Online Free MCQs on Mechanical Properties of Solids for JEE Mains – High-Level Practice

MCQs on Mechanical Properties of Solids for JEE Mains

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Time: 02:00

Q1: A body of mass \( m = 10 \, \text{kg} \) is attached to one end of a wire of length \(0.3\,\text{m}\). In a space station the wire’s breaking stress is \(4.8\times10^7\,\text{N/m}^2\) and its cross‐sectional area is \(10^{-2}\,\text{cm}^2\). The maximum angular speed (in rad/s) with which it can be rotated about the other end is _______.

Correct Answer: A) \( 4 \, \text{rad/s} \)

Explanation: The maximum tension a wire can sustain is \( \sigma_{\text{max}}A \). Here, \(A=10^{-2}\,\text{cm}^2 = 10^{-2}\times10^{-4}\,\text{m}^2 = 10^{-6}\,\text{m}^2\). Equating the centripetal force required (\(m\omega^2L\)) to the maximum tension, we have:
\[ m\omega^2L = \sigma_{\text{max}}A \quad \Longrightarrow \quad \omega = \sqrt{\frac{\sigma_{\text{max}}A}{mL}} = \sqrt{\frac{(4.8\times10^7)(10^{-6})}{10\times0.3}} = \sqrt{16} = 4\,\text{rad/s}. \]

Time: 02:00

Q2: A uniform cylindrical rod of length \(L\) and radius \(r\) is made from a material whose Young’s modulus is \(Y\). When heated by a temperature \(T\) and simultaneously subjected to a compressional force \(F\) so that its length remains unchanged, the coefficient of volume expansion is (nearly) equal to:

Correct Answer: C) \( \frac{3F}{\pi r^2YT} \)

Explanation: With no net change in length, the compressive shortening \( \Delta L_{\text{mech}} = \frac{FL}{\pi r^2Y} \) cancels the thermal expansion \( \Delta L_{\text{th}} = \alpha L T \). Hence, the linear expansion coefficient is \( \alpha = \frac{F}{\pi r^2YT} \). Since the coefficient of volume expansion is approximately \(3\alpha\) (for isotropic materials), we get:
\[ \beta \approx 3\alpha = \frac{3F}{\pi r^2YT}. \]

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Q3: Brass and steel wires of length \(1\,\text{m}\) each and cross-sectional area \(1\,\text{mm}^2\) (i.e. \(1\times10^{-6}\,\text{m}^2\)) are connected in series. One end of the combined wire is fixed while the other is pulled so that the net elongation is \(0.2\,\text{mm}\). Given that \(Y_{\text{brass}}=60\times10^9\,\text{N/m}^2\) and \(Y_{\text{steel}}=120\times10^9\,\text{N/m}^2\), the stress required is:

Correct Answer: E) None of the above

Explanation: For each wire, the extension is given by \(\Delta L = \frac{FL}{AY}\). Since the wires are in series, the total extension is:
\[ \Delta L_{\text{total}} = \frac{F}{A}\left(\frac{1}{60\times10^9} + \frac{1}{120\times10^9}\right) = F \times 2.5\times10^{-5}. \] Setting \(\Delta L_{\text{total}}=0.0002\,\text{m}\) gives: \[ F = \frac{0.0002}{2.5\times10^{-5}} = 8\,\text{N}. \] The stress is then: \[ \sigma = \frac{F}{A} = \frac{8}{1\times10^{-6}} = 8.0\times10^6\,\text{N/m}^2. \] Since \(8.0\times10^6\,\text{N/m}^2\) is not among the provided options, the correct choice is “None of the above.”

Time: 02:00

Q4: The elastic limit of brass is \(379\,\text{MPa}\). What should be the minimum diameter of a brass rod to support a \(400\,\text{N}\) load without exceeding this limit?

Correct Answer: B) \( 1.16\,\text{mm} \)

Explanation: The maximum stress must not exceed \(379\,\text{MPa}\). Thus, the minimum cross‐sectional area required is: \[ A_{\min} = \frac{400}{379\times10^6} \approx 1.055\times10^{-6}\,\text{m}^2. \] The diameter \(d\) is given by: \[ d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4(1.055\times10^{-6})}{\pi}} \approx 1.16\,\text{mm}. \]

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Q5: A steel wire of radius \(2.0\,\text{mm}\) supports a load of \(4\,\text{kg}\). (Assume \(g=9.81\,\text{m/s}^2\) for calculation.) What is the tensile stress developed in the wire?

Correct Answer: C) \( 3.1\times10^6\,\text{N/m}^2 \)

Explanation: The load is \(4\,\text{kg}\) so the weight is \(F = 4\times9.81 \approx 39.24\,\text{N}\). The cross-sectional area of the wire is: \[ A = \pi (2.0\times10^{-3})^2 \approx 1.257\times10^{-5}\,\text{m}^2. \] Thus, the tensile stress is: \[ \sigma = \frac{F}{A} \approx \frac{39.24}{1.257\times10^{-5}} \approx 3.12\times10^6\,\text{N/m}^2. \]

Time: 02:00

Q6: (Repeated) A steel wire of radius \(2.0\,\text{mm}\) supports a load of \(4\,\text{kg}\). With \(g=9.81\,\text{m/s}^2\), what is the tensile stress in the wire?

