SATHEE: Gravitation Question 112

Gravitation Question 112

Question: A simple pendulum has a time period $[T_{1}]$ when on the earth’s surface and $[T_{2}]$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of $[T_{2} / T_{1}]$ is[IIT-JEE 2001]

Options:

A) 1

B) $[\sqrt{2}]$

C) 4

D) 2

Show Answer

Answer:

Correct Answer: D

Solution:

If acceleration due to gravity is g at the surface of earth then at height R it value becomes $[g^{'} = g {(\frac{R}{R + h})}^{2} = \frac{g}{4}]$ $[T_{1} = 2 π \sqrt{\frac{l}{g}} and, T_{2} = 2 π \sqrt{\frac{l}{g / 4}}]$ $[\frac{T_{2}}{T_{1}} = 2]$

Gravitation Question 112

Question: A simple pendulum has a time period $[T_{1}]$ when on the earth’s surface and $[T_{2}]$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of $[T_{2} / T_{1}]$ is[IIT-JEE 2001]

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

Important Resources

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NCERT Exercise Video Solution

Gravitation Question 112

Question: A simple pendulum has a time period [T1] when on the earth’s surface and [T2] when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of [T2/T1] is[IIT-JEE 2001]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: A simple pendulum has a time period $[T_{1}]$ when on the earth’s surface and $[T_{2}]$ when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of $[T_{2} / T_{1}]$ is[IIT-JEE 2001]