 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

An ammeter reads upto $1$ ampere. Its internal resistance is $0.81$ $ohm$. To increase the range to $10$ $A$ the value of the required shunt is
A
$0.03\,\Omega$
B
$0.3\,\Omega$
C
$0.9\,\Omega$
D
$0.09\,\Omega$

Explanation

${i_g} \times G = \left( {i - {i_g}} \right)S$

$\therefore$ $S = {{{i_g} \times G} \over {i - {i_g}}} = {{1 \times 0.81} \over {10 - 1}} = 0.09\Omega$
2

AIEEE 2003

The thermo $e.m.f.$ of a thermo -couple is $25$ $\mu V/{}^ \circ C$ at room temperature. A galvanometer of $40$ $ohm$ resistance, capable of detecting current as low as ${10^{ - 5}}\,A,$ is connected with the thermo couple. The smallest temperature difference that can be detected by this system is
A
${16^0}C$
B
${12^0}C$
C
${8^0}C$
D
${20^0}C$

Explanation

Let $\theta$ be the smallest temperature difference that can be detected by the thermocouple, then

$I \times R = \left( {25 \times {{10}^{ - 6}}} \right)\theta$

where ${\rm I}$ is the smallest current which can be detected by the galvanometer of resistance $R.$

$\therefore$ ${10^{ - 5}} \times 40 = 25 \times {10^{ - 6}} \times \theta$

$\therefore$ $\theta = {16^ \circ }C.$
3

AIEEE 2003

The length of a wire of a potentiometer is $100$ $cm$, and the $e.$ $m.$ $f.$ of its standard cell is $E$ volt. It is employed to measure the $e.m.f.$ of a battery whose internal resistance in $0.5\Omega .$ If the balance point is obtained at $1=30$ $cm$ from the positive end, the $e.m.f.$ of the battery is

where $i$ is the current in the potentiometer wire.

A
${{30E} \over {100.5}}$
B
${{30E} \over {\left( {100 - 0.5} \right)}}$
C
${{30\left( {E - 0.5i} \right)} \over {100}}$
D
${{30E} \over {100}} - 0.5i$, where i is the current in the potentiometer wire

Explanation

Potential gradient along wire, K = ${E \over {100}}$ volt/cm

For battery V = E' – ir, where E' is emf of battery.

or K × 30 = E' – ir, where current i is drawn from battery

or ${{E \times 30} \over {100}}$ = E' + 0.5i

or E' = ${{30E} \over {100}} - 0.5i$
4

AIEEE 2002

If in the circuit, power dissipation is $150W,$ then $R$ is A
$2\,\Omega$
B
$6\,\Omega$
C
$5\,\Omega$
D
$4\,\Omega$

Explanation

The equivalent resistance is ${{\mathop{\rm R}\nolimits} _{eq}} = {{2 \times R} \over {2 + R}}$

$\therefore$ Powder dissipation $P = {{{V^2}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$

$\therefore$ $150 = {{15 \times 15} \over {{R_{eq}}}}$

$\therefore$ ${{\mathop{\rm R}\nolimits} _{eq}} = {{15} \over {10}} = {3 \over 2}$

$\Rightarrow {{2R} \over {2 + R}} = {3 \over 2}$

$\Rightarrow 4R = 6 + 3R$

$\Rightarrow R = 6\Omega$