Mechanical Properties of Solids Previous Year Questions — JEE Main Physics
21 past-year Mechanical Properties of Solids questions from JEE Main (Physics) — page 3.
Q43 (2024)
A wire of length $L$ and radius $r$ is clamped at one end. If its other end is pulled by a force $F$, its length increases by $l$. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:
- $3\text{ times}$
- $\frac{3}{2}\text{ times}$
- $4\text{ times}$
- $2\text{ times}$
### Related Formula
Young's modulus $Y$ is defined as:
$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{l / L} = \frac{F L}{A l}$$
Solving for the extension $l$:
$$l = \frac{F L}{A Y} = \frac{F L}{\pi r^2 Y}$$
Since the material does not change, $Y$ remains constant.
### Core Logic
Let the original parameters be $F$, $r$, $L$, $l$. The extension is:
$$l = \frac{F L}{\pi r^2 Y} \quad \dots (i)$$
When the force and radius are both halved, keeping original length $L$ and Young's Modulus $Y$ constant:
* New Force, $F' = \frac{F}{2}$
* New Radius, $r' = \frac{r}{2}$
The new extension $l'$ is:
$$l' = \frac{F' L}{\pi (r')^2 Y} = \frac{(F/2) L}{\pi (r/2)^2 Y}$$
$$l' = \frac{F L / 2}{\pi r^2 Y / 4} = 2 \left( \frac{F L}{\pi r^2 Y} \right)$$
### Step 1: Calculate Final Extension
Comparing this with equation (i):
$l' = 2l$
Thus, the increase in length becomes $2\text{ times}$ its original value.
### Pattern Recognition
From the formula $l \propto \frac{F}{r^2}$, if $F$ is scaled by $\frac{1}{2}$ and $r$ by $\frac{1}{2}$, the scaling factor for $l$ is $\frac{1/2}{(1/2)^2} = 2$ directly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q51 (2024)
Two metallic wires $P$ and $Q$ have same volume and are made up of same material. If their area of cross sections are in the ratio $4:1$ and force $F_1$ is applied to $P$, an extension of $\Delta l$ is produced. The force which is required to produce same extension in $Q$ is $F_2$. The value of $\frac{F_1}{F_2}$ is:
### Related Formula
The Young's modulus $Y$ is:
$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta l / l} = \frac{F l}{A \Delta l}$$
Solving for extension $\Delta l$:
$$\Delta l = \frac{F l}{A Y}$$
Since volume $V = A l$, we substitute $l = \frac{V}{A}$:
$$\Delta l = \frac{F V}{A^2 Y}$$
### Core Logic
We are given that the wires are made of the same material ($Y$ is constant) and have the same volume ($V$ is constant). Also, we want to achieve the same extension ($\Delta l$ is constant).
Thus, from the formula:
$$\Delta l \propto \frac{F}{A^2} \implies F \propto A^2$$
### Step 1: Calculate the Force Ratio
Using the proportionality for both wires:
$$\frac{F_1}{F_2} = \left( \frac{A_1}{A_2} \right)^2$$
Given the cross-sectional area ratio is $4:1$:
$$\frac{A_1}{A_2} = \frac{4}{1}$$
Substitute this into the ratio equation:
$$\frac{F_1}{F_2} = \left( \frac{4}{1} \right)^2 = 16$$
### Pattern Recognition
For a constant volume and material, extension $\Delta l \propto \frac{F}{A^2}$. Since area ratio is $4$, the required force ratio is $4^2 = 16$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q1560 (2024)
If the average depth of an ocean is $4000\text{ m}$ and the bulk modulus of water is $2 \times 10^{9}\text{ N}\cdot\text{m}^{-2}$, then the fractional compression $\frac{\Delta V}{V}$ of water at the bottom of the ocean is $\alpha \times 10^{-2}$. The value of $\alpha$ is ______.
(Given, $g = 10\text{ m}\cdot\text{s}^{-2}$, $\rho = 1000\text{ kg}\cdot\text{m}^{-3}$).
### Related Formula
$$B = -\frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} \implies \left|\frac{\Delta V}{V}\right| = \frac{\Delta P}{B}$$
Where the hydrostatic pressure difference at depth is $\Delta P = \rho g h$.
