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JEE • PHYSICS • MECHANICAL PROPERTIES OF SOLIDS

Mechanical properties of solids practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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241 Medium
241 of 251
Application +4 | -1
An elastic material of Young’s modulus YY is subjected to a stress SS. The elastic energy stored per unit volume of the material is
View answer and explanation
Correct option: B
S22Y\frac{{{S^2}}}{{2Y}}
Option B (S22Y\frac{{{S^2}}}{{2Y}}) is correct because the elastic energy stored per unit volume is half the square of the stress divided by Young's modulus.
242 Medium
242 of 251
Understanding +4 | -1
What is the Young's modulus of elasticity for a perfectly rigid body?
View answer and explanation
Correct option: A
Infinity
Option A (Infinity) is correct because a perfectly rigid body has zero strain for any amount of stress, leading to an infinite Young's modulus.
243 Medium
243 of 251
Application +4 | -1
The pressure of a medium is changed from 1.01×1051.01 \times {10^5} Pa to 1.165  ×1051.165\,\, \times {10^5} Pa and change in volume is 10  %10\,\,\% keeping temperature constant. The Bulk modulus of the medium is
View answer and explanation
Correct option: D
1.55  ×  1051.55\,\, \times \,\,{10^5} Pa
Option D (1.55 × 1051.55\, \times \,{10^5} Pa) is correct because it's the bulk modulus calculated using the given pressure and volume changes.
244 Medium
244 of 251
Understanding +4 | -1
Statement I: A hollow shaft is found to be stronger than a solid shaft made of same material.
Statement II: The torque required to produce a given twist in hollow cylinder is greater than that required to twist a solid cylinder of same size and material.
View answer and explanation
Correct option: A
Statement I is true and statement II is true
Option A (Statement I is true and statement II is true) is correct because a hollow shaft is stronger and requires more torque than a solid shaft of the same size and material.
245 Medium
245 of 251
Understanding +4 | -1
A: Diamond is more elastic than rubber.
R: When same deforming force is applied diamond deforms less than rubber.
View answer and explanation
Correct option: A
If both A &\& R are true and the R is the correct explanation of the A
Option A is correct because diamond is more elastic than rubber, and this is because diamond deforms less than rubber when the same deforming force is applied.
246 Medium
246 of 251
Application +4 | -1
A mass of 100 gram100 \mathrm{~gram} is attached to the end of a rubber string 49 cm49 \mathrm{~cm} long and having an area of cross-section 20 mm220 \mathrm{~mm}^2. The string is whirled round, horizontally at a constant speed of 40 rps40 \mathrm{~rps} in a circle of radius 51 cm51 \mathrm{~cm}. Find Young's modulus of rubber.
View answer and explanation
Correct option: A
3.95×109 Nm−23.95 \times 10^9 \mathrm{~Nm}^{-2}
Option A (3.95×109 Nm−23.95 \times 10^9 \mathrm{~Nm}^{-2}) is correct because it is calculated using the formula for Young's modulus, considering the centripetal force.
247 Medium
247 of 251
Application +4 | -1
A wire of length L and cross-section A is made of material of Young’s modulus Y. It is stretched by an amount x, the work done is
View answer and explanation
Correct option: C
Yx2A2L\frac{{Y{x^2}A}}{{2L}}
Option C (Yx2A2L\frac{{Y{x^2}A}}{{2L}}) is correct because it represents the work done in stretching the wire, derived from the formula relating Young's modulus, stress, and strain.
248 Medium
248 of 251
Application +4 | -1
A metal rod of Young’s modulus 2×1010 Nm−22 \times {10^{10}}\,N{m^{ - 2}} undergoes an elastic strain of 0.2%0.2\%. The energy per unit volume stored in the rod in joule/m3joule/{m^3} is
View answer and explanation
Correct option: A
400
Option A (400) is correct because it is the result of substituting the given values into the strain energy per unit volume formula.
249 Medium
249 of 251
Application +4 | -1
Cross sectional area of a wire (Y=5×1011\left(\mathrm{Y}=5 \times 10^{11}\right. N/m2\mathrm{N} / \mathrm{m}^2 ) is 0.1 cm20.1 \mathrm{~cm}^2. Find the force required to make its length double.
View answer and explanation
Correct option: B
5×106 N5 \times 10^6 \mathrm{~N}
Option B (5×106 N5 \times 10^6 \mathrm{~N}) is correct because it is the force calculated using Young's modulus formula to double the length of the wire.
250 Medium
250 of 251
Application +4 | -1
A 14.5 kg14.5 \mathrm{~kg} mass, fastened to the end of a steel wire of unstretched length 1 m1 \mathrm{~m}, is whirled in a vertical circle with an angular velocity of 2 rev/s2 \mathrm{~rev} / \mathrm{s} at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm20.065 \mathrm{~cm}^2. The elongation of the wire when the mass is at the lowest point of its path is;

[ Ysteel =2×1011 N/m2 ] \text {[ } Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2 \text { ] }
View answer and explanation
Correct option: C
1.87 mm1.87 \mathrm{~mm}
Option C (1.87 mm1.87 \mathrm{~mm}) is correct because it is the elongation calculated using Young's modulus, the force due to gravity and centripetal acceleration, and the given dimensions of the wire.
251 Medium
251 of 251
Application +4 | -1
A wire increases by 10−310^{-3} of its length when a stress of 1×108 Nm−21 \times 10^8 \mathrm{~Nm}^{-2} is applied to it. What is the Young's modulus of the material of the wire?
View answer and explanation
Correct option: A
1011 Nm−210^{11} \mathrm{~Nm}^{-2}
Option A (1011 Nm−210^{11} \mathrm{~Nm}^{-2}) is correct because Young's modulus is calculated by dividing stress by strain.
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