NEET UG PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Easy

1 of 1

Application +4 | -1

A wire of length '

2 \mathrm{~m}

' is clamped horizontally between two fixed support. A mass

\mathrm{m}=5 \mathrm{~kg}

is hanged from middle of wire. The vertical and depression in wire (in

\mathrm{cm}

) in equilibrium is (Young modulus of wire

=2.4 \times 10^9 \mathrm{~N} / \mathrm{m}^2

, cross-sectional area

=1 \mathrm{~cm}^2

)

View answer and explanation

Correct option: A

Option A is correct. The vertical depression in the wire is calculated using the formula for the sag of a wire under its own weight, considering Young's modulus and the wire's properties.

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A 20 kg load is suspended by a wire of area of cross section is

1\,m{m^2}

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\left[ {g = 10\,m{s^{ - 2}}\,} \right]\,

In magnitude hydraulic stress is equal to: A wire extends by

\mathrm{1 mm}

when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be Stress to strain ratio is equivalent to A

2\,\,m

long rod of radius

1\,\,cm

which is fixed from one end is given a twist of

0.8

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Y

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\mathrm{K}

and modulus of rigidity

\eta