 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2006

In a common base mode of a transistor, the collector current is $5.488$ $mA$ for an emitter current of $5.60mA.$ The value of the base current amplification factor $\left( \beta \right)$ will be
A
$49$
B
$50$
C
$51$
D
$48$

Explanation

${I_C} = 5.488\,mA,\,\,{I_e} = 5.6\,mA,\,{I_B} = {I_E} - {I_C}$

$\beta = {{{I_c}} \over {{I_B}}} = {{5.488} \over {5.6 - 5.485}} = 49$
2

AIEEE 2005

In a common base amplifier, the phase difference between the input signal voltage and output voltage is
A
$\pi$
B
${\pi \over 4}$
C
${\pi \over 2}$
D
$0$

Explanation

Zero; In common base amplifier circuit, input and output voltage are in the same phase.
3

AIEEE 2005

In a full wave rectifier circuit operating from $50$ $Hz$ mains frequency, the fundamental frequency in the ripple would be
A
$25$ $Hz$
B
$50$ $Hz$
C
$70.7$ $Hz$
D
$100$ $Hz$

Explanation

Input frequency, $f = 50\,Hz \Rightarrow T = {1 \over {50}}$

For full wave rectifier, ${T_1} = {T \over 2} = {1 \over {100}} \Rightarrow {f_1} = 100\,Hz.$
4

AIEEE 2005

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $2480$ $nm$ is incident on it. The band gap in $(eV)$ for the semiconductor is
A
$2.5$ $eV$
B
$1.1$ $eV$
C
$0.7$ $eV$
D
$0.5$ $eV$

Explanation

Band gap $=$ energy of photon of wavelength $2480$ $nm.$ So,

$\Delta E = {{hc} \over \lambda }$

$= \left( {{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {2480 \times {{10}^{ - 9}}}}} \right) \times {1 \over {1.6 \times {{10}^{ - 19}}}}eV$

$= 0.5\,eV$