Each of three blocks P, Q and R shown in figure has a mass of 3 ~kg$3 \mathrm{~kg}$. Each of the wire A and B has cross-sectional area 0.005 ~cm²$0.005 \mathrm{~cm}^2$ and Young's modulus 2 × 10¹¹ Nm⁻²$2 \times 10^{11} \mathrm{Nm}^{-2}$. Neglecting friction, the longitudinal strain on wire B is _ _ _ _ × 10⁻⁴$\_ \_ \_ \_ \times 10^{-4}$. (Take g = 10 m / s²$g = 10 \mathrm{m / s^2}$)
Three 3kg blocks hanging and resting on table attached via wires A and B through a pulley.
Numerical Answer Type:
Enter a numerical valueAnswer: 2 to 2+4 marks
Solution & Explanation
Related Formula
a = Net Pulling ForceTotal Mass$$a = \frac{\text{Net Pulling Force}}{\text{Total Mass}}$$Strain = StressY = (T)/(AY)$$\text{Strain} = \frac{\text{Stress}}{Y} = \frac{T}{AY}$$
Core Logic
Three 3kg blocks hanging and resting on table attached via wires A and B through a pulley.
First, calculate the common acceleration of the system of three blocks. Block R provides the driving gravitational force (3g$3g$), pulling blocks P and Q. Then isolate the tension in wire B (T₁$T_1$) pulling block P. Finally, apply Young's modulus to find the strain.
Step 1: Find Acceleration
Total driving force = Weight of block R = 3 × 10 = 30 ~N$3 \times 10 = 30 \mathrm{~N}$.
Total mass being accelerated = mP + mQ + mR = 3 + 3 + 3 = 9 ~kg$m_P + m_Q + m_R = 3 + 3 + 3 = 9 \mathrm{~kg}$.
(Wait, the pdf solution says 30 - T₁ = 3 × a ⇒ T₁ = 20 ~N$30 - T_1 = 3 \times a \Rightarrow T_1 = 20 \mathrm{~N}$. 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 - T₁ = 3a$30 - T_1 = 3a$, then T₁$T_1$ 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 - T₁ = 3 × (10/3) ⇒ T₁ = 20 N$30 - T_1 = 3 \times (10/3) \Rightarrow T_1 = 20 \mathrm{N}$. This means wire B is the vertical wire supporting block R.)
In multi-block connected systems, always find common 'a' first using Σ F / Σ m$\Sigma F / \Sigma m$. Then isolate the single block the wire directly pulls to find tension. Strain acts statically under that tension.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Laws of Motion
More Mechanical Properties of Solids Previous-Year Questions — Page 3
Q24jee_main_2025_24_jan_eveningBulk Modulus
The increase in pressure required to decrease the volume of a water sample by 0.2% is P × 10⁵$P \times 10^{5}$ Nm ⁻²$^{-2}$ . Bulk modulus of water is 2.15 × 10⁹$2.15 \times 10^{9}$ Nm ⁻²$^{-2}$ . The value of P is ____.
Numerical Answer.Answer: 43 to 43
Solution
Related Formula
Bulk Modulus formula:
B = - (Δ P)/((Δ V)/(V)) Δ P = B ( (-Δ V)/(V) )$$B = - \frac{\Delta P}{\frac{\Delta V}{V}} \implies \Delta P = B \left( \frac{-\Delta V}{V} \right)$$
Comparing with P × 10⁵$P \times 10^{5}$, we get P = 43$P = 43$.
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.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q23jee_main_2025_28_jan_eveningBulk Modulus
The volume contraction of a solid copper cube of edge length 10cm$10\mathrm{cm}$ , when subjected to a hydraulic pressure of 7 × 10⁶Pa$7 \times 10^{6}\mathrm{Pa}$ , would be ____________ mm³$mm^3$ .
