heavily
(d) vary continuously.
Answer:
(c) increases heavily

Question 6.
State whether the following statements are true or false.
(а) An electric motor converts mechanical energy into electrical energy.
(b) An electric generator works on the principle of electromagnetic induction.
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines.
(d) A wire with a green insulation is usually the live wire of an electric supply.
Answer:
(a) False
(b) True
(c) True
(d) False.

Question 7.
List two methods of producing magnetic fields.
Answer:

1. Passing electric current through a straight conductor
2. Passing electric current through a circular loop
3. Passing electric current through a solenoid

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.
Answer:
The pattern of the magnetic field around a current-carrying solenoid is same as that of a bar magnet. Yes, we can determine the poles of a solenoid by placing a bar magnet near it.

If north pole is placed near side B and repulsion is felt them B is North pole and vice versa.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
Answer:
When the direction of current is perpendicular to the direction of magnetic field in a current-carrying conductor.

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:
The direction of current (I) is opposite to the direction of the electron beam. Since the electron beam is deflected to the right side, applying, Fleming’s left hand rule, it is found that the direction of the magnetic field is vertically downwards.

Question 11.
Draw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a spilt ring in an electric motor?
Answer:
An electric motor is a rotating device that converts electrical energy to mechanical energy.
Principle: It works on the principle that when electric current is passed through a conductor placed perpendicularly in a magnetic field, a force acts on it as a result of which the conductor begins to move. An electrical motor, is shown in the figure given on the next page.

Current in the coil ABCD enters from the source battery through conducting brush X and flows back to the battery through brush Y. The current in the arm AB of the coil flows from A to B. In arm CD it flows from C to D, that is, opposite to that through arm AB. On applying Fleming’s left hand rule for the direction of force on a current-carrying conductor in a magnetic field, we find that the force acting on arm AB pushes it downwards while the force acting on arm CD pushes it upwards. Thus the coil and the axle O rotate anti-clockwise. At half rotation, Q makes contact with the brush X and P with brush Y. Therefore the current in the coil gets reversed and flows along the path DCBA. The reversal of current also reverses the direction of force acting on the two arms AB and CD. Thus the arm AB of the coil that was earlier pushed down is now pushed up and the arm CD previously pushed up is now pushed down. Therefore the coil and the axle rotate half a turn more in the same direction. The reversing of the current is repeated at each half rotation, giving rise to a continuous rotation of the coil and to the axle.
The split ring reverses the direction of flow of current in an electric motor.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric fans, refrigerators, mixers, washing machines, computers and CD players.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is (i) pushed into the coil, (ii) withdrawn from inside the coil, (iii) held stationary inside the coil?
Answer:
(i) The needle of the galvanometers moves momentarily in one direction
(ii) The needle of the galvanometers moves momentarily but in the opposite direction to (i)
(iii) No deflection of needle.

Question 14.
Two circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
By changing current in the coil A, a current will be induced in coil B. This is because magnetic field associated with A changes when current in coil A is changed. This changes the magnetic field lines associated with coil B and thus a current is induced in it.

Question 15.
State the rule to determine the direction of a (i) magnetic field produced around a straight conductor-carrying current, (ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and (iii) current induced in a coil due to its rotation in a magnetic field.
Answer:
(i) Imagine that you are holding a current-carrying straight conductor in your right hand such that thumb points towards the direction of current, then your fingers will wrap around the conductor in the direction of the field lines of the magnetic field, as shown in the figure. This is known as the right hand thumb rule.

(ii) Fleming’s left-hand rule determines the force on a conductor: According to this rule, stretch the thumb, forefinger and middle finger of your left hand such that they are mutually perpendicular.

If the fore-finger points in the direction of the magnetic field and the middle-finger in the direction of current, then the thumb will point in the direction of motion or force acting on the conductor.

(iii) Fleming’s right hand rule determines the direction of the induced current: Stretch the thumb, forefinger and middle finger of the right hand so that they are perpendicular to each other, as shown in figure. If the forefinger indicates the direction of the magnetic field and the thumb shows the direction of motion of the conductor, then the middle finger will show the direction of induced current. This simple rule is called Fleming’s right-hand rule.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes?
Answer:
In an electric generator, mechanical energy is used to rotate a conductor in a magnetic field to produce electricity.
Principle: It works on the principle that when magnetic lines of force linked with a coil changes, a current is induced in the coil.

An electric generator is shown in the figure below.

When the coil ABCD is rotated clockwise, by applying Fleming’s right-hand rule, induced currents are set up in the these arms along the directions AB and CD. Thus an induced current flows in the direction ABCD. If there are larger number of turns in the coil, the current generated in each turn adds up to give a large current through the coil.

After half a rotation, arm CD starts moving up and AB moving down. As a result, the direction of the induced currents in both the arms change, giving rise to the net induced current in the direction DCBA. The current in the external circuit now flows from B₁ and B₂. Thus after every half rotation the polarity of the currnet in the respective arms changes. Such a current, which changes direction after equal intervals of time, is called an alternating current (abbreviated as AC). This device is called an AC generator.

The function of the brushes is to make contact with the rotating rings and through them supply current to the external circuit.

Question 17.
When does an electric short circuit occur?
Answer:
Electric short-circuit occurs when live wire comes in direct contact with the neutral wire due to some fault in an appliance or damaged insulation of the wires.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
It is used as a safety measure to ensure that any leakage of current to a metallic body does not give severe shock to the user.

Earth wire provides a low resistance conducting path for electric current to the earth which may have leaked from an appliance.

Bihar Board Class 10 Science Chapter 13 Magnetic Effects of Electric Current Textbook Activities

Activity 13.1 (NCERT Textbook Page 223)

Discussion and conclusion:
The needle of the compass is deflected when electric current is passed through it since electric current produces a magnetic field.

Activity 13.2 (NCERT Textbook Page 224)

Discussion and conclusion:
The iron filings arrange themselves in a pattern. The magnet exerts its influence in the region surrounding it. This region surrounding the magnet is called magnetic field.

Activity 13.4 (NCERT Textbook Page 226)

Discussion and conclusion:
The direction of magnetic field due to a current carrying straight conductor depends upon the direction of current flowing through it. If we reverse the direction of current the direction of magnetic field is also reversed.

