Mechanical Properties of Solids JEE Main previous year questions with solutions

5 solved JEE Main questions on Mechanical Properties of Solids, free to read — no sign-in needed. The full chapter has 115 questions; sign in to attempt the remaining 110 in the exam simulator.

1. Q1JEE Main 2026 (04 Apr, Shift 1)Bulk Modulus
   The increase in the pressure required to decrease the volume $(\Delta V)$ of water is $6.3 \times 10^7$ N/m$^2$. The percentage decrease in the volume is _____. (Bulk modulus of water $= 2.1 \times 10^9$ N/m$^2$.)
   1. A.$2\%$
   2. B.$3\%$
   3. C.$6\%$
   4. D.$4\%$
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   Answer: (B)

   Bulk modulus is given by $B = \dfrac{\Delta P}{-\dfrac{\Delta V}{V}}$ The fractional decrease in volume is $-\dfrac{\Delta V}{V} = \dfrac{\Delta P}{B}$ Substituting the given values: $-\dfrac{\Delta V}{V} = \dfrac{6.3 \times 10^7}{2.1 \times 10^9} = 3 \times 10^{-2}$ The percentage decrease in volume is $\left(-\dfrac{\Delta V}{V}\right) \times 100$ Percentage decrease $= 3 \times 10^{-2} \times 100 = 3\%$ Answer: $3\%$
2. Q2JEE Main 2025 (02 Apr, Shift 1)Work Done in Stretching a Wire
   A steel wire of length 2 m and Young's modulus $2.0 \times 10^{11} \mathrm{Nm}^{-2}$ is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and $10^{-3}$ respectively, then the elastic potential energy density of the wire is _______ $\times 10^5$ (in SI units)
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   Answer: 25

   $\begin{aligned} & \ell=2 \mathrm{~m} ; \mathrm{Y}=2 \times 10^{11} \frac{\mathrm{~N}}{\mathrm{~m}^2} \\ & \begin{aligned} \mu=-\frac{\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right)}{\left(\frac{\Delta \ell}{\ell}\right)} \Rightarrow \frac{\Delta \ell}{\ell} & =\frac{1}{\mu} \times\left(\frac{\Delta \mathrm{r}}{\mathrm{r}}\right) \\ & =\frac{1}{0.2} \times\left(10^{-3}\right)\end{aligned}\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{\Delta \ell}{\ell}=5 \times 10^{-3} \\ & \begin{aligned} \mathrm{u}=\frac{1}{2} \mathrm{y} \varepsilon_{\ell}^2 & =\frac{1}{2} \times 2 \times 10^{11} \times\left[5 \times 10^{-3}\right]^2 \\ & =25\end{aligned}\end{aligned}$
3. Q3JEE Main 2024 (27 Jan, Shift 2)Elasticity
   Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A) : The property of body, by virtue of which it tends to regain its original shape when the external force is removed, is Elasticity. Reason (R) : The restoring force depends upon the bonded inter atomic and inter molecular force of solid. In the light of the above statements, choose the correct answer from the options given below :
   1. A.(A) is false but (R) is true
   2. B.(A) is true but (R) is false
   3. C.Both (A) and (R) are true and (R) is the correct explanation (A)
   4. D.Both (A) and (R) are true but (R) is not the correct explanation of (A)
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   Answer: (C)

   Elasticity is the ability of a material to deform under stress and then return to its original shape when the stress is removed. The restoring force in elastic materials is associated with the inter-atomic or inter-molecular forces that act to bring the atoms or molecules back to their original positions when the material is deformed. When an external force is applied to a material, it causes a displacement of atoms or molecules from their equilibrium positions. The restoring force is a result of the potential energy stored in these inter-atomic or inter-molecular bonds, and it works to bring the material back to its original state. Hence, both (A) and (R) are true and (R) is the correct explanation (A)
4. Q4JEE Main 2023 (15 Apr, Shift 1)Young's Modulus and Breaking Stres
   A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force$f$, its length increases by $l$. Another wire of same material of length $2L$ and radius $2r$ is pulled by a force $2f$. Then the increase in its length will be:
   1. A.$4l$
   2. B.$\dfrac{l}{2}$
   3. C.$2l$
   4. D.$l$
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   Answer: (D)

   Young's modulus is same for both the wires since the material is same. For the first wire, $Y=\dfrac{fL}{\pi {r}^{2}l}...(i)$ For the second wire, $Y=\dfrac{2f(2L)}{4\pi {r}^{2}l'}...(ii)$ Dividing (i) by (ii) $\dfrac{fL}{\pi {r}^{2}l}\times \dfrac{4\pi {r}^{2}l'}{4fL}=1 \Rightarrow l'=l$
5. Q5JEE Main 2022 (27 Jul, Shift 1)Shearing stress
   A square aluminium (shear modulus is $25\times {10}^{9}{Nm}^{-2}$) slab of side $60cm$ and thickness $15cm$ is subjected to a shearing force (on its narrow face) of $18.0\times {10}^{4}N$. The lower edge is riveted to the floor. The displacement of the upper edge is _____ $\mu m$.
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   Answer: 48

   Shearing stress is given by $\dfrac{F}{A}=\eta \dfrac{x}{l}$ Thus, displacement is $x=\dfrac{Fl}{A\eta }$ $\Rightarrow x=\dfrac{18\times {10}^{4}\times 60\times {10}^{-2}}{60\times {10}^{-2}\times 15\times {10}^{-2}\times 25\times {10}^{9}}$ $=48\times {10}^{-6}m=48\mu m$