Verified Solution • 2025–26 Session moderate

QJEEM22Ph19a1: JEE Main 2022 Physics - Class 11 JEEM22Ph19a1 Solution: A stone tide to a spring of length L is whirled in a vertical…

A stone tide to a spring of length L is whirled in a vertical circle with the other end of the spring at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is x(u2−gL)\sqrt {x({u^2} - gL)} . The value of x is -

Multiple Choice Options

  • A 3
  • B 2
  • C 1
  • D 5

Step-by-Step Verified Solution

v→=u2−2gLj^\overrightarrow v = \sqrt {{u^2} - 2gL} \widehat j

JEE Main 2022 (Online) 27th June Evening Shift Physics - Circular Motion Question 43 English Explanation

u→=ui^\overrightarrow u = u\widehat i

∴\therefore ∣v→−u→∣=(u2−2gL)+u2\left| {\overrightarrow v - \overrightarrow u } \right| = \sqrt {({u^2} - 2gL) + {u^2}}

=2u−2gL= \sqrt {2u - 2gL}

∴\therefore x=2x = 2

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