Activity 13.7 (NCERT Textbook Page 230)

Discussion and conclusion:
Current carrying wire placed in magnetic field experience force. On reversing the direction of current, the direction of force is reversed.

Activity 13.8 (NCERT Textbook Page 233-234)

Discussion and conclusion:

• An emf and current is induced in the circuit when magnetic field linked with the coil changes.
• When the magnet is stationary, magnetic field linked with the coil does not change. Hence there is no induced emf and current in the coil.
• The direction of induced current depends on the direction of motion of the magnet.

Activity 13.9 (NCERT Textbook Page 235)

Discussion and conclusion:

• Whenever electric current through coil-1 is changed (either increased or decreased), a pd is induced in coil-2.
• The coil-1 (containing battery) is called the primary coil whereas coil-2 is called the secondary coil.
• The induced pd in coil-2 is due to change in magnetic field lines in it which in turn is due to change in magnetic field lines in coil-1 through which current is changed.

Bihar Board Class 10 Science Solutions Chapter 11 Human Eye and Colourful World

Bihar Board Class 10 Science Chapter 11 Human Eye and Colourful World InText Questions and Answers

In-text Questions (Page 190)

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye lens to adjust its focal length is called power of accommodation.

Question 2.
A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
Person should use concave lens.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
Near point of normal human eye is 25 cm and far point is infinity.

Question 4.
A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
The student is suffering from myopia. It can be corrected by using concave lens of suitable focal length.

Bihar Board Class 10 Science Chapter 11 Human Eye and Colourful World Textbook Questions and Answers

Question 1.
The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia
(b) accommodation
(c) near-sightedness
(d) far-sightedness
Answer:
(b) accommodation

Question 2.
The human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina
Answer:

Question 3.
The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m
Answer:
(c) 25 cm

Question 4.
The change in focal length of an eye lens is caused by the action of the
(a) pupil
(b) retina
(c) ciliary muscles
(d) iris
Answer:
(c) ciliary muscles

Question 5.
A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?
Answer:
(i) For distant vision P = -5.5D
f = \(\frac{100}{P}=\frac{100}{5.5}\) = -18.2 cm

(ii) For near vision P = 1.5 D
f = \(\frac{100}{P}=\frac{100}{1.5}\) = 66.7 cm

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer:
Distance of far point, x = 80 cm, P = ?
The lens will form image at 80 cm of the object lying at ∞
Here, υ = – 80 cm, u = ∞ cm and \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answer:
For diagram Refer Q No. 1 (Long Answer Questions)
Distance of near point = 1 m = 100 cm
Here u = – 1 m = – 100 cm, υ = – 25 cm

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
This is because the focal length of the eye lens cannot be decreased below a certain minimum limit i.e., 25 cm.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
The image distance in the eye is fixed and equal to the distance of the retina from the eye lens. When the object distance increases focal length of the eye lens increases.

Question 10.
Why do stars twinkle?
Answer:
The twinkling of a star is due to atmospheric refraction of starlight. The starlight, on entering the earth’s atmosphere undergoes refraction continuously before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. Since the physical conditions of the earth’s atmosphere are not stationary. Apparent position changes and star appears twinkling.

Question 11.
Explain why the planets do not twinkle.
Answer:
The planets are much closer to the earth, and are thus seen as extended sources. If we consider a planet as a collection of a large number of point-sized sources of light, the total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying the twinkling effect.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
In the morning, the Sun lies near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelength. This gives rise to the reddish appearance of the Sun.

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
This is because at a large height, there is no atmosphere and hence no refraction. Therefore the sky appears dark.

Bihar Board Class 10 Science Chapter 11 Human Eye and Colourful World Textbook Activities

Activity 11.2 (NCERT Textbook Page 193)

Discussion and Conclusion:
We observe a beautiful band of seven colours on a screen. This is called visible spectrum. The activity shows that white light is composed of seven colours. The phenomenon of splitting of white light into seven colours on passing through a glass prism is called dispersion of light.

Activity 11.3 (NCERT Textbook Page 196)

Discussion and Conclusion:
In the arrangement we find that fine microscopic sulphur particles precipitate in water in a couple of minutes. Light gets scattered from the minute sulphur particles and we observe blue colour from the three sides of the tank. From the fourth side of the tank facing the hole in a cardboard, we observe transmitted light. As blue colour has been scattered, we observe first the orange red colour and then bright crimson red colour on the screen.

Bihar Board Class 10 Science Solutions Chapter 10 Light Reflection and Refraction

Bihar Board Class 10 Science Chapter 10 Light Reflection and Refraction InText Questions and Answers

In-text Questions (Page 168)

Question 1.
Define the principal focus of a concave mirror.
Answer:
A point on the principal axis of a spherical mirror where the rays of light parallel to the principal axis meet after reflection from the spherical mirror is called the principal focus.

Question 2.
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
Focal Length f = \(\frac{\mathrm{R}}{2}\)
Here, R = 20 cm
therefore, f = 10 cm

Question 3.
Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave mirror

Question 4.
Why do we prefer a convex mirror as a rearview mirror in vehicles?
Answer:
A convex mirror always gives an erect though diminished image for an object placed at any distance from it. Moreover, it gives a wider field view so that the driver can see a lot more traffic behind him than that possible by a plane mirror. This enables him to drive safely.

In-text Questions (Page 171)

Question 1.
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Given, Radius of curvature, R = 32 cm
Focal length f = \(\frac{\mathrm{R}}{2}\)
∴ f = \(\frac{32}{2}\) = 16 m
Hence, focal length is 16 m.

Question 2.
A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Answer:
Given, m = -3, u = -10 cm
Magnification, m = \(\frac{-v}{u}\) ⇒ -3 = \(\frac{-v}{-10}\)
υ = -30 cm
Image is located 30 cm in front of mirror.

In-text Questions (Page 176)

Question 1.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
The ray of light will bend towards the normal as it is entering in denser medium (water) from rarer medium (air).

Question 2.
Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ ms^-1.
Answer:
Given, n_g = 1.50, e = 3 × 10⁸

Speed of light in glass is 2 × 10⁸

Question 3.
Find out, from table 10.3 (NCERT Textbook page 175), the medium having higher optical density. Also find the medium with lowest optical density.
Answer:
Minimum optical density: Air
Maximum optical density: Diamond

Question 4.
You are given kerosene, turpentine and water. In which of these does the light travels fastest? Use the information given in table 10.3.
Answer:
Light travels fastest in a medium having minimum optical density or refractive index. Since water has the least refractive index, the speed of light is fastest in water.

Question 5.
The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
According to this statement light travels 2.42 times slower in diamond as compared to vacuum or air.

In-text Questions (Page 184)

Question 1.
Define 1 dioptre of power of a lens.
Answer:
1 dioptre is the power of a lens whose focal length is 1 metre.

Question 2.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also find out the power of the lens.
Answer:
Here υ = 50 cm, m = -1
Magnification m = \(\frac{v}{u}\) ⇒ -1 = \(\frac{(50)}{u}\)
∴ u = -50 cm
Object is placed 50 cm in front of the mirror.
For power

Question 3.
Find the power of concave lens of focal length 2 m.
Answer:
Here, f= 2 m
p = \(\frac{1}{f}\) ⇒ P = \(\frac {1}{2}\) = 0.5D
Power of lens is 0.5D.

Bihar Board Class 10 Science Chapter 10 Light Reflection and Refraction Textbook Questions and Answers

Question 1.
Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay cannot be used to make a lens as it is not transparent.

Question 2.
The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principal focus.
Answer:
(d) Between the pole of the mirror and its principal focus.

Question 3.
Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(а) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer:
(b) At twice the focal length

Question 4.
A spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be
(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex, but the lens is concave.
Answer:
(a) both concave.

Question 5.
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane
(b) concave
(c) convex
(d) either plane or convex.
Answer:
(d) either plane or convex.

Question 6.
Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50 cm.
(c) A convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Answer:
(a) A convex lens of focal length 5 cm.

Question 7.
We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Answer:

Object should be placed between pole and principal focus of the mirror. Range of distance is less than 15 cm. The image will be virtual, erect, large sized and behind the mirror.

Question 8.
Name the type of mirror used in the following situations:
(a) Headlights of a car
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace.
Support your answer with reason.
Answer:
(a) Concave mirror: for getting powerful beam of light.
(b) Convex mirror: for getting erect and small sized image.
(c) Concave mirror: for concentrating sunlight to produce heat.

Question 9.
One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes, it will produce a complete image of the object. But the image will be less bright as some amount of light will be blocked by the black paper.

Question 10.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
u = -25 cm, f = 10 cm, h = 5 cm


Image is real and inverted.

Question 11.
A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
Here, υ = -10 cm, f = 15 cm, u = ?

Question 12.
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
Here, u = -10 cm, f = 15 cm, υ = ?

Question 13.
The magnification produced by a plane mirror is +1. What does this mean?
Answer:
m = \(\frac{h^{\prime}}{h}\) = 1
h’ = h
Image produced by plane mirror is same sized, virtual and erect.

Question 14.
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
Height of object (h) = 5 cm
u = -20 cm; f = +15 cm, R = +30 cm

∴ h’ = \(\frac{15}{7}\) = 2.14 cm
Since h’ is positive, and m < 1 image is virtual, erect and diminished.

Question 15.
An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Answer:
Size of object (h) = 7 cm, u = -21 cm, f = -18 cm

At a distance of 54 cm sharp image can be obtained.

Image is real, inverted and enlarged.

Question 16.
Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
f = \(\frac{1}{\mathrm{P}}\), f = \(\frac{1}{-2}\) = 0.5 cm, f = -50 cm
Focal length is negative, lens is concave.

Question 17.
A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
f = \(\frac{1}{\mathrm{P}}\), f = \(\frac{1}{1.5}\) = = 0.67 m = + 67 cm
Focal length is positive, lens is converging.

Bihar Board Class 10 Science Chapter 10 Light Reflection and Refraction Textbook Activities

Activity 10.1 (Textbook page 161)

Discussion and Conclusion:
The inner curved surface of a large shining spoon acts as a concave mirror. Our face is the object. When the spoon is at a small distance from the mirror, the object lies between pole P and focus F of concave mirror. We observe enlarged, erect image of our face behind the spoon. When we move the spoon slowly away from our face, we observe that

• When face is at focus of mirror/spoon, highly enlarged, inverted image of our face is seen at infinity.
• When face is between F and C of mirror/spoon, an enlarged, inverted image of our faces is seen.
• When face is at the centre of curvature of the mirror/spoon, an inverted image of same size is seen.
• When face is beyond C, an inverted and diminished image of our face is seen.
• When the face is very far off, highly diminished, inverted image of our face is seen at the focus of spoon/mirror.

When we reverse the spoon, its curved portion is bulged out. It behaves as a convex mirror. The image of our face is virtual, erect and diminished. As we move the spoon away from our face, image moves away till it is at the focus of the mirror. The image, however, continues to be virtual, erect and finally, it is diminished to almost point size.

Activity 10.2 (Textbook page 162)

Discussion and Conclusion:
The source of light, the sun in this case is at infinite distance from the concave mirror. Therefore, on reflection from the concave mirror, the sunlight collects at the focus of the mirror. By moving the sheet of paper back and forth gradually, we find a bright, sharp spot of light on the paper. This spot is the real, point size image of the Sun at the focus of the concave mirror (see figure).

When we hold the mirror and the paper in the same position for a few minutes, the paper may burn out. This is because the Sun rays are concentrated on the bright, sharp spot of light on the paper.

Activity 10.5 (Textbook page 167)

Discussion and Conclusion:
When we hold a pencil in the upright position in front of a convex mirror, we observe the image of the pencil in the mirror, at the back of the mirror. The image is erect, virtual and smaller in size than the object.

As the pencil is moved away from the mirror, the image becomes smaller and smaller, moving away from the mirror.

On repeating the activity, we find that as the object is moved away from the mirror, the image would move closer to the focus of the mirror.

Activity 10.6 (Textbook page 167)

Discussion and Conclusion:
When we observe the image of a distant object, say a distant tree, in a plane mirror, we may not see a full length image. When we try with plane mirrors of different sizes, we find that the image of the entire object is seen when size of plane mirror is atleast half the size of the object.

When we repeat this activity with a concave mirror, we do see full length image of the object for certain locations of the object.

