A body of mass 200 g is tied to a spring of spring constant 12 . 5 N m - 1 , while the other end of spring is fixed at point O . If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad s - 1 , then the ratio of extension in the spring to its natural length will be :

Question

A body of mass 200 g is tied to a spring of spring constant 12 . 5 N m - 1 , while the other end of spring is fixed at point O . If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad s - 1 , then the ratio of extension in the spring to its natural length will be :, 1 : 2, 1 : 1, 2 : 3, 2 : 5

MARKS App · Accepted Answer

Let the extension in spring be x and the natural length of spring be l . Here the centrifugal force will stretch the spring. So, centrifugal force = spring force k x = m l + x ω 2 ⇒ 12 . 5 x = 1 5 l + x 25 ⇒ 1 . 5 x = l ⇒ x l = 1 1 . 5 = 2 3 ⇒ x : l = 2 : 3

Search any question & find its solution

Looking for more such questions to practice?