Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in…

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Q22 jee_main_2025_02_april_morning Elastic Moduli

A steel wire of length

2~m

and Young's modulus

2.0\times 10¹¹~N/m²

is stretched by a force. If Poisson's ratio and transverse strain for the wire are 0.2 and

10⁻³

respectively, then the elastic potential energy density of the wire is

[value] \times 10⁵

(in SI units)

Numerical Answer. Answer: 25 to 25

Solution

Related Formula

σ = εtransverseεlongitudinal

u = (1)/(2) Y εlongitudinal²

Core Logic

Given parameters:

Young's modulus, $Y = 2.0 \times 10¹¹~N/m²$ $Y = 2.0 \times 10^{11}\mathrm{~N/m}^2$
Poisson's ratio, $σ = 0.2$ $\sigma = 0.2$
Transverse strain, $εₜᵣₐₙₛ = 10⁻³$ $\epsilon_{\text{trans}} = 10^{-3}$

First, calculate the longitudinal strain $ε$ $\epsilon$:

σ = εₜᵣₐₙₛε 0.2 = 10⁻³ε

ε = 10⁻³0.2 = 5 \times 10⁻³

Now, compute the elastic potential energy density $u$ $u$:

u = (1)/(2) Y ε² = (1)/(2) \times (2.0 \times 10¹¹) \times (5 \times 10⁻³)²

u = 10¹¹ \times 25 \times 10⁻⁶ = 25 \times 10⁵~J/m³

Expressing as $x \times 10⁵$ $x \times 10^5$, we have $x = 25$ $x = 25$.

Step 1: Final Conclusion

The value of the coefficient is 25.

Pattern Recognition

Poisson's ratio is defined as the ratio of lateral/transverse strain to longitudinal strain. Use this to determine the longitudinal stretch, then apply the basic potential energy density formula $(1)/(2) Y ε²$ $\frac{1}{2} Y \epsilon^2$ directly.

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids

Q19 jee_main_2025_07_april_morning Elastic Behaviour

Two wires A and B are made of same material having ratio of lengths

LALB = (1)/(3)

and their diameters ratio

dAdB = 2

. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

A. $1:6$ $1:6$
B. $1:12$ $1:12$
C. $3:4$ $3:4$
D. $1:3$ $1:3$

Solution

Related Formula

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

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

Area of cross-section of wire with diameter $d$ $d$ is:

A = (π d²)/(4) Δ L = (4 F L)/(π d² Y)

Core Logic Since both wires are made of the same material ($ Core Logic Since both wires are made of the same material ($Y_A = Y_B) and stretched with the same force ($) and stretched with the same force ($ F_A = F_B):

Δ L \propto (L)/(d²)

Set up the ratio for wires A and B:

(Δ LA)/(Δ LB) = ( (LA)/(LB) ) \times ( (dB)/(dA) )²

Step 1: Substitute Given Ratios Substitute the ratios$ Step 1: Substitute Given Ratios Substitute the ratios $\frac{L_A}{L_B} = \frac{1}{3}and$ and $\frac{d_A}{d_B} = 2 \implies \frac{d_B}{d_A} = \frac{1}{2}:

(Δ LA)/(Δ LB) = ( (1)/(3) ) \times ( (1)/(2) )² = (1)/(3) \times (1)/(4) = (1)/(12)

Therefore, the ratio of their elongations is $ Therefore, the ratio of their elongations is $1:12 . Pattern Recognition Sees: Same material, same stretching force. Shortcut: Elongation scales directly with length and inversely with the square of the diameter (radius). Thus, ratio is $. Pattern Recognition Sees: Same material, same stretching force. Shortcut: Elongation scales directly with length and inversely with the square of the diameter (radius). Thus, ratio is $(1/3) / (2^2) = 1/12$. Chapter Mix Class 11 Physics: Mechanical Properties of Solids

Q3 jee_main_2025_08_april_evening Young's Modulus

A

3~m

long wire of radius

3~mm

shows an extension of

0.1~mm

when loaded vertically by a mass of

50~kg

in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is

P \times 10¹¹~N ⁻²

, where the value of

P

is: (Take

g = 3π~m/s²

)

A. $5$ $5$
B. $10$ $10$
C. $25$ $25$
D. $2.5$ $2.5$

Solution

Related Formula

Y = StressStrain = (F / A)/(Δ L / L) = (mg L)/(π r² Δ L)

where, $Y$ $Y$ = Young's modulus $F = mg$ $F = mg$ = stretching force (load) $A = π r²$ $A = \pi r^2$ = cross-sectional area of the wire $L$ $L$ = original length of the wire $Δ L$ $\Delta L$ = extension produced

Core Logic

Given parameters:

Original length of wire, $L = 3~m$ $L = 3\mathrm{~m}$
Radius of wire, $r = 3~mm = 3 \times 10⁻³~m$
Mass loaded, $m = 50~kg$ $m = 50\mathrm{~kg}$
Acceleration due to gravity, $g = 3π~m/s²$ $g = 3\pi\mathrm{~m/s}^{2}$
Extension, $Δ L = 0.1~mm = 10⁻⁴~m$

Substitute the values into the Young's modulus formula:

Y = 50 \times (3π) \times 3π \times (3 \times 10⁻³)² \times 10⁻⁴

Step 1: Detailed Computation

Simplify the equation:

Y = 450ππ \times 9 \times 10⁻⁶ \times 10⁻⁴

Cancelling $π$ $\pi$ from numerator and denominator:

Y = 4509 \times 10⁻¹⁰ = 50 \times 10¹⁰ = 5 \times 10¹¹~N/m²

Comparing this with $P \times 10¹¹~N/m²$ $P \times 10^{11}\mathrm{~N/m}^{2}$:

$P = 5$ $P = 5$

Pattern Recognition

Sees: Standard Young's modulus measurement. Trap: Don't forget that the radius is in millimeters and the extension is in millimeters. Convert all units to SI systematically first. Calculation Tip: The variable $g = 3π$ $g = 3\pi$ is chosen deliberately to cancel the $π$ $\pi$ in the area formula. Keep your eye open for these mathematical shortcuts! ✓

Chapter Mix

Class 11 Physics: Elasticity

Q5 jee_main_2025_04_april_evening Elastic Moduli

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 top. Considering infinitesimally small displacement in the upper edge, the angular displacement

θ

of the rod axis from its original position would be : (shear moduli,

G=10¹⁰ N/m²

)

A. $1/160π$ $1/160\pi$
B. $1/4π$ $1/4\pi$
C. $1/40π$ $1/40\pi$
D. $1/2π$ $1/2\pi$

Solution

Related Formula

G = σshearθ = (F)/(A \cdot θ)

where $A = π r²$ $A = \pi r^2$ is the cross-sectional area.

Core Logic

Rearranging the formula for $θ$ $\theta$:

θ = (F)/(A \cdot G) = (F)/(π r² \cdot G)

Substituting the values:

F = 10⁵ N

r = 4 cm = 4 \times 10⁻² m

G = 10¹⁰ N/m²

Step 1: Calculate Angular Displacement

θ = 10⁵π \times (4 \times 10⁻²)² \times 10¹⁰

θ = 10⁵π \times 16 \times 10⁻⁴ \times 10¹⁰ = (10⁵)/(16π \times 10⁶)

θ = (1)/(160π) radians

Pattern Recognition

Shear strain definition is straightforwardly matching the geometry angle of shift. Avoid radius-to-diameter or unit mismatches.

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids

Q jee_main_2025_04_april_morning Shear Modulus of Elasticity

Two slabs with square cross section of different materials (1, 2) with equal sides (

l

) and thickness

d₁

and

d₂

such that

d₂ = 2d₁

and

l > d₂

. Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is

θ₂ = 2θ₁

. If the shear moduli of material 1 is

4 \times 10⁹~N/m²

, then shear moduli of material 2 is

x \times 10⁹~N/m²

, where value of x is

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

Shear modulus definition:

η = Shear StressShear Strain = (F / A)/(θ) = (F)/(Aθ)

where $A = l \cdot d$ $A = l \cdot d$ is the area of the narrow face parallel to the shearing force $F$ $F$.

Core Logic

We are given:

$θ₂ = 2θ₁$ $\theta_2 = 2\theta_1$
Shearing forces are equal ( $F₁ = F₂ = F$ $F_1 = F_2 = F$).
Thickness relationship: $d₂ = 2d₁$ $d_2 = 2d_1$
Resisting face area: $A₁ = l d₁$ $A_1 = l d_1$ and $A₂ = l d₂ = 2l d₁ = 2A₁$ $A_2 = l d_2 = 2l d_1 = 2A_1$

Shear strain deformation slab configuration layout for Q22 - JEE Main 2025 Morning

Step 1: Link Modulus Terms

Express the angular deformation for each slab:

θ₁ = (F)/(l d₁ η₁)

θ₂ = (F)/(l d₂ η₂) = (F)/(2l d₁ η₂)

Substitute these expressions into $θ₂ = 2θ₁$ $\theta_2 = 2\theta_1$:

(F)/(2l d₁ η₂) = 2 \cdot ((F)/(l d₁ η₁))

(1)/(2η₂) = (2)/(η₁) 4η₂ = η₁ η₂ = (η₁)/(4)

Step 2: Solve for x

Given $η₁ = 4 \times 10⁹ N/m²$ $\eta_1 = 4 \times 10^{9}\text{ N/m}^2$:

η₂ = (4 \times 10⁹)/(4) = 1 \times 10⁹ N/m²

Comparing with $η₂ = x \times 10⁹ N/m²$ $\eta_2 = x \times 10^9\text{ N/m}^2$: $x = 1$ $x = 1$

Pattern Recognition

Shear stress is inversely proportional to the face area parallel to the applied shearing force ( $A = l \cdot d$ $A = l \cdot d$). Doubling the slab thickness doubles the resisting contact area, halving the applied shear stress under identical lateral loading.

Evaluation Rubric / Model Answer

1

Chapter Mix

Class 11 Physics: Mechanical Properties of Solids

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate the Force Ratio

Pattern Recognition

Chapter Mix

Reference Study Guides

More Mechanical Properties of Solids Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Substitute Given Ratios

Step 1: Substitute Given Ratios

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Detailed Computation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Angular Displacement

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Link Modulus Terms

Step 2: Solve for x

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

More Mechanical Properties of Solids Questions — jee_main_2024_29_january_evening

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