(Given bulk modulus of copper = 1.4 × 10¹¹ Nm⁻²$= 1.4 \times 10^{11} \mathrm{Nm}^{-2}$ )
Numerical Answer.Answer: 50
Solution
Related Formula
Bulk modulus B$B$ measures a material's resistance to uniform compression and is defined as the ratio of hydraulic pressure to volumetric strain:
B = (Δ P)/(((Δ V)/(V))) Δ V = (Δ P · V)/(B)$$B = \frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} \implies \Delta V = \frac{\Delta P \cdot V}{B}$$
Core Logic
Given parameters [cite: 192, 193]:
Edge length of the cube, a = 10 cm = 0.1 m$a = 10 \text{ cm} = 0.1 \text{ m}$
Always perform unit conversions carefully at the final step to avoid handling complex decimals during calculations. Converting 1 m³ = 10⁹ mm³$1 \text{ m}^3 = 10^9 \text{ mm}^3$ ensures a clean, error-free conversion.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Qjee_main_2025_29_jan_morningElasticity
The fractional compression ( V)$\left(\frac{\Delta\mathrm{V}}{\mathrm{V}}\right)$ of water at the depth of 2.5km$2.5\mathrm{km}$ below the sea level is %$\%$ Given, the Bulk modulus of water = 2× 10⁹Nm⁻²$= 2\times 10^{9}\mathrm{Nm}^{-2}$ , density of water = 10³kgm⁻³$= 10^{3}\mathrm{kg}\mathrm{m}^{-3}$ , acceleration due to gravity = g = 10ms⁻²$= \mathrm{g} = 10\mathrm{ms}^{-2}$ .
A. 1.75
B. 1.0
C. 1.5
D. 1.25
Solution
Related Formula
B = (Δ P)/(((Δ V)/(V))) = (ρ g h)/(((Δ V)/(V)))$$B = \frac{\Delta P}{\left(\frac{\Delta V}{V}\right)} = \frac{\rho g h}{\left(\frac{\Delta V}{V}\right)}$$
Core Logic
Calculate the hydrostatic pressure change at depth h = 2.5 km = 2500 m$h = 2.5\text{ km} = 2500\text{ m}$ :
Δ P = ρ g h = 10³ · 10 · 2500 = 2.5 × 10⁷ ~N· m⁻²$$\Delta P = \rho g h = 10^3 \cdot 10 \cdot 2500 = 2.5 \times 10^7 \mathrm{~N\cdot m^{-2}}$$
Now compute the percentage fractional compression :
Class 11 Physics: Mechanical Properties of Solids
Class 11 Physics: Mechanical Properties of Fluids
Qjee_main_2024_01_february_morningElastic Moduli
With rise in temperature, the Young's modulus of elasticity:
A. changes erratically
B. decreases
C. increases
D. remains unchanged
Solution
Related Formula
Intermolecular forces weaken as thermal agitation increases:
Y = StressStrain$$Y = \frac{\text{Stress}}{\text{Strain}}$$
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 arrow$\rightarrow$ Thermal expansion up arrow$\rightarrow$ Intermolecular bonds weaken arrow$\rightarrow$ Modulus of elasticity down.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
Q43jee_main_2024_29_january_eveningHooke's Law and Young's Modulus
A wire of length L$L$ and radius r$r$ is clamped at one end. If its other end is pulled by a force F$F$, its length increases by l$l$. 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:
Thus, the increase in length becomes 2 times$2\text{ times}$ its original value.
Pattern Recognition
From the formula l ∝ (F)/(r²)$l \propto \frac{F}{r^2}$, if F$F$ is scaled by (1)/(2)$\frac{1}{2}$ and r$r$ by (1)/(2)$\frac{1}{2}$, the scaling factor for l$l$ is (1/2)/((1/2)²) = 2$\frac{1/2}{(1/2)^2} = 2$ directly.
Chapter Mix
Class 11 Physics: Mechanical Properties of Solids
More Mechanical Properties of Solids Questions — jee_main_2024_30_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
