Verified Solution • 2025–26 Session moderate

QJEEM23Ph12ab: JEE Main 2023 Physics - Class 11 JEEM23Ph12ab Solution: A body of mass 200g is tied to a spring of spring constant 12.5 …

A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s. Then the ratio of extension in the spring to its natural length will be :

Multiple Choice Options

  • A 1:02
  • B 2:03
  • C 2:05
  • D 1:01

Step-by-Step Verified Solution

JEE Main 2023 (Online) 24th January Evening Shift Physics - Circular Motion Question 24 English Explanation

∵kx=mω2(ℓ+x)\because k x=m \omega^{2}(\ell+x)

12.5(x)=15(5)2(ℓ+x)12.5(x)=\frac{1}{5}(5)^{2}(\ell+x)

⇒52x=ℓ+x\Rightarrow \frac{5}{2} x=\ell+x

⇒32x=ℓ\Rightarrow \frac{3}{2} x=\ell

⇒xℓ=23\Rightarrow \frac{x}{\ell}=\frac{2}{3}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring