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Class 12-science H C VERMA Solutions Physics Chapter 11 - Thermal and Chemical Effect of Electric Current

Thermal and Chemical Effect of Electric Current Exercise 218

Solution 1

Heat=   

   

   

Solution 2

Heat produced by coil=heat to increase temperature of coil

  

  

 min

Solution 3

  

  

  

Now, P=1000W

  

  

  

Thermal and Chemical Effect of Electric Current Exercise 219

Solution 4

(a)   

  

  

(b)   

  

 m

(c) Number of turns=  

  

Solution 5

Resistance of the bulb

  

Resistance of copper wire

  

Total resistance of copper wire(both side connecting wire) =   

The effective resistance

  

The current supplied by the power station

  

Power supplied/consumed by both side connecting wire

  

 mW

Solution 6

Power=  

  

  

(a)   

  W

(b)   

  W

Solution 7

Resistance of the bulb

  

For minimum power, voltage should be minimum

  

Power,  W

For maximum power, voltage should be maximum

  

 W

Solution 8

Resistance of the bulb

  

Let maximum voltage bulb can withstand

  

 volt

Solution 9

Power supplied by the heater=1000W

Power consumed by water= W

  

  

 minutes

Solution 10

(a) volume of water boiled= cc

Mass of water boiled=  

 Kg

Heat required for boiling water

  

  

 J

We know

1000 watt-hour=  

 cost of boiling 4 cups of water=  

 (b) If room temperature drops to   

Heat required,   

  

 J

 cost of boiling 4 cups of water=  

Solution 11

Resistance of the bulb,   

Case 1: When 220V supply is used.

Excess power=100-40=60W

Power converted into light= W

Case 2: When 200V supply is used.

Power consumed by bulb= W

Excess power=82.64-40=42.64W

Power converted into light= W

Now, percentage drop in intensity of light=  

Solution 12

Equivalent resistance of the circuit

  

Current through battery

  

  

 A

6  and 2  resistance are in parallel combination so current will be distributed in inverse ratio of resistance i.e., in ratio of 1:3.

Current in 2  resistance= A

(a) Heat generated by 2  resistance= heat taken by water and calorimeter

  

  

  

(b) Equivalent resistance of the circuit

  

Current in 2  resistance

  

  

I=2A

Heat generated by 2  resistance=heat taken by water and calorimeter

  

   

  

Solution 13

  

  

  

Emf;   

  

 V

Solution 14

Difference in temperature   ;   

  

  

Emf   

  

 V

Solution 15

Here,

  

  

  

  

  

  

Thus, the neutral temperature

  

So, inversion temperature  

Solution 16

One equivalent mass of the substance requires 96500 coulombs.

(a) For a monovalent material

Equivalent mass=molecular mass

So,   atoms require 96500C

1 atom requires=  

(b) For a divalent material

Equivalent mass=  

So,   atoms require 96500C

1 atom requires=  

Solution 17

Since Ag is monoatomic.

Equivalent mass of silver=molecular mass

 gm

The ECE of silver

  

Using,

  

 gm

Solution 18

  

  

 A

Solution 19

Mass of 1 liter hydrogen,  gm

Now,

  

  

 Minutes.

Solution 20

(a) For the trivalent metal salt

  

Now,   

  

Atomic weight=  

(b)   

  

 gm

Solution 21

 A

Surface area,   

Thickness, t=0.1mm= cm

Volume of Ag deposited=  for one side

For both sides, Mass of Ag= gm

Using,   

  

 minutes 

Solution 22

 gm

 sec

Now,   

  

 A

Heat developed in the 20  resistor

  

  

  

Solution 23

Terminal potential energy, TPD=10V

Emf=12V

We know that, while discharging

  

  

 A

Using,   

 gm

Solution 24

Surface area   

Thickness   

Volume of Ag  

  

  

Mass of Ag=density volume

  

 kg

Using,   

  

 C

Energy spent by cell= work done by the cell

  

 kJ

When this energy is used to heat 100gm of water, rise in temperature.

  

  

  

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