Home
Class 12
PHYSICS
STATEMENT -1 : The time period of oscial...

STATEMENT -1 `:` The time period of osciallation of a simple pendulum of constant length is more at a place inside a mine than on the surface of the earth.
and
STATEMENT -2 `:` The frequency of osciallations ofa simple pendulum is more at aplace inside a mine than on the surface of the earth .

Text Solution

Verified by Experts

The correct Answer is:
3
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    AAKASH INSTITUTE|Exercise ASSIGNMENT ( SECTION-F( MATRIX MATCH TYPE QUESTIONS) )|3 Videos

  • OSCILLATIONS

    AAKASH INSTITUTE|Exercise ASSIGNMENT ( Section-G( Integers Answer Type Question) )|4 Videos

  • OSCILLATIONS

    AAKASH INSTITUTE|Exercise ASSIGNMENT (SECTION -D( Comprehension) )|1 Videos

  • NUCLEI

    AAKASH INSTITUTE|Exercise ASSIGNMENT (SECTION-D)|10 Videos

  • PHYSICAL WORLD

    AAKASH INSTITUTE|Exercise ASSIGNMENT (Section-B)|5 Videos

Similar Questions

Explore conceptually related problems

The period of oscillation of a simple pendulum of constant length at earth surface is T. Its period inside a mine is C

The time period of a simple pendulum of infinite length is (R=radius of earth).

The period of oscillation of a simple pendulum of constant length at surface of the earth is T. Its time period inside mine will be

Time period of a simple pendulum inside a satellite orbiting earth is

The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is

If T_(1) and T_(2) are the time-periods of oscillation of a simple pendulum on the surface of earth (of radius R) and at a depth d, the d is equal to

STATEMENT-1 : Time period of a simple pendulum changeswhen the solid bob is replaced by a hollow sphere of same radius but difference mass and STATEMENT-2 : The time period of a simple pendulum depends on force acting on bob per unit mass dueto the earth.

If the length of a simple pendulum is equal to the radius of the earth, its time period will be