Again, when we use a convex mirror, the full length image of an object is seen, wherever the object may be located. The reasons behind these observations are that

• in a plane mirror, size of image is always equal to the size of the object,
• in a concave mirror, the image is equal in size or smaller in size to the size of the object only for certain locations of the object, and
• in a convex mirror, the image is always virtual, erect and shorter than the object, wherever the object maybe located.

Activity 10.7 (Textbook page 172)

Discussion and Conclusion:
Figure shows a coin O at the bottom of a bucket filled with water. When we view this coin with our eye to a side above water, we observe the image I of the coin, which is above O. When we try to pick up this coin in one go, we do not succeed in picking up the coin. This is because we move our hand upto I, where the coin is being observed, but actually the coin lies below at O, the bottom of the bucket. When our friends try the same way, they also fail to pick up the coin. However, if some one understands that the coin is at the bottom of the bucket, he/she can pick up the coin.

Activity 10.8 (Textbook page 172)

Discussion and Conclusion:
This activity is an extension of the last activity 10.7. In a large shallow bowl filled with water, the coin is at the bottom at O. On account of refraction of light, image I of the coin appears to be raised. When we move our eye slowly away from the bowl, the coin disappears from our sight. This happens when rays starting from coin O fail to enter our eye, after refraction at the water surface.

When a friend pours more water gently into the bowl without disturbing the coin, the coin becomes visible again from the same position of our eye. This happens because on adding water, real depth of the coin increases. The apparent depth of the coin is equal to real depth divided by refractive index of water. Therefore, apparent depth of the coin also increases. The apparent position of the coin rises slightly above I and it becomes visible from the same position of our eye.

Activity 10.9 (Textbook page 172)

Discussion and Conclusion:
Figure shows a thick straight line ABCD in ink drawn over a sheet of white paper. A glass slab PQRS is placed over the line such that edge PQ of the slab makes an acute angle with the line ABCD.

When we look at the portion BC of the line under the slab, from the sides, careful observation shows that the line is bent at the edges B and C. This is because of refraction of light. At B, light enters from air to glass. Therefore, it bends towards the normal. At C, light enters from glass to air. Therefore, it bends away from the normal. The emergent ray CD, after two refractions, is parallel to AB, but is slightly displaced laterally. When the glass slab is so placed that one of its edges PQ is normal to the line ABCD, the part BC of the line under the slab does not appear to be bent. This is because for the incident ray AB, angle of incidence i = 0. Therefore, angle of refraction r is also zero. Therefore, no refraction occurs at B, and there is no bending. Similarly, no refraction occurs at C and hence there is no bending at C.

When we look at the portion of the line BC from the top of the glass slab, it appears to be raised. This is also an account of refraction of light. Rays of light starting from BC pass from glass to air. When viewed from the top of slab, these rays bend away from the normal. Therefore, the image of portion BC of the line appears to be raised.

Activity 10.11 (Textbook page 177)

Discussion and Conclusion:
When we direct a convex lens in our hand towards the Sun, the light from the Sun is focussed on a sheet of paper. By moving the sheet of paper back and forth, we obtain a sharp, bright image of the Sun on the paper.

When the paper and lens are held in the same position for the some time, the paper may burn out. This is because of heat produced by the Sun rays concentrate on the spot where the image of Sun is formed. This is what we observed in Activity 10.2.

Bihar Board Class 10 Science Solutions Chapter 9 Heredity and Evolution

Bihar Board Class 10 Science Chapter 9 Heredity and Evolution InText Questions and Answers

In-text Questions (Page 143)

Question 1.
If a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait B is present in a greater percentage of the population, so trait B is likely to have arisen earlier.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
Variations arise in the offsprings due to the inaccuracies which occur during the process of DNA replication. The favourable variations enable the organism to survive under adverse conditions and these variations are passed on to their progeny which promotes the survival of the species.

Example: The in-built variations present in the colony of bacterial cells enable them to develop antibiotic resistance in some cases.

In-text Questions (Page 147)

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel crossed pure tall pea plant (TT) with pure dwarf plant (tt) and observed in F1 generation a progeny which was hybrid tall (Tt). This showed that only one of the traits was able to express itself in the F1 generation i.e., tallness. The recessive dwarf trait was not able to express itself in the F1 generation but reappeared in the F2 generation on selfing the hybrid F1 plants.

Question 2.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
When Mendel crossed pea plant with round-yellow seeds (RRYY) with wrinkled-green seeds (rryy), all the progeny obtained in F1 generation was RrYy (round and yellow seeds). On self-pollination of F1 plants (RrYy), a phenotypic ratio of 9 : 3 : 3 : 1 for round-yellow, round-green, wrinkled-yellow and wrinkled-green pea seeds respectively were obtained. On the basis of the above experiments, Mendel concluded that round and yellow seeds have dominant characters. Occurrence of new phenotypic combinations shows that genes for round and yellow seeds are inherited independently of each other.

Question 3.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
The information is not sufficient to tell whether trait A or trait 0 is dominant.

Case I: If the blood group A is considered dominant over O, then the combinations possible are given alongside. The males are AO and females are OO. So, the daughters will have blood group O due to the presence of OO in them.

Case II: If the blood group O is considered dominant over A, then the combinations possible again show that child may have blood group O.
Since, in both the combinations it is possible for the child to have blood group O, it becomes difficult to establish which trait is dominant.

Question 4.
How is the sex of the child determined in human beings?
Answer:
In human beings the sex is genetically determined. Human male produces half of the gametes with X chromosome and half with Y chromosome. Human female produces all the gametes having X chromosome. When a sperm carrying X chromosome fertilises an ovum carrying X chromosome, the child born will be a girl (XX). When a sperm carrying Y chromosome fertilises an ovum carrying X chromosome, the child born will be a boy (XY).

In-text Questions (Page 150)

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
The individuals with a particular trait may increase in a population by

• Inheritance
• Natural selection

Question 2.
Why are traits acquired during the lifetime of an individual not inherited?
Answer:
The traits acquired during the lifetime of an individual are not inherited because they occur in non- reproductive cells of the organism. No change occurs in the DNA of the organism when the traits are acquired, so such traits are not passed on to the progeny.

Question 3.
Why are the small numbers of surviving tigers a cause of worry from the point of view of genetics?
Answer:
The small number of tigers is a cause of worry because if they die and become extinct then their genes will be lost forever. This will lead to loss of biodiversity and can create an imbalance in nature and its processes too.

