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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
An $$LCR$$ circuit is equivalent to a damped pendulum. In an $$LCR$$ circuit the capacitor is charged to $${Q_0}$$ and then connected to the $$L$$ and $$R$$ as shown below :

If a student plots graphs of the square of maximum charge $$\left( {Q_{Max}^2} \right)$$ on the capacitor with time$$(t)$$ for two different values $${L_1}$$ and $${L_2}$$ $$\left( {{L_1} > {L_2}} \right)$$ of $$L$$ then which of the following represents this graph correctly ?
$$\left( {plots\,\,are\,\,schematic\,\,and\,\,niot\,\,drawn\,\,to\,\,scale} \right)$$

A
B
C
D

Explanation

From $$KVL$$ at any time $$t$$



$${q \over c} - iR - L{{di} \over {dt}} = 0$$

$$i = - {{dq} \over {dt}} \Rightarrow {q \over c} + {{dq} \over {dt}}R + {{L{d^2}q} \over {d{t^2}}} = 0$$

$${{{d^2}q} \over {d{t^2}}} + {R \over L}{{dq} \over {dt}} + {q \over {Lc}} = 0$$

From damped harmonic oscillator, the amplitude is

given by $$A = {A_0}e - {{dt} \over {2m}}$$

Double differential equation

$${{{d^2}x} \over {d{t^2}}} + {b \over m}{{dx} \over {dt}} + {k \over m}x = 0$$

$${Q_{\max }} = {Q_0}e - {{Rt} \over {2L}} \Rightarrow Q_{\max }^2 = Q_0^2e - {{Rt} \over L}$$

Hence damping will be faster for lesser self inductance.
2

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
An arc lamp requires a direct current of $$10$$ $$A$$ at $$80$$ $$V$$ to function. If it is connected to a $$220$$ $$V$$ $$(rms),$$ $$50$$ $$Hz$$ $$AC$$ supply, the series inductor needed for it to work is close to :
A
$$0.044$$ $$H$$
B
$$0.065$$ $$H$$
C
$$80$$ $$H$$
D
$$0.08$$ $$H$$

Explanation

Here

$$i = {e \over {\sqrt {{R^2} + X_L^2} }}$$

$$ = {e \over {\sqrt {{R^2} + {\omega ^2}{L^2}} }}$$

$$ = {e \over {\sqrt {{R^2} + 4{\pi ^2}{v^2}{L^2}} }}$$

$$10 = {{220} \over {\sqrt {64 + 4{\pi ^2}{{\left( {50} \right)}^2}L} }}$$

$$\left[ {\,\,\,} \right.$$ as $$\left. {R = {V \over {\rm I}} = {{80} \over {10}} = 8\,\,\,} \right]$$

On solving we get

$$L = 0.065\,H$$
3

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Two coaxial solenoids of different radius carry current $$I$$ in the same direction. $$\overrightarrow {{F_1}} $$ be the magnetic force on the inner solenoid due to the outer one and $$\overrightarrow {{F_2}} $$ be the magnetic force on the outer solenoid due to the inner one. Then :
A
$$\overrightarrow {{F_1}} $$ is radially in wards and $$\overrightarrow {{F_2}} = 0$$
B
$$\overrightarrow {{F_1}} $$ is radially outwards and $$\overrightarrow {{F_2}} = 0$$
C
$$\overrightarrow {{F_1}} = \overrightarrow {{F_2}} = 0$$
D
$$\overrightarrow {{F_1}} $$ is radially inwards and $$\overrightarrow {{F_2}} $$ is radially outards

Explanation

$$\mathop {F{}_1}\limits^ \to = \mathop {F{}_2}\limits^ \to = 0$$

because of action and reaction pair
4

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
Match List - $$1$$ (Electromagnetic wave type ) with List - $$2$$ (Its association/application) and select the correct option from the choices given below the lists:
A
$$1 - iv,\,\,2 - iii,\,\,3 - ii,\,\,4 - i$$
B
$$1 - i,\,\,2 - ii,\,\,3 - iv,\,\,4 - iii$$
C
$$1 - iii,\,\,2 - ii,\,\,3 - i,\,\,4 - iv$$
D
$$1 - i,\,\,2 - ii,\,\,3 - iii,\,\,4 - iv$$

Explanation

$$\left( 1 \right)\,\,\,\,\,\,$$ Infrared rays are used to treat muscular strain because these are heat rays.

$$\left( 2 \right)\,\,\,\,\,\,$$ Radio waves are used for broadcasting because these waves have very long wavelength runging from few centimeters to few hundred kilometers

$$\left( 3 \right)\,\,\,\,\,\,$$ $$X$$-rays are used to detect fracture of bones because they have high penetrating power but they can't penetrate through denser medium like bones.

$$\left( 4 \right)\,\,\,\,\,\,$$ Ultraviolet rays are absorbed by ozone of the atmosphere.

Questions Asked from Alternating Current and Electromagnetic Induction

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
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