# Thermal and Chemical Effects of Electric Current Exercise HC Verma Solutions Ch-33 Vol-2

Thermal and Chemical Effects of Electric Current Exercise HC Verma Solutions Ch-33 Vol-2 Concept of Physics Vol-2 for ISC Class-12. Step by Step Solution of Exercise Questions of Ch-33 Thermal and Chemical Effects of Electric Current HC Verma Concept of Physics. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## Thermal and Chemical Effects of Electric Current Exercise HC Verma Solutions Ch-33 Vol-2

 Board ISC and other board Publications Bharti Bhawan Publishers Ch-33 Thermal and Chemical Effects of Electric Current Class 12 Vol 2nd writer H C Verma Book Name Concept of Physics Topics Solution of Exercise Questions Page-Number 218, 219, 220

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Thermal and Chemical Effects of Electric Current Exercise HC Verma Questions Solutions

### (HC Verma Ch-33 Concept of Physics Vol-2 for ISC Class-12)

(Page-218)

#### Question-1 :-

An electric current of 2.0 A passes through a wire of resistance 25 Ω. How much heat will be developed in 1 minute?

Given:-

Current through the wire, i = 2 A

Resistance of the wire, R = 25 Ω

Time taken, t = 1 min = 60 s

Heat developed across the wire,

H = i2Rt
= 2 × 2 × 25 × 60
= 100 × 60 J = 6000 J

#### Question-2 :-

A coil of resistance 100 Ω is connected across a battery of emf 6.0 V. Assume that the heat developed in the coil is used to raise its temperature. If the heat capacity of the coil is 4.0 J K−1, how long will it take to raise the temperature of the coil by 15°C?

Given:-

Resistance of the coil, = 100 Ω,

Emf of the battery, V = 6 V,

Change in temperature, ∆T = 15°C

Heat produced across the coil,

This heat produced is used to increase the temperature of the coil.

#### Question-3 :-

The specification on a heater coil is 250 V, 500 W. Calculate the resistance of the coil. What will be the resistance of a coil of 1000 W to operate at the same voltage?

Let R be the resistance of the coil.

The power P consumed by a coil of resistance R when connected across a supply V is given by

### Thermal and Chemical Effects of Electric Current Exercise HC Verma Questions Solutions

(HC Verma Ch-33 Concept of Physics Vol-2 for ISC Class-12)

(Page-219)

#### Question-4 :-

A heater coil is to be constructed with a nichrome wire (ρ = 1.0 × 10−6 Ωm) that can operate at 500 W when connected to a 250 V supply.

(a) What would be the resistance of the coil?

(b) If the cross-sectional area of the wire is 0.5 mm2, what length of the wire will be needed?

(c) If the radius of each turn is 4.0 mm, how many turns will be there in the coil?

(a) Let R be the resistance of the coil.

The power P consumed by a coil of resistance R when connected across a supply V is given by

(b) We know:-

(c) Let n be the number of turns in the coil. Then,

#### Question-5 :-

A bulb with rating 250 V, 100 W is connected to a power supply of 220 V situated 10 m away using a copper wire of area of cross-section 5 mm2. How much power will be consumed by the connecting wires? Resistivity of copper = 1.7 × 10−8 Ωm

Let R be the resistance of the bulb. If P is the power consumed by the bulb when operated at voltage V, then

The effective resistance,

Reff = R + Rc = 625.034 Ω

The current supplied by the power station,

The power supplied to one side of the connecting wire,

#### Question-6 :-

An electric bulb, when connected across a power supply of 220 V, consumes a power of 60 W. If the supply drops to 180 V, what will be the power consumed? If the supply is suddenly increased to 240 V, what will be the power consumed?

The resistance of a bulb that consumes power P and is operated at voltage V is given by

(a) Now the supply drops to V’ = 180 V.

So, the power consumed,

(b) Now the supply increases to V” = 240 V. Therefore,

#### Question-7 :-

A servo voltage stabiliser restricts the voltage output to 220 V ± 1%. If an electric bulb rated at 220 V, 100 W is connected to it, what will be the minimum and maximum power consumed by it?

Output voltage, V = 220 V ± 1%

= 220 V ± 2.2 V

The resistance of a bulb that is operated at voltage V and consumes power P is given by

(a) For minimum power to be consumed, output voltage should be minimum. The minimum output voltage,

V’ = (220 − 2.2) V

= 217.8 V

The current through the bulb,

i′=V′R=217.8484=0.45A

Power consumed by the bulb, P’ = i’ × V’

= 0.45 × 217.8 = 98.0 W

(b) For maximum power to be consumed, output voltage should be maximum. The maximum output voltage,

V” = (220 + 2.2) V

= 222.2 V

The current through the bulb,

Power consumed by the bulb,

P” = i” × V”

= 0.459 × 222.2

= 102 W

#### Question-8 :- (Thermal and Chemical Effects of Electric Current Exercise HC Verma )

An electric bulb marked 220 V, 100 W will get fused if it is made to consume 150 W or more. What voltage fluctuation will the bulb withstand?

