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NEET • PHYSICS • MECHANICAL PROPERTIES OF SOLIDS

Mechanical properties of solids practice · Easy

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Showing 1-30 of 43 questions

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1 Easy
1 of 43
Application +4 | -1
A 20 kg load is suspended by a wire of area of cross section is 1 mm21\,m{m^2}. The stress produced in N/m2N/{m^2} is [g=10 ms−2 ] \left[ {g = 10\,m{s^{ - 2}}\,} \right]\,
View answer and explanation
Correct option: B
2×1082 \times {10^8}
Option B (2×1082 \times {10^8}) is correct because stress is calculated as force divided by area, where the force is the weight of the load (20 kg * 10 m/s²) and the area is 1 mm² converted to m².
2 Easy
2 of 43
Application +4 | -1
A wire extends by 1mm\mathrm{1 mm} when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm will be
View answer and explanation
Correct option: A
8\mathrm{8}
Option A (8) is correct because elongation is inversely proportional to the area and directly proportional to the force.
3 Easy
3 of 43
Application +4 | -1
A cube of side 40  mm\mathrm{40}\,\,\mathrm{mm} has its upper face displaced by 0.1 mm\mathrm{0}.\mathrm{1}\,\mathrm{mm} by a tangential force of 8  kN\mathrm{8}\,\,\mathrm{kN}. The shearing modulus of cube is
View answer and explanation
Correct option: A
2  ×109 Nm−22\,\, \times {10^9}\,\mathrm{N}{\mathrm{m}^{ - 2}}
Option A (2×109 Nm−22 \times {10^9}\,\mathrm{N}{\mathrm{m}^{ - 2}}) is correct because it is the result of calculating the shearing modulus using the given values of tangential force, area, and displacement.
4 Easy
4 of 43
Understanding +4 | -1
The elastic potential energy of a stretched wire is given by
View answer and explanation
Correct option: B
U=AY2Ll2U=\frac{A Y}{2 L} l^2
Option B (U=AY2Ll2U=\frac{A Y}{2 L} l^2) is correct because it represents the formula for elastic potential energy stored in a stretched wire.
5 Easy
5 of 43
Application +4 | -1
A metal cube of side 8 cm. If the top of the cube is displaced by 0.04 cm with respect to its bottom. What is the shearing stress if the rigidity modulus is 4×106 N/m2  [in  N/m2]4 \times {10^6}\,N/{m^2}\,\,\left[ {\mathrm{in}\,\,N/{m^2}} \right]
View answer and explanation
Correct option: B
2×1042 \times {10^4}
Option B (2×1042 \times {10^4}) is correct because shearing stress is the product of the rigidity modulus and the shearing strain.
6 Easy
6 of 43
Application +4 | -1
A wire of 4 m length 16 mm radius is clamped at upper end. The lower end is twisted by an angle of 30∘30^\circ. The angle of shear is
View answer and explanation
Correct option: C
0.12∘0.12^\circ
Option C (0.12∘0.12^\circ) is correct because the angle of shear is calculated by dividing the arc length at the wire's surface by the wire's length.
7 Easy
7 of 43
Application +4 | -1
The interatomic distance for a metal is 3×10−10 m3 \times 10^{-10} \mathrm{~m}. If the interatomic force constant is 3.6×10−9 N/A∘3.6 \times 10^{-9} \mathrm{~N} / \mathop {\text{A}}\limits^ \circ, then the Young's modulus (in Nm−2\mathrm{Nm}^{-2} ) will be
View answer and explanation
Correct option: A
1.2×10111.2 \times 10^{11}
Option A (1.2×10111.2 \times 10^{11}) is correct because Young's modulus is calculated from the interatomic force constant and distance.
8 Easy
8 of 43
Application +4 | -1
Young's modulus of rubber is 104 N/m210^4 \mathrm{~N} / \mathrm{m}^2 and area of cross-section is 2 cm22 \mathrm{~cm}^2. If force of 2×1052 \times 10^5 dynes is applied along its length, then its initial length LL becomes
View answer and explanation
Correct option: C
2L
Option C (2L) is correct because the final length is twice the initial length based on the given Young's modulus, area, and force.
9 Easy
9 of 43
Remembering +4 | -1
For most of the material, Young's modulus (Y)(Y) and rigidity modulus (G)(G) are related as
View answer and explanation
Correct option: B
G=Y3G=\frac{Y}{3}
Option B (G=Y3G=\frac{{Y}}{{3}}) is correct because the relationship between Young's modulus (Y) and rigidity modulus (G) typically falls within the range of G = Y/3 to Y/2 for most materials.
10 Easy
10 of 43
Understanding +4 | -1
The length of an iron wire is LL and area of cross-section is AA. The increase in length is ll on applying the force FF on its two ends. Which of the following statement(s) is/are correct?
View answer and explanation
Correct option: A
Increase in length is inversely proportional to its length LL.
Option C (Increase in length is inversely proportional to A) is correct because according to Young's modulus formula, the increase in length is inversely proportional to the area of cross-section.
11 Easy
11 of 43
Understanding +4 | -1
Assertion(A): Bulk modulus of an incompressible fluid is infinite.

