Mechanical Properties of Solids Previous Year Questions — JEE Main Physics
21 past-year Mechanical Properties of Solids questions from JEE Main (Physics) — page 4.
Q54 (2024)
Two blocks of mass $2 \text{ kg}$ and $4 \text{ kg}$ are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is $4.0 \times 10^{-5} \text{ m}$ and Young's modulus of the metal is $2.0 \times 10^{11} \text{ N/m}^2$. The longitudinal strain developed in the wire is $\frac{1}{\alpha \pi}$. The value of $\alpha$ is _______. [Use $g = 10 \text{ m/s}^2$]
{{IMG1}}
### Related Formula
$$T = \frac{2 m_1 m_2}{m_1 + m_2} g$$
$$\text{Strain} = \frac{\Delta \ell}{\ell} = \frac{F}{A Y} = \frac{T}{A Y}$$
### Core Logic
First, determine the tension $T$ in the wire produced by the accelerating system (Atwood machine). Then, apply the formula for longitudinal strain using the calculated tension, cross-sectional area, and Young's Modulus.
### Step 1: Calculate Tension
$$T = \left( \frac{2 \times 2 \times 4}{2 + 4} \right) g = \left( \frac{16}{6} \right) \times 10 = \frac{160}{6} = \frac{80}{3} \text{ N}$$
### Step 2: Calculate Area
Given radius $r = 4.0 \times 10^{-5} \text{ m}$.
$$A = \pi r^2 = \pi (4 \times 10^{-5})^2 = 16\pi \times 10^{-10} \text{ m}^2$$
### Step 3: Calculate Strain
$$\text{Strain} = \frac{T}{A Y} = \frac{80/3}{(16\pi \times 10^{-10}) \times (2 \times 10^{11})}$$
$$\text{Strain} = \frac{80/3}{32\pi \times 10} = \frac{80}{3 \times 320\pi} = \frac{1}{12\pi}$$
### Step 4: Extract Alpha
Comparing with $\frac{1}{\alpha \pi}$:
$\alpha = 12$
### Pattern Recognition
Linkage problem: Mechanics (Atwood tension) $\to$ Solids (Strain). Memorize Atwood tension $T = 2m_1m_2g/(m_1+m_2)$ to skip drawing free body diagrams during exams.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Laws of Motion
Q58 (2024)
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02\%$ is ________ $\mathrm{m}$.
(Take density of sea water $= 10^3\mathrm{\ kgm^{-3}}$, Bulk modulus of rubber $= 9 \times 10^8\mathrm{\ Nm^{-2}}$, and $g = 10\mathrm{\ ms^{-2}}$)
### Related Formula
$$\beta = \frac{-\Delta P}{\frac{\Delta V}{V}}$$
$$\Delta P = \rho g h$$
### Core Logic
The change in pressure $\Delta P$ is the hydrostatic pressure at depth $h$.
$$\Delta P = -\beta \frac{\Delta V}{V}$$
$$\rho g h = -\beta \frac{\Delta V}{V}$$
### Step 2: Calculation
Given values:
$\rho = 10^3\mathrm{\,kg/m^3}$
$g = 10\mathrm{\,m/s^2}$
$\beta = 9 \times 10^8\mathrm{\,N/m^2}$
$\frac{\Delta V}{V} = -0.02\% = -\frac{0.02}{100}$
Substitute into the equation:
$$10^3 \times 10 \times h = - (9 \times 10^8) \times \left(-\frac{0.02}{100}\right)$$
$$10^4 \times h = 9 \times 10^8 \times 2 \times 10^{-4}$$
$$10^4 h = 18 \times 10^4$$
$$h = 18\mathrm{\,m}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties Of Solids
Class 11 Physics: Mechanical Properties Of Fluids
Q35 (2026)
The strain-stress plot for materials A, B, C and D is shown in the figure. Which material has the largest Young's modulus?
