JEE Advanced Previous Year Questions with Solutions on Elasticity

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

Download JEE Advanced Previous Year Questions with Solutions on Elasticity PDF

Question 1) An elastic string of length 42 cm and cross-section area 10^–4 m² is attached between two pegs at a distance of 6 mm, as shown in the figure. A particle of mass m is kept at the midpoint of the string stretched as shown in the figure by 20 cm and released. As the string attains natural length, the particle attains a speed of 20 m/s. Then young modulus Y of the string is of order.

(A) 10⁸

(B)10¹²

(C)10⁶

(D) 10⁴

Answer:(C)

Solution:

Elastic stream energy × volume = kinetic energy

Y = [(0.05 × 400 × 0.42)/(0.2)² × 10^-4]

Y = 2.1 × 10⁶ N/m²

Question 2) A spring with natural length l₀ has a tension T₁ when its length is l_1, and the tension is T₂ when its length is l₂. The natural length of spring will be:

Solution:

Let the natural length be L₀

Using hook’s law ,Y= TL/AdL , where dL = L – L₀

Case 1: When tension is T₁ length of wire = L₁

L₁ – L₀ = T₁L₀/AY ——-(1)

Case 2: Tension is T₂ and length of wire = L₂

L₂ – L₀ = T₂L₀/AY ——-(2)

Dividing both equations :

Question 3) One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is:

(A) 0.25

(B) 0.50

(C) 2.00

(D) 4.00

Answer: (C) 2.00

Solution:

For thick wire,

For thin wire,

Δl₂/Δl₁ = 2

Question 4) In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are)

(A) P has more tensile strength than Q

(B) P is more ductile than Q

(C) P is more brittle than Q

(D) The Young’s modulus of P is more than that of Q

Answer: (A) and (B)

Solution:

P has more tensile strength than Q. From graph slope tanθ=strain/stress, more will be slope, more will be the tensile strength.

Therefore, graph P has more tensile strength than Q. As P has more tensile strength, P is more ductile than Q. As P elongates more before it breaks. Thus, options (a) and (b) are true.

Question 5) A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross-sectional area is 4.9 × 10^-7 m². If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s^–1. If Young’s modulus of the material of the wire is n × 10⁹ Nm^-2, the value of n is?

(A) 4

(B) 2

(C) 8

(D) 1

Answer: (A) 4

Solution:

The frequency of the oscillation can be written as

ω = √(YA/mL)

⇒ Y = ω²mL/A

Question 6) A 20 cm long string, having a mass of 0.1 g, is fixed at both ends. The tension in the string is 0.5 N. The string is set into vibration using an external vibrator of frequency 100 Hz. Find the separation between the successive nodes on a string.

Answer: 5 cm

Solution:

The distance between two successive nodes

Question 7) A wire of density 9 × 10^-3 kg cm^-3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9 × 10^-4. The lowest frequency of the transverse vibrations in the wire is (Young’s modulus of wire Y = 9 × 10¹⁰ N/m²), to the nearest integer is _________

Answer: 35 Hz

Solution:

Y = Stress/Strain

= (T/A)/Strain

T/A = Y × Strain = 9 × 10⁹ × 4.9 × 10^-4

Also, mass of the wire, m = Alσ

Mass per unit length, μ = m/l = Aσ

Fundamental frequency in the string

f = (1/2l)√(YΔl/ρl)

Question 8) The adjacent graph shows the extension (Δl) of a wire of length 1m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is 10^-6 m², Calculate Young’s modulus of the material of the wire.

(A) 2 × 10¹¹ N/m²

(B) 2 × 10^-11 N/m²

(C) 3 × 10^-12 N/m²

(D) 2 × 10^-13 N/m²

Answer: (A) 2 × 10¹¹ N/m²

Solution:

Using, Y = (F/A)/ (△l/l)

= (20 × 1)/(10^-6 × 10^-4)

= 2 × 10¹¹ N/m²

Question 9) Two steel wires having the same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1:4, the ratio of their diameters is

(A) √2: 1

(B) 1:2

(C) 2:1

(D) 1:√2

Answer: (A) √2: 1

Solution:

If force F acts along the length L of the wire of cross-section A, then energy stored in the unit volume of wire is given by

Energy density = (½) stress × strain

= (½) × (F/A) × (F/AY)

= (½) × (F²/A²Y)

= (½) × (F² × 16/(πd²)²Y)

If u₁ and u₂ are the densities of the two wires, then

Question 10) A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. If the wire is stretched by the amount x, the work done is____

Answer:

Solution:

Work done, W = (½) Kx²

Where, K = YA/L and x = extension in wire

Therefore, W = (½)(YA/L)x²

Also Read:

Elasticity JEE Main Previous Year Questions With Solutions