The increase in pressure required to decrease the volume of a water sample by 0.2% is P x 105 Nm -2 . Bulk modulus of…

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Q35 jee_main_2026_23_january_morning Stress-Strain Curve

The strain-stress plot for materials A, B, C and D is shown in the figure. Which material has the largest Young's modulus?

Stress-Strain Curve diagram for Q35 - JEE Main 2026 Morning — A graph plotting Strain on the y-axis vs Stress on the x-axis for four materials.

A. C
B. D
C. A
D. B

Solution

Related Formula

Y = StressStrain

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 = StrainStress = (1)/(Y)

Since $Y = (1)/(Slope)$ $Y = \frac{1}{Slope}$, the material with the largest Young's modulus $Y$ $Y$ 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.

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids

Q39 jee_main_2026_24_january_morning Thermal Stress

A brass wire of length 2m and radius 1mm at

27°

C is held taut between two rigid supports. Initially it was cooled to a temperature of

-43°

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 ____

°

C

A. $-86$ $-86$
B. $-71$ $-71$
C. $-65$ $-65$
D. $-80$ $-80$

Solution

Related Formula

Thermal Tension, T = Y A α Δ θ

Core Logic

For the first cooling, the tension $T$ $T$ generated is:

T = Y A α Δ θ₁

T = Y A α (27 - (-43))

T = Y A α (70) (i)

Let $θ$ $\theta$ be the final temperature required for a tension of $1.4T$ $1.4T$.

1.4T = Y A α (27 - θ) (ii)

Step 1: Ratio of Tensions

Dividing equation (ii) by equation (i):

(1.4T)/(T) = (27 - θ)/(70)

1.4 = (27 - θ)/(70)

27 - θ = 1.4 \times 70 = 98

θ = 27 - 98 = -71° C

Pattern Recognition

Thermal tension strictly scales with the absolute difference in temperature from the unstressed state, $Δ θ$ $\Delta \theta$. Multiplying tension by a factor simply scales $Δ θ$ $\Delta \theta$ by that identical factor.

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids Class 11 Physics: Thermal Properties of Matter

Q39 jee_main_2026_28_january_morning Young's Modulus

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 ~m²

and

4.5 \times 10⁻⁵ ~m²

, 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 ____.

A. $1$ $1$
B. $4$ $4$
C. $2$ $2$
D. $5$ $5$

Solution

Related Formula

Y = StressStrain = (F / A)/(Δ / ) = (F )/(A Δ )

Core Logic

Since the wires are stretched by the same magnitude ( $Δ A = Δ B$ $\Delta \ell_A = \Delta \ell_B$) under the same load ( $FA = FB$ $F_A = F_B$), the Young's Modulus $Y$ $Y$ is proportional to $( )/(A)$ $\frac{\ell}{A}$.

Step 1: Setting up the Ratio

(YA)/(YB) = (( A)/( B)) ((AB)/(AA))

Substitute the given values: $A = 6.0 ~cm$ $\ell_A = 6.0 \mathrm{~cm}$ $B = 5.4 ~cm$ $\ell_B = 5.4 \mathrm{~cm}$ $AA = 3.0 \times 10⁻⁵ ~m²$ $AB = 4.5 \times 10⁻⁵ ~m²$

Step 2: Calculation

(YA)/(YB) = ((6.0)/(5.4)) ( 4.5 \times 10⁻⁵3.0 \times 10⁻⁵)

= ((60)/(54)) ((4.5)/(3.0))

= ((10)/(9)) ((1.5)/(1)) = (15)/(9) = (5)/(3)

Step 3: Finding x

Given ratio is $x : 3$ $x : 3$.

(x)/(3) = (5)/(3) \Rightarrow x = 5

Pattern Recognition

Whenever "stretched by same magnitude under same load" appears, strip out $F$ $F$ and $Δ$ $\Delta \ell$ to leave $Y \propto / A$ $Y \propto \ell / A$.

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids

Q34 jee_main_2026_28_january_evening Speed of Longitudinal Wave

The speed of a longitudinal wave in a metallic bar is

400 m/s

. If the density and Young's modulus of the bar material are increased by 0.5% and 1% respectively then the speed of the wave is changed approximately to ____ m/s.

A. $399$ $399$
B. $398$ $398$
C. $402$ $402$
D. $401$ $401$

Solution

Related Formula

Vsound = \sqrt((Y)/(ρ))

where $Y$ $Y$ is Young's modulus and $ρ$ $\rho$ is density.

