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Page No 90:

Question 1:

A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure12.1. The current recorded in the ammeter will be

(a) maximum in (i)
(b) maximum in (ii)
(c) maximum in (iii)
(d) the same in all the cases

Answer:

Since the resistance is same in all the three circuit diagrams so the current will be same in all the three cases.
Hence, the correct answer is option D. 

Page No 90:

Question 2:

In the following circuits (Figure 12.2), heat produced in the resistor or combination of resistors connected to a 12 V battery will be

(a) same in all the cases
(b) minimum in case (i)
(c) maximum in case (ii)
(d) maximum in case (iii)

Answer:

Heat produced in time (t) is given by
Heat (H) = V2R×t
where, V = potential difference
R = Net resistance of the circuit
t = time

R1net= 2 ΩR2net=2 Ω+ 2 Ω = 4 ΩR3net=2×22+2=1 Ω

Since, R3net is minimum so heat produced will be maximum in case (iii)
Hence, the corrrect answer is option D.



Page No 91:

Question 3:

Electrical resistivity of a given metallic wire depends upon
(a) its length
(b) its thickness
(c) its shape
(d) nature of the material

Answer:

Electrical resistivity of a given metallic wire depends upon the nature of the material.
Hence, the correct answer is option D

Page No 91:

Question 4:

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly
(a) 1020
(b) 1016
(c) 1018
(d) 1023

Answer:

Current, i = 1 A
Time, t = 16 s
We know
i=qtq=i×t

q=1×16q= 16 C

Now, 
q=n×e
where, n is number of electrons lost or gained
e = charge on 1 electron (1.6×10-19 C)

n=qen= 161.6×10-19n= 1020 

Hence, the correct answer is option A

Page No 91:

Question 5:

Identify the circuit (Figure 12.3) in which the electrical components have been properly connected.

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)

Answer:

All the electrical components have been correctly connected in case (ii). This is because ammeter is connected in series and voltmeter is connected parallel to the resistor. Moreover, terminal of both the devices connected to positive terminal of the cell must be positive. 
Hence, the correct answer is option B. 



Page No 92:

Question 6:

What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 10 Ω
(c) 5 Ω
(d) 1 Ω

Answer:

Resistance will be maximum when all the five resistors are connected in series. 
Rmax=15+15+15+15+15Rmax=1 Ω
Hence, the correct answer is option D. 

Page No 92:

Question 7:

What is the minimum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 1/25 Ω
(c) 1/10 Ω
(d) 25 Ω

Answer:

Resistance will be minimum when all the five resistors are in parallel connection. Since all the five resistors are identical so the formula will be 
Rmin=R5Rmin=155Rmin=125 Ω

Hence, the correct answer is option B. 
 

Page No 92:

Question 8:

The proper representation of series combination of cells (Figure 12.4) obtaining maximum potential is

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)

Answer:

The positive terminal of a cell should be connected to the negative terminal of the adjacent cell. Case (i) represents the correct combination of cells. 
Hence, the correct answer is option A. 

Page No 92:

Question 9:

Which of the following represents voltage?

(a) Work doneCurrent × Time

(b) Work done × Charge

(c) Work done×TimeCurrent

(d) Work done × Charge × Time

Answer:

Potential difference (V) is defined as the ratio of work done (W) and charge (Q).
V=WQ

Charge is defined as 
Q=i×t
where i = current 
          t = time  

So, 
V=Wi×t
Voltage=Work doneCurrent×time

Hence, the correct answer is option A. 


 

Page No 92:

Question 10:

A cylindrical conductor of length l and uniform area of cross section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A

Answer:

Let the resistivity of the material be ρ
Resistance (R) of the first cylindrical conductor = ρ×lA
where, l = length of conductor
           A = area of cross section

Now, another cylindrical conductror has double length but same resistance. 
Let its area of cross section be A'
Its resistance (R') will be = ρ×2lA'
Since resistance is same
R = R'
ρ×lA=ρ×2lA'

Solving it, we get A' = 2A

Hence, the correct answer is option C. 

 



Page No 93:

Question 11:

A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?

