JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

In an $LCR$ series $a.c.$ circuit, the voltage across each of the components, $L,C$ and $R$ is $50V$. The voltage across the $L.C$ combination will be
A
$100V$
B
$50\sqrt 2$
C
$50$ $V$
D
$0$ $V$ (zero)

Explanation

Since the phase difference between $L$ & $C$ is $\pi ,$

$\therefore$ net voltage difference across $LC=50-50=0$
2

AIEEE 2004

Alternating current can not be measured by $D.C.$ ammeter because
A
Average value of current for complete cycle is zero
B
$A.C.$ Changes direction
C
$A.C.$ can not pass through $D.C.$ Ammeter
D
$D.C.$ Ammeter will get damaged.

Explanation

$D.C.$ ammeter measure average current in $AC$ current, average current is zero for complete cycle. Hence reading will be zero.
3

AIEEE 2003

The core of any transformer is laminated so as to
A
reduce the energy loss due to eddy currents
B
make it light weight
C
make it robust and strong
D
increase the secondary voltage

Explanation

Laminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.
4

AIEEE 2003

When the current changes from $+ 2A$ to $-2A$ in $0.05$ second, an $e.m.f.$ of $8$ $V$ is inducted in a coil. The coefficient of self- induction of the coil is
A
$0.2H$
B
$0.4H$
C
$0.8$ $H$
D
$0.1$ $H$

Explanation

$e = - {{\Delta \phi } \over {\Delta t}} = {{ - \Delta \left( {LI} \right)} \over {\Delta t}} = - L{{\Delta I} \over {\Delta t}}$

$\therefore$ $\left| e \right| = L{{\Delta I} \over {\Delta t}} \Rightarrow 8 = L \times {4 \over {0.05}}$

$\Rightarrow L = {{8 \times 0.05} \over 4} = 0.1H$