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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Medium

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Application +4 | -1

The area of cross section of a steel wire

\left( {Y = 2.0\,\, \times {{10}^{11}}\,N/{m^2}} \right)

is

0.1\,\,c{m^2}

. The force required to double its length will be

View answer and explanation

Correct option: D

2 \times {10^6}N

Option D (

2 \times {10^6}N

) is correct because it correctly applies Young's modulus formula to calculate the force required to double the wire's 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.