question_answer

If an engine lifts \[4000\,\,{{m}^{3}}\] of water from a depth of 500 m in 40 minutes, then calculate. (i) Work done by the motor and (ii) Power of the motor [\[\text{g}=\text{1}0\text{ m}/{{\text{s}}^{\text{2}}}\], density of water\[=\text{1}{{0}^{\text{3}}}\text{kg }{{\text{m}}^{-\text{3}}}\]]

Answer:

Volume \[(v)=4000\,{{m}^{3}}\] depth \[(h)=500\,m\] time \[(t)=40\,\min =40\times 60\,\sec \] \[g=10\,m/{{s}^{2}}\] density \[(d)=100\,\,kg/{{m}^{3}}\] \[W=mgh=vdgh\] \[=4000\times 10\times 1000\times 500\] \[=20\times {{10}^{9}}\] \[=20\times {{10}^{10}}J\] \[P=\frac{w}{t}=\frac{2\times {{10}^{10}}}{40\times 60}=\frac{100}{2}\times {{10}^{6}}=8.33\times {{10}^{6}}w\]                                                 \[=8330\,kw\]