A simple pendulum has a time period $T_{1}$ when on 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 $\frac{T_{2}}{T_{1}}$ is

Q. A simple pendulum has a time period

T_{1}

, when on the earth's surface ; and

T_{2}

, when takes to height

R

above the earth's surface (

R

is the radius of the earth ). The value of

T_{2} / T_{1}

is:

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A simple pendulum has a time period T1 when on the earth's surface and T2 when taken to a height 2R above the earth's surface, where R is the radius of the earth. The value of (T1T2) is:

The correct option is B 13T1=2π√lg g′=gR2(R+h)2=gR2(R+2R)2 ∵h=2R g′=g9 T2=2π ⎷lg9 T2=2π√9l3=3T1, T1T2=13

A simple pendulum has a time period $T_{1}$ when on the earth's surface and $T_{2}$ when taken to a height $2 R$ above the earth's surface, where $R$ is the radius of the earth. The value of $(\frac{T_{1}}{T_{2}})$ is: