Question

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 ($g=10m/s^2$):

• A. $900$ joule
• B. $1500$ joule
• C. $1800$ joule
• D. $2100$ joule

Answer 1

The correct option is C $1800$ joule

Given, $mass=15kg,h=15m$ ,mass escaped=$6kg$

Let's first find out the rate (say R) at which mass of bucket ( filled with water ) changes as it is lifted up ( due to leakage ) with respect to height ( to which it has been pulled ):

Total Change in Mass $=6Kg$

Total height it being pulled up $=15m$

Then;

$R=\frac{6}{15}Kg/m$.

Assume it be constant throughout the process !!

Further ;

Let's suppose the bucket is lifted by a very small height $\left(ds\right)$

Then ;

Mass of bucket at that instant $=15-[\frac{6}{15}\times ds]$

Then work done is given by :

(using $W=mgh$ ) $\Rightarrow dW=15-[c\times ds]\times g\times s$

On integrating the above equation we get:

$W=[15\times g\times s]$- On integrating the above equation we get :

$W=[15\times g\times s]-[\frac{6}{15}\times (\frac{s^2}{2})\times g]$

Now $s$ ranges from $0$ to $15m$ & taking $g=10m/s^2$ . using it in above equation we get,

$W=(15\times 10\times 15)-[\frac{6}{15}\times [\frac{(15{)}^2}{2}]\times 10]$

On solving we get The Work done in the whole process which is

$W=1800J$