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The bodies whose masses are in the ratio $2:1$ are dropped simultaneously at two places $A$ and $B$ where the acceleration due to gravity are ${g_A}$ and ${g_B}$ respectively. If they reach the ground simultaneously, the ratio of the heights from which they are dropped is
1) ${g_A}:{g_B}$
2) $2{g_A}:{g_B}$
3) ${g_A}:2{g_B}$
4) $\sqrt {{g_A}} :\sqrt {{g_B}} $

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Last updated date: 28th Feb 2024
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IVSAT 2024
Answer
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Hint: Acceleration due to gravity is the gravitational force acting on a body of mass$1Kg$. It changes with the change in the planet as the mass and radius of the planet changes. From the given data in the question we can say that to solve it we need a formula that connects distance, velocity, acceleration, and time.
Formula used:
$s = ut + \dfrac{1}{2}a{t^2}$
Where is the $u$ initial velocity, $a$is the acceleration, $s$ is the displacement, $t$ is the time.

Complete Step-by-step answer:
Let the mass of $A$ be $m_A$, the mass of $B$ be $m_B$, the time taken by $A$ be
$t_A$, the time taken by $B$ be $t_B$, the height from which $A$ is dropped be $h_A$, the height from which $B$ is dropped be ${h_B}$
From the question, we came to know that,
${m_A}:{m_B} = 2:1$
and,
${t_A} = {t_B}$
We know that,
$s = ut + \dfrac{1}{2}a{t^2}$
Where is the $u$ initial velocity, $a$is the acceleration, $s$ is the displacement, $t$ is the time.
 From the given data and the known formula we can state that,
$ \Rightarrow {h_A} = u{t_A} + \dfrac{1}{2}{g_A}{t_A}^2$
$ \Rightarrow {h_B} = u{t_B} + \dfrac{1}{2}{g_B}{t_B}^2$
As both the balls are released from rest i.e. $u = 0$
Hence,
$ \Rightarrow {h_A} = \dfrac{1}{2}{g_A}{t_A}^2$
And
$ \Rightarrow {h_B} = \dfrac{1}{2}{g_B}{t_B}^2$
After taking the ratio of both this we get
$ \Rightarrow \dfrac{{{h_A}}}{{{h_B}}} = \dfrac{{{g_A}{t^2}_A}}{{{g_B}{t^2}_B}}$
As ${t_A} = {t_B}$
$ \Rightarrow \dfrac{{{h_A}}}{{{h_B}}} = \dfrac{{{g_A}}}{{{g_B}}}$

Hence the answer to the given question is (1) ${g_A}:{g_B}$

Note:
There is one more way to solve this by calculating the final velocity by using $v = u + at$ (where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time) of the ball then applying conservation of energy i.e. $mgh = \dfrac{1}{2}m{v^2}$ where $m$ is the mass, $g$ is the acceleration due to gravity, $v$ is the velocity, $h$is the distance ball covered. Using this we will get the value of $h$in the terms of $g$.