A copper wire of negligible mass, ${\rm{1 m}}$ length and cross-sectional area ${10^{ - 6}}$ is kept on a smooth horizontal table with one end fixed. A ball of mass ${\rm{1}}$ kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rad ${s^{ - 1}}$. If the elongation in the wire is ${10^{ - 3}}{\rm{m}}$, then the Young’s modulus is

$4\,\, \times {10^{11}}\,{\rm{N}}{{\rm{m}}^{ - 2}}$

$6 \times {10^{11}}\,{\rm{N}}{{\rm{m}}^{ - 2}}$

$8\, \times {10^{11}}\,{\rm{N}}{{\rm{m}}^{ - 2}}$

$10\,\, \times {10^{11}}\,{\rm{N}}{{\rm{m}}^{ - 2}}$

JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Medium

1 of 1

Application +4 | -1

A copper wire of negligible mass,

\mathrm{1 m}

length and cross-sectional area

{10^{ - 6}}

is kept on a smooth horizontal table with one end fixed. A ball of mass

\mathrm{1}

kg is attached to the other end. The wire and the ball are rotated with an angular velocity 20 rad

{s^{ - 1}}

. If the elongation in the wire is

{10^{ - 3}}\mathrm{m}

, then the Young’s modulus is

View answer and explanation

Correct option: A

4\,\, \times {10^{11}}\,\mathrm{N}{\mathrm{m}^{ - 2}}

Option A (

4 \times {10^{11}}\,\mathrm{N}{\mathrm{m}^{ - 2}}

) is correct because it is derived from the formula relating Young's modulus to the force, length, area, and elongation.

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l

L

l

\mathrm{A}

1 \mathrm{~cm}

Y=\frac{F l}{A \Delta l},