Mechanical Properties of Solids Previous Year Questions — JEE Main Physics
21 past-year Mechanical Properties of Solids questions from JEE Main (Physics) — page 2.
Q922 (2025)
Two slabs with square cross section of different materials (1, 2) with equal sides ($l$) and thickness $d_{1}$ and $d_{2}$ such that $d_{2} = 2d_{1}$ and $l > d_{2}$. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is $\theta_{2} = 2\theta_{1}$. If the shear moduli of material 1 is $4 \times 10^{9}\mathrm{~N/m}^{2}$, then shear moduli of material 2 is $x \times 10^{9}\mathrm{~N/m}^{2}$, where value of x is
### Related Formula
Shear modulus definition:
$$\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F / A}{\theta} = \frac{F}{A\theta}$$
where $A = l \cdot d$ is the area of the narrow face parallel to the shearing force $F$.
### Core Logic
We are given:
* $\theta_2 = 2\theta_1$
* Shearing forces are equal ($F_1 = F_2 = F$).
* Thickness relationship: $d_2 = 2d_1$
* Resisting face area: $A_1 = l d_1$ and $A_2 = l d_2 = 2l d_1 = 2A_1$
{{SOL_IMG1}}
### Step 1: Link Modulus Terms
Express the angular deformation for each slab:
$$\theta_1 = \frac{F}{l d_1 \eta_1}$$
$$\theta_2 = \frac{F}{l d_2 \eta_2} = \frac{F}{2l d_1 \eta_2}$$
Substitute these expressions into $\theta_2 = 2\theta_1$:
$$\frac{F}{2l d_1 \eta_2} = 2 \cdot \left(\frac{F}{l d_1 \eta_1}\right)$$
$$\frac{1}{2\eta_2} = \frac{2}{\eta_1} \implies 4\eta_2 = \eta_1 \implies \eta_2 = \frac{\eta_1}{4}$$
### Step 2: Solve for x
Given $\eta_1 = 4 \times 10^{9}\text{ N/m}^2$:
$$\eta_2 = \frac{4 \times 10^9}{4} = 1 \times 10^{9}\text{ N/m}^2$$
Comparing with $\eta_2 = x \times 10^9\text{ N/m}^2$:
$x = 1$
### Pattern Recognition
Shear stress is inversely proportional to the face area parallel to the applied shearing force ($A = l \cdot d$). Doubling the slab thickness doubles the resisting contact area, halving the applied shear stress under identical lateral loading.
### Evaluation Rubric / Model Answer
1
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q24 (2025)
The increase in pressure required to decrease the volume of a water sample by 0.2% is $P \times 10^{5}$ Nm $^{-2}$ . Bulk modulus of water is $2.15 \times 10^{9}$ Nm $^{-2}$ . The value of P is ____.
### Related Formula
Bulk Modulus formula:
$$B = - \frac{\Delta P}{\frac{\Delta V}{V}} \implies \Delta P = B \left( \frac{-\Delta V}{V} \right)$$
### Core Logic
Given specifications:
- Fractional volume drop, $\frac{-\Delta V}{V} = 0.2\% = \frac{0.2}{100} = 2 \times 10^{-3}$
- Bulk modulus value, $B = 2.15 \times 10^{9}\ \mathrm{N/m^2}$
Calculating excess pressure:
$$\Delta P = 2.15 \times 10^{9} \times 2 \times 10^{-3}$$
$$\Delta P = 4.3 \times 10^{6}\ \mathrm{N/m^2} = 43 \times 10^{5}\ \mathrm{N/m^2}$$
Comparing with $P \times 10^{5}$, we get $P = 43$.
### Pattern Recognition
Bulk modulus measures a fluid's resistance to compression. A tiny percentage reduction in volume requires a massive amount of pressure due to water's near-incompressibility.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q23 (2025)
The volume contraction of a solid copper cube of edge length $10\mathrm{cm}$ , when subjected to a hydraulic pressure of $7 \times 10^{6}\mathrm{Pa}$ , would be ____________ $mm^3$ .
(Given bulk modulus of copper $= 1.4 \times 10^{11} \mathrm{Nm}^{-2}$ )
### Related Formula
Bulk modulus $B$ measures a material's resistance to uniform compression and is defined as the ratio of hydraulic pressure to volumetric strain:
$$B = \frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} \implies \Delta V = \frac{\Delta P \cdot V}{B}$$
### Core Logic
Given parameters [cite: 192, 193]:
* Edge length of the cube, $a = 10 \text{ cm} = 0.1 \text{ m}$
* Initial volume, $V = a^3 = (0.1)^3 = 10^{-3} \text{ m}^3 = 10^6 \text{ mm}^3$
* Hydraulic pressure increase, $\Delta P = 7 \times 10^6 \text{ Pa}$
* Bulk Modulus, $B = 1.4 \times 10^{11} \text{ N/m}^2$
Substitute these values into the volume change equation :
$$\Delta V = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}}$$
$$\Delta V = \frac{7 \times 10^3}{1.4 \times 10^{11}} = 5 \times 10^{-7} \text{ m}^3 \quad \text{}$$
Convert the volume contraction into $\text{mm}^3$:
$$\Delta V = 5 \times 10^{-7} \times (10^3)^3 \text{ mm}^3 = 5 \times 10^{-7} \times 10^9 \text{ mm}^3 = 50 \text{ mm}^3$$
### Pattern Recognition
Always perform unit conversions carefully at the final step to avoid handling complex decimals during calculations. Converting $1 \text{ m}^3 = 10^9 \text{ mm}^3$ ensures a clean, error-free conversion.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q1288 (2025)
The fractional compression $\left(\frac{\Delta\mathrm{V}}{\mathrm{V}}\right)$ of water at the depth of $2.5\mathrm{km}$ below the sea level is $\%$ Given, the Bulk modulus of water $= 2\times 10^{9}\mathrm{Nm}^{-2}$ , density of water $= 10^{3}\mathrm{kg}\mathrm{m}^{-3}$ , acceleration due to gravity $= \mathrm{g} = 10\mathrm{ms}^{-2}$ .
- 1.75
- 1.0
- 1.5
- 1.25
### Related Formula
$$B = \frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} = \frac{\rho g h}{\left(\frac{\Delta V}{V}\right)}$$
### Core Logic
Calculate the hydrostatic pressure change at depth $h = 2.5\text{ km} = 2500\text{ m}$ :
$$\Delta P = \rho g h = 10^3 \cdot 10 \cdot 2500 = 2.5 \times 10^7 \mathrm{~N\cdot m^{-2}}$$
Now compute the percentage fractional compression :
$$\frac{\Delta V}{V} \times 100 = \frac{\Delta P}{B} \times 100 = \frac{2.5 \times 10^7}{2 \times 10^9} \times 100\% = 1.25\%$$
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Mechanical Properties of Fluids
Q1351 (2024)
With rise in temperature, the Young's modulus of elasticity:
- changes erratically
- decreases
- increases
- remains unchanged
### Related Formula
Intermolecular forces weaken as thermal agitation increases:
$$Y = \frac{\text{Stress}}{\text{Strain}}$$
### Core Logic
When temperature rises, the mean separation between atoms or molecules increases due to thermal expansion. This weakens the intermolecular binding forces, making the material easier to deform for the same amount of applied stress. Consequently, the value of Young's modulus decreases with a rise in temperature.
### Step 1: Final Conclusion
Thus, Young's modulus of elasticity decreases with the increase in temperature.
### Pattern Recognition
Temperature up $\rightarrow$ Thermal expansion up $\rightarrow$ Intermolecular bonds weaken $\rightarrow$ Modulus of elasticity down.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
