Question

A body of mass '$m$' is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1cm$. If the angular velocity is doubled, the elongation in the spring is $5cm$. The original length of the spring is

• A. $15cm$
• B. $12cm$
• C. $16cm$
• D. $10cm$

Answer 1

The correct option is A $15cm$

Given,

Centrifugal force will stretch the string

$m(l+1)\omega {}^2=kx$

At elongation $(x=1)$

$m(l+1)\omega {}^2=k\times 1......\left(1\right)$

At elongation $(x=5)$

$m(l+5){\left(2\omega \right)}^2=K\times 5......\left(2\right)$
From (1) and (2)
$l=15cm$