T is the time period of simple pendulum on the earth's surface. Its time period becomes x T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:

Question

T is the time period of simple pendulum on the earth's surface. Its time period becomes x T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be: (A) (1/4) (B) 4 (C) (1/2) (D) 2

Tardigrade Admin · Accepted Answer

At surface of earth time period T=2 π √(ℓ/ g ) At height h = R g prime=( g /(1+( h ) R )2)=( g /4) ∴ xT =2 π √(ℓ/( g / 4)) ⇒ xT =2 × 2 π √(ℓ/ g ) ⇒ xT =2 T ⇒ x =2

Q. $T$ is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:

$\frac{1}{4}$

4

$\frac{1}{2}$

2

At surface of earth time period
$T = 2 π \frac{ℓ}{g}$
At height $h = R$
$g^{'} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}} = \frac{g}{4}$
$∴ x T = 2 π \frac{ℓ}{( g /4 )}$
$\Rightarrow x T = 2 \times 2 π \frac{ℓ}{g}$
$\Rightarrow x T = 2 T \Rightarrow x = 2$

Q. T is the time period of simple pendulum on the earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:

41​

4

21​

2

At surface of earth time period T=2πgℓ​​ At height h=R g′=(1+Rh​)2g​=4g​ ∴xT=2π(g/4)ℓ​​ ⇒xT=2×2πgℓ​​ ⇒xT=2T⇒x=2

Q. $T$ is the time period of simple pendulum on the earth's surface. Its time period becomes $x T$ when taken to a height $R$ (equal to earth's radius) above the earth's surface. Then, the value of $x$ will be:

$\frac{1}{4}$

$\frac{1}{2}$

At surface of earth time period
$T = 2 π \frac{ℓ}{g}$
At height $h = R$
$g^{'} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}} = \frac{g}{4}$
$∴ x T = 2 π \frac{ℓ}{( g /4 )}$
$\Rightarrow x T = 2 \times 2 π \frac{ℓ}{g}$
$\Rightarrow x T = 2 T \Rightarrow x = 2$