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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
The bob of a simple pendulum executes simple harmonic motion in water with a period $$t,$$ while the period of oscillation of the bob is $${t_0}$$ in air. Neglecting frictional force of water and given that the density of the bob is $$\left( {4/3} \right) \times 1000\,\,kg/{m^3}.$$ What relationship between $$t$$ and $${t_0}$$ is true
A
$$t = 2{t_0}$$
B
$$t = {t_0}/2$$
C
$$t = {t_0}$$
D
$$t = 4{t_0}$$

Explanation

$$t = 2\pi \sqrt {{\ell \over {{g_{eff}}}}} ;\,{t_o}\,\, = 2\pi \sqrt {{\ell \over g}} $$



$$m{g_{eff}} = mg - B = my - V \times 100 \times g$$

$$\therefore$$ $${g_{eff}} = g - {{100} \over {\left( {m/v} \right)}}g$$

$$ = g - {{1000} \over {{4 \over 3} \times 1000}}g = {g \over 4}$$

$$\therefore$$ $$t = 2\pi \sqrt {{\ell \over {g/4}}} \,\,\,\,\,\,\,\,\,\,\,t = 2{t_0}$$
2

AIEEE 2003

MCQ (Single Correct Answer)
The displacement of particle varies according to the relation
$$x=4$$$$\left( {\cos \,\pi t + \sin \,\pi t} \right).$$ The amplitude of the particle is
A
$$-4$$
B
$$4$$
C
$$4\sqrt 2 $$
D
$$8$$

Explanation

$$x = 4\left( {\cos \pi t + \sin \pi t} \right)$$

$$ = \sqrt 2 \times 4\left( {{{\sin \pi t} \over {\sqrt 2 }} + {{\cos \pi t} \over {\sqrt 2 }}} \right)$$

$$x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right)$$
3

AIEEE 2003

MCQ (Single Correct Answer)
A body executes simple harmonic motion. The potential energy $$(P.E),$$ the kinetic energy $$(K.E)$$ and total energy $$(T.E)$$ are measured as a function of displacement $$x.$$ Which of the following statements is true ?
A
$$K.E$$ is maximum when $$x=0$$
B
$$T.E$$ is zero when $$x=0$$
C
$$K.E$$ is maximum when $$x$$ is maximum
D
$$P.E$$ is maximum when $$x=0$$

Explanation

$$K.E. = {1 \over 2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)$$

When $$x=0,$$ $$K.E$$ is maximum and is equal to $${1 \over 2}m{\omega ^2}{a^2}.$$
4

AIEEE 2003

MCQ (Single Correct Answer)
The length of a simple pendulum executing simple harmonic motion is increased by $$21\% $$. The percentage increase in the time period of the pendulum of increased length is
A
$$11\% $$
B
$$21\% $$
C
$$42\% $$
D
$$10\% $$

Explanation

$$T = 2\pi \sqrt {{\ell \over g}} $$ and $$T' = 2\pi \sqrt {{{1.21\ell } \over g}} $$

$$\left( \, \right.$$ as $$\ell ' = \ell + 21\% $$ of $$\left. \ell \right)$$

$$\% $$ increase $$ = {{T' - T} \over T} \times 100$$

$$ = {{\sqrt {1.21\ell } - \sqrt \ell } \over {\sqrt \ell }} \times 100$$

= $${{\sqrt {{{121} \over {100}}l} - \sqrt l } \over {\sqrt l }} \times 100$$

= $${{{{11} \over {10}}\sqrt l - \sqrt l } \over {\sqrt l }} \times 100$$

= $${{{{11} \over {10}} - 1} \over 1} \times 10$$

$$ = 10\% $$

Questions Asked from Simple Harmonic Motion

On those following papers in MCQ (Single Correct Answer)
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