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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

1 Medium

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Application +4 | -1

A steel wire of length

3.2 \mathrm{~m}\left(Y_s=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right)

and a copper wire of length

4.4 \mathrm{~m}\left(Y_c=1.1 \times 10^{11}\right.

\mathrm{Nm}^{-2}

), both of radius

1.4 \mathrm{~mm}

are connected end to end. When stretched by a load, the net elongation is found to be

1.4 \mathrm{~mm}

. The load applied, in Newton, will be: (Given

\pi=\frac{22}{7}

)

View answer and explanation

Correct option: D

154

Option D (154) is correct because it is the load calculated using the elongation formula considering both steel and copper wires connected in series.

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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.