JEE Main 2021 (Online) 18th March Evening Shift | Simple Harmonic Motion Question 82 | Physics

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 :-

JEE Main 2021 (Online) 17th March Evening Shift

MCQ (Single Correct Answer)

-1

Two particles A and B of equal masses are suspended from two massless springs of spring constants K₁ and K₂ respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is

$${{{K_1}} \over {{K_2}}}$$

$$\sqrt {{{{K_1}} \over {{K_2}}}} $$

$${{{K_2}} \over {{K_1}}}$$

$$\sqrt {{{{K_2}} \over {{K_1}}}} $$

JEE Main 2021 (Online) 17th March Evening Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion . (take ln 2 = 0.693)

0.69 $$\times$$ 10² kg s^$$-$$1

3.3 $$\times$$ 10² kg s^$$-$$1

1.16 $$\times$$ 10^$$-$$2 kg s^$$-$$1

5.7 $$\times$$ 10^$$-$$3 kg s^$$-$$1

PREVIOUS NEXT

Questions Asked from Simple Harmonic Motion (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions