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JEE • PHYSICS • MECHANICAL PROPERTIES OF SOLIDS

Mechanical properties of solids practice · Medium

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181 Medium
181 of 251
Application +4 | -1
Equal torsional torques act on two rods xandyx\mathrm{ }and\mathrm{ }y having equal length. The diameter of rod yy is twice the diameter of rod x.   If  θx   and   θyx.{\mkern 1mu} \,{\mkern 1mu} \mathrm{If}\,\,{\theta _x}{\mkern 1mu} {\mkern 1mu} \,and{\mkern 1mu} \,{\mkern 1mu} {\theta _y} are the angles of twist, then θxθy=\frac{{{\mathrm{\theta }_x}}}{{{\mathrm{\theta }_y}}} =
View answer and explanation
Correct option: D
16
Option D (16) is correct because the angle of twist is inversely proportional to the fourth power of the diameter, so doubling the diameter reduces the twist by a factor of 16.
182 Medium
182 of 251
Application +4 | -1
The upper end of a wire of diameter 12 mm12 \mathrm{~mm} and length 1 m1 \mathrm{~m} is clamped and its other end is twisted through an angle of 30∘30^{\circ}. The angle of shear is
View answer and explanation
Correct option: B
0.18∘0.18^{\circ}
Option B (0.18∘0.18^{\circ}) is correct because it is calculated using the formula relating the angle of twist, radius, length, and angle of shear.
183 Medium
183 of 251
Application +4 | -1
The upper end of a wire of radius 4  mm\mathrm{4}\,\,\mathrm{mm} and length 100  cm\mathrm{100}\,\,\mathrm{cm} is clamped and its other end is twisted through and angle of 300{30^0}. Then angle of shear is
View answer and explanation
Correct option: B
0.12∘0.12^\circ
Option B (0.12∘0.12^\circ) is correct because the angle of shear is calculated using the formula ϕ=rlθ\phi = \frac{r}{l}\theta, where r is the radius, l is the length, and θ\theta is the angle of twist.
184 Medium
184 of 251
Application +4 | -1
The bulk modulus of a spherical object is ′B′.'B'.If it is subjected to uniform pressure ′p′,'p',the fractional decreases in radius is
View answer and explanation
Correct option: A
p3B\frac{p}{{3B}}
Option A (p3B\frac{p}{{3B}}) is correct because it is derived from the definition of bulk modulus and the relationship between volume and radius for a sphere.
185 Medium
185 of 251
Application +4 | -1
If a bar is made of copper whose coefficient of linear expansion is one and a half times that of iron, the ratio of the force developed in the copper bar to the iron bar of identical lengths and cross-sections, when heated through the same temperature range (Young’s modulus for copper may be taken equal to that of iron) is
View answer and explanation
Correct option: A
32\frac{3}{2}
Option A (32\frac{3}{2}) is correct because the force developed is proportional to the coefficient of linear expansion, and the coefficient for copper is 1.5 times that of iron.
186 Medium
186 of 251
Application +4 | -1
A wire of length L\mathrm{L}, area of cross section A\mathrm{A} is hanging from a fixed support. The length of the wire changes to L1{\mathrm{L}_\mathrm{1}} when mass M\mathrm{M} is suspended from its free end. The expression for Young’s modulus is:
View answer and explanation
Correct option: D
MgLA (L1−L)\frac{{\mathrm{MgL}}}{{\mathrm{A}\,\mathrm{(}{\mathrm{L}_\mathrm{1}} - \mathrm{L)}}}
Option D (MgLA (L1−L)\frac{{\mathrm{MgL}}}{{\mathrm{A}\,\mathrm{(}{\mathrm{L}_\mathrm{1}} - \mathrm{L)}}}) is correct because it is the correct formula for Young's modulus, which is stress divided by strain.
187 Medium
187 of 251
Understanding +4 | -1
The general form of potential energy curve for atoms or molecules can be represented by the following equation U(R)=ARn−BRmU\left( R \right) = \frac{A}{{{R^n}}} - \frac{B}{{{R^m}}} . Here, R is the interatomic or molecular distance, A and B are coefficients, nn and mm are the exponents. In the above equation
View answer and explanation
Correct option: B
Second term represents the attractive part of the potential
Option B (Second term represents the attractive part of the potential) is correct because the negative sign in front of the second term indicates an attractive potential, as the potential energy decreases with increasing distance.
188 Medium
188 of 251
Analyzing +4 | -1
Which of the following relations is true
View answer and explanation
Correct option: D
σ=0.5Y−ηη\sigma = \frac{{0.5Y - \eta }}{\eta }
Option D is correct because the correct relation between Young's modulus (Y), shear modulus (η\eta), bulk modulus (K), and Poisson's ratio (σ\sigma) is Y=2η(1+σ)Y = 2\eta(1 + \sigma) which can be rearranged to σ=0.5Y−ηη\sigma = \frac{{0.5Y - \eta}}{\eta}.
