(C) According to the law of conservation of energy,the potential energy at the top is converted into kinetic energy at the bottom.$mgh = \frac{1}{2}mv

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(C) According to the law of conservation of energy,the potential energy at the top is converted into kinetic energy at the bottom.$mgh = \frac{1}{2}mv^2$Here,$m$ is the mass of the object,$g$ is the acceleration due to gravity,$h$ is the height,and $v$ is the final velocity.Canceling $m$ from both sides,we get $gh = \frac{1}{2}v^2$,which simplifies to $v = \sqrt{2gh}$.Since all three objects fall from the same height $h$ and are subject to the same acceleration due to gravity $g$,their final speeds will be independent of their masses.Therefore,the ratio of their speeds is $v_1 : v_2 : v_3 = \sqrt{2gh} : \sqrt{2gh} : \sqrt{2gh} = 1 : 1 : 1$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

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Three different objects of masses $m_1, m_2$ and $m_3$ are allowed to fall from rest and from the same point $O$ along three different frictionless paths. The speeds of the three objects,on reaching the ground,will be in the ratio of

AIIMS 2002 All AIIMS

A
$m_1 : m_2 : m_3$
B
$m_1 : 2m_2 : 3m_3$
C
$1 : 1 : 1$
D
$\frac{1}{m_1} : \frac{1}{m_2} : \frac{1}{m_3}$

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

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MediumMHT CET 2025

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EasyAIIMS 2016

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Three different objects of masses $m_1, m_2$ and $m_3$ are allowed to fall from rest and from the same point $O$ along three different frictionless paths. The speeds of the three objects,on reaching the ground,will be in the ratio of

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$A$ particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $v = \alpha \sqrt{x}$,where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x = 0$ to $x = d$ will be:

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