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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF SOLIDS

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The elastic limit and ultimate strength for steel wire is

2.48 \times 10^8 \mathrm{~Pa}

and

4.89 \times 10^8 \mathrm{~Pa}

. A steel wire of length

10 \mathrm{~m}

and diameter of cross-section

2 \mathrm{~mm}

is given, then find maximum force that can be applied to ends of wire without breaking the wire

\left(\gamma_{\text {steel }}=2 \times 10^{11}\right)

View answer and explanation

Correct option: B

1535 \mathrm{~N}

Option B (

1535 \mathrm{~N}

) is correct because it is calculated using the breaking stress and the cross-sectional area of the wire.

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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.