{{IMG1}}
- C
- D
- A
- B
### Related Formula
$$Y = \frac{\text{Stress}}{\text{Strain}}$$
### Core Logic
In a plot where Strain is on the y-axis and Stress is on the x-axis, the slope of the line equals Strain/Stress.
$$Slope = \frac{\text{Strain}}{\text{Stress}} = \frac{1}{Y}$$
Since $Y = \frac{1}{Slope}$, the material with the largest Young's modulus $Y$ will correspond to the line with the minimum slope (least steep line).
### Step 1: Final Conclusion
From the given graph, curve C is closest to the x-axis, meaning it has the smallest slope. Therefore, material C has the largest Young's modulus.
### Pattern Recognition
Sees: "strain-stress plot" instead of standard "stress-strain plot" → The axes are swapped! Always double check axis labels. Strain on Y means Slope = 1/Y.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q39 (2026)
A brass wire of length 2m and radius 1mm at $27^{\circ}$ C is held taut between two rigid supports. Initially it was cooled to a temperature of $-43^{\circ}$ C creating a tension T in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to 1.4T, is ____$^{\circ}$C
- $-86$
- $-71$
- $-65$
- $-80$
### Related Formula
$$\text{Thermal Tension, } T = Y A \alpha \Delta \theta$$
### Core Logic
For the first cooling, the tension $T$ generated is:
$$T = Y A \alpha \Delta \theta_1$$
$$T = Y A \alpha (27 - (-43))$$
$$T = Y A \alpha (70) \quad \dots \text{(i)}$$
Let $\theta$ be the final temperature required for a tension of $1.4T$.
$$1.4T = Y A \alpha (27 - \theta) \quad \dots \text{(ii)}$$
### Step 1: Ratio of Tensions
Dividing equation (ii) by equation (i):
$$\frac{1.4T}{T} = \frac{27 - \theta}{70}$$
$$1.4 = \frac{27 - \theta}{70}$$
$$27 - \theta = 1.4 \times 70 = 98$$
$$\theta = 27 - 98 = -71^{\circ} \text{C}$$
### Pattern Recognition
Thermal tension strictly scales with the absolute difference in temperature from the unstressed state, $\Delta \theta$. Multiplying tension by a factor simply scales $\Delta \theta$ by that identical factor.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Thermal Properties of Matter
Q39 (2026)
Two wires A and B made of different materials of length 6.0 cm and 5.4 cm, respectively and area of cross sections $3.0 \times 10^{-5} \mathrm{~m}^{2}$ and $4.5 \times 10^{-5} \mathrm{~m}^{2}$ , respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of A to that of B is $x : 3$. The value of x is ____.
- $1$
- $4$
- $2$
- $5$
### Related Formula
$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta \ell / \ell} = \frac{F \ell}{A \Delta \ell}$$
### Core Logic
Since the wires are stretched by the same magnitude ($\Delta \ell_A = \Delta \ell_B$) under the same load ($F_A = F_B$), the Young's Modulus $Y$ is proportional to $\frac{\ell}{A}$.
### Step 1: Setting up the Ratio
$$\frac{Y_A}{Y_B} = \left(\frac{\ell_A}{\ell_B}\right) \left(\frac{A_B}{A_A}\right)$$
Substitute the given values:
$\ell_A = 6.0 \mathrm{~cm}$
$\ell_B = 5.4 \mathrm{~cm}$
$A_A = 3.0 \times 10^{-5} \mathrm{~m}^2$
$A_B = 4.5 \times 10^{-5} \mathrm{~m}^2$
### Step 2: Calculation
$$\frac{Y_A}{Y_B} = \left(\frac{6.0}{5.4}\right) \left(\frac{4.5 \times 10^{-5}}{3.0 \times 10^{-5}}\right)$$
$$= \left(\frac{60}{54}\right) \left(\frac{4.5}{3.0}\right)$$
$$= \left(\frac{10}{9}\right) \left(\frac{1.5}{1}\right) = \frac{15}{9} = \frac{5}{3}$$
### Step 3: Finding x
Given ratio is $x : 3$.
$$\frac{x}{3} = \frac{5}{3} \Rightarrow x = 5$$
### Pattern Recognition
Whenever "stretched by same magnitude under same load" appears, strip out $F$ and $\Delta \ell$ to leave $Y \propto \ell / A$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