(a) R1 = R2 = R3
(b) R1 > R2 > R3
(c) R3 > R2 > R1
(d) R2 > R3 > R1

Answer:

Since the current (I) is plotted on y-axis and potential difference (V) is plotted on x-axis, so the slope of the graph is 1R. It means higher the slope the lesser is the resistance. So, R1 will be minimum and R3 will be maximum. 
Hence, the correct answer is option C. 

Page No 93:

Question 12:

If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be
(a) 100 %
(b) 200 %
(c) 300 %
(d) 400 %

Answer:

If the current is increased by 100% then the current will become twice i.e 2I. The power dissipated through a resistor is given by 
P=I2R
If the current is made twice then power dissipated will become four times i.e. power will become 400%. 
Hence, the correct answer is option D. 
 

Page No 93:

Question 13:

The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed

Answer:

Resistivity of a material depends only on the temperature and the nature of material. Shape of the resistor has no effect on the resistivity. 
Hence, the correct answer is option C. 

Page No 93:

Question 14:

In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(a) Brightness of all the bulbs will be the same
(b) Brightness of bulb A will be the maximum
(c) Brightness of bulb B will be more than that of A
(d) Brightness of bulb C will be less than that of B

Answer:

Since all the three bulbs are connected in parallel so potential difference (V) across all the bulbs is same. Power dissipated is given by 
P=V2R

Lower the resistance, higher the power dissipation. Hence higher the brightness. 
Out of all the three bulbs, 100 W bulb will have minimum resistance because rated voltage of all the bulbs will be same  and its power dissipation is maximum. 
Thus, 100 W will have maximum brightness and 40 W bulb will have minimum brightness. 
Hence, the correct answer is option C.

 

Page No 93:

Question 15:

In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J

Answer:

Since 2 Ω and 4 Ω are connected in series so their net resistance is 
Rnet=R1+R2

Rnet=2+4Rnet=6 Ω

Current (I) in the circuit is given by 
I=VR1+R2I=66I=1 A

Heat dissipated through 4 Ω resistor is given by 
H=I2×R×tH=12×4×5H=20 J

Hence, the correct answer is option C. 
 

Page No 93:

Question 16:

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A

Answer:

Power (P) = 1 kW = 1000 W
Potential difference (V) = 220 V
We know, 
P = VI
I=PVI=1000220I=4.54 A

From all the given options, 5 A fuse wire is correct
Hence, the correct answer is option D. 



Page No 94:

Question 17:

Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have
(a) same current flowing through them when connected in parallel
(b) same current flowing through them when connected in series
(c) same potential difference across them when connected in series
(d) different potential difference across them when connected in parallel

Answer:

When two resistors are connected in series with a battery then current passing through them is same. 
Hence, the correct answer is option B. 

Page No 94:

Question 18:

Unit of electric power may also be expressed as
(a) volt ampere
(b) kilowatt hour
(c) watt second
(d) joule second

Answer:

Power is defined as 
P = VI
where, V = Potential difference (Volt) 
            I = Current (Ampere)

So, unit of electric power may also be expressed as volt ampere.
Hence, the correct answer is option A. 

Page No 94:

Question 19:

A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

Answer:

The correct circuit diagram is shown as below

Page No 94:

Question 20:

Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

Answer:

P = I2R
where, 
P = Potential difference
I = Current
R = Resistance 

18=I2×2I2=182I2=9 I=3 A

This is the maximum current that can flow through the resistor. 

Page No 94:

Question 21:

Should the resistance of an ammeter be low or high? Give reason.

Answer:

Resistance of an ammeter should ideally be zero so that it does not affect the flow of current in the circuit. Hence, resistance of an ammeter should be very low as zero resistance is not possible in real life. 

Page No 94:

Question 22:

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2Ω in series with a combination of two resistors (4Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason.

Answer:



Net resistance of two 4 Ω resistors connected in parallel is 2 Ω. Since we have two resistors of same resistance in series so the potential drop across each one of them will be same. 

Page No 94:

Question 23:

How does use of a fuse wire protect electrical appliances?

Answer:

Fuse wire has low melting point and high resistance. So, whenever there is surge in the electric current, the fuse wire melts and breaks the circuit. This prevents any damage to the electrical appliances. Thus, a fuse wire protects the electrical appliances. 

Page No 94:

Question 24:

What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Answer:

Electrical resistivity is the inherent property of the resistor because of which it opposes the flow of electric current. It depends only on the nature and temperature of the resistor.
Resistance varies as directly proportional with the length. When the length is doubled then resistance also become double. Current is inversely proportional to resistance according to Ohm's law. If the resistance becomes double then current will become half i.e 5 A to 2.5 A.
This is why the reading of the ammeter decreases to half.   



Page No 95:

Question 25:

What is the commercial unit of electrical energy? Represent it in terms of joules.

Answer:

The commercial unit of electrical energy is Kilowatt hour (kWh)

1 kWh=1000 W×1 h1 kWh=1000 Js×3600 s1 kWh=3.6××106 J

Page No 95:

Question 26:

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.

Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Answer:

Let the resistance of the electric lamp be Rlamp
Current (I) = 1 A
Resistance of conductor (Rconductor) = 5 Ω
Potential difference of battery (V) = 10 V

Using Ohm's law, 

Rnet=VIRnet=101Rnet=10 Ω

We know in series connection
Rnet=Rlamp+Rconductor10=Rlamp+5Rlamp=5 Ω

When 10 Ω resistor is connected with the series combination of lamp and conductor then the net resistance of the circuit becomes 5 Ω
Using Ohm's law
I=VRnetI=105I=2 A

Current passing through 5 Ω conductor will remain same i.e 1 A as the 2 A current coming from the battery will equally divide between the two parallel branches of the circuit. Moreover, potential difference across the lamp will remain same because current passing through it remains same. 

Page No 95:

Question 27:

Why is parallel arrangement used in domestic wiring?

Answer:

Parallel arrangement is used in domestic wiring because of the following reasons
1. Net resistance of the electric circuit decreases in parallel connection as compared to series connection due to which heat loss is reduced.
2. Potential difference across each electrical device remains same i.e 220 V due to which all the electrical devices work upto their maximum potential. 
3. Each electrical device can be operated independently.
4. If one electrical device stops working still other electrical devices continue to work properly. 

Page No 95:

Question 28:

B1, B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused?
(ii) What happens to the reading of A1, A2, A3 and A when the bulb B2 gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Answer:

 Since all the three bulbs are identical so current through each bulb will remain same i.e 1 A

(i) Even when bulb B1 gets fused, other two bulbs will continue to glow with same brightness because all the three bulbs are connected in parallel so potential difference across each bulb remains same. 

(ii) When bulb B2 gets fused then the reading A2 will become zero. The reading of A1 and A3 will remain same as 1 A because potential difference across them remains same. However, the reading of A will decrease and will become 2 A. 

(iii). Power dissipated by bulb B1 will be
P1=VIP1=4.5×1P1=4.5 W

Similarly, 
P2=4.5 WP3= 4.5 W

Hence total power dissipated by the circuit will be 
Pnet= P1+P2+P3Pnet= 4.5+4.5+4.5Pnet =13.5 W

Page No 95:

Question 29:

Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.
(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

Answer:

(a) In series combination, potential difference get divided whereas in parallel potential difference across each bulb remains full. Hence in series bulbs glow with less brightness as compared to the bulbs in parallel connection. 

(b) Now if one bulb gets fused in series connection then all the other bulbs will stop glowing whereas in parallel connection other bulbs will continue to glow. This is because in series combination even a single faulty component break the circuit whereas this does not happen in parallel circuit. 
 

Page No 95:

Question 30:

State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Answer:

Ohm's Law states that at constant temperature, potential difference across a conductor is directly proportional to the current passing through it. 
Mathematically
V α IVI=R

where, 
V = potential difference
I = current
R = Constant of proportionality and it is called Resistance

Experimental verification
1. Take a nichrome wire, number of electric cells, ammeter, voltmeter and key. 
2. First, connect one electric cell in the circuit and note down the ammeter reading corresponding to it. 
3. Then, connect one more electric cell in the circuit and note down the ammeter reading. Continue this process until all the electric cells are connected in the circuit. 

