JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Medium

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Application +4 | -1

A wire of length

4 \mathrm{~m}

is hanged vertically from ceiling if it is stretched due to its own weight, then find out the energy stored per unit volume.

\left[Y=2 \times 10^7 \mathrm{~N} / \mathrm{m}^2\right.

and

r=10^3

\mathrm{kg} / \mathrm{m}^3

]

View answer and explanation

Correct option: D

10 \mathrm{~J} / \mathrm{m}^3

Option D (10 J/m³) is correct because it is calculated using the formula for potential energy density in terms of Young's modulus, density, gravity, and length.

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A wire is stretched under a force. If the wire suddenly snaps the temperature of the wire The increases in length is

l

of a wire of length

L

by the longitudinal stress. Then the stress is proportional to A steel wire of length

l

and cross sectional area

\mathrm{A}

is stretched by

1 \mathrm{~cm}

under a given load. When the same load is applied to another steel wire of double its length and half of its cross section area, the amount of stretching (extension) is Assertion: From the relation

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

we can say that, if length of a wire is doubled, its Young's modulus of elasticity will also becomes two times.

Reason: Modulus of elasticity is a material property. A wire can be broken by applying a load of 200 N. The force required to break another wire of the same length and same material, but double in diameter, is Assertion(A): If radius of cylindrical wire is doubled, then this wire can bear four times stress.

Reason(R): By doubling the radius, area of cross-section will become four times.