Elasticity | JEE Main previous year questions with Solutions

Q 1 :
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 [2024]

3 times
4 times
$\frac{3}{2}$ times
2 times

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(D) Young's modulus, $Y = \frac{s t r e s s}{s t r a i n} = \frac{F / A}{∆ l / L} = \frac{F L}{A ∆ l}$

In case I : $Y = \frac{F L}{{πr}^{2} l}$ ...(i)

In case II : $Y = \frac{(F / 2) L}{π {(\frac{r}{2})}^{2} ∆ l} = \frac{2 F L}{π r^{2} ∆ l}$ ...(ii)

From (i) and (ii), $\frac{F L}{π r^{2} l} = \frac{2 F L}{π r^{2} ∆ l} \Rightarrow ∆ l = 2 l$

Q 2 :
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 [2024]

$Y / 4$
$4 Y$
$Y$
$2 Y$

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4

(C) Young's modulus depends on the material not on length and cross-sectional area. So young's modulus remains same.

Q 3 :
With rise in temperature, the Young's modulus of elasticity [2024]

changes erratically
decreases
increases
remains unchanged

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4

(B) $Y = \frac{F / A}{∆ l / l} = \frac{F / A}{α ∆ T}$ as $∆ T$ increases, $Y$ decreases

Q 4 :
A wire of cross-sectional area A, modulus of elasticity $2 \times 10^{11}$ ${Nm}^{- 2}$ and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle, it sags lower from its original position, making an angle $θ = \frac{1}{100}$ radian on the points of support. The value of A is _____ $\times 10^{- 4}$ $m^{2}$ (consider $x < < L$ ). [2024]

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(1)

$\tan θ = \frac{x}{L} \approx θ, As θ is small$

$2 T \sin θ = 20 \Rightarrow T θ = 10$

$T \cdot \frac{1}{100} = 10 \Rightarrow T = 1000 N$

Change in length, $Δ L = 2 \sqrt{x^{2} + L^{2}} - 2 L$

$\Rightarrow Δ L = 2 L [1 + \frac{x^{2}}{2 L^{2}} - 1] = \frac{x^{2}}{L}$

$Stress = Y strain \Rightarrow \frac{T}{A} = Y \frac{Δ L}{(2 L)}$

$\frac{T}{A} = Y \cdot \frac{x^{2}}{L (2 L)} = \frac{Y}{2} \cdot (\frac{x^{2}}{L^{2}}) = \frac{Y}{2} \cdot θ^{2}$

$\frac{1000}{A} = \frac{Y θ^{2}}{2} = \frac{2 \times 10^{11}}{2} \times \frac{1}{100 \times 100}$

$\Rightarrow A = 1 \times 10^{- 4} m^{2}$

Q 5 :
Two persons pull a wire towards themselves. Each person exerts a force of 200 N on the wire.

Young’s modulus of the material of the wire is $1 \times 10^{11}$ ${Nm}^{- 2}$ . Original length of the wire is 2 m and the area of cross-section is 2 ${cm}^{2}$ . The wire will extend in length by _____ μm. [2024]

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(20)

$\frac{F}{A} = Y \frac{Δ ℓ}{ℓ} \Rightarrow Δ ℓ = \frac{F ℓ}{A Y}$

$Δ ℓ = \frac{200 \times 2}{2 \times 10^{- 4} \times 10^{11}} = 2 \times 10^{- 5} = 20 μ m$

Q 6 :
Two metallic wires P and Q have the same volume and are made up of the 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 $∆ λ$ is produced. The force required to produce the same extension in Q is $F_{2}$ . The value of $\frac{F_{1}}{F_{2}}$ is _____. [2024]

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(16)

$Y = \frac{Stress}{Strain} = \frac{F / A}{Δ ℓ / ℓ} = \frac{F ℓ}{A Δ ℓ}$

$Δ ℓ = \frac{F ℓ}{A Y}$

$V = A ℓ \Rightarrow ℓ = \frac{V}{A}$

$Δ ℓ = \frac{F V}{A^{2} Y}$

Y and V is same for both the wires

$Δ ℓ \propto \frac{F}{A^{2}}$

$\frac{Δ ℓ_{1}}{Δ ℓ_{2}} = \frac{F_{1}}{A_{1}^{2}} \times \frac{A_{2}^{2}}{F_{2}}$