Tabulate all the readings in the form of a table having column for potential difference and current. Then plot the graph between potential difference and the current in which a straight line passing through origin will be obtained. The slope of V-I graph indicates the resistance of the circuit. 
Ohm's law does not hold good under all conditions. It holds good only if temperature remains constant. 

 

Page No 95:

Question 31:

What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

 

Answer:

Electrical resistivity is the inherent property of a conductor due to which it opposes the flow of current in the wire. It does not depend on the length and area of cross section of wire. It depends only on the temperature and nature of the material of the wire.
Its SI unit is Ωm (Ohm metre). 
Experiment to study the factors on which the resistance of conducting wire depends:
1. Take a battery, ammeter, voltmeter, nichrome wires of different lengths and area of cross sections, copper wire. 
2. Connect the nichrome wire in the circuit and measure the ammeter reading.
3. Now change the length of the nichrome wire and note down the ammeter reading. 
4. Now change the area of cross section of nichrome wire and note down the ammeter reading.
5. Now replace the nichrome wire with copper wire of same length and area of cross section and note down the ammeter reading. 
6. Tabulate all the readings and calculate the resistance from the ratio of potential difference and current. 

We will observe that resistance of wire increases with the increase in length of the nichrome wire and decreases with the increase in area of cross section. Moreover, when we replace the nichrome wire with copper wire of same length and area of cross section even then resistance will be different. 
These observations show that resistance of wire depends on the length, area of cross section and nature of the material. 

Page No 95:

Question 32:

How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Answer:

Take three resistances R1, R2, and R3
Connect them in series combination using battery, ammeter and key. First of all, connect the ammeter between battery and R1 and note down the reading. After that, connect the ammeter between R1 and R2 and note down the reading. After that, connect the ammeter between R3 and battery. 
We will observe that ammeter shows the same reading in all the three situations. This shows that same current flows through all the three resistors. 

Page No 95:

Question 33:

How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?

Answer:

1. Take three resistors R1, R2 and R3 and connect them in parallel arrangement.
2. Connect voltmeter in parallel to measure the potential difference across the combination.
3. Now, remove the resistor R1 and measure the potential difference across the remaining resistor's combination. 
4. Now, remove the resistor R2 and measure the potential difference across the remaining resistor's combination.
Potential difference reading will be same in all the three cases. This shows that the same potential difference exists across three resistors connected in parallel arrangement to a battery. 
 



Page No 96:

Question 34:

What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

Answer:

Joule's law of heating states that heat produced in the resistor is 
1. directly proportional to the square of the current.
2. directly proportional to the resistance. 
3. directly proportional to the current.

Mathematically, 
H=i2Rt
where, 
H = Heat produced
I = Current
R = Resistance
t = time

Experimental demonstration
1. Take a water heating immersion rod, water, bucket, thermometer, regulator.
2. Immerse the water heating immersion rod in the water and measure the time taken to heat the water upto a certain temperature.
3. Now increase the current to double with the help of regulator and now measure the time taken to heat the water upto the same temperature. 
Then, we will observe that time taken in second case will be one fourth the time taken in earlier case. Hence, it verifies the Joule's law of heating.

Electric fuse, toaster, oven, kettle,  are the four applications of Joule's law of heating in daily life. 

Page No 96:

Question 35:

Find out the following in the electric circuit given in Figure 12.9
(a) Effective resistance of two 8 Ω resistors in the combination
(b) Current flowing through 4 Ω resistor
(c) Potential difference across 4 Ω resistance
(d) Power dissipated in 4 Ω resistor
(e) Difference in ammeter readings, if any.

Answer:

(a) Both 8 Ω resistors are connected in parallel so the effective resistance of the two will be 
1Rp=1R1+1R21Rp=18+181Rp=281Rp=14Rp=4 Ω

(b) Net resistance of the circuit will be 
Rnet=4+4Rnet=8 Ω

So, current in 4 Ω resistor using Ohm's law is
I=VRnetI=88I=1 A

(c) Potential difference across 4 Ω resistor is
V=IRV=1×4V=4 V

(d) Power dissipated through 4 Ω is 

P=i2RP=12×4P=4 W

(e) Both ammeters A1 and A2 will record same readings because they are connected in series. Hence difference in their readings will be zero. 



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