A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of 105 N at the…

Year	2026	2025	2024	Total
Questions	4	9	8	21

Q24 jee_main_2025_24_jan_evening Bulk Modulus

The increase in pressure required to decrease the volume of a water sample by 0.2% is

P \times 10⁵

Nm

⁻²

. Bulk modulus of water is

2.15 \times 10⁹

Nm

⁻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) )

Core Logic

Given specifications:

Fractional volume drop, $(-Δ V)/(V) = 0.2% = (0.2)/(100) = 2 \times 10⁻³$
Bulk modulus value, $B = 2.15 \times 10⁹ N/m²$

Calculating excess pressure:

Δ P = 2.15 \times 10⁹ \times 2 \times 10⁻³

Δ P = 4.3 \times 10⁶ N/m² = 43 \times 10⁵ N/m²

Comparing with $P \times 10⁵$ $P \times 10^{5}$, 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

Q23 jee_main_2025_28_jan_evening Bulk Modulus

The volume contraction of a solid copper cube of edge length

10cm

, when subjected to a hydraulic pressure of

7 \times 10⁶Pa

, would be ____________

mm³

. (Given bulk modulus of copper

= 1.4 \times 10¹¹ 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 \cdot V)/(B)

Core Logic

Given parameters [cite: 192, 193]:

Edge length of the cube, $a = 10 cm = 0.1 m$ $a = 10 \text{ cm} = 0.1 \text{ m}$
Initial volume, $V = a³ = (0.1)³ = 10⁻³ m³ = 10⁶ mm³$
Hydraulic pressure increase, $Δ P = 7 \times 10⁶ Pa$ $\Delta P = 7 \times 10^6 \text{ Pa}$
Bulk Modulus, $B = 1.4 \times 10¹¹ N/m²$ $B = 1.4 \times 10^{11} \text{ N/m}^2$

Substitute these values into the volume change equation :

Δ V = 7 \times 10⁶ \times 10⁻³1.4 \times 10¹¹

Δ V = 7 \times 10³1.4 \times 10¹¹ = 5 \times 10⁻⁷ m³

Convert the volume contraction into $mm³$ $\text{mm}^3$:

Δ V = 5 \times 10⁻⁷ \times (10³)³ mm³ = 5 \times 10⁻⁷ \times 10⁹ mm³ = 50 mm³

Pattern Recognition

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

Q jee_main_2025_29_jan_morning Elasticity

The fractional compression

( V)

of water at the depth of

2.5km

below the sea level is

%

Given, the Bulk modulus of water

= 2\times 10⁹Nm⁻²

, density of water

= 10³kgm⁻³

, acceleration due to gravity

= g = 10ms⁻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)))

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 \cdot 10 \cdot 2500 = 2.5 \times 10⁷ ~N\cdot m⁻²

Now compute the percentage fractional compression :

(Δ V)/(V) \times 100 = (Δ P)/(B) \times 100 = (2.5 \times 10⁷)/(2 \times 10⁹) \times 100% = 1.25%

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2024_01_february_morning Elastic 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

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

Q43 jee_main_2024_29_january_evening Hooke's Law and Young's Modulus

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:

A. $3 times$ $3\text{ times}$
B. $(3)/(2) times$ $\frac{3}{2}\text{ times}$
C. $4 times$ $4\text{ times}$
D. $2 times$ $2\text{ times}$

Solution

Related Formula

Young's modulus $Y$ $Y$ is defined as:

Y = StressStrain = (F / A)/(l / L) = (F L)/(A l)

Solving for the extension $l$ $l$:

l = (F L)/(A Y) = (F L)/(π r² Y)

Since the material does not change, $Y$ $Y$ remains constant.

Core Logic

Let the original parameters be $F$ $F$, $r$ $r$, $L$ $L$, $l$ $l$. The extension is:

l = (F L)/(π r² Y) (i)

When the force and radius are both halved, keeping original length $L$ $L$ and Young's Modulus $Y$ $Y$ constant:

New Force, $F' = (F)/(2)$ $F' = \frac{F}{2}$
New Radius, $r' = (r)/(2)$ $r' = \frac{r}{2}$

The new extension $l'$ $l'$ is:

l' = (F' L)/(π (r')² Y) = ((F/2) L)/(π (r/2)² Y)

l' = (F L / 2)/(π r² Y / 4) = 2 ( (F L)/(π r² Y) )

Step 1: Calculate Final Extension

Comparing this with equation (i):

$l' = 2l$ $l' = 2l$

Thus, the increase in length becomes $2 times$ $2\text{ times}$ its original value.

Pattern Recognition

From the formula $l \propto (F)/(r²)$ $l \propto \frac{F}{r^2}$, 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

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Angular Displacement

Pattern Recognition

Chapter Mix

Reference Study Guides

More Mechanical Properties of Solids Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Final Extension

Pattern Recognition

Chapter Mix

More Mechanical Properties of Solids Questions — jee_main_2025_04_april_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back