Reason(R): Density of incompressible fluid remains constant.
View answer and explanation
Correct option: B
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.) is correct because both the assertion and reason are true, but the constant density doesn't directly explain why the bulk modulus is infinite.
12 Easy
12 of 43
Understanding +4 | -1
Assertion(A): If length of a wire is halved, its Young's modulus of elasticity will become two times.

Reason(R): The ratio of longitudinal stress and longitudinal strain is called Young's modulus of elasticity.
View answer and explanation
Correct option: D
If Assertion is false but Reason is true.
Option D (If Assertion is false but Reason is true) is correct because Young's modulus is a material property and does not depend on the length of the wire.
13 Easy
13 of 43
Application +4 | -1
Two wires of same material and length but radii in the ratio 2:3 are stretched by two forces to produce equal elongation. The ratio of two forces is
View answer and explanation
Correct option: D
0.172916667
The question is incomplete and does not provide enough information to determine the ratio of forces. It needs additional information such as the Young's modulus, length, and elongation to calculate the forces and their ratio.
14 Easy
14 of 43
Application +4 | -1
The Young's modulus of a wire is YY, if the energy per unit volume is EE, then the strain will be
View answer and explanation
Correct option: A
2EY\sqrt{\frac{2 E}{Y}}
Option A (2EY\sqrt{\frac{2 E}{Y}}) is correct because the strain is derived from the energy per unit volume formula.
15 Easy
15 of 43
Application +4 | -1
Two wires of copper having the length in the ratio 4: 1 and their radii ratio as 1: 8 are stretched by the same force. The ratio of longitudinal strain in the two will be
View answer and explanation
Correct option: D
2.667361111
Option D (2.667361111) is correct because longitudinal strain is proportional to length and inversely proportional to the square of the radius when the force is the same.

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16 Easy
16 of 43
Application +4 | -1
If the ratio of lengths, radii and Young's moduli of steel and copper are 4, 3 and 2 respectively. The ratio of their increase in lengths would be (same force FF is applied to them)
View answer and explanation
Correct option: A
29\frac{2}{9}
Option A (29\frac{2}{9}) is correct because the ratio of increase in lengths is derived from Young's modulus formula, considering the ratios of lengths, radii, and Young's moduli.
17 Easy
17 of 43
Understanding +4 | -1
Given below are two statements: one is labelled as
Assertion (A) and the other is labelled as Reason
(R).

Assertion (A): The property of body, by virtue of
which it tends to regain its original shape when the
external force is removed, is Elasticity.

Reason (R): The restoring force depends upon the
bonded inter atomic and inter molecular force of
solid.

