$A$ simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4 \ m$,then the time period of small oscillations will be . . . . . . $s$. [Take $g = \pi^2 \ ms^{-2}$]

  • A
    $5$
  • B
    $6$
  • C
    $7$
  • D
    $8$

Explore More

Similar Questions

The value of acceleration due to gravity at a height of $10 \,km$ from the surface of the Earth is $x$. At what depth inside the Earth is the value of the acceleration due to gravity the same value $x$ (in $\,km$)?

The height in terms of radius of the earth $(R)$, at which the acceleration due to gravity becomes $g/9$, where $g$ is acceleration due to gravity on earth's surface, is . . . . . . .

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

Given below are two statements:
Statement $I$: The law of gravitation holds good for any pair of bodies in the universe.
Statement $II$: The weight of any person becomes zero when the person is at the centre of the earth.
In the light of the above statements,choose the correct answer from the options given below.

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo