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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Medium

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Application +4 | -1

An aluminium rod has a breaking strain 0.1%. The minimum cross-sectional area of the rod in

{m^2}

in order to support a load of

{10^6}\,N

is if (Young’s modulus is

7 \times {10^9}\,\,N{m^{ - 2}})

View answer and explanation

Correct option: C

0.14

Option C (0.14) is correct because the minimum cross-sectional area is calculated using the formula

A = \frac{{Fl}}{{eY}}

, where

F

is the force,

l

is the original length,

e

is the extension, and

Y

is Young's modulus.

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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.