A string of area of cross-section $4 \mathrm{~mm}^{2}$ and length 0.5 is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of $5 \mathrm{~m} / \mathrm{s}$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is $...... \times 10^{-5}$. (Use Young's modulus $10^{11} \mathrm{~N} / \mathrm{m}^{2}$ and $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Elasticity & SHM

A string of area of cross-section

4 mm^{2}

and length 0.5 is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of

5 m / s

at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is

...... \times 1 0^{- 5}

. (Use Young's modulus

1 0^{11} N / m^{2}

and

g = 10 m / s^{2}

)

Answer: 30

Solution

At the bottom of the circular path, the tension in the string is:

T = m g + \frac{m v ^{2}}{R} = 2 \times 10 + \frac{2 \times 25}{0.5} = 20 + 100 = 120

N.

Strain is defined as

ϵ = \frac{stress}{Y} = \frac{F / A}{Y} = \frac{F}{A Y}

.

Given

A = 4

mm²

= 4 \times 1 0^{- 6}

m² and

Y = 1 0^{11}

N/m²:

ϵ = \frac{120}{4 \times 1 0 ^{- 6} \times 1 0 ^{11}} = \frac{120}{4 \times 1 0 ^{5}} = 30 \times 1 0^{- 5}

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Elasticity & SHM →

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Elasticity & SHM

Solution

More JEE Main questions on Elasticity & SHM