Chapter 10, Problem 62A

a.

To determine

Work done by pump.

Answer to Problem 62A

Work done by pump is $\text{1}\text{.1}\times {\text{10}}^5\text{J}$

Explanation of Solution

Given:

Volume of oil is $V=0.55{\text{m}}^3$

Height of platform is $h=25\text{m}$

Density of oil is $\rho =0.82{\text{g/cm}}^3=0.82\times {10}^3{\text{kg/m}}^3$

Time is $t=35\text{s}$

Formula used:

The work done by pump is given by,

  $\begin{array}{l}W=F_{g}d=mgh=\left(V\times \rho \right)gh\mathrm{.....}\left(1\right)\\ \left(\because F_g=mg\text{and}m=V\rho \right)\end{array}$

Calculation:

Putting the given values in equation (1)

  $\begin{array}{l}W=\left(0.55\times 0.82\times {10}^3\right)\times 9.8\times 25\\ =110495\text{J}\\ \text{=1}\text{.1}\times {\text{10}}^5\text{J}\\ W\text{=1}\text{.1}\times {\text{10}}^5\text{J}\mathrm{......}(2)\end{array}$

b.

To determine

Power developed by pump.

Answer to Problem 62A

Power developed by pump is $P=3.14\text{kJ}$

Explanation of Solution

Given:

Volume of oil is $V=0.55{\text{m}}^3$

Height of platform is $h=25\text{m}$

Density of oil is $\rho =0.82{\text{g/cm}}^3=0.82\times {10}^3{\text{kg/m}}^3$

Time is $t=35\text{s}$

Formula used:

Power developed is given by

  $P=\frac{W}{t}$ …. (3)

Calculation:

From previous part, $W=\text{1}\text{.1}\times {\text{10}}^5\text{J}$

  $\begin{array}{l}P=\frac{1.1\times {10}^5}{35}=3.14\times {10}^3\\ =3.14\text{kJ}\end{array}$