Two similar springs P and Q have spring constants K\(_P\) and K \(_Q\) , such that K\(_P\) > K \(_Q\) . They are stretched first by the same amount (case a), then by the same force (case b). The work done by the springs \(W_P\) and W\(_Q\) are related as, in case (a) and case (b) respectively
- $W_P > W_Q ; W_Q > W_P$
- $W_P < W_Q ; W_Q < W_P$
- $W_P = W_Q ; W_P > W_Q$
- $W_P = W_Q ; W_P = W_Q$
The Correct Option is A
Solution and Explanation
Here, \(K_P > K_Q\)
Case (a): Elongation (x) in each spring is same.
\(W_P=\frac{1}{2} K_P x^2,W_Q=\frac{1}{2}K_Qx^2\)
\(\therefore \, \, \, \, \, \, W_P > W_Q\)
Case (b) : Force of elongation is same.
So, \(x_1=\frac{F}{K_p}\) and \(x_2=\frac{F}{K_Q}\)
\(W_P=\frac{1}{2}K_P x_2^1=\frac{1}{2} \frac{F^2}{K_P}\)
\(W_Q=\frac{1}{2}K_Q x_2^2=\frac{1}{2} \frac{F^2}{K_Q}\)
\(\therefore \, \, \, \, \, \, W_P < W_Q\)
Top Questions on Hooke's Law
- Three forces are acting on a particle in which \( F_1 \) and \( F_2 \) are perpendicular. If \( F_1 \) is removed, find the acceleration of the particle.
- KEAM - 2025
- Physics
- Hooke's Law
A 4 kg mass is suspended as shown in the figure. All pulleys are frictionless and spring constant \( K \) is \( 8 \times 10^3 \) Nm\(^{-1}\). The extension in spring is ( \( g = 10 \) ms\(^{-2}\) )
- AP EAPCET - 2024
- Physics
- Hooke's Law
A 3 kg block is connected as shown in the figure. Spring constants of two springs \( K_1 \) and \( K_2 \) are 50 Nm\(^{-1}\) and 150 Nm\(^{-1}\) respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is ( \( g = 10 \) ms\(^{-2}\) )
- AP EAPCET - 2024
- Physics
- Hooke's Law
- Hooke’s law holds good up to
- AP PGECET - 2024
- Metallurgy
- Hooke's Law
- Three identical masses are at the three corners of the triangle, connected by massless identical springs (rest length \( l_0 \)) forming an isosceles right angle triangle. If the two sides of equal length (of length \( 2l_0 \)) lie along the positive x-axis and positive y-axis, then the force on the mass that is not at the origin but on the x-axis is given by \( \hat{i} + \hat{j} \) with \( a \) and \( b \).
- IIITH UGEE - 2024
- Physics
- Hooke's Law
Questions Asked in NEET exam
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): A typical unfertilized, angiosperm embryo sac at maturity is 8-nucleate and 7-celled.
Reason (R): The egg apparatus has 2 polar nuclei.In the light of the above statements, choose the correct answer from the options given below:
- NEET (UG) - 2025
- Angiosperms
- A ball of mass 0.5 kg is dropped from a height of 10 m. The ball hits the ground and rises to a height of 1.5 m. The impulse imparted to the ball during its collision with the ground is : (Take \(g = 9.8 \, \text{m/s}^2\))
- NEET (UG) - 2025
- Collision and Impulse
- Consider the diameter of a spherical object being measured with the help of a Vernier Callipers. Suppose its 10 Vernier Scale Divisions (V.S.D.) are equal to its 9 Main Scale Divisions (M.S.D.). The least count in the M.S. is 0.1 cm and the zero of V.S. is at -0.1 cm when the jaws of Vernier callipers are closed. If the main scale reading for the diameter is \(M = 5\) cm and the number of coinciding vernier division is 8, the measured diameter after zero error correction, is:
- NEET (UG) - 2025
- Measurement of length
- A container has two chambers of volumes \(V_1 = 2\) litres and \(V_2 = 3\) litres separated by a partition made of a thermal insulator. The chambers contain \(n_1 = 2\) moles and \(n_2 = 3\) moles of ideal gas at pressures \(p_1 = 1\) atm and \(p_2 = 2\) atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of :
- NEET (UG) - 2025
- Ideal Gases
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :
- NEET (UG) - 2025
- Moment Of Inertia
Concepts Used:
Hooke’s Law
Hooke’s Law states that for small deformities, the stress and strain are proportional to each other. Thus,
Stress ∝ Strain
Stress = k × Strain … where k is the Modulus of Elasticity.
When a limited amount of Force or deformation is involved then concept of Hooke’s Law is only applicable . If we consider the fact, then we can deviate from Hooke's Law. This is because of their extreme Elastic limits.