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JEE Main 2021 (Online) 20th July Evening Shift
MCQ (Single Correct Answer)  +4  -1
A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as :
A
$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 - v_2^2}}} $$
B
$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 + v_2^2}}} $$
C
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 + v_2^2}}} $$
D
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}} $$
2
JEE Main 2021 (Online) 20th July Evening Shift
MCQ (Single Correct Answer)  +4  -1
Consider a binary star system of star A and star B with masses mA and mB revolving in a circular orbit of radii rA an rB, respectively. If TA and TB are the time period of star A and star B, respectively,

Then :
A
$${{{T_A}} \over {{T_B}}} = {\left( {{{{r_A}} \over {{r_B}}}} \right)^{{3 \over 2}}}$$
B
$${T_A} = {T_B}$$
C
$${T_A} > {T_B}$$ (if $${m_A} > {m_B}$$)
D
$${T_A} > {T_B}$$ (if $${r_A} > {r_B}$$)
3
JEE Main 2021 (Online) 18th March Evening Shift
MCQ (Single Correct Answer)  +4  -1
The function of time representing a simple harmonic motion with a period of $${\pi \over \omega }$$ is :
A
cos($$\omega$$t) + cos(2$$\omega$$t) + cos(3$$\omega$$t)
B
sin2($$\omega$$t)
C
sin($$\omega$$t) + cos($$\omega$$t)
D
3cos$$\left( {{\pi \over 4} - 2\omega t} \right)$$
4
JEE Main 2021 (Online) 18th March Morning Shift
MCQ (Single Correct Answer)  +4  -1
The time period of a simple pendulum is given by $$T = 2\pi \sqrt {{l \over g}} $$. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-
A
2%
B
3%
C
5%
D
4%
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