189 Medium
189 of 251
Application +4 | -1
Find the density of the metal under a pressure of 20,000 N cm−220,000 \mathrm{~N} \mathrm{~cm}^{-2}. Given density of the metal =11 g cm−3=11 \mathrm{~g} \mathrm{~cm}^{-3}, bulk modulus of the metal =8×109 Nm−2=8 \times 10^9 \mathrm{~Nm}^{-2}.
View answer and explanation
Correct option: A
11.28 g cm−311.28 \mathrm{~g} \mathrm{~cm}^{-3}
Option A (11.28 g cm−311.28 \mathrm{~g} \mathrm{~cm}^{-3}) is correct because it is calculated by considering the bulk modulus and the pressure applied to the metal.
190 Medium
190 of 251
Understanding +4 | -1
If the tension on a wire is removed at once, then:
View answer and explanation
Correct option: D
Its temperature increases
Option D (Its temperature increases) is correct because the potential energy stored in the wire due to tension is converted into heat when the tension is released.
191 Medium
191 of 251
Application +4 | -1
A wire of length 4 m4 \mathrm{~m} is hanged vertically from ceiling if it is stretched due to its own weight, then find out the energy stored per unit volume. [Y=2×107 N/m2\left[Y=2 \times 10^7 \mathrm{~N} / \mathrm{m}^2\right. and r=103r=10^3 kg/m3\mathrm{kg} / \mathrm{m}^3 ]
View answer and explanation
Correct option: D
10 J/m310 \mathrm{~J} / \mathrm{m}^3
Option D (10 J/m³) is correct because it is calculated using the formula for potential energy density in terms of Young's modulus, density, gravity, and length.
192 Medium
192 of 251
Application +4 | -1
When change in volume of a cube under a pressure at atm of 10 is 1% ; compressibility of material of black in m2/N{m^2}/N is (Take 1atm =105Pa= {10^5}Pa)
View answer and explanation
Correct option: B
10−8{10^{ - 8}}
Option B (10−8{10^{ - 8}}) is correct because compressibility is the reciprocal of the bulk modulus, which relates pressure to the fractional change in volume.
193 Medium
193 of 251
Application +4 | -1
Two exactly similar wires of steel and copper are stretched by equal forces. If the difference in their elongations is 0.5  cm0.5\,\,cm, the elongation (l)\left( l \right) of each wire is Ys(steel)=2.0×1011N/m2{Y_s}\left( {\mathrm{steel}} \right) = 2.0 \times {10^{11}}N/{m^2}, Yc(copper)=1.2×1011N/m2{Y_c}\left( {\mathrm{copper}} \right) = 1.2 \times {10^{11}}N/{m^2}
View answer and explanation
Correct option: A
ls=0.75  cm,lc=1.25  cm{l_s} = 0.75\,\,cm,{l_c} = 1.25\,\,cm
Option A (ls=0.75  cm,lc=1.25  cm{l_s} = 0.75\,\,cm,{l_c} = 1.25\,\,cm) is correct because it satisfies both the elongation ratio based on Young's modulus and the given difference in elongations.
194 Medium
194 of 251
Analyzing +4 | -1
Assertion: If length of a rod is doubled, the breaking load remains the same.
Reason: Breaking load is equal to the elastic limit.
View answer and explanation
Correct option: C
If assertion is true but reason is false
The correct answer is C (If assertion is true but reason is false) because breaking load depends on the cross-sectional area, not the length, and it's related to breaking stress, not the elastic limit.
195 Medium
195 of 251
Understanding +4 | -1
Young’s modulus of perfectly rigid body material is
View answer and explanation
Correct option: A
Infinite
Option A (Infinite) is correct because a perfectly rigid body experiences no deformation under stress, resulting in an infinite Young's modulus.
196 Medium
196 of 251
Application +4 | -1
A wire of Young’s modulus 1.5×1012  Nm−21.5 \times {10^{12}}\;\mathrm{N}{\mathrm{m}^{ - 2}} is stretched by a force so as to produce a strain of 2×104.2 \times {10^4}. The energy stored per unit volume is
View answer and explanation
Correct option: D
3  ×104  Jm−33\mathrm{\;} \times {10^4}\mathrm{\;J}{\mathrm{m}^{ - 3}}
Option D (3  ×104  Jm−33\mathrm{\;} \times {10^4}\mathrm{\;J}{\mathrm{m}^{ - 3}}) is correct because it is the energy stored per unit volume calculated using the given Young's modulus and strain.
197 Medium
197 of 251
Remembering +4 | -1
Which is the most elastic
View answer and explanation
Correct option: C
Quartz
Option C (Quartz) is correct because quartz exhibits a higher resistance to deformation under stress compared to the other options, making it more elastic.
198 Medium
198 of 251
Application +4 | -1
If longitudinal strain for a wire is 0.03\mathrm{0}\mathrm{.03} and its Poisson’s ratio is 0.5\mathrm{0}\mathrm{.5}, then its lateral strain is
View answer and explanation
Correct option: C
0.015
Option C (0.015) is correct because lateral strain is calculated by multiplying Poisson's ratio by longitudinal strain.