In-text Questions (Page 151)

Question 1.
What factors could lead to the rise of a new species?
Answer:
The factors which lead to the rise of a new species are:

• Natural selection
• genetic drift
• reproductive isolation.

Question 2.
Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not? [CBSE 2012, 2013]
Answer:
A self-pollinating species does not depend on other plant for getting pollinated, so geographical isolation will not be a major factor in its speciation.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
An asexually reproducing organism does not depend on other organism for its reproductive process, so geographical isolation will not be a major factor in its speciation.

In-text Questions (Page 156)

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
The homologous organs help to determine the closeness of species in evolutionary terms as they have a common ancestor. For example, the structural similarity in the forelimbs of frogs, lizards, birds and humans points out to their relationship during evolution.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
Homologous organs have a similar structure and a common ancestor but are different in their functions. Analogous organs have different structure and ancestors but perform a similar function. Since, the wing of a bat and wing of butterfly are different in structure but perform the same function of flying, they are said to be analogous organs and not homologous organs.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
The remains or impressions of the dead animals or plants that lived in the remote past are called fossils. The fossils provide an evidence for evolution and help to establish the evolutionary relationships among the various groups of organisms.

In-text Questions (Page 158)

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Answer:
Human beings who look so different from each other in terms of size, colour and looks are said to belong to the same species as they can interbreed among themselves and produce fertile offsprings.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
Though the more complex body designs have evolved over time, it does not mean that the older designs are inefficient. The older forms still survive along with the newer forms. For example, bacteria are most primitive forms which are able to survive even in harsh habitats like hot springs or polar ice caps. So, it is not possible to say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design. All of them have evolved a body design which is best suited according to the environmental conditions and the mode of life they have adopted during the course of evolution.

Bihar Board Class 10 Science Chapter 9 Heredity and Evolution Textbook Questions and Answers

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW

Question 2.
An example of homologous organs is
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer:
(d) all of the above.

Question 3.
In evolutionary terms, we have more in common with
(a) a Chinese school-boy.
(b) a chimpanzee.
(c) a spider.
(d) a bacterium.
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light- coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
Since, children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis we can say that light eye colour trait is dominant.

Question 5.
How are the areas of study-evolution and classification-interlinked?
Answer:
Classification is the grouping of organisms on the basis of their similarities and differences. Organisms which share similar characteristics are placed in one group and considered to be more closely related evolutionary than the members of the other groups. So, evolution and classification are interlinked.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:
Homologous organs: Organs having the same basic structure but different functions are called as homologous organs. They have a common origin (common ancestor).
Example: Forelimbs of man, a lizard, a frog and a bird have the same basic structure and design but perform different functions.

Analogous organs: Organs which perform the same function but have different basic structure are called analogous organs: They have a different origin (different ancestors).
Example: The wings of insect and the wings of bird have different structure but perform the same function of flying.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
The experiment can be designed in which dogs having pure black colour coat (BB) are crossed with dogs having pure white colour coat (bb) to obtain the F1 progeny. If all the dogs obtained in the F1 generation are black in colour, we can conclude that black coat colour is dominant in dogs. But, if all the dogs in F1 generation are white in colour, then white coat colour will be considered as dominant.

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
The remains or impressions of the dead animals or plants that lived in the remote past are called fossils. The fossils provide an evidence for evolution and help to establish the evolutionary relationships among the various groups of organisms. On digging into the Earth, the fossils we find closer to the surface are more recent than the fossils we find in deeper layers, this helps us to ascertain the time in history when different species were formed or became extinct.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
The experiment carried out by Stanley L. Miller and Harold C. Urey provides an evidence for the origin of life from inanimate matter.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:
Asexual reproduction produces a progeny which are genetically identical to their parents. No variations occur in them due to which they are more vulnerable under adverse conditions. Sexual reproduction is a process in which fusion of gametes which are genetically dissimilar, takes place. Such a fusion of gametes results in variations which enable the organism to withstand adverse conditions. This also causes accumulation of variations over a period of time which consequently result in the evolution of new species.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Gametes are produced by cell division called meiosis in the male and female reproductive cells. The gametes have equal genetic material. Both male and female gametes contribute equal genetic content during their fusion to form a zygote. Zygote develops into an embryo and later into an individual. Hence, equal genetic contribution of male and female parents are ensured in the progeny.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Yes, only variations that confer an advantage to an individual organism will survive in a population. The favourable variations enable the organism to withstand the adverse conditions, ensure its survival and help to transfer them to their progeny. The variations which do not confer an advantage will not be able to help the organism to survive under adverse conditions, and would ultimately get eliminated from the population along with the organism.

Bihar Board Class 10 Science Chapter 9 Heredity and Evolution Textbook Activities

Activity 9.1 (Textbook page 143)

The presence of attached ear lobe is a recessive trait. The organisms with free ear lobe show the dominant trait. The recessive traits express themselves only in homozygous condition. So, the Law of Dominance given by Mendel will apply and the offsprings will have the traits as shown below:

Activity 9.2 (Textbook page 144)

The experiment done to confirm will be called as the test cross. It involves crossing the individuals obtained in the F2 generation with the parent with ‘tt’ as trait combination. The individuals having tt trait combination would be visible as dwarfs. The confusion of the tall plants- TT and Tt trait combination can be resolved by the test cross. The individuals having TT trait combination will give rise to all tall progeny after crossing with tt trait combination. The individuals with Tt trait combination will produce 50% progeny tall and 50% progeny dwarf after the cross with tt trait combination.

Bihar Board Class 10 Science Solutions Chapter 8 How do Organisms Reproduce?

Bihar Board Class 10 Science Chapter 8 How do Organisms Reproduce? InText Questions and Answers

In-text Questions (Page 128)

Question 1.
What is the importance of DNA copying in reproduction?
Answer:
The DNA copying is important because:

• it ensures the transmission of characteristics from parents to their offsprings.
• variations are created during DNA copying which increases the chances of survival of the species and in the evolution of new life forms.

Question 2.
Why is variation beneficial to the species but not necessarily for the individual?
Answer:
The variations which arise during the process of reproduction enable the organisms to survive under adverse conditions. The variations which arise in an individual are beneficial only if they are inherited by the offsprings of the individual for several generations. So, we can say that variations are not necessarily beneficial for the individual unless they are transmitted to the other members of the species.

