d
\(\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,Here,\,volume\,of\,blood\,pumped\\
by\,man's\,heart\\
V = 5\,litres\, = 5 \times {10^{ - 3}}\,{m^3}\,\,\left( {1\,lite = {{10}^{ - 3}}\,{m^3}} \right)\\
Time\,in\,which\,this\,volume\,of\,blood\,\\
pumps,\,\\
t = 1\,\min \, = 60\,s\\
pressure\,at\,which\,the\,blood\,pumps,\\
p = 150\,mm\,of\,Hg\, = 0.15\,m\,of\,Hg\\
\,\,\, = \,\left( {0.15\,m} \right)\left( {13.6 \times {{10}^3}\,kg/{m^3}} \right)\left( {10\,m/{s^2}} \right)
\end{array}\)

\(\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {p = h\rho g} \right)\\
\,\,\, = 20.4 \times {10^3}\,N/{m^2}\\
\therefore \,\,power\,of\,the\,heart = \frac{{PV}}{t}\\
 = \frac{{\left( {20.4 \times {{10}^3}\,N/{m^2}} \right)\left( {5 \times {{10}^{ - 3}}{m^{ - 3}}} \right)}}{{60\,s}}\\
 = 1.70\,W
\end{array}\)