Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The mass of a bucket full of water is 15kg. It is being pulled up from a 15m deep well. Due to a hole in the bucket 6kg water flows out of the bucket. The work done in drawing the bucket out of the well will be: A. 900 joule B. 1500 joule C. 1800 joule D. 2100 joule

Last updated date: 17th Apr 2024
Total views: 323.4k
Views today: 5.23k
Verified
323.4k+ views
Hint: Work is defined as the force which causes a body to move. This is the fundamental idea on which we can rely to find out the solution. We must apply conservation of energy to solve the given problem.

Complete step by step answer:
In the question it is given that,
Mass = 15kg
Height of the deep well or h = 15m
Mass escaped = 6kg
Let us now find out the rate at which the mass of the bucket, which is filled with water changes as it is lifted up, due to the presence of the leakage, in accordance to the height, which will be considered when the bucket will be pulled.
Let us consider that the rate is R.
So, we know that the total mass is 6kg
Total height it being pulled up is 15m
So, now we can write:
R = $\dfrac{\text{6}}{\text{15}}$kg/m
We have assumed that R is constant throughout the process.
Now let us consider that the bucket is lifted by a very small height. Let the height be ds.
Then, the mass of the bucket at that instant will be $15-\left[ \dfrac{6}{15}\times ds \right]$
Then we have to find the work done. So, the work done will be given as:
(using W = mgh)
$\Rightarrow dW=15-\left[ c\times ds \right]\times g\times s$
Then we have to integrate the above equation.
Now on integrating the above equation we get:
W = [$15\times g\times s$]
So now, $\left[ 15\times g\times s \right]-\left[ \dfrac{6}{5}\times (\dfrac{{{s}^{2}}}{2})\times g \right]$
We know that the range of s is from 0 to 15m and we have taken the value of g as 10 $m/{{s}^{2}}$. So, using the above equation we get the following expression:
$W=(15\times 10\times 15)-\left[ \dfrac{6}{15}\times \left[ \dfrac{{{15}^{2}}}{2} \right]\times 10 \right]$
This gives the value as 1800 J.
Hence the correct answer is Option C.

Note: In the given question, the conservation of energy is maintained, so the rate of expulsion of water is kept into consideration with the work done on pulling it upwards. Therefore, basically the weight of the container is constantly reducing with the upward motion of the container. Relative work also comes into play as the water flows down under the influence of gravity while the container is pulled against gravity. The energy of the system is constant at every moment.