Core Logic

Taking natural logarithm on both sides:

ln V = (1)/(2) ln Y - (1)/(2) ln ρ

Differentiating to find fractional errors/changes:

(Δ V)/(V) = (1)/(2) (Δ Y)/(Y) - (1)/(2) (Δ ρ)/(ρ)

Step 1: Calculate Percentage Change

Multiply by 100 to get percentages:

(Δ V)/(V) \times 100 = (1)/(2) ( (Δ Y)/(Y) \times 100 ) - (1)/(2) ( (Δ ρ)/(ρ) \times 100 )

Given $(Δ Y)/(Y) \times 100 = 1%$ $\frac{\Delta Y}{Y} \times 100 = 1\%$ and $(Δ ρ)/(ρ) \times 100 = 0.5%$ .

(Δ V)/(V) \times 100 = (1)/(2)(1%) - (1)/(2)(0.5%) = 0.5% - 0.25% = 0.25%

Step 2: Calculate New Speed

(Δ V)/(V) = (0.25)/(100) = (1)/(400)

Δ V = (1)/(400) \times 400 = 1 m/s

The final speed is:

Vfinal = V + Δ V = 400 + 1 = 401 m/s

Pattern Recognition

For products and quotients of the form $A^a / B^b$ $A^a / B^b$, the small percentage change is $\pm a(%Δ A) \mp b(%Δ B)$ $\pm a(\%\Delta A) \mp b(\%\Delta B)$. Here, velocity scales with $Y1/2 ρ-1/2$ $Y^{1/2} \rho^{-1/2}$, so change is $(1)/(2)(%Δ Y) - (1)/(2)(%Δ ρ)$ .

Chapter Mix

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

Q24 jee_main_2025_02_april_evening Hooke's Law and Elongation of String

The length of a light string is

1.4m

when the tension on it is

5N

. If the tension increases to

7N

, the length of the string is

1.56m

. The original length of the string is ______

m

.

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

By Hooke's Law, the tension ( $T$ $T$) in a stretched elastic string is proportional to its change in length:

T = k \cdot Δ L = k(L - L₀)

where: $k$ $k$ = spring constant of the string $L$ $L$ = current stretched length $L₀$ $L_0$ = natural original length

Core Logic

We can set up two algebraic equations using the given conditions:

Case 1: Tension $T₁ = 5 N$ $T_1 = 5 \ \mathrm{N}$ produces stretched length $L₁ = 1.4 m$ $L_1 = 1.4 \ \mathrm{m}$:

5 = k(1.4 - L₀) (1)

Case 2: Tension $T₂ = 7 N$ $T_2 = 7 \ \mathrm{N}$ produces stretched length $L₂ = 1.56 m$ $L_2 = 1.56 \ \mathrm{m}$:

7 = k(1.56 - L₀) (2)

Step 1: Solve for natural length

Divide equation (1) by equation (2) to eliminate the spring constant $k$ $k$:

(5)/(7) = (1.4 - L₀)/(1.56 - L₀)

Cross-multiply and solve for $L₀$ $L_0$:

5(1.56 - L₀) = 7(1.4 - L₀)

7.8 - 5 L₀ = 9.8 - 7 L₀

Rearrange to isolate the variable:

2 L₀ = 9.8 - 7.8 = 2

L₀ = 1 m

Thus, the original length of the string is $1 ~m$ $1 \mathrm{~m}$.

Pattern Recognition

Sees: Elastic stretching of a string/wire under variable load. Trap: Attempting to resolve details like Young's modulus or cross-sectional area. Taking ratios allows the spring constant $k$ $k$ to cancel cleanly, saving computational effort. Shortcut: Use the ratio equation $(T₁)/(T₂) = (L₁ - L₀)/(L₂ - L₀)$ directly. Solving the linear equation yields $L₀ = 1 m$ $L_0 = 1\text{ m}$.

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids

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 ____.

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Mechanical Properties of Solids Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Ratio of Tensions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the Ratio

Step 2: Calculation

Step 3: Finding x

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Percentage Change

Step 2: Calculate New Speed

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for natural length

Pattern Recognition

Chapter Mix

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The increase in pressure required to decrease the volume of a water sample by 0.2% is $P \times 10⁵$ $P \times 10^{5}$ Nm $⁻²$ $^{-2}$ . Bulk modulus of water is $2.15 \times 10⁹$ $2.15 \times 10^{9}$ Nm $⁻²$ $^{-2}$ . The value of P is ____.