### Core Logic
Calculate the gauge pressure $\Delta P$ at the ocean floor ($h = 4000\text{ m}$):
$$\Delta P = 1000 \times 10 \times 4000 = 4 \times 10^7\text{ N/m}^2$$
### Step 1: Compute fractional volume change
$$\left|\frac{\Delta V}{V}\right| = \frac{4 \times 10^7}{2 \times 10^9} = 2 \times 10^{-2}$$
### Step 2: Match parameters to find alpha
Comparing $2 \times 10^{-2}$ to $\alpha \times 10^{-2}$ gives:
$\alpha = 2$
### Pattern Recognition
Hydrostatic scaling expressions yield precise decade cancellations when tracked directly relative to high exponent elastic constants like Bulk Moduli.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids / Fluids
Q1824 (2024)
Each of three blocks P, Q and R shown in figure has a mass of $3 \mathrm{~kg}$. Each of the wire A and B has cross-sectional area $0.005 \mathrm{~cm}^2$ and Young's modulus $2 \times 10^{11} \mathrm{Nm}^{-2}$. Neglecting friction, the longitudinal strain on wire B is $\_ \_ \_ \_ \times 10^{-4}$. (Take $g = 10 \mathrm{m / s^2}$)
{{IMG1}}
### Related Formula
$$a = \frac{\text{Net Pulling Force}}{\text{Total Mass}}$$
$$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{T}{AY}$$
### Core Logic
{{SOL_IMG1}}
First, calculate the common acceleration of the system of three blocks. Block R provides the driving gravitational force ($3g$), pulling blocks P and Q. Then isolate the tension in wire B ($T_1$) pulling block P. Finally, apply Young's modulus to find the strain.
### Step 1: Find Acceleration
Total driving force = Weight of block R = $3 \times 10 = 30 \mathrm{~N}$.
Total mass being accelerated = $m_P + m_Q + m_R = 3 + 3 + 3 = 9 \mathrm{~kg}$.
$$a = \frac{30}{9} = \frac{10}{3} \mathrm{~m/s^2}$$
### Step 2: Find Tension in Wire B
Wire B connects block P to block Q. The only horizontal force accelerating block P is the tension $T_1$ in wire B.
$$T_1 = m_P a = 3 \times \frac{10}{3} = 10 \mathrm{~N}$$
*(Wait, the pdf solution says $30 - T_1 = 3 \times a \Rightarrow T_1 = 20 \mathrm{~N}$. Let's examine the arrangement from the image context. If block R hangs, wire A connects R to Q, and wire B connects Q to P. If B connects R and Q, it bears more tension. Let's trust the PDF calculation: The solution implies block P is the hanging block driving it? No, if $30 - T_1 = 3a$, then $T_1$ is the tension supporting a hanging block. That means wire B is supporting a 3kg hanging block against gravity, or wire B connects Q and R. According to PDF logic: $30 - T_1 = 3 \times (10/3) \Rightarrow T_1 = 20 \mathrm{N}$. This means wire B is the vertical wire supporting block R.)*
### Step 3: Calculate Strain
Area $A = 0.005 \mathrm{~cm}^2 = 0.005 \times 10^{-4} \mathrm{~m}^2 = 5 \times 10^{-7} \mathrm{~m}^2$.
$$Y = 2 \times 10^{11} \mathrm{~N/m^2}$$
$$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{T_1}{A Y}$$
$$\text{Strain} = \frac{20}{5 \times 10^{-7} \times 2 \times 10^{11}}$$
$$\text{Strain} = \frac{20}{10 \times 10^4} = \frac{20}{10^5} = 20 \times 10^{-5} = 2 \times 10^{-4}$$
### Pattern Recognition
In multi-block connected systems, always find common 'a' first using $\Sigma F / \Sigma m$. Then isolate the single block the wire directly pulls to find tension. Strain acts statically under that tension.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Laws of Motion
Q34 (2024)
Young's modulus of material of a wire of length 'L' and cross-sectional area A is Y. If the length of the wire is doubled and cross-sectional area is halved then Young's modulus will be :
- $\frac{Y}{4}$
- $4Y$
- $Y$
- $2Y$
### Related Formula
$$Y = \frac{\text{Stress}}{\text{Strain}}$$
(However, Y is an intrinsic property of the material.)
### Core Logic
Young's modulus ($Y$) is a fundamental material property (intensive property). It depends exclusively on the nature of the material and its temperature, not on the macroscopic physical dimensions such as length or cross-sectional area of the specific specimen.
### Step 1: Final Conclusion
Since the material remains unchanged, the Young's modulus remains the same.
Therefore, $Y_{new} = Y$.
### Pattern Recognition
Whenever dimensions (length, area, radius) change but the material is constant, intrinsic properties like density, resistivity, and elastic moduli remain strictly invariant.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
