JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

MCQ (Single Correct Answer)
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' = ${{30E} \over {100}} - 0.5i$
2

AIEEE 2002

MCQ (Single Correct Answer)
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 $P = {{{V^2}} \over {{{\mathop{\rm R}\nolimits} _{eq}}}}$

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

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

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

$\Rightarrow 4R = 6 + 3R$

$\Rightarrow R = 6\Omega$
3

AIEEE 2002

MCQ (Single Correct Answer)
If a current is passed through a spring then the spring will
A
expand
B
compress
C
remains same
D
none of these

Explanation

When current is passed through a spring then current flows parallel in the adjacent turns.

NOTE : When two wires are placed parallel to each other and current flows in the same direction, the wires attract each other.

Similarly here the various turns attract each other and the spring will compress.
4

AIEEE 2002

MCQ (Single Correct Answer)
The mass of product liberated on anode in an electrochemical cell depends on (where $t$ is the time period for which the current is passed).
A
${\left( {It} \right)^{1/2}}$
B
$It$
C
$I/t$
D
${I^2}t$

Explanation

According to Faraday's first law of electrolysis

$m = ZIt \Rightarrow m \propto It$

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