If effective length of a simple pendulum is equal to radius of earth (R), time period will be
$T = π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{2 R}{g}}$
$T = 2 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{R}{2 g}}$

Question

If effective length of a simple pendulum is equal to radius of earth (R), time period will be
$T = π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{2 R}{g}}$
$T = 2 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{R}{2 g}}$

Toppr Admin · Accepted Answer

Formula-T-2-x3C0-x221A-LRg-R-L-If the effective length of a simple pendulum is equal to the radius of the earth -R-then-T-2-x3C0-x221A-R-xD7-Rg-R-R-2-x3C0-x221A-R2g-2R-x2234-T-2-x3C0-x221A-R2g

If effective length of a simple pendulum is equal to radius of earth (R), time period will beT=π√RgT=2π√2RgT=2π√RgT=2π√R2g

formula,T=2π√LRg(R+L)If the effective length of a simple pendulum is equal to the radius of the earth (R),then,T=2π√R×Rg(R+R)=2π√R2g(2R)∴T=2π√R2g

If effective length of a simple pendulum is equal to radius of earth (R), time period will be
$T = π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{2 R}{g}}$
$T = 2 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{R}{2 g}}$

formula,

$T = 2 π \sqrt{\frac{L R}{g (R + L)}}$

If the effective length of a simple pendulum is equal to the radius of the earth (R),

then,

$T = 2 π \sqrt{\frac{R \times R}{g (R + R)}}$

$= 2 π \sqrt{\frac{R^{2}}{g (2 R)}}$

$∴ T = 2 π \sqrt{\frac{R}{2 g}}$