$Assertion:$ The time period of a simple pendulum on a satellite orbiting the Earth is infinity.
$Reason:$ The time period of a simple pendulum is inversely proportional to $\sqrt{g}$.

  • A
    If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
  • B
    If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.
  • C
    If the Assertion is correct but Reason is incorrect.
  • D
    If both the Assertion and Reason are incorrect.

Explore More

Similar Questions

Which vector in the figures best represents the acceleration of a pendulum mass at the intermediate point in its swing?

$A$ simple pendulum of length $l_1$ has a periodic time of $2.4 \text{ s}$. Another simple pendulum of length $l_2 < l_1$ has a periodic time of $1.8 \text{ s}$. The periodic time of a simple pendulum of length $(l_1 - l_2)$ is nearly: (in $\text{ s}$)

$A$ heavy small-sized sphere is suspended by a string of length $l$. The sphere rotates uniformly in a horizontal circle with the string making an angle $\theta$ with the vertical. Then the time period of this conical pendulum is

Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing $SHM$ with frequency relation $f_{A}=2 f_{B}$,then the relation is:

In a simple pendulum,the period of oscillation $T$ is related to the length of the pendulum $l$ as

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