In-text Questions (Page 133)

Question 1.
How does binary fission differ from multiple fission?
Answer:
Binary Fission:

1. The parent cell divides/splits equally into two daughter cells.
2. It occurs under favourable conditions. For example, Amoeba

Multiple Fission:

1. The parent cell splits into many daughter cells.
2. It occurs under unfavourable conditions.
   For example, Plasmodium (Malarial parasite)

Question 2.
How will an organism be benefited if it reproduces through spores?
Answer:
The spores have a thick wall which help them to tide over and survive under unfavourable conditions. Spores germinate on the return of favourable conditions. So, spores benefit the organism to ensure its survival.

Question 3.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:
Multicellular organisms consist of specialised cells which form the various tissues and the organs of the body of the organism present at different locations in the body of the organism. This makes it difficult for cell by cell division of the organism. So, multicellular animals need a complex method of reproduction and cannot give rise to new individuals through regeneration.

Question 4.
Why is vegetative propagation practised for growing some types of plants?
Answer:
Vegetative reproduction is practised for growing some types of plants because

• some plants have lost the ability to form seeds and can be propagated by vegetative reproduction. For example, Banana, rose, orange, jasmine, etc.
• it helps in creating genetically identical plants.
• plants which take a longer time to grow and mature can be grown by this method.
• it can be used to grow such plants which require more care during early stages of their development.

Question 5.
Why is DNA copying an essential part of the process of reproduction?
Answer:
DNA copying is an essential part of the process of reproduction because:

• it helps in the transmission of characteristics from the parents to their offsprings.
• the variations are created during copying which increases the chances of the survival of the members of the species under adverse conditions.
• it helps in the evolution of the species over time by creating variations during copying of DNA.

In-text Questions (Page 140)

Question 1.
How is the process of pollination different from fertilisation?
Answer:
Pollination:

1. The transfer of pollen grains from the anther to stigma of a flower.
2. It is of two types-Self-pollination and cross pollination
3. It is a physical process.
4. It occurs only in the plants.

Fertilisation:

1. The process of fusion of male and female gamete to form a zygote is called fertilisation.
2. It is of two types-Internal fertilisation and external fertilisation.
3. It is a bio-chemical process.
4. It takes place both in the plants as well as in animals.

Question 2.
What is the role of the seminal vesicles and the prostate gland?
Answer:
Seminal vesicles- The sperms are stored in the seminal vesicles. It also provides nutrition to the sperms.
Prostate gland- The secretions of this gland make the transport of the sperms easier.

Question 3.
What are the changes seen in girls at the time of puberty?
Answer:
The changes which occur in the girls at the time of puberty are:

• Increase in the breast size.
• Growth of hair in the armpits and the pubic region.
• Menstruation and ovulation begin.
• Excess fat gets deposited in various parts of the body like hips and thighs.

Question 4.
How does the embryo get nourishment inside the mother’s body?
Answer:
The embryo develops inside the mother’s body for about nine months. Inside the uterus, the outer tissue surrounding the embryo develops figner-like projections called villi. These villi are surrounded by uterine tissue and maternal blood. They provide a large surface area for exchange of oxygen and nutrients. Also, there is special tissue Placenta is nutrients. Also, there is special tissue placenta is embedded in the uterine wall. The embryo receives oxygen and nutrients from the mother’s blood through the placenta. The waste materials produced by the embryo are also removed through the placenta.

Question 5.
If a woman is using a copper-T, will it help in protecting her from sexually transmitted diseases?
Answer:
No, copper-T are placed in the uterus to prevent pregnancy. They can cause side effects due to irritation of the uterus.

Bihar Board Class 10 Science Chapter 8 How do Organisms Reproduce? Textbook Questions and Answers

Question 1.
Asexual reproduction takes place through budding in
(a) Amoeba
(b) yeast
(c) Plasmodium
(d) leishmania
Answer:
(b) A sexually reproduction takes place through budding in yeast.

Question 2.
Which of the following is not a part of the female reproductive system in human beings?
(a) Ovary
(b) Uterus
(c) vas deferens
(d) Fallopian tube
Answer:
(c) vas deferens is not a part of the female reproductive system in human beings.

Question 3.
The anther contains
(a) sepals
(b) ovules
(c) carpel
(d) pollen grains
Answer:
(d) The anther contains pollen grains.

Question 4.
What are the advantages of sexual reproduction over asexual reproduction?
Answer:
Sexual reproduction is more advantageous than asexual reproduction because:

• variations are caused during sexual reproduction which increases the chances of survival of an organism under adverse conditions.
• the progeny obtained by asexual reproduction is genetically identical to the parents and does not lead to evolution of the organism. The variations caused during the sexual reproduction leads to accumulation of variations which can lead to the evolution of new species.

Question 5.
What are the functions performed by the testis in human beings?
Answer:
Testis perform the following functions in the human beings:

• They are the sites for sperm production.
• They provide a suitable temperature required for the sperm formation.
• Male sex hormones are released by the testes.

Question 6.
Why does menstruation occur?
Answer:
Menstruation is a process which occurs when the egg is not fertilised by the sperm. The lining of the uterus prepares itself by becoming thick and spongy every month for receiving the fertilised egg and nourish the growing embryo. If the egg does not get fertilised then this lining breaks and comes out through the vagina along with blood by the process called menstruation.

Question 7.
Draw a labelled diagram of the longitudinal section of a flower.
Answer:

Question 8.
What are the different methods of contraception?
Answer:
The methods to avoid pregnancy are called as contraceptive methods. The different methods are:
Physical Methods: Prevent union of sperm and egg. For example, Condoms, diaphragm and cervical caps.

Chemical Methods: Act by changing the hormonal balance of the body so that eggs are not released and fertilisation cannot occur. They may have side-effects too. For example, Oral pills.

Surgical Methods:

• Vasectomy: The vas deferens of male is blocked to prevent sperm transfer.
• Tubectomy: The fallopian tube of female is blocked to prevent egg to reach uterus.
• Copper-T or loop is placed in uterus to prevent pregnancy.

Question 9.
How are the modes for reproduction different in unicellular and multicellular organisms? [CBSE2016]
Answer:
Unicellular organisms usually reproduce by fission or budding as they have only one cell and there is no separate tissue for reproduction.

Multicellular organisms have different modes of reproduction like sexual and asexual methods because they have special reproductive tissues and systems for the process of reproduction.

Question 10.
How does reproduction help in providing stability to populations of species?
Answer:
Reproduction provides stability to population of species as it helps in maintaining a balance between the birth rate and death rate in a population.

Bihar Board Class 10 Science Chapter 8 How do Organisms Reproduce? Textbook Activities

Activity 8.1 (Textbook Page 129)

The budding process of yeast may be seen as the yeast multiply by the process of budding.

Activity 8.2 (Textbook Page 129)

The slice of bread would be seen covered with a white cottony mass of fungal threads called hyphae, belonging to the fungus Rhizopus. Rhizopus reproduces by spores formed in the sporangia borne on the hyphae.

Activity 8.3 (Textbook Page 129)

Amoeba has no definite shape, has cytoplasm and a nucleus in a flexible cell membrane. During binary fission Amoeba undergoes nuclear division and forms two nuclei. The two nuclei formed by division move to opposite directions in the parent cell. Ultimately, the cell divides and forms two daughter cells.

Activity 8.4 (Textbook Page 130)

Spirogyra is a filamentous alga found as slimy filamentous green masses in clean water, ditches and ponds. Its filaments come to the surface of water and form dense green coloured structures on it. It has unbranched chains of cylindrical cells having spiral chloroplasts. It occurs as a slimy, green ‘blanket weed’ in ponds.

Activity 8.5 (Textbook Page 132)

Only the pieces of potato having the notch or the bud give rise to fresh green shoots whereas those pieces which do not have the notch or bud are not able to reproduce. This is so because the reproductive buds are present in the potato at the regions called as eyes form where the buds arise to form new plants. Potato is a stem.

Activity 8.6 (Textbook Page 132)

Only the portions which have at least one leaf will be able to form new plants because the leaves arise from the nodes. Nodes are the regions where reproductive buds are present in most of the plants. So, new plants will arise from the reproductive buds present at nodes in plants having at least one leaf, whereas the portion between two leaves known as the internode will not be able to form new plants.

Activity 8.7 (Textbook Page 135)

The seeds when cut open will show two large cotyledons attached to the middle of the embryonal axis. The cotyledons have reserve food material to nourish the developing embryo. The portion above the cotyledon is the future shoot called as plumule and the portion below the cotyledons is the future root called as the radicle.

Bihar Board Class 10 Science Solutions Chapter 7 Control and Coordination

Bihar Board Class 10 Science Chapter 7 Control and Coordination InText Questions and Answers

In-text Questions (Page 119)

Question 1.
What is the difference between a reflex action and walking?
Answer:

Question 2.
What happens at the synapse between two neurons?
Answer:
Nerve impulses are carried from one neuron to another neuron which lie end to end but are not connected. The small gap between the neurons is called as synapse. It is filled with a chemical (neurotransmitter substance). The electrical impulse from the axon of one neuron sets off the release of these chemicals into the synapse. The chemicals released transmit the impulse to the dendrite of the next neuron to create an electrical impulse in it. This presence of synapse between the junctions of axon endings and dendrites helps in transmission of impulse in one direction.

Question 3.
Which part of the brain maintains posture and equilibrium of the body?
Answer:
Cerebellum region of hind brain.

Question 4.
How do we detect the smell of an agarbatti (incense stick)?
Answer:
Olfactory receptors present in ours nose detect smell and transmit the information to forebrain which interprets the message.

Question 5.
What is the role of the brain in reflex action?
Answer:
Reflexes are of two types: Spinal reflexes and cerebral reflexes. Spinal reflexes are controlled by spinal cord but the message is also sent to brain for interpretation of the action. For example, removing hands on touching a hot object. Cerebral reflexes are controlled by the brain. For example, production of saliva on sight of spicy food.

In-text Questions (Page 122)

Question 1.
What are plant hormones?
Answer:
Chemical substances which help in the proper growth and development of the plants and also enable them to respond to the environment are called plant hormones. For example, auxin, cytokinin, etc.

Question 2.
How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?
Answer:
The leaves of the sensitive plants like ‘Touch-me-not’ lose turgor pressure in cells and cause drooping of its leaves/leaflets. The movement of the shoot towards light occurs due to more growth of shoot on the part away from light as auxin concentration gets increased at this part.

Question 3.
Give an example of a plant hormone that promotes growth.
Answer:
Auxin.

Question 4.
How do auxins promote the growth of a tendril around a support?
Answer:
The concentration of auxin gets increased on the free side of the tendril as compared to the part in contact with the support. So, more growth occurs on the free side and enables the tendril to coil around a support.

Question 5.
Design an experiment to demonstrate hydrotropism.
Answer:
Hydrotropism is a plant growth in which the direction of growth is determined by the stimulus or gradient in water concentration.

Take a sieve, put some sawdust in it and make it moist by adding some water. Now place a few germinating seeds on moist sawdust. After some days, the radicles will be seen growing downwards under the influence of gravity. But after a further few days, it can be observed that the radicles bend back and enter the moist sawdust due to the hydrotropic response of the roots.

In-text Questions (Page 125)

Question 1.
How does chemical coordination take place in animals?
Answer:
Chemical coordination is achieved by the hormones secreted by the endocrine glands (ductless glands) in the animals. Hormones are secreted in very small amounts, secreted directly into the blood and act only at the target organ. They help in the coordination of various activities of the body. For example, stress hormone called adrenaline is secreted to respond during emergency situations.

Question 2.
Why is the use of iodised salt advisable?
Answer:
Iodised salt is a source of iodine which is a constituent of hormone called thyroxine. Due to reduced synthesis of iodine the fat, carbohydrate and protein metabolism gets disturbed which leads to a disease called goitre. So, in order to avoid goitre, the use of iodised salt is advisable.

Question 3.
How does our body respond when adrenaline is secreted into the blood?
Answer:
Adrenaline is secreted in higher amounts in our body during stress or emergency situations to respond to the situation in an effective manner. It increases the rate of heart beat, breathing rate and blood pressure which causes release of more glucose in the blood to provide energy. In this way adrenaline helps the organism to overcome the stressful situation. Hence, its also called as stress hormone or emergency hormone.

Question 4.
Why are some patients of diabetes treated by giving injections of insulin?
Answer:
Sometimes due to insufficient production of insulin by the pancreas, the level of glucose in the blood rises. The excess glucose gets excreted by the person in urine and leads to the disorder called diabetes. So, in order to supplement the requirement of insulin, the patients are treated by giving insulin injections.

Bihar Board Class 10 Science Chapter 7 Control and Coordination Textbook Questions and Answers

Question 1.
Which of the following is a plant hormone?
(a) Insulin
(b) Thyroxin
(c) Oestrogen
(d) Cytokinin
Answer:
(d) Cytokinin

Question 2.
The gap between two neurons is called a
(a) dendrite
(b) synapse
(c) axon
(d) impulse
Answer:
(b) synapse

Question 3.
The brain is responsible for
(a) thinking
(b) regulating the heartbeat
(c) balancing the body
(d) all of the above
Answer:
(d) all of the above

Question 4.
What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?
Answer:
Receptors are present in our sense organs like nose, ear, tongue, skin, etc. They help in receiving the stimulus and transmitting it through the sensory nerves to the central nervous system for generating a response.

If receptors do not work properly then we will not be able to perceive the information and this can lead to many problems. For example, if the receptors present on the skin do not work then we will not be able to sense the objects properly. Similarly, if receptors of the nose do not work properly then we will not be able to sense the smell properly.

Question 5.
Draw the structure of a neuron and explain its function.
Answer:
The basic structural and functional unit of the nervous system is called as a nerve cell or neuron. A neuron has three parts-Cell body, dendrites and axon.

The dendrites help to acquire the information from the receptors on the sense organs and transmit it in form of electrical impulse to the axon through the cell body of the neuron. At the end of the axon, the electrical impulse leads to the release of chemicals at the synapse which starts a similar electrical impulse in the next neuron on being received by its dendrites. This enables transmission of nerve impulse in one direction from one neuron to another till it reaches the target organ, muscles, gland or brain.

Question 6.
How does phototropism occur in plants?
Answer:
The directional movement of a plant or its part in response to unidirectional exposure to light is called phototropism. The site of perception of light is the shoot/ root apex where auxin is synthesised. The shoot apex bends towards light while the root apex bends away from light. Auxin is a shade loving hormone which is uniformly spread in the stem. But, due to unidirectional exposure to light the auxin moves away from light and moves towards the shaded part of the stem. More amount of auxin on the shaded side of the plant causes more growth on the dark (shaded) side of stem, thereby bending the stem towards the source of light.

But in the case of the root an increase in the auxin on the side away from light causes a slower growth on that side. So, the root bends away from light.
Thus, the effect of auxin on growth of stem is opposite to its effect on the growth of root.

Question 7.
Which signals will get disrupted in case of a spinal cord injury?
Answer:
Reflex actions will get disrupted along with the involuntary actions of the nervous system.

Question 8.
How does chemical coordination occur in plants?
Answer:
Plant hormones (phytohormones) help in chemical coordination in the plants. These are of two types: Growth promoters like auxin, gibberelin, cytokinin; and growth inhibitors like abscissic acid and ethylene. Auxins cause apical growth, cytokinin causes cell division, and gibberelin causes growth of stem. Abscissic acid causes closing of stomata and wilting of leaves, ethylene causes fruit ripening.

Question 9.
What is the need for a system of control and coordination in an organism?
Answer:
Multicellular organisms have different tissues and organs which are specialised for a particular function. There is a need of these organs to act in a coordinated manner for the proper functioning of the body. So, the nervous system and the endocrine system are needed for effective control and coordination in an organism.

Question 10.
How are involuntary actions and reflex actions different from each other?
Answer:
Involuntary actions:

• The actions of the organism which are not in control of its will are called as involuntary actions.
• It is controlled by the hindbrain region of the brain.
• They are comparatively slower than the reflex actions.
• Stimulus in these actions is usually internal.
• For example, movements in muscular wall of oesophagus, beating movement of heart muscles.

Reflex actions:

• An automatic, immediate, involuntary response of an organism.
• It is controlled by the spinal cord.
• It is a very fast response.
• Stimulus in these responses is generally external.
• For example, knee jerk movement or moving the hand away on touching a hot object.

Question 11.
Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.
Answer:
Nervous Control:

• Information is transmitted in the form of electrochemical impulse/reaction.
• Transmission of impulse is very fast.
• The impulse through nerve is not specific.
• It produces short lasting responses.

Hormonal Control:

• Information is carried in the form of chemicals called hormones through the bloodstream.
• Transmission of impulse is slower.
• The action of hormones on the target organ is highly specific.
• It produces responses which are longer lasting.

Question 12.
What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?
Answer:
Movement in Sensitive Plant:

1. The response is produced due to the change in amount of water in the plant cells.
2. No nerve or specialised tissue involved in the conduction of information.
3. It occurs due to plant hormones and not due to specialised proteins.

Movement in Legs:

1. The response occurs due to contraction and relaxation of the muscle cells.
2. Nerves are involved along with muscle cells in conduction of information.
3. It occurs due to specialised proteins which help in contraction and relaxation of muscles.

Bihar Board Class 10 Science Chapter 7 Control and Coordination Textbook Activities

Activity 7.1 (Textbook Page 115)

Sugar has a sweet taste. No difference in taste will be observed if we eat sugar while pressing our nose between thumb and index finger, because the smell of sugar does not influence its taste. The smell of food perceived by the receptors of smell (olfactoreceptors) helps to stimulate the receptors of taste (gustatoreceptors). We will not be able to fully appreciate the taste of the food we are eating if we press our nose between thumb and index finger while eating because we won’t be able to smell the food.

Activity 7.2 (Textbook Page 121)

Yes, the old parts of the shoot and root changed direction. When the flask is placed in the cardboard which is open from one side which faces light coming from a window, the shoots of freshly germinated seeds show growth by bending towards light and roots show growth by bending away from light.

But, when the flask is turned so that the shoots are away from light and the roots towards light, we observe after few days that the shoots again grow by bending towards light and the roots grow by bending away from light. This experiment shows that the shoots of plants respond by showing growth movement towards light (i.e, positive phototropism) and roots of plants respond by showing growth movement away from light (i.e., negative phototropism)

Activity 7.3 (Textbook Page 123)