Question

Define practical simple pendulum. Show that motion of bob of pendulum with small amplitude is linear S.H.M. Hence obtain an expression for its period. What are the factors on which its period depends ?
The total free surface energy of a liquid drop is ` pi sqrt 2 ` times the surface tensin of the liquid. Calculate the diameter of the drop in S.I unit.

Answer 1

Factors on which its period depends are :
(a) Period of simple pendulum is directly proportional to the square root of its length
i.e., ` " " T prop sqrt L `
( b ) Period of simple pendulum is inversely proportional to the square root of acceleration due to gravity.
i.e., ` T prop ( 1 ) /( sqrt g ) `
Numerical :
Given : ` " "E = pi sqrt 2 T `
` E = T Delta A `
` E = 4pi r ^ 2 T" " [ because Delta A = 4pi r ^ 2 ] `
` pi sqrt 2 T = 4pi r ^ 2 T `
` 2 sqrt 2 r^ 2 = 1 `
` therefore " " r ^ 2 = ( 1 ) /( 2 sqrt 2 ) `
` rArr r = (( 1 )/ ( 2 sqrt 2 ))^ ( 1//2) = 0.5946 `
` therefore ` Diametre , ` d = 2 r `
` = 2 xx 0.5946 `
` = 1.1893 m `