JEE Main Elasticity Previous Year Questions with Solutions

There is a change of length, volume or shape when an external force is applied. The property of the body by virtue of which a body tends to regain its original shape (or) size when external forces are removed is called elasticity.

The internal restoring forces cause the elastic bodies to regain their original shape. This internal restoring force, acting per unit area of the deformed body is called stress. The ratio of change of any dimension to its original dimension is called strain. The law that relates stress and strain is called Hooke’s law. Hooke’s law states that for a small deformation, the stress caused in the body is proportional to the strain.

Within the elastic limit, Stress ∝ Strain ⇒ Stress/Strain = constant

This constant is known as modulus of elasticity (or) coefficient of elasticity. A higher value of modulus of elasticity shows that the material is harder to deform. The unit of modulus is Pascal.

There are different types of modulus of elasticity like Young’s modulus, Bulk modulus and Sheer modulus.

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JEE Main Past Year Questions With Solutions on Elasticity

Q1: A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by ΔT. The net change in its length is zero. Let L be the length of the rod, A is its area of cross-section. Y is Young’s modulus, and α is its coefficient of linear expansion. Then, F is equal to

a. L²YαΔT

b. AY/αΔT

c. AYαΔT

d. LAYαΔT

Solution:

Thermal expansion, ΔL = LαΔT ——-(1)

Let ΔL’  be the compression produced by applied force

Y = FL/AΔL’ ⇒ F = YAΔL’ /L————-(2)

Net change in length = 0 ⇒ΔL’ = ΔL ——(3)

From (1),(2) and (3)

F = YA x (LαΔT)/L = YAαΔT

Answer: (c) AYαΔT

Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is

a. 0.2 J

b. 10 J

c. 20 J

d. 0.1 J

Solution

Elastic energy per unit volume = ½ x stress x strain

Elastic Energy = ½ x stress x strain x volume

= ½ x F/A x (ΔL /L) x (AL)

= ½ x FΔL

= ½ x 200 x 10^-3

Elastic Energy = 0.1 J

Answer: (d) 0.1 J

Q3: A rod of length L at room temperature and uniform area of cross-section A, Is made of a metal having a coefficient of linear expansion α. It is observed that an external compressive force F is applied to each of its ends, prevents any change in the length of the rod when its temperature rises by ΔT K. Young’s modulus, Y for this metal is

a.F/A αΔT

b.F/Aα(ΔT – 273)

c. F/2AαΔ

d. 2F/AαΔT

Solution:

Young’s Modulus Y = stress/strain = (F/A)/(Δl/l)

Substituting the coefficient of linear expansion

α =Δl /(lΔT)

Δl /l= αΔT

Y= (F/AαΔT)

Answer: (a) F/AαΔT

Q4: Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and has radius R. Wire B is 1.5 m long and has a radius of 2mm. If the two wires stretch by the same length for a given load, then the value of R is close t

a. 1.5 mm

b. 1.9 mm

c. 1.7 mm

d. 1.3 mm

Solution:

Δ₁= Δ₂

(Fl₁/πr₁²y₁) = (Fl₂/πr₂²y₂)

2/(R² x 7)= 1.5/(2²x 4)

R= 1.75 mm

Answer: (c) 1.7 mm

Q5: The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

a. 1 mm

b. 1.15 mm

c. 0.90 mm

d. 1.36 mm

Solution

Stress = F/A

Stress = 400 x 4/πd²

= 379 x 10⁶ N/m²

d² = (400 x 4)/(379 x 10⁶π)

d = 1.15 mm

Answer: (b) 1.15 mm

Q6: A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod is nearly equal to

a. 9F/(πr²YT)

b. 6F/(πr²YT)

c. 3F/(πr²YT)

d. F/(3πr²YT)

Solution

Y = (F/πr²) x L/ΔL

ΔL = Fl/πr²Y——–(1)

Change in length due to temperature change

ΔL=LαΔT————(2)

From equa (1) and (2)

L αΔT = FL/AY

α= F/AYΔT

α = F/πr²YT

Coefficient of volume expansion

3∝ = 3F/πr²YT

Answer: (c) 3F/πr²YT

Q7: The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?

(a) length = 200 cm, diameter = 2 mm

(b) length = 300 cm, diameter = 3 mm

(c) length = 50 cm, diameter = 0.5 mm

(b) length = 100 cm, diameter = 1 mm

Solution:

Since all four wires are made from the same material Young’s modulus will be the same.

ΔL ∝ L/D²

In (a) L/D² = 200/(0.2)² = 5 x 10³ cm^-1

In (b) L/D² = 300/(0.3)² = 3.3 x 10³ cm^-1

In (c) L/D² = 50/(0.5)² = 20 x 10³ cm^-1

In (d) L/D² = 100/(0.1)² = 10 x 10³ cm^-1

Answer: (c) length = 50 cm, diameter = 0.5 mm

Q8: A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of

(a) 1/9

(b) 81

(c) 1/81

(d) 9

Solution

Stress = Force/Area

Stress = Force/L²

Now, dimensions increases by a factor of 9

Now, S = (volume x density) x g /L²

S = L³ x ρ g /L² = L ρ g

Stress S ∝ L

S₂/S₁ = L_2/L₁ = 9L₁/L₁ = 9

Answer: (d) 9

Q9. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of the area a floats on the surface of the liquid, covering an entire cross-section of the cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere is Mg/αAB. Find the value of α.

a. 4

b. 5

c. 3

d. 2

Solution:

Increase in pressure is Δp = Mg/A

Bulk modulus is B = Δp /(ΔV/V)

ΔV/V = Δp / B = Mg/AB——(1)

The volume of the sphere is V = (4/3)πR³

ΔV/V = 3(ΔR/R)

From equation (1) we get

Mg/AB = 3(ΔR/R)

ΔR/R = Mg/3AB

Therefore α = 3

Answer: (c) 3

Q10: A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g = 3.1πm/s², what will be the tensile stress that would be developed in the wire?

a. 4.8 x 10⁶ N/m²

b. 3.1 x 10⁶ N/m²

c. 6.2 x 10⁶ N/m²

d. 5.2 x 10⁶ N/m²

Solution:

Tensile stress = Force/Area

Tensile stress = (4)(3.1π)/π(2 x 10^-3)²

Tensile stress = 3.1 x 10⁶ Nm^-2

Answer: (b) 3.1 x 10⁶ N/m²

Q11: A steel rail of length 5m and area of cross-section 40cm² is prevented from expanding along its length while the temperature rises by 10⁰C. If the coefficient of linear expansion and Young’s modulus of steel is 1.2 x 10^-5 K^-1 and 2 x 10¹¹ Nm^-2 respectively, the force developed in the rail is approximately

a. 2 x 10⁹ N

b. 3 x 10^-5 N

c. 2 x 10⁷ N

d. 1 x 10⁵ N

Solution:

A = 40 cm² = 4 x 10^-3 m²

ΔT = 10⁰C

Y = 2 x 10¹¹ Nm^-2

α = 1.2 x 10^-5 K^-1

Force = YAαΔT

Force = (2 x 10¹¹ )(4 x 10^-3)(1.2 x 10^-5)(10) = 9.6 x 10⁴ N

Force ≈ 1 x 10⁵ N

Answer: (d) 1 x 10⁵ N

Q12: If S is the stress and Y is Young’s Modulus of the material of the wire, the energy stored in the wire per unit volume is

a. 2Y/S

b. S/2Y

c. 2S²Y

d. S²/2Y

Solution:

Young’s modulus, Y = Stress/Strain

⇒ Strain = Stress/Y = S/Y

Energy stored per unit volume = ½ x stress x strain

= Stress x Stress/2Y = S²/2Y (since Young’s modulus, Y = Stress/Strain)

Answer: (d) S²/2Y

Q13: A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is

a. F/2l

b. Fl

c. 2Fl

d. Fl/2

Solution

Young’s Modulus Y = FL/Al

Therefore, F = YAl/L

dW = Fdl = YAl(dl)/L

Work done = YAl²/2L

Work done = Fl/2

Answer: (d) Fl/2

Q14: Two wires are made of the same material and have the same volume. However, wire 1 has a cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of the wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?

a. F

b. 4F

c. 6F

d. 9F

Solution

For the same material, Young’s modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire L₁ is and that of L₂ is 3A

V = V₁ =V₂

V = A x L₁ = 3A x L₂ ⇒ L₂ = L₁/3

Y = (F/A)/(ΔL/L)

F₁ = YA(ΔL₁/L₁)

F₂ = Y3A(ΔL_2/L₂)

Given ΔL₁ = ΔL₂= Δx (for the same extension)

F₂ = Y3A(Δx/(L₁/3)) = 9. (YAΔx/L₁) = 9F₁ = 9F

Answer: (d) 9F

Q15: A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weighs W each is hung at the two ends, the elongation of the wire will be (in mm)

a. l/2

b.l

c. 2

d. Zero

Solution

Y = (Force x L)/(A x l) = WL/A l

l = WL/AY

Due to the arrangement of the pulley, the length of wire is L/2 on each side and so the elongation will be l/2. For both sides, elongation = l

Answer: (b) l

Also Read:

Elasticity JEE Advanced Previous Year Questions With Solutions

Hooke’s Law