199 Medium
199 of 251
Application +4 | -1
If a spring extends by xx on loading, then the energy stored by the spring is (if TT is tension in the spring and kk is spring constant)
View answer and explanation
Correct option: B
T22k\frac{{{T^2}}}{{2k}}
Option B (T22k\frac{{{T^2}}}{{2k}}) is correct because the energy stored in a spring is related to the tension and spring constant by this formula.
200 Medium
200 of 251
Application +4 | -1
A 0.10.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m1\mathrm{ }m and its cross-sectional area is 4.9×10−7m2.4.9 \times {10^{ - 7}}{\mathrm{m}^{2.}} If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140  rads−1140\,\,\mathrm{rad}{\mathrm{s}^{ - 1}}. If the Young’s modulus of the material of the wire is n×109 Nm−2n \times {10^9}\,\mathrm{N}{\mathrm{m}^{ - 2}}, the value of nn is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the Young's modulus is calculated using the formula for the angular frequency of SHM in a wire, and solving for n.
201 Medium
201 of 251
Understanding +4 | -1
What happens when the applied load increases and upto breaking stress in the experiment to determine the Young's modulus of elasticity?
View answer and explanation
Correct option: A
The area of wire goes on decreasing and wire extends and breaks.
Option A is correct because as the load increases, the wire stretches (extends), and its cross-sectional area decreases until it reaches the breaking stress and fractures.
202 Medium
202 of 251
Understanding +4 | -1
A: Strain is a dimensionless quantity.
R: Strain is internal force per unit area of a body
View answer and explanation
Correct option: C
If A is true but R is false
Option C is correct because strain is dimensionless (change in dimension/original dimension), and the reason is false as strain is not internal force per unit area.
203 Medium
203 of 251
Understanding +4 | -1
In Searle's apparatus, used for study of extension of wires verses load, we have two wires. During the experiment, we study the extension in one wire. The use of second wire is
View answer and explanation
Correct option: B
To compensate for the change in length caused by changes in temperature of atmosphere during experimentation.
Option B is correct. The second wire in Searle's apparatus compensates for length changes due to temperature variations during the experiment.
204 Medium
204 of 251
Understanding +4 | -1
If G={G}= shear modulus and Y=Y= Young's modulus, then in general:
View answer and explanation
Correct option: A
Y>G{Y} > {G}
Option A (Y>G{Y} > {G}) is correct because Young's modulus (Y) typically represents resistance to tensile stress, while shear modulus (G) represents resistance to shear stress, and Y is generally larger than G for most materials.
205 Medium
205 of 251
Understanding +4 | -1
Statement I: With increase of temperature, elastic property of a substance decreases.
Statement II: Elasticity is due to intermolecular forces which decreases with the increase of inter molecular distance.
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 increased temperature weakens intermolecular forces, thus decreasing elasticity.
206 Medium
206 of 251
Application +4 | -1
A 5 m long aluminium wire Y=7×1010  Nm−2  Y = 7 \times {10^{10}}\;\mathrm{N}{\mathrm{m}^{ - 2\;}} of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in the copper wire (Y=12×102Nm  −2)  \left( {Y = 12 \times {{10}^2}\mathrm{Nm}{\mathrm{\;}^{ - 2}}} \right)\; of the same length under the same weight, the diameter should now be (in mm)
View answer and explanation
Correct option: C
2.3
Option C (2.3) is correct because the diameter is inversely proportional to the square root of Young's modulus for the same elongation, length, and weight. Calculating the new diameter using the given Young's moduli results in approximately 2.3 mm.
207 Medium
207 of 251
Understanding +4 | -1
An elastic metal rod will change its length when it
View answer and explanation
Correct option: D
Both (2) and (3)
Option D (Both (2) and (3)) is correct because stress is developed in both cases when the metal rod is pulled or stretched.
208 Medium
208 of 251
Analyzing +4 | -1
A: It is the breaking stress and not the breaking strength which depends on the material
R: Breaking strength =Breaking stressArea=\frac{\text {Breaking stress}}{\text {Area}}
View answer and explanation
Correct option: D
If both A and R are false
Option D (If both A and R are false) is correct because both the assertion and the reason are incorrect statements about breaking stress and breaking strength.
209 Medium
209 of 251
Understanding +4 | -1
A: Adiabatic elasticity of a gas is greater than isothermal elasticity
R: EadiabaticEisothermal =γ\frac{E_{\text {adiabatic}}}{E_{\text {isothermal }}}=\gamma
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. Adiabatic elasticity is greater than isothermal elasticity because the ratio is \γ\gamma, which is always greater than 1, and the reason correctly explains the assertion.
210 Medium
210 of 251
Application +4 | -1
The breaking stress of a wire is 6×108 N/m26 \times 10^8 \mathrm{~N} / \mathrm{m}^2 and density is 3×103Kg/m33 \times 10^3 \mathrm{Kg} / \mathrm{m}^3. What should be the maximum length of the wire so that it may not break under its own weight?
View answer and explanation
Correct option: A
20 km20 \mathrm{~km}
Option A (20 km) is correct because it is the maximum length calculated using the breaking stress, density, and gravity.
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