Correct Answer: C) \( 3.1\times10^6\,\text{N/m}^2 \)

Explanation: (As in Q5) With a load of \(4\,\text{kg}\) and using \(g=9.81\,\text{m/s}^2\), the tensile stress is calculated as approximately \(3.1\times10^6\,\text{N/m}^2\).

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Q7: Young’s moduli of two wires A and B are in the ratio \(7:4\). Wire A is \(2\,\text{m}\) long with radius \(R\), and wire B is \(1.5\,\text{m}\) long with radius \(2\,\text{mm}\). If both wires stretch by the same amount under a given load, the value of \(R\) is close to:

Correct Answer: C) \( 1.7\,\text{mm} \)

Explanation: For a wire, the extension is \( \Delta L = \frac{FL}{\pi R^2 Y} \). Equal extension for both wires under the same load gives:
\[ \frac{2}{\pi R^2 (7k)} = \frac{1.5}{\pi (2\times10^{-3})^2 (4k)} \] Canceling common factors and solving for \(R\) yields \(R\approx1.75\times10^{-3}\,\text{m}\), i.e. about \(1.7\,\text{mm}\).

Time: 02:00

Q8: A copper wire of length \(1.0\,\text{m}\) and a steel wire of length \(0.5\,\text{m}\) (with equal cross-sectional areas) are joined end-to-end. If a load stretches the copper wire by \(1\,\text{mm}\), what is the total extension of the composite wire? (Given: \(Y_{\text{Cu}}=1.0\times10^{11}\,\text{N/m}^2\) and \(Y_{\text{steel}}=2.0\times10^{11}\,\text{N/m}^2\))

Correct Answer: D) \( 1.25\,\text{mm} \)

Explanation: Let \(F\) be the applied force and \(A\) the cross-sectional area. For the copper part:
\[ \Delta L_{\text{Cu}} = \frac{F}{A\,Y_{\text{Cu}}}\times 1.0, \] and since \(\Delta L_{\text{Cu}}=1\,\text{mm}\), we have \(F=\;1\times10^{-3}A\,Y_{\text{Cu}}\). For the steel part:
\[ \Delta L_{\text{steel}} = \frac{F}{A\,Y_{\text{steel}}}\times 0.5 = \frac{1\times10^{-3}A\,Y_{\text{Cu}}\times0.5}{A\,Y_{\text{steel}}} = \frac{0.5\times10^{-3}\,Y_{\text{Cu}}}{Y_{\text{steel}}}. \] Plugging in the given values: \[ \Delta L_{\text{steel}} = \frac{0.5\times10^{-3}\times10^{11}}{2.0\times10^{11}} = 0.25\,\text{mm}. \] Hence, total extension = \(1.0 + 0.25 = 1.25\,\text{mm}\).

Time: 02:00

Q9: A uniform wire (with \(Y=2\times10^{11}\,\text{N/m}^2\)) is subjected to a tensile stress of \(5\times10^7\,\text{N/m}^2\). If the overall volume change is \(0.02\%\), the fractional decrease in its radius is approximately:

Correct Answer: C) \( 0.25\times10^{-4} \)

Step 1: Calculate the axial strain using Hooke’s law: \[ \epsilon_{\text{axial}} = \frac{\sigma}{Y} = \frac{5\times10^7}{2\times10^{11}} = 2.5\times10^{-4}. \]
Step 2: For small deformations, the total fractional change in volume is approximately: \[ \frac{\Delta V}{V} \approx \epsilon_{\text{axial}} + 2\epsilon_{\text{lateral}}, \] where \(\epsilon_{\text{lateral}}\) is the strain in the lateral (radius) direction.
Step 3: Given that the overall volume change is \(0.02\%\), we write: \[ \frac{\Delta V}{V} = 2\times10^{-4}. \]
Step 4: Substitute the axial strain: \[ 2\times10^{-4} \approx 2.5\times10^{-4} + 2\epsilon_{\text{lateral}}. \]
Step 5: Solve for \(\epsilon_{\text{lateral}}\): \[ 2\epsilon_{\text{lateral}} \approx 2\times10^{-4} - 2.5\times10^{-4} = -0.5\times10^{-4}, \] \[ \epsilon_{\text{lateral}} \approx -0.25\times10^{-4}. \]
The negative sign indicates a decrease. Hence, the fractional decrease in the radius is: \[ 0.25\times10^{-4}. \]

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Q10: A steel wire can sustain a \(100\,\text{kg}\) load without breaking. If the same wire is cut into two equal parts, each part can sustain a load of:

Correct Answer: C) \( 100\,\text{kg} \)

Explanation: The breaking load of a wire depends on the tensile strength (stress limit) and cross‐sectional area. Cutting the wire does not change its material strength or cross‐sectional area. Therefore, each half can sustain the same \(100\,\text{kg}\) load individually.