Question

A mass of 100g is attached to the end of a rubber string 49cm long and

Answer 1

The answer is "100 x 51 x $(80 π)^{2}$".

Explanation:

A mass of 100 gram is attached to the end of a rubber string 49 cm long and having an area of cross section 20 sq. mm. the string is whirled round, horizontally at a constant speed of 40 r.p.s. in a circle of radius 51 cm. find the Young's modulus of rubber. Explain the answer also.

Given ,

L = 49 cm

l = 51 cm - 49 c = 2 cm

Cross sectional area =  20 $mm^{2}$ = 0.2 $cm^{2}$,

mass (m) = 100 g , R = 51 cm

ω = 40 r.p.s = 80 π rad·$sec^{- 1}$

When the mass attached to the string is whirled, the length of the string increases under the effect of centrifugal force.

The centrifugal force, F = mR$ω^{2}$

= 100 x 51 x $(80 π)^{2}$

Answer 2

Thus the value of Young's modulus is Y=3.95×10^9 N.m−2

Explanation:

Complete statement:

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

F= m r ω^2

F = 100 × 51 × (2 × π × 40)^2 dyne  

Y = F× l / A × Δl

Y  = ( 100 × 51 × 4π^2 × 1600) × 49 / ( 20×10−2) × (51 − 49)

Y = 3.95 × 10^10 dyne cm−2

Y =3.95×10^9 N.m−2