 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will
A
flow from Antimony to Bismuth at the hot junction
B
flow from Bismuth to Antimony at the cold junction
C
now flow through the thermocouple
D
flow from Antimony to Bismuth at the cold junction

Explanation

At cold junction, current flows from Antimony to Bismuth (because current flows from metal occurring later in the series to metal occurring earlier in the thermoelectric series).
2

AIEEE 2005

An energy source will supply a constant current into the load if its internal resistance is
A
very large as compared to the load resistance
B
equal to the resistance of the load
C
non-zero but less than the resistance of the load
D
zero

Explanation

$I = {E \over {R + r}},\,$ Internal resistance $\left( r \right)$ is

zero, $I = {E \over R} =$ constant.
3

AIEEE 2005

The resistance of hot tungsten filament is about $10$ times the cold resistance. What will be resistance of $100$ $W$ and $200$ $V$ lamp when not in use ?
A
$20\Omega$
B
$40\Omega$
C
$200\Omega$
D
$400\Omega$

Explanation

$P = Vi = {{{V_2}} \over R}$

${R_{hot}} = {{{V^2}} \over P} = {{200 \times 200} \over {100}} = 400\Omega$

${R_{cold}} = {{400} \over {10}} = 40\Omega$
4

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)

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