In the light of the above statements, choose the
correct answer from the options given below:
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct
explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because elasticity is the property to regain original shape, and this property arises from interatomic forces.
18 Easy
18 of 43
Understanding +4 | -1
The breaking stress for a wire of radius r\mathrm{r} of given material is F N/m2\mathrm{F} \mathrm{~N} / \mathrm{m}^2. The breaking stress for the wire of same material of radius 2r2 \mathrm{r} is:
View answer and explanation
Correct option: C
F\mathrm{F}
Option C (F) is correct because breaking stress is a material property and depends only on the material, not the dimensions of the wire.
19 Easy
19 of 43
Application +4 | -1
For steel Y=2×1011Nm−2Y=2 \times 10^{11} \mathrm{Nm}^{-2}. The force required to double the length of a steel wire of area 1 cm21 \mathrm{~cm}^2 is
View answer and explanation
Correct option: A
2×107 N2 \times 10^7 \mathrm{~N}
Option A (2×107 N2 \times 10^7 \mathrm{~N}) is correct because doubling the length means strain is 1, and force is calculated using Young's modulus formula.
20 Easy
20 of 43
Application +4 | -1
An elongation of 0.4% in a wire of cross-sectional area is 10−6 m2{10^{ - 6}}\,{m^2} causes a tension of 100 N. Y for the wire is [in  N/m2][\mathrm{in }\,\,N/{m^2}]
View answer and explanation
Correct option: D
25×10925 \times {10^9}
Option D (25×10925 \times {10^9}) is correct because Young's modulus is calculated by dividing stress by strain.
21 Easy
21 of 43
Application +4 | -1
What is the formula for force when a rod fixed between two rigid support at both end. A temperature is changed by ΔT,Y\Delta {T}, {Y}-young modulus, α=\boldsymbol{\alpha}= coefficient of linear expension :
View answer and explanation
Correct option: A
F=YAαΔTF=Y A \alpha \Delta T
Option A (F=YAαΔTF=Y A \alpha \Delta T) is correct because it represents the thermal stress formula for a rod fixed between two rigid supports.
22 Easy
22 of 43
Application +4 | -1
The length of a wire is 4 m. Its length is increased by 2 mm when a force acts on it. The strain is
View answer and explanation
Correct option: A
5×10−45 \times {10^{ - 4}}
Option A (5×10−45 \times {10^{ - 4}}) is correct because strain is the change in length divided by the original length.
23 Easy
23 of 43
Application +4 | -1
A steel wire of 6 mm diameter is stretched by applying a stress in the wire is 3×106 N/m2.3 \times {10^6}\,N/{m^2}. Find the applied force in stretched wire
View answer and explanation
Correct option: D
84.8 N
Option D (84.8 N) is correct because the applied force is calculated by multiplying the stress by the area of the wire's cross-section.
24 Easy
24 of 43
Application +4 | -1
A cube is subjected to a uniform volume compression. If the side of the cube decreases by 1  %\mathrm{1}\,\,\mathrm{\% } the bulk strain is
View answer and explanation
Correct option: C
0.03\mathrm{0}\mathrm{.03}
Option C (0.03\mathrm{0}\mathrm{.03}) is correct because bulk strain is the volumetric strain, which is three times the linear strain for a cube.
25 Easy
25 of 43
Application +4 | -1
A ball is falling in a lake to a depth 200 m shows a decreases 0.1 % in its volume at the bottom. The bulk modulus of the ball is [in  N/m2][g=10 ms−2 ] [\mathrm{in }\,\,N/{m^2}]\left[ {g = 10\,m{s^{ - 2}}\,} \right]\,
View answer and explanation
Correct option: C
2×1092 \times {10^9}
Option C (2×1092 \times {10^9}) is correct because bulk modulus is the ratio of pressure to volumetric strain.
26 Easy
26 of 43
Application +4 | -1
The force constant of a wire is k and that of another wire of the same material is 2k. When both the wires are stretched, then work done is
View answer and explanation
Correct option: B
W2=2 W1{W_2} = 2\,{W_1}
Option B (W2=2 W1{W_2} = 2\,{W_1}) is correct because the work done is proportional to the force constant, given the same stretch.
27 Easy
27 of 43
Application +4 | -1
A uniform steel bar of cross-sectional area AA and length LL is suspended, so that it hangs vertically. The stress at the middle point of the bar is (ρ(\rho is the density of steel)
View answer and explanation
Correct option: B
Lρg2\frac{L \rho g}{2}
Option B (Lρg2\frac{L \rho g}{2}) is correct because the stress at the middle point depends on the weight of the lower half of the bar.
28 Easy
28 of 43
Application +4 | -1
An increase in pressure of 100 kPa100 \,\mathrm{kPa} causes a certain volume of water to decrease to 5×10−35 \times 10^{-3} percent of its original volume. The speed of sound (compressional waves) in water is close to
View answer and explanation
Correct option: C
1400 m/s1400 \mathrm{~m} / \mathrm{s}
Option C (1400 m/s1400 \mathrm{~m} / \mathrm{s}) is correct because it is the approximate speed of sound in water, calculated using the bulk modulus and density of water.
29 Easy
29 of 43
Understanding +4 | -1
Statement-I: A metal wire held vertically is longer than when it placed on a horizontal table.
Statement-II: Due to its own weight, some elongation is produced when it is held vertically.
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 the wire stretches due to its own weight when held vertically.
30 Easy
30 of 43
Analyzing +4 | -1
Assertion(A): If a wire is stretched, only half of the work done in stretching the wire remains stored as elastic potential energy.

Reason(R): Potential energy stored in the wire is (1/2)×(1 / 2) \times stress ×\times strain.
View answer and explanation
Correct option: C
If Assertion is true but Reason is false.
Option C (If Assertion is true but Reason is false) is correct because while the assertion about half the work being stored is true, the reason provides an incorrect formula for potential energy.
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