Exams › JEE Main › Physics
The acceleration due to gravity is 0.5% larger at the poles than at the equator. A pendulum that beats seconds (period exactly 2 s) at the poles is taken to the equator. What is its period there?
- 1.950 s
- 1.995 s
- 2.050 s
- 2.005 s
Correct answer: 2.005 s
Solution
Smaller g at the equator lengthens the period. Use the ratio of periods with the 0.5% difference in g.
Related JEE Main Physics questions
- A block of mass M is attached to a horizontal spring and oscillates in simple harmonic motion with amplitude A1. When the block crosses the equilibrium position, a smaller mass m is gently placed on it, and thereafter the two move together. The new amplitude is A2. What is the value of the ratio A1/A2?
- An ideal gas is trapped in a vertical cylinder of cross-sectional area A by a frictionless piston of mass M. When the gas volume is V0 and its pressure is P0, the piston is in equilibrium. If the piston is given a small displacement from this position and then let go, and the entire setup is thermally isolated, the piston performs SHM with frequency
- A mass m is connected to a spring with spring constant k, giving it a natural angular frequency ω0. If an external driving force varying as cos(ωt) acts on the system, the oscillator’s displacement will be proportional to
- A particle undergoing simple harmonic motion has kinetic energy K0. The maximum potential energy and the total energy are, respectively,
- A simple pendulum is suspended from the roof of a lift that is initially at rest, and its period is T. The lift then moves upward such that the distance it covers is given by y = t², where y is in metres and t is in seconds. Taking g = 10 m/s², what will be the new time period of the pendulum?
- A damped oscillator’s amplitude falls to 0.9 of its initial value in 5 s. After an additional 10 s, what fraction of the initial amplitude will it have, if the factor is denoted by α?
⚔️ Practice JEE Main Physics free + battle 1v1 →