$Δ ℓ_{1} = Δ ℓ_{2}$

$\frac{F_{1}}{F_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}} = {(\frac{4}{1})}^{2} = 16$

Q 7 :
Each of three blocks P, Q, and R shown in the figure has a mass of 3 kg. Each of the wires A and B has a cross-sectional area of 0.005 ${cm}^{2}$ and Young’s modulus $2 \times 10^{11}$ $N m^{2}$ . Neglecting friction, the longitudinal strain on wire B is _____ $\times 10^{- 4}$ . (Take g = 10 $m / s^{2}$ ). [2024]

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(2)

$3 g = (3 + 3 + 3) a \Rightarrow a = \frac{3}{9} g = \frac{1}{3} g$

$Now, 3 g - T = 3 a$

$\Rightarrow T = 3 (g - a) = 3 (g - \frac{g}{3}) = 2 g = 20 N$

$Y = \frac{longitudinal stress}{longitudinal strain}$

$longitudinal strain = \frac{longitudinal stress}{Y}$

$= \frac{T}{A Y} = \frac{20}{5 \times 10^{- 7} \times 2 \times 10^{11}} = 2 \times 10^{- 4}$

Q 8 :
Two blocks of mass 2 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The radius of the wire is $4.0 \times 10^{- 5}$ m and Young’s modulus of the metal is $2.0 \times 10^{11}$ $N / m^{2}$ . The longitudinal strain developed in the wire is $\frac{1}{α π}$ . The value of $α$ is _______ .

[Use g = 10 $m / s^{2}$ ] [2024]

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(12)

$T = (\frac{2 m_{1} m_{2}}{m_{1} + m_{2}}) g = \frac{80}{3} N$

$A = π r^{2} = 16 π \times 10^{- 10} m^{2}$

$Strain = \frac{Δ ℓ}{ℓ} = \frac{F}{A Y} = \frac{T}{A Y}$

$= \frac{\frac{80}{3}}{16 π \times 10^{- 10} \times 2 \times 10^{11}} = \frac{1}{12 π}$

$α = 12$

Q 9 :
One end of a metal wire is fixed to a ceiling, and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load, and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of the upper wire to that of the lower wire will be _____.

[Area of cross-section of wire = 0.005 ${cm}^{2}$ , Y = $2 \times 10^{11}$ ${Nm}^{- 2}$ and g = 10 ${ms}^{- 2}$ ] [2024]

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(3)

$\sum f_{net} = 0$

$T_{2} = T_{1} + 2 g = g + 2 g = 3 g$

$T_{1} = g$ ...(i)

${(stress)}_{L} = Y \times {(strain)}_{L}$

${(strain)}_{L} = \frac{stress}{Y} = \frac{F}{Y A}$

${(strain)}_{upper} = \frac{T_{2}}{Y A}$

${(strain)}_{lower} = \frac{T_{1}}{Y A}$

$\frac{{(strain)}_{upper}}{{(strain)}_{lower}} = \frac{T_{2}}{T_{1}} = \frac{3 g}{g} = 3$

Topic Question Set

Q 2 :
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 [2024]

Q 3 :
With rise in temperature, the Young's modulus of elasticity [2024]

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Topic Question Set

Q 2 : 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 [2024]

Q 3 : With rise in temperature, the Young's modulus of elasticity [2024]

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Q 5 : Two persons pull a wire towards themselves. Each person exerts a force of 200 N on the wire. Young’s modulus of the material of the wire is 1×1011 Nm-2. Original length of the wire is 2 m and the area of cross-section is 2 cm2. The wire will extend in length by _____ μm. [2024]

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Q 7 : Each of three blocks P, Q, and R shown in the figure has a mass of 3 kg. Each of the wires A and B has a cross-sectional area of 0.005 cm2 and Young’s modulus 2×1011 Nm2. Neglecting friction, the longitudinal strain on wire B is _____ ×10-4. (Take g = 10 m/s2). [2024]

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Q 2 :
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 [2024]

Q 3 :
With rise in temperature, the Young's modulus of elasticity [2024]