Question

Water is drawn from a well in a 5kg drum of capacity 55L by two ropes connected to the top of the drum. The linear mass density of each rope is $0.5kg{m^{ - 1}}$. The work done in lifting water in the well 20m below is $[g = 10m{s^{ - 2}}]$
\eqalign{ & {\text{A}}{\text{. }}1.4 \times {10^4}J \cr & {\text{B}}{\text{. }}1.5 \times {10^5}J \cr & {\text{C}}{\text{. }}1.4 \times 10 \times 6J \cr & {\text{D}}{\text{. }}18J \cr}

Answer 1

The work done in lifting water from the well is the sum of work done in lifting the drum along with water and work done in lifting the two ropes. Work done can be calculated using the formula,$W = mgh$, where m is the total mass of the system, h is the height or displacement of the system, and g is the acceleration due to gravity. While pulling the rope, its center of gravity will be at its center, therefore, consider its height from this center of gravity or center of the rope.

Formula used:
$W = mgh$
where,
$m$ represents the mass of the body,
$g$ represents the acceleration due to gravity,
$h$ represents the height at which the body is kept.

Complete step by step answer:
Given,
Mass of the drum=5kg
Capacity of drum=55L
Linear mass density of the rope=$0.5kg{m^{ - 1}}$
Depth of the well=20m
Acceleration due to gravity = g= $10m{s^{ - 2}}$
Since water and drums are lifted from the well the length of the rope must be equal to the depth of the well.
Therefore, the length of the rope=20m.

\eqalign{ & {\text{Mass of the rope = Linear mass density of the rope }} \times {\text{ length of the rope}} \cr & \Rightarrow {\text{Mass of the rope }} = 0.5 \times 20 \cr & \Rightarrow {\text{Mass of the rope}} = 10kg \cr}

Therefore,

\eqalign{ & {\text{The mass of two ropes(}}{{\text{m}}_{tworope}}{\text{) = 2 x (mass of each rope)}} \cr & {{\text{m}}_{tworope}}{\text{ = 2 x 10kg}} \cr & {{\text{m}}_{tworope}}{\text{ = 20kg}} \cr}

The work done in lifting the drums along with water can be calculated as

$W_1$ = (mass of the drums + mass of water inside drums) $\times$ g $\times$ height from which it is drawn
$W_1$ = (5 + 55) $\times$ 10 $\times$ 20
$W_1 = 12000 J$

The work done in lifting the rope can be calculated as
${{\text{W}}_{\text{2}}}{\text{ = mass of the ropes}} \times {\text{g}} \times {\text{height}}$
As the center of gravity will be at its center of the length of the rope, height=10m
Thus,
${{\text{W}}_{\text{2}}}{\text{ = 20}} \times 10 \times 10 = 2000J$
Total work done in lifting the water to the ground is the sum of work done in lifting the drum along with water and work done in lifting the rope.

Therefore,
\eqalign{ & {\text{Total work done = }}{{\text{W}}_1} + {{\text{W}}_2} \cr & {\text{Total work done = }}12000 + 2000 \cr & {\text{Total work done = }}14000J \cr & {\text{Total work done = }}1.4 \times {10^4}J \cr}
Hence, the work done in lifting water in the well 20m below is $1.4 \times {10^4}J$
Hence, the correct answer is option A.

Note:
We have considered that the mass of the rope is evenly distributed and the rope is inextensible. Work done is nothing but the product of force acting on a body and its displacement. When you lift any object or body, the force will be its weight given by mg, where m is the mass of the body. An important thing here to note is that we assume the drum with water filled in it as a point mass, therefore, we take its whole displacement. But the rope is an extended object, therefore, we assume that a load is attached to it at its center of gravity and we take its displacement from this point which is the midpoint of the rope. We again ignore the part of the rope which is tied to the drum.