Question

A mass of 100 gm is attached to end of rubber string 49 cm long and having an area of cross section 20 square mm. The string is whirled round horizontally at a constant speed of 40 r.p.s. in circle of radius 51 cm. Find the Young modulus of rubber?

Answer 1

When the mass attached at the end of rubber string is whirled into a horizontal circle, the restoring force in the rubber string is equal to centripetal force.

F = mrω² = 100 x 51 x (2π x 40)² dyne

\(\therefore\ Y = \frac{F\times l}{A\times\Delta l} = \frac{100\times51\times4\pi ^2\times40\times40\times49}{20\times10^{-2}(51 - 49)}\)

= 3.95 x 10¹¹ dyne/cm²

= 3.95 x 10⁹ N/m²

Bulk Modulus of Elasticity