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NEET • PHYSICS • MECHANICAL PROPERTIES OF SOLIDS

Mechanical properties of solids practice · Easy

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Showing 31-43 of 43 questions
31 Easy
31 of 43
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.
32 Easy
32 of 43
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.
33 Easy
33 of 43
Application +4 | -1
On all the six surfaces of a unit cube, equal tensile force of FF is applied. The increase in length of each side will be (Y=(Y = Young’s modulus, σ=\sigma = Poisson’s ratio)
View answer and explanation
Correct option: C
F(1−2σ)Y\frac{{F\left( {1 - 2\sigma } \right)}}{Y}
Option C (F(1−2σ)Y\frac{{F\left( {1 - 2\sigma } \right)}}{Y}) is correct because it represents the increase in length of each side when equal tensile force is applied on all six surfaces of a unit cube.
34 Easy
34 of 43
Remembering +4 | -1
Rubber is an example of
View answer and explanation
Correct option: C
Elastomer
Option C (Elastomer) is correct because rubber exhibits high elasticity and can undergo large deformations under stress, characteristic of elastomers.
35 Easy
35 of 43
Understanding +4 | -1
When a force is applied at one end of an elastic wire, it produces a strain EE in the wire. If YY is the Young's modulus of the material of the wire, the amount of energy stored per unit volume of the wire is given by
View answer and explanation
Correct option: D
12(Y×E2)\frac{1}{2}\left(Y \times E^2\right)
Option D, 12(Y×E2)\frac{1}{2}(Y \times E^2), is correct because it represents the energy stored per unit volume in an elastic material, derived from the area under the stress-strain curve.
36 Easy
36 of 43
Remembering +4 | -1
The value of modulus of rigidity for ideal liquid is
View answer and explanation
Correct option: A
0
Option A (0) is correct because an ideal liquid cannot resist shear stress, hence its modulus of rigidity is zero.
37 Easy
37 of 43
Understanding +4 | -1
A long piece of rubber is wider than it is thick. When is is stretched in length by some amount:
View answer and explanation
Correct option: D
Both its thickness and width decrease
Option D (Both its thickness and width decrease) is correct because when a rubber piece is stretched, its dimensions perpendicular to the direction of the force decrease due to Poisson's effect.
38 Easy
38 of 43
Understanding +4 | -1
According to Hooke’s law of elasticity, if stress is increased, them the ratio of stress to strain
View answer and explanation
Correct option: B
Remains constant
Option B (Remains constant) is correct because Hooke's law states that the ratio of stress to strain is constant within the elastic limit.
39 Easy
39 of 43
Understanding +4 | -1
The Young's modulus of a wire is numerically equal to the stress which will
View answer and explanation
Correct option: B
double the length of the wire
Option B (double the length of the wire) is correct because when stress equals Young's modulus, strain equals 1, meaning the length doubles.
40 Easy
40 of 43
Application +4 | -1
A steel wire is 1 m1 \mathrm{~m} long and 1 mm21 \mathrm{~mm}^2 in area of crosssection. If it takes 200 N200 \mathrm{~N} to stretch this wire by 1 mm1 \mathrm{~mm}, how much force will be required to stretch a wire of the same material as well diameter from its normal length of 10 m10 \mathrm{~m} to a length of 1002 cm1002 \mathrm{~cm} ?
View answer and explanation
Correct option: A
400
The correct answer is 400 because the force required is proportional to the change in length and inversely proportional to the original length.
41 Easy
41 of 43
Application +4 | -1
Two forces F1F_1 and F2F_2 are applied on two rods PP and QQ of same materials such that elongation in rods are same. If ratio of their radii is xx : yy and ratio of length is m:nm: n, then ratio of F1:F2F_1: F_2 is
View answer and explanation
Correct option: B
(xy)2⋅nm\left(\frac{x}{y}\right)^2 \cdot \frac{n}{m}
Option B is correct. The ratio of forces is derived from the Young's modulus formula, considering the same elongation and different radii and lengths of the rods.
42 Easy
42 of 43
Application +4 | -1
If to break a wire of 1 m1 \mathrm{~m} length, minimum 40 kg40 \mathrm{~kg}-wt is required, then to break a wire of same material 6 m6 \mathrm{~m} in length and double in radius, the breaking weight required will be
View answer and explanation
Correct option: C
160 kg160 \mathrm{~kg}-wt
Option C (160 kg160 \mathrm{~kg}-wt) is correct because the breaking force is proportional to the cross-sectional area, which is proportional to the square of the radius.
43 Easy
43 of 43
Application +4 | -1
A wire of length '2 m2 \mathrm{~m}' is clamped horizontally between two fixed support. A mass m=5 kg\mathrm{m}=5 \mathrm{~kg} is hanged from middle of wire. The vertical and depression in wire (in cm\mathrm{cm}) in equilibrium is (Young modulus of wire =2.4×109 N/m2=2.4 \times 10^9 \mathrm{~N} / \mathrm{m}^2, cross-sectional area =1 cm2=1 \mathrm{~cm}^2)
View answer and explanation
Correct option: A
5
Option A is correct. The vertical depression in the wire is calculated using the formula for the sag of a wire under its own weight, considering Young's modulus and the wire's properties.
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