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NEET PHYSICSWORK, ENERGY AND POWERMedium

Question

Two similar springs PP and QQ have spring constants kPk_P and kQk_Q, such that kP>kQk_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 WPW_P and WQW_Q are related as, in case (a) and case (b), respectively:

A

WP=WQ;WP>WQW_P = W_Q; W_P > W_Q

B

WP=WQ;WP=WQW_P = W_Q; W_P = W_Q

C

WP>WQ;WP<WQW_P > W_Q; W_P < W_Q

D

WP<WQ;WP<WQW_P < W_Q; W_P < W_Q

Step-by-Step Solution

The work done (WW) in stretching a spring is stored as elastic potential energy, given by the formula W=12kx2W = \frac{1}{2}kx^2, where kk is the spring constant and xx is the displacement .

  1. Case (a): Same displacement (xP=xQ=xx_P = x_Q = x) Using W=12kx2W = \frac{1}{2}kx^2, since xx is constant, WkW \propto k. Given kP>kQk_P > k_Q, it follows that WP>WQW_P > W_Q.

  2. Case (b): Same force (FP=FQ=FF_P = F_Q = F) We know F=kxF = kx, so x=F/kx = F/k . Substituting this into the work formula: W=12k(F/k)2=F22kW = \frac{1}{2}k(F/k)^2 = \frac{F^2}{2k}. Here, since FF is constant, W1/kW \propto 1/k. Given kP>kQk_P > k_Q, it follows that WP<WQW_P < W_Q.

Thus, the correct relationships are WP>WQW_P > W_Q for case (a) and WP<WQW_P < W_Q for case (b).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERsimilarspringsspringconstantsstretched

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