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NEET • PHYSICS • MECHANICAL PROPERTIES OF SOLIDS

Mechanical properties of solids practice · Mixed difficulty

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Showing 91-120 of 177 questions

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91 Medium
91 of 177
Application +4 | -1
A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is 4.8×107N/m24.8 \times {10^7}N/{m^2}. The area of cross section of the wire is 10−6m2{10^{ - 6}}{m^2}. The maximum angular velocity with which it can be rotated in a horizontal circle
View answer and explanation
Correct option: B
4 rad/secrad/sec
Option A (4 rad/sec) is correct because the maximum angular velocity is calculated using the breaking stress, area, mass, and length.
92 Easy
92 of 177
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.
93 Easy
93 of 177
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.
94 Medium
94 of 177
Remembering +4 | -1
The restoring force per unit area is known as:
View answer and explanation
Correct option: C
Stress
Option C (Stress) is correct because stress is defined as the restoring force per unit area within a deformable body.
95 Medium
95 of 177
Understanding +4 | -1
If Y,KY,K and η\eta are the values of Young’s modulus, bulk modulus and modulus of rigidity of any material, respectively. Choose the correct relation for these parameters.
View answer and explanation
Correct option: C
K=Yη9η−3YN/m2K = \frac{{Y\eta }}{{9\eta - 3Y}}N/{m^2}
Option C (K=Yη9η−3YN/m2K = \frac{{Y\eta }}{{9\eta - 3Y}}N/{m^2}) is the correct relation between Young’s modulus, bulk modulus and modulus of rigidity.
96 Medium
96 of 177
Remembering +4 | -1
Stress is a _____ quantity.
View answer and explanation
Correct option: D
Tensor
The correct answer is D (Tensor) because stress is a tensor quantity that describes forces acting on a cross-sectional area.
97 Easy
97 of 177
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.
98 Medium
98 of 177
Application +4 | -1
Compressibility of water is 5×1010  m2N−1.  5 \times {10^{10}}\;{\mathrm{m}^2}{\mathrm{N}^{ - 1}}.\; The change in volume of 100 mL water subjected 15×106  Pa15 \times {10^6}\mathrm{\;Pa} pressure will
View answer and explanation
Correct option: D
Decrease by 0.75 mL
Option D (Decrease by 0.75 mL) is correct because the volume of water decreases under pressure, and the calculated decrease is 0.75 mL based on the given compressibility and pressure.
99 Hard
99 of 177
Application +4 | -1
If Youngs moduls of a material is 108 N/m210^8 \mathrm{~N} / \mathrm{m}^2 and coefficient of Rigidity is 3×108 N/m23 \times 10^8 \mathrm{~N} / \mathrm{m}^2 then its bulk modulus will be-
View answer and explanation
Correct option: A
12.5×106 N/m212.5 \times 10^6 \mathrm{~N} / \mathrm{m}^2
Option A (12.5×106 N/m212.5 \times 10^6 \mathrm{~N} / \mathrm{m}^2) is correct because it is the calculated bulk modulus using the given Young's modulus and coefficient of rigidity.
100 Medium
100 of 177
Understanding +4 | -1
If the length of a wire is reduced to half, then it can hold the Load
View answer and explanation
Correct option: B
Same
Option B (Same) is correct because the load a wire can hold depends on the breaking stress of the material and the cross-sectional area, not the length of the wire.
101 Medium
101 of 177
Application +4 | -1
A wire can be broken by a load of 20 kg-wt. The force required to break wire of same material with twice the diameter will be
View answer and explanation
Correct option: D
80 kg-wt
Option D (80 kg-wt) is correct because the breaking force is proportional to the cross-sectional area, which is proportional to the square of the diameter.
102 Medium
102 of 177
Application +4 | -1
When a pressure of 100 atmosphere is applied on a spherical ball, then its volume reduces to 0.01%0.01\%. The bulk modulus of the material of the rubber in dyne/cm2dyne/c{m^2} is
View answer and explanation
Correct option: C
1×10121 \times {10^{12}}
Option C (1×10121 \times {10^{12}}) is correct because it is the calculated bulk modulus of the rubber based on the given pressure and volume reduction.
103 Medium
103 of 177
Understanding +4 | -1
The length of a wire under stress changes by 0.01%0.01\%. The strain produced is
View answer and explanation
Correct option: A
1×10−41 \times {10^{ - 4}}
Option A (1×10−41 \times {10^{ - 4}}) is correct because strain is defined as the change in length divided by the original length, which is 0.01% or 0.0001.
104 Medium
104 of 177
Analyzing +4 | -1
Assertion:- Elastic restoring forces may be conservative
Reason:- The value of strain for same stress are different while increasing the load and while decreasing the load.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because elastic restoring forces can be conservative, and hysteresis explains the difference in strain for the same stress during loading and unloading, but hysteresis doesn't explain why restoring forces *may* be conservative.
105 Medium
105 of 177
Remembering +4 | -1
Dimensions of stress are
