Question

A mass 'm' kg is whirled in a vertical plane by
tying it at the end of a flexible wire of length 'L'
and area of cross-section 'A' such that it just
completes the vertical circle. When the mass is at
its lowest position, the strain produced in the wire is
(Young's modulus of the wire is 'Y')

1)AY/6 mg
2)6mgAY
3)5mgAY
4)AY/5mg​

Answer 1

Solution :

At the lowest point of journey net force (T) at lowest point is:

$\implies$ $\sf{mg - T = \frac{mv^{2}}{l}}$

$\implies$ $\sf{T =\frac{mv^{2}}{l} +mg}$ - - - - (1)

This tension produce extension in string, hence using Hooks law:

$\implies$ $\boxed{\sf{Y =\frac{Stress}{Strain}}}$

We get:

$\implies$ $\sf{Y =\frac{\frac{T}{A}}{Strain}}$

$\implies$ $\sf{Strain= \frac{T}{AY}}$ - - - - (2)

Now substituting the value of T from eq.(1 ), we get:

$\implies$ $\sf{Strain = \frac{ \frac{ {mv}^{2} }{I} + mg}{AY}}$

$\implies$ $\sf{Strain = \frac{ \frac{m( \sqrt{5g \times 1}) ^{2} }{I} + mg}{AY}}$

Note:

Putting value of $\sf{v = \sqrt{5gl} = \sqrt{5g \times 1}}$

$\implies$ $\sf{Strain = \frac{ \frac{m5g}{1} + mg }{AY}}$

$\implies$ $\sf{Strain = \frac{6mg}{AY}}$

Therefore,

Correct option: (2) 6mg/AY

________________

Answered by: Niki Swar, Goa❤️