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A mass of 100 grams is attached to the e...

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.

Text Solution

Verified by Experts

The correct Answer is:
3.95
When the mass is rotated at the end of the rubber string, the restoring force in the string is equal to the centripetal force.
`therefore F= mr omega^(2) = mr(2pi v)^(2)`
`=100 xx 51 xx (2 xx pi xx 40)^(2)` dyne
Also l= 49cm, `Delta l= 51- 49= 2cm`
`A = 20 m m^(2) = 20 xx 10^(-2) cm^(2)`
Hence `Y= (F)/(A).(l)/(Delta l)`
`= (100 xx 51 xx 4 xx 9.87 xx 1600 xx 49)/(20 xx 10^(-2) xx 2) [pi^(2) = 9.87]`
`= 3.95 xx 10^(10)` dyne `cm^(-2)`
`=3.95 xx 10^(9) Nm^(-2)`
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