(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy between $A$ and $B$:$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$$v^2 = u^2 - 2gL$$v = \sqrt{u^2 - 2gL}$.At position $B$, the velocity is directed vertically upward, so $\vec{v} = v \hat{j} = \sqrt{u^2 - 2gL} \hat{j}$.The change in velocity is $\Delta \vec{v} = \vec{v} - \vec{u} = v \hat{j} - u \hat{i}$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v^2 + u^2} = \sqrt{(u^2 - 2gL) + u^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

Difficult

$A$ stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

IIT 1998 All IIT

AIPMT 2004 All AIPMT

A
$\sqrt{u^2 - 2gL}$
B
$\sqrt{2gL}$
C
$\sqrt{u^2 - gL}$
D
$\sqrt{2(u^2 - gL)}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Work, Energy, Power and Collision

Subtopic

Vertical Circular Motion

Previous Year

IIT 1998 Paper All IIT years →

Previous Year

AIPMT 2004 Paper All AIPMT years →

$A$ stone is tied to a string of length $L$ and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

Difficult

View Solution

The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

EasyMHT CET 2024

View Solution

$A$ stone of mass $40$ g is tied to a light inextensible string of length $50$ cm and whirled in a vertical circle at the rate of $30$ revolutions per minute. The tension in the string when it is at the lowest point of the circle is (Take gravitational acceleration, $g = 10 \text{ m/s}^2$ and $\pi^2 = 10$) (in $N$)

DifficultMHT CET 2026

View Solution

$A$ ball attached to one end of a string swings in a vertical plane such that its acceleration at point $A$ (extreme position) is equal to its acceleration at point $B$ (mean position). The angle $\theta$ is :-

Difficult

View Solution

In a circus,a stuntman rides a motorbike on a circular track of radius $R$ in the vertical plane. The minimum speed at the highest point of the track will be:

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

$A$ ball attached to one end of a string swings in a vertical plane such that its acceleration at point $A$ (extreme position) is equal to its acceleration at point $B$ (mean position). The angle $\theta$ is :-

In a circus,a stuntman rides a motorbike on a circular track of radius $R$ in the vertical plane. The minimum speed at the highest point of the track will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module