Given that the operating voltage is V and power consumed is P.

Therefore, the resistance of the bulb,

The power fluctuation, p = 150 W. So, the voltage fluctuation that the bulb can withstand,

The bulb will withstand up to 270 V.

⇒ t = 29.17 minutes

The coil of an electric bulb takes 40 watts to start glowing. If more than 40 W are supplied, 60% of the extra power is converted into light and the remaining into heat. The bulb consumes 100 W at 220 V. Find the percentage drop in the light intensity at a point if the supply voltage changes from 220 V to 200 V.

In the Case-I :

When the supply voltage is 220 V.

Power consumed by the bulb = 100 W

Excess power = 100 − 40 = 60 W

Power converted to light = 60% of 60 W = 36 W

In the Case-II :

When the supply voltage is 200 V.

Power consumed

= (200/220)×100 = 82.64W

Excess power = 82.64 − 40 = 42.64 W

Power converted to light = 60% of 42.64 W = 25.584 W

Percentage drop in light intensity,

p =  (36−25.584)/36 × 100

⇒ p = 28.93 ≈ 29

The effective resistance of the circuit,

2000J of heat raises the temp. by 1K

5832J of heat raises the temp. by 2.916K.

(b) When 6Ω resistor get burnt

Reff= 1 + 2 = 3Ω

i = 6/3 = 2Amp. Heat = 2 × 2 × 2 ×15 × 60 = 7200J

2000J raises the temp. by 1K

7200J raises the temp by 3.6k

#### Question-13 :- (Thermal and Chemical Effects of Electric Current Exercise HC Verma )

The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0°C and 0.001°C. Find the thermo-emf (Seebeck emf) developed. For bismuth-silver, a = − 46 × 10−6 V°C−1 and b = −0.48 × 10−6 V°C−2.

Given:-

Difference in temperature, θ = 0.001°C,

a = − 46 × 10−6 V °C−1

b = − 0.48 × 10−5 V °C−2

Emf, E = aθ + (1/2)bθ2

⇒ E = (− 46 × 10−6) × (0.001)

−(1/2) × (0.48×10−6) × (0.001)2

= − 46 × 10−9 − 0.24 × 10−12

= − 46.0024 × 10−9

= − 4.6 × 10−8 V

#### Question-14 :-

Find the thermo-emf developed in a copper-silver thermocouple when the junctions are kept at 0°C and 40°C. Use the data given in the table no (33.1).

 Metal with lead (Pb) a μV/°C b μV/(°C) Aluminium -0.47 0.003 Bismuth -43.7 -0.47 Copper 2.76 0.012 Gold 2.90 0.0093 Iron 16.6 -0.030 Nickel 19.1 -0.030 Platinum -1.79 -0.035 Silver 2.50 0.012 Steel 10.8 -0.016

Difference in temperature, θ = 40°C

Emf, Ecs = acsθ + (1/2) bcsθ2 ………..(1)

acs = [2.76 − (−43.7) μV

= 46.46 μV/°C

bcs = [0.012 − (−0.47) μV/°C

= 0.482 μV/°C2

Putting this value in eq. (1), we get:-

Ecs = 46.46 × 10−6 × 40 + (1/2) × 0.482 × 10−6 × (40)2

= 1.04 × 10−5 V

#### Question-15 :-

Find the neutral temperature and inversion temperature of a copper-iron thermocouple if the reference junction is kept at 0°C. Use the data given in the table no (33.1).

 Metal with lead (Pb) a μV/°C b μV/(°C) Aluminium -0.47 0.003 Bismuth -43.7 -0.47 Copper 2.76 0.012 Gold 2.90 0.0093 Iron 16.6 -0.030 Nickel 19.1 -0.030 Platinum -1.79 -0.035 Silver 2.50 0.012 Steel 10.8 -0.016

Neutral temperature, θn= −(a/b)

acu Fe = acu Pb − aFePb

= 2.76−16.6 = 13.84μV°C−1

bcu Fe= bcu Pb − bFePb

= 0.012 + 0.030 = 0.042μV°C−2

Thus, the neutral temperature,

The inversion temperature is double the neutral temperature, i.e. 659 °C.