### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2005

A moving coil galvanometer has $150$ equal divisions. Its current sensitivity is $10$- divisions per milliampere and voltage sensitivity is $2$ divisions per millivolt. In order that each division reads $1$ volt, the resistance in $ohms$ needed to be connected in series with the coil will be -
A
${10^5}$
B
${10^3}$
C
$9995$
D
$99995$

## Explanation

KEY CONCEPT : Resistance of Galvanometer,

$G = {{Current\,\,\,sensitivity} \over {Voltage\,\,\,sensityvity}} \Rightarrow G = {{10} \over 2} = 5\Omega$

Here ${i_g} =$ Full scale deflection current $= {{150} \over {10}} = 15\,\,mA$

$V=$ voltage to be measured $=150$ volts

(such that each division reads $1$ volt)

$\Rightarrow R = {{150} \over {15 \times {{10}^{ - 3}}}} - 5 = 9995\Omega$
2

### AIEEE 2005

Two sources of equal $emf$ are connected to an external resistance $R.$ The internal resistance of the two sources are ${R_1}$ and ${R_2}\left( {{R_1} > {R_1}} \right).$ If the potential difference across the source having internal resistance ${R_2}$ is zero, then
A
$R = {R_2} - {R_1}$
B
$R = {R_2} \times \left( {{R_1} + {R_2}} \right)/\left( {{R_2} - {R_1}} \right)$
C
$R = {R_1}{R_2}/\left( {{R_2} - {R_1}} \right)$
D
$R = {R_1}{R_2}/\left( {{R_1} - {R_2}} \right)$

## Explanation

${\rm I} = {{2\varepsilon } \over {R + {R_1} + {R_2}}}$

Potential difference across second cell

$= V = \varepsilon - {\rm I}{R_2} = 0$

$\varepsilon - {{2\varepsilon } \over {R + {R_1} + {R_2}}}.{R_2} = 0$

$R + {R_1} + {R_2} - 2{R_2} = 0$

$R + {R_1} - {R_2} = 0$

$\therefore$ $R = {R_2} - {R_1}$
3

### AIEEE 2005

In the circuit, the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance, the value of the resistor $R$ will be -
A
$100\Omega$
B
$200\Omega$
C
$1000\Omega$
D
$500\Omega$

## Explanation

$iR = 2 = 12 - 500i$

$\therefore$ $i = {1 \over {50}}$

$\therefore$ ${1 \over {50}} \times R = 2$

$R = 100\Omega$
4

### AIEEE 2005

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be
A
four times
B
doubled
C
halved
D
one fourth

## Explanation

$H = {{{V^2}t} \over R}$

Resistance of half the coil $= {R \over 2}$

$\therefore$ As $R$ reduces to half, $'H'$ will be doubled.