View answer and explanation
Correct option: D
[ M L−1 T−2 ]\mathrm{[}\,\mathrm{M}\,{\mathrm{L}^{ - 1}}\,{\mathrm{T}^{ - 2}}\,]
Option D ([ M L−1 T−2 ]\mathrm{[}\,\mathrm{M}\,{\mathrm{L}^{ - 1}}\,{\mathrm{T}^{ - 2}}\,\mathrm{]}) is correct because stress is force per unit area, and the dimensions of force are [MLT−2][MLT^{-2}] and area are [L2][L^2], so stress is [ML−1T−2][ML^{-1}T^{-2}].
106 Medium
106 of 177
Understanding +4 | -1
A wire fixed at the upper end stretches by length ll by applying a force FF. The work done in stretching is
View answer and explanation
Correct option: D
Fl2\frac{{Fl}}{2}
Option D (Fl2\frac{{Fl}}{2}) is correct because the work done in stretching a wire is given by half the product of force and extension.
107 Medium
107 of 177
Understanding +4 | -1
Write copper, steel, glass and rubber in order of increasing coefficient of elasticity.
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Correct option: D
Rubber, glass, copper, steel
Option D (Rubber, glass, copper, steel) is correct because it lists the materials in order of increasing Young's modulus (coefficient of elasticity).
108 Medium
108 of 177
Understanding +4 | -1
The Poisson’s ratio cannot have the value
View answer and explanation
Correct option: A
0.7\mathrm{0}\mathrm{.7}
Option A (0.7) is the correct answer because Poisson's ratio is generally between 0 and 0.5 for most materials, and 0.7 exceeds this range.
109 Medium
109 of 177
Application +4 | -1
The temperature of a wire of length 1m1\mathrm{ }m and area of cross section 1  cm21\,\,c{m^2} is increased from 00C  to  1000C{0^0}\mathrm{C}\,\,\mathrm{to}\,\,{100^0}C. If the rod is not allowed to increased in length, the force required will be (α=10−5/0C(\alpha = {10^{ - 5}}{/^0}C and Y=1011N/m2)Y = {10^{11}}\mathrm{N}/{\mathrm{m}^2})
View answer and explanation
Correct option: B
104 N{10^4}\,\mathrm{N}
Option B (104 N{10^4}\,\mathrm{N}) is correct because the force required is calculated using the formula F=YAαΔTF = YA\alpha \Delta T, where Y is Young's modulus, A is the area, α\alpha is the coefficient of linear expansion, and ΔT\Delta T is the change in temperature.
110 Medium
110 of 177
Remembering +4 | -1
Young's modulus is defined as the ratio of
View answer and explanation
Correct option: C
Tensile stress and longitudinal strain
Option C (Tensile stress and longitudinal strain) is correct because Young's modulus is defined as the ratio of tensile stress to longitudinal strain.
111 Medium
111 of 177
Remembering +4 | -1
The Young’s modulus of the material of a wire is equal to the
View answer and explanation
Correct option: B
Stress required to produce unit strain
Option B is correct because Young's modulus is defined as the stress required to produce unit strain.
112 Easy
112 of 177
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.
113 Medium
113 of 177
Remembering +4 | -1
To keep constant time, watches are fitted with balance wheel made of
View answer and explanation
Correct option: A
Invar
Option A is correct because Invar has a very low coefficient of thermal expansion, making it suitable for balance wheels in watches to maintain constant time.
114 Easy
114 of 177
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.
115 Easy
115 of 177
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.
116 Easy
116 of 177
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.
117 Easy
117 of 177
Application +4 | -1
The ratio of radii of two wires of same material is 1:3. Then the ratio of their elongations on applying the same load is
View answer and explanation
Correct option: C
0.375694444
The question is incomplete and does not provide enough information to determine the ratio of elongations. It needs additional information such as the applied load, length, and Young's modulus to calculate the elongations and their ratio.
118 Medium
118 of 177
Application +4 | -1
When a certain weight is suspended from a long uniform wire, its length increases by 1 cm. If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one, the increase in length will be
View answer and explanation
Correct option: C
4 cm
Option C (4 cm) is correct because elongation is inversely proportional to the square of the diameter.
119 Easy
119 of 177
Understanding +4 | -1
Young's modulus of the material of a wire of length LL and radius rr is YNm−2Y \mathrm{Nm}^{-2}. If the length is reduced to L2\frac{L}{2} and radius to r2\frac{r}{2}, the Young's modulus will be
View answer and explanation
Correct option: A
YY
Option A (YY) is correct because Young's modulus is a material property and does not depend on the dimensions of the wire.
120 Medium
120 of 177
Application +4 | -1
A force FF is needed to break a copper wire having radius RR. The force needed to break a copper wire of radius 2R2R will be
View answer and explanation
Correct option: C
4 F4\,F
Option C (4 F4\,F) is correct because the force needed to break a wire is proportional to the cross-sectional area, which is proportional to the square of the radius.
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