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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2003

MCQ (Single Correct Answer)
The displacement $$y$$ of a wave travelling in the $$x$$-direction is given by $$$y = {10^{ - 4}}\,\sin \left( {600t - 2x + {\pi \over 3}} \right)\,\,metres$$$
where $$x$$ is expressed in metres and $$t$$ in seconds. The speed of the wave - motion, in $$m{s^{ - 1}}$$, is
A
$$300$$
B
$$600$$
C
$$1200$$
D
$$200$$

Explanation

$$y = {10^{ - 4}}\sin \left( {600t - 2x + {\pi \over 3}} \right)$$

But $$y = A\sin \left( {\omega t - kx + \phi } \right)$$

On comparing we get $$\omega = 600;\,k = 2$$

$$v = {\omega \over k} = {{600} \over 2} = 300\,m{s^{ - 1}}$$
2

AIEEE 2003

MCQ (Single Correct Answer)
A metal wire of linear mass density of $$9.8$$ $$g/m$$ is stretched with a tension of $$10$$ $$kg$$-$$wt$$ between two rigid supports $$1$$ metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency $$n.$$ The frequency $$n$$ of the alternating source is
A
$$50$$ $$Hz$$
B
$$100$$ $$Hz$$
C
$$200$$ $$Hz$$
D
$$25$$ $$Hz$$

Explanation

KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by

$$n = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz$$

As the string is vibrating in resonance to a.c of frequency $$n,$$ therefore both the frequencies are same.
3

AIEEE 2002

MCQ (Single Correct Answer)
When temperature increases, the frequency of a tuning fork
A
increases
B
decreases
C
remains same
D
increases or decreases depending on the material

Explanation

KEY CONCEPT : The frequency of a tuning fork is given by the expression

$$f = {{{m^2}k} \over {4\sqrt 3 \pi {\ell ^2}}}\sqrt {{Y \over \rho }} $$

As temperature increases, $$\ell $$ increases and therefore $$f$$ decreases.
4

AIEEE 2002

MCQ (Single Correct Answer)
A wave $$y=a$$ $$\sin \left( {\omega t - kx} \right)$$ on a string meets with another wave producing a node at $$x=0.$$ Then the equation of the unknown wave is
A
$$y = \alpha \,\sin \,\left( {\omega t + kx} \right)$$
B
$$y = - \alpha \,\sin \,\left( {\omega t + kx} \right)$$
C
$$y = \alpha \,\sin \,\left( {\omega t - kx} \right)$$
D
$$y = - \alpha \,\sin \,\left( {\omega t - kx} \right)$$

Explanation

To form a node there should be superposition of this wave with the reflected wave. The reflected wave should travel in opposite direction with a phase change of $$\pi $$. The equation of the reflected wave will be

$$y = a\sin \left( {\omega t + kx + \pi } \right)$$

$$ \Rightarrow y = - a\sin \left( {\omega t + kx} \right)$$

Questions Asked from Waves

On those following papers in MCQ (Single Correct Answer)
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AIEEE 2002 (5)
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