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QUESTION: 1

A cell of constant emf first connected to a resistance ** R_{1}** and then connected to resistance

Solution:

We know, Power delivered = ** i^{2}R** where

Power delivered in both cases is equal then

QUESTION: 2

The seals of a galvanometer of resistance 100Ω contains 25 division. It gives a deflection of one division of passing a current of 4 × 10^{–4}** A**. The resistance in ohm to be added to it, so, that it may become a voltmeter of range 2.5

Solution:

The deflection of one division of galvanometer is given by the current = 4 × 10^{–4} **A**

or the current sensitivity of galvanometer is

= 4 × 10^{–4} **A/div**

** i_{g}** = full scale deflection

= current sensitivity × total number of turns

= 4 × 10

Resistance of galvanometer = 100Ω

The value of resistance required,

The correct answer is: 150

QUESTION: 3

The resistance of the filament of a lamp increase with the increase in temperature. A lamp rated 100** W** and 200

Solution:

Let the resistance of the lamp filament be * R. *Then when the voltage drops expected power is

Here

But it will not be 90% of earlier value, because fall in temperature is smaller.

The correct answers are: 81 ** W**, between 81 and 90

QUESTION: 4

The figure shows the variation of ** v** with

Solution:

R = R_{0} + R_{0}α(T_{1}-T_{0})

tan3α = R_{0} + R_{0}α(T_{1}-T_{0})...............(1)

tan2α = R_{0} + R_{0}α(T_{2}-T_{0})...............(2)

tan3α - tan2α = R_{0}α(T_{1}-T_{2})

(T_{1}-T_{2}) α tan(3α) - tan(2α)

tan(3α-2α) = (tan3α - tan2α)/(1 + tan3αtan2α)

(T_{1} - T_{2}) α tanα(1 + tan3αtan2α)

So, tanα is the right answer.

QUESTION: 5

The potential difference across the terminal of a battery is 50 * V* when 11

Solution:

For a closed circuit cells supplies a constant current in the circuit.

Equation of cells *E* = *V* + *ir*

For ** V** = 50

Similarly for

From Equation (i) and (ii), we get

r = 1Ω

Substituting the value of

The correct answer is: 61V, 1Ω

QUESTION: 6

Masses of three wire of copper are in ratio 1:3:5, and their length are in ratio 5:3:1. The ratio of their electrical resistance is

Solution:

Ratio of masses of wires

(** m_{1}, m_{2}, m_{3}**) = 1 : 3 : 5

Ratio of length of wires

Electrical resistance is given by

The correct answer is: 125 : 15 : 1

QUESTION: 7

When a galvanometer is shorted by resistance ** S**, its current capacity increase

Solution:

Using the relation

QUESTION: 8

A wire of resistance 5Ω is drown out so that its new length is 3 times its original length. What is the resistance of the new wire?

Solution:

Let ** L_{1}** is the original length and

Let

Volume is constant

QUESTION: 9

The resistance of a wire at 300K is found to be 0.3Ω. If the temperature coefficient of resistance of wire is 1.5 × 10^{–3} K^{–1}, the temperature at which resistance become 0.6Ω is :

Solution:

Given,

** T** = 300K = 27°

Temperature coefficient of resistance

Dividing Equation (ii) by equation (i) we get

The correct answer is: 993K

QUESTION: 10

An electric immersion heater of 1.08KW is immersed in water. After the water has reached a temperature of 100°C, how much time will be required to produced 100* g* of steam?

Solution:

** L** is the latent heat of vaporisation of water, the heat required to produced 1

** L** = 540

Energy supplied = 1080

Time required to boil 100

The correct answer is: 210

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