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

Question

A simple pendulum has a time period T1 when on the earths surface and T2 when taken to a height 2R above the earths surface where R is the radius of the earth. The value of (T1/T2) is: (A)  (1/9)  (B)  (1/3)  (C)  √3  (D)  9

Tardigrade Admin · Accepted Answer

The periodic time of a simple pendulum is given by, T=2π √(l/g) Where taken to height 2R. g=g( 1+(h/Re) )2=g( 1+(2R/R) )-2=g(3)-2 ∴ (T1/T2)=√(1/32) ⇒ T2=3T1 ⇒ (T2/T1)=3

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

$\frac{1}{9}$

$\frac{1}{3}$

$3$

$9$

3

Q. A simple pendulum has a time period T1​ when on the earths surface and T2​ when taken to a height 2R above the earths surface where R is the radius of the earth. The value of (T1​/T2​) is:

91​

31​

3​

9

3

The periodic time of a simple pendulum is given by, T=2πgl​​ Where taken to height 2R. g=g(1+Re​h​)2=g(1+R2R​)−2=g(3)−2 ∴ T2​T1​​=321​​ ⇒ T2​=3T1​ ⇒ T1​T2​​=3

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

$\frac{1}{9}$

$\frac{1}{3}$

$3$

$9$