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JEE Main 2021 (Online) 27th August Evening Shift

Numerical
Two simple harmonic motion, are represented by the equations $${y_1} = 10\sin \left( {3\pi t + {\pi \over 3}} \right)$$ $${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$$ Ratio of amplitude of y1 to y2 = x : 1. The value of x is ______________.
Your Input ________

Answer

Correct Answer is 1

Explanation

$${y_1} = 10\sin \left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ Amplitude = 10

$${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$$

$${y_2} = 10\left( {{1 \over 2}\sin 3\pi t + {{\sqrt 3 } \over 2}\cos 3\pi t} \right)$$

$${y_2} = 10\left( {\cos {\pi \over 3}\sin 3\pi t + \sin {\pi \over 3}\cos 3\pi t} \right)$$

$${y_2} = 10\left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ Amplitude = 10

So ratio of amplitudes = $${{10} \over {10}}$$ = 1
2

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
Two simple harmonic motions are represented by the equations

$${x_1} = 5\sin \left( {2\pi t + {\pi \over 4}} \right)$$ and $${x_2} = 5\sqrt 2 (\sin 2\pi t + \cos 2\pi t)$$. The amplitude of second motion is ................ times the amplitude in first motion.
Your Input ________

Answer

Correct Answer is 2

Explanation

$${x_2} = 5\sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2\pi t + {1 \over {\sqrt 2 }}\cos 2\pi t} \right)\sqrt 2 $$

$$ = 10\sin \left( {2\pi t + {\pi \over 4}} \right)$$

$$\therefore$$ $${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$$
3

JEE Main 2021 (Online) 27th July Evening Shift

Numerical
A particle executes simple harmonic motion represented by displacement function as

x(t) = A sin($$\omega$$t + $$\phi$$)

If the position and velocity of the particle at t = 0 s are 2 cm and 2$$\omega$$ cm s$$-$$1 respectively, then its amplitude is $$x\sqrt 2 $$ cm where the value of x is _________________.
Your Input ________

Answer

Correct Answer is 2

Explanation

x(t) = A sin($$\omega$$t + $$\phi$$)

v(t) = A$$\omega$$ cos ($$\omega$$t + $$\phi$$)

2 = A sin$$\phi$$ ...... (1)

2$$\omega$$ = A$$\omega$$ cos$$\phi$$ ....... (2)

From (1) and (2)

tan$$\phi$$ = 1

$$\phi$$ = 45$$^\circ$$

Putting value of $$\phi$$ in equation (1),

$$2 = A\left\{ {{1 \over {\sqrt 2 }}} \right\}$$

$$A = 2\sqrt 2 $$

$$ \therefore $$ x = 2
4

JEE Main 2021 (Online) 25th July Morning Shift

Numerical
A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60$$^\circ$$ to the vertical will be (g = 10 m/s2) ____________ m/s.
Your Input ________

Answer

Correct Answer is 2

Explanation



Applying work energy theorem :

wg + wT = $$\Delta$$K

$$-$$mgl(1 $$-$$ cos60$$^\circ$$) = $${1 \over 2}$$mv2 $$-$$ $${1 \over 2}$$mu2

v2 = u2 $$-$$ 2gl(1 $$-$$ cos60$$^\circ$$)

v2 = 9 $$-$$ 2 $$\times$$ 10 $$\times$$ 0.5$$\left( {{1 \over 2}} \right)$$

v2 = 4

v = 2 m/s

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