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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Question for Short Answer

Objective-I

Objective-II

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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 :-

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 :-

Given:-

Resistance of the coil, R = 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 :-

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

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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-219)

Question-4 :-

(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 :-

Answer-5 :-

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,

R_eff = R + R_c = 625.034 Ω

The current supplied by the power station,

The power supplied to one side of the connecting wire,

Question-6 :-

Answer-6 :-

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 :-

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 )

Answer-8 :-

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,

Answer-12 :-

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

R_eff= 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 )

Answer-13 :-

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).

Answer-14 :-

Difference in temperature, θ = 40°C

Emf, E_cs = a_csθ + (1/2) b_csθ² ………..(1)

a_cs = [2.76 − (−43.7) μV

= 46.46 μV/°C

b_cs = [0.012 − (−0.47) μV/°C

= 0.482 μV/°C²

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

E_cs = 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).

Answer-15 :-

Neutral temperature, θ_n= −(a/b)

a_{cu Fe} = a_{cu Pb} − a_FePb

= 2.76−16.6 = 13.84μV°C⁻¹

b_{cu Fe}= b_{cu Pb} − b_FePb

= 0.012 + 0.030 = 0.042μV°C⁻²

Thus, the neutral temperature,