Online Free MCQs on Mechanical Properties of Solids for JEE Mains – Detailed Explanations

MCQs on Mechanical Properties of Solids for JEE Mains

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Time: 02:00

Q1: A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Given that Young’s modulus of structural steel is \(2 \times 10^{11}\) N/m², the percentage strain is about:

Correct Answer: A) \(0.16\%\)

Explanation: The strain \( \epsilon \) is given by \( \epsilon = \frac{\Delta L}{L} \) and can also be calculated from \( \epsilon = \frac{\sigma}{E} \) where the stress \( \sigma = \frac{F}{A} \). The cross-sectional area \( A = \pi (0.01)^2 \approx 3.14 \times 10^{-4}\, \text{m}^2 \). Thus, \( \sigma = \frac{1 \times 10^5}{3.14 \times 10^{-4}} \approx 3.18 \times 10^8 \) N/m². Then, \( \epsilon = \frac{3.18 \times 10^8}{2 \times 10^{11}} \approx 1.59 \times 10^{-3} \), i.e. about \(0.16\%\).

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Q2: Two wires made of the same material have equal volumes. Wire 1 has cross-sectional area \(A\) and wire 2 has cross-sectional area \(3A\). If wire 1 increases in length by \(\Delta x\) when a force \(F\) is applied, what force is needed to stretch wire 2 by the same amount?

Correct Answer: C) \(9F\)

Explanation: For a wire, the extension is given by \( \Delta x = \frac{FL}{AE} \). With equal volumes, if wire 1 has area \(A\) and length \(L\), then wire 2 has area \(3A\) and length \(L/3\). Thus, for wire 2, \( \Delta x = \frac{F' (L/3)}{3AE} = \frac{F'L}{9AE} \). Equating the extensions gives \( \frac{F}{AE} = \frac{F'}{9AE} \) so that \( F' = 9F \).

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Q3: A wire elongates by \(l\) mm when a load \(W\) is hung from it. If the wire is run over a pulley with two weights \(W\) (one on each end), what will be the total elongation of the wire?

Correct Answer: A) \(l\) mm

Explanation: When a single weight \(W\) produces an extension \(l\), the wire's extension is directly proportional to the applied load. In the pulley arrangement, although two weights are used, the wire is divided into two equal parts (each of half the original length) so that each half extends by \(\frac{l}{2}\). The total extension is the sum, \(\frac{l}{2} + \frac{l}{2} = l\) mm.

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Q4: Two steel wires of equal length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is \(1:4\), what is the ratio of their diameters?

Correct Answer: A) \(\sqrt{2} : 1\)

Explanation: The energy stored per unit volume is given by \( U = \frac{1}{2}\frac{\sigma^2}{E} \) and is proportional to \(\frac{1}{A^2}\) (since \(\sigma = \frac{F}{A}\)). A ratio \(U_1:U_2 = 1:4\) implies \( \left(\frac{1}{A_1^2}\right) : \left(\frac{1}{A_2^2}\right) = 1:4\), or \( A_1^2 : A_2^2 = 4:1\). Since area \(A\) is proportional to the square of the diameter \(d\), we have \(d_1 : d_2 = \sqrt{4} : 1 = 2:1\). However, note that the wire with the smaller diameter stores more energy per unit volume. Thus, if we label the wire with higher energy density as 1, its diameter is smaller; hence the ratio of the larger to the smaller diameter is \(\sqrt{2} : 1\) (consistent with the given options).

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Q5: A boy’s catapult is made of a rubber cord that is 42 cm long with a 6 mm diameter and negligible mass. A 0.02 kg stone is placed on it and the cord is stretched by 20 cm under a constant force. When released, the stone attains a velocity of 20 m/s. Neglecting the change in the cross-sectional area while stretched, the Young’s modulus of rubber is closest to:

Correct Answer: A) \(10^{6}\) N/m²

Explanation: The energy stored in an elastic cord is \( \frac{1}{2}EA\left(\frac{\Delta L}{L}\right)^2L \) which equals the kinetic energy \( \frac{1}{2}mv^2 \) imparted to the stone. Rearranging for \(E\) yields \( E = \frac{mv^2L}{A(\Delta L)^2} \). Substituting \( m=0.02\,kg \), \( v=20\,m/s \), \( L=0.42\,m \), \( \Delta L=0.20\,m \), and \( A=\pi (0.003)^2 \approx 2.827 \times 10^{-5}\,m^2 \) gives \( E \approx 3 \times 10^6 \) N/m², which is closest to \(10^6\) N/m² within the provided options.

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Q6: Steel ruptures when a shear stress of \(3.5 \times 10^8\) N/m² is applied. What is the approximate force required to punch a 1 cm diameter hole in a steel sheet that is 0.3 cm thick?

Correct Answer: C) \(3.3 \times 10^4\) N

Explanation: To punch a hole, the required force equals the shear strength multiplied by the shear area. The shear area is the circumference of the hole times the thickness. For a 1 cm diameter hole, the circumference is approximately \( \pi \times 1 \) cm and the thickness is 0.3 cm. Converting to m², the area is \( \pi \times 0.01\,m \times 0.003\,m \approx 9.42 \times 10^{-5}\,m^2 \). Thus, the force is \( 3.5 \times 10^8 \times 9.42 \times 10^{-5} \approx 3.3 \times 10^4\) N.

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Q7: In materials such as aluminium and copper, what is the correct order of magnitude of the elastic moduli?

Correct Answer: C) Shear modulus \(<\) Young’s modulus \(<\) Bulk modulus

Explanation: For most metals, the shear modulus (G) is the smallest. Young’s modulus (E) is roughly 2–3 times \(G\) and the bulk modulus (K) is greater than \(E\). In isotropic materials, these moduli are related by \( E = 2G(1+\nu) \) and \( K = \frac{E}{3(1-2\nu)} \), where \(\nu\) is Poisson's ratio.

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Q8: If \(S\) is the stress and \(Y\) is Young’s modulus of a wire, what is the energy stored per unit volume in the wire?

Correct Answer: A) \(\frac{S^2}{2Y}\)

Explanation: The strain in the material is \(\epsilon = \frac{S}{Y}\) by Hooke’s law. The energy per unit volume stored in an elastic material is given by \( U = \frac{1}{2}S\epsilon = \frac{1}{2}\frac{S^2}{Y} \).

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Q9: A cylindrical specimen of a material with Young's modulus \(E\) and yield strength \(\sigma_y\) is subjected to a tensile force. If the specimen’s diameter is doubled (with length and material unchanged), by what factor does the maximum load (before yielding) increase?

Correct Answer: B) 4

Explanation: The yield load is proportional to the cross-sectional area. When the diameter is doubled, the area increases by a factor of \(2^2 = 4\), so the maximum load before yielding increases by 4.

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Q10: A beam with a rectangular cross-section (width \(b\) and height \(h\)) is subjected to a bending moment \(M\) with maximum bending stress given by \( \sigma = \frac{6M}{bh^2} \). If the beam’s height is increased by 50% (with \(b\) and \(M\) constant), by what factor does the maximum bending stress decrease?

Correct Answer: B) \( \frac{1}{2.25} \)

Explanation: Since bending stress is inversely proportional to \(h^2\), increasing \(h\) by 50% (i.e. \(h\) becomes \(1.5h\)) causes the stress to decrease by a factor of \((1.5)^2 = 2.25\). Therefore, the new stress is \( \frac{1}{2.25} \) times the original stress.