The mass of a bucket containing water is 10 kg. What is the work done in pulling up the bucket from a well of depth 10 m if water is pouring out at a uniform rate from a hole in it and there is loss of 2kg of water from it while it reaches the top (g= 10 / sec² )

The correct answer/option is 900 J.

Let's break down the physical situation step-by-step to understand how to compute the total work done.

1. Identify the given parameters:
- Initial mass of the bucket containing water, $m_0=10\text{kg}$
- Total depth of the well, $H=10\text{m}$
- Total loss of water during the ascent, $\mathrm{\Delta m}=2\text{kg}$
- Acceleration due to gravity, $g=10{\text{m/s}}^2$
- Water pours out at a uniform rate with respect to height/distance.

2. Express mass as a function of height:
Since the water leaks at a uniform rate as the bucket is pulled up, the mass of the bucket at any height $y$ (where $y$ ranges from $0$ at the bottom to $H=10\text{m}$ at the top) decreases linearly.
The mass $m\left(y\right)$ at height $y$ can be written as:
$m\left(y\right)=m_0-\frac{\mathrm{\Delta m}}{H}y$

Substituting the given values:
$m\left(y\right)=10-\frac{2}{10}y=10-0.2y$

3. Set up the force function:
The upward force required to pull the bucket at a constant speed is equal to the weight of the bucket at height $y$:
$F\left(y\right)=m\left(y\right)\cdot g$
$F\left(y\right)=(10-0.2y)\cdot 10=100-2y$

4. Calculate the work done by integration:
Work done $W$ is the integral of the force over the distance from $y=0$ to $y=10\text{m}$:
$W={\int }_0^{10}F\left(y\right)dy$
$W={\int }_0^{10}(100-2y)dy$

Integrating the terms:
$W={[100y-y^2]}_0^{10}$
$W=(100\cdot 10-{10}^2)-0$
$W=1000-100=900\text{J}$

Alternative Method (Average Mass):
Since the leakage rate is uniform, the average mass of the bucket during the displacement is simply the arithmetic mean of its initial and final masses:
$m_{\text{avg}}=\frac{m_{\text{initial}}+m_{\text{final}}}{2}=\frac{10+8}{2}=9\text{kg}$

The work done is then:
$W=m_{\text{avg}}\cdot g\cdot H=9\cdot 10\cdot 10=900\text{J}$