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(A) The work done against gravity is given by the change in gravitational potential energy,which is $W = mgh$,where $m$ is the mass,$g$ is the acceler

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$1: 1$

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(A) The work done against gravity is given by the change in gravitational potential energy,which is $W = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the vertical displacement.In both cases,the mass $m = 50 \ kg$ and the vertical height $h = 20 \ m$ are the same.Since gravity is a conservative force,the work done against it depends only on the initial and final vertical positions,not on the path taken.Therefore,the work done in Case-$1$ is $W_1 = mgh = 50 	imes g 	imes 20 = 1000g \ J$.The work done in Case-$2$ is also $W_2 = mgh = 50 	imes g 	imes 20 = 1000g \ J$.The ratio of work done is $W_1 : W_2 = 1000g : 1000g = 1: 1$.Thus,option $A$ is correct.

$A$ body of mass $50 \ kg$ is lifted to a height of $20 \ m$ from the ground in two different ways as shown in the figure. The ratio of work done against gravity in both the respective cases will be:

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