A wire of length ${\rm{L}}$, area of cross section ${\rm{A}}$ is hanging from a fixed support. The length of the wire changes to ${{\rm{L}}_{\rm{1}}}$ when mass ${\rm{M}}$ is suspended from its free end. The expression for Young’s modulus is:

$\frac{{{\rm{MgL}}}}{{{\rm{A}}\,{\rm{(}}{{\rm{L}}_{\rm{1}}} - {\rm{L)}}}}$

$\frac{{{\rm{Mg}}{{\rm{L}}_{\rm{1}}}}}{{{\rm{AL}}}}$

$\frac{{{\rm{Mg}}\,{\rm{(}}{{\rm{L}}_{\rm{1}}}{\rm{ - L)}}}}{{{\rm{AL}}}}$

$\frac{{{\rm{MgL}}}}{{{\rm{A}}{{\rm{L}}_{\rm{1}}}}}$

JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Medium

1 of 1

Application +4 | -1

A wire of length

\mathrm{L}

, area of cross section

\mathrm{A}

is hanging from a fixed support. The length of the wire changes to

{\mathrm{L}_\mathrm{1}}

when mass

\mathrm{M}

is suspended from its free end. The expression for Young’s modulus is:

View answer and explanation

Correct option: D

\frac{{\mathrm{MgL}}}{{\mathrm{A}\,\mathrm{(}{\mathrm{L}_\mathrm{1}} - \mathrm{L)}}}

Option D (

\frac{{\mathrm{MgL}}}{{\mathrm{A}\,\mathrm{(}{\mathrm{L}_\mathrm{1}} - \mathrm{L)}}}

) is correct because it is the correct formula for Young's modulus, which is stress divided by strain.

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l

L

l

\mathrm{A}

1 \mathrm{~cm}

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