Question

In a horizontal pipe line of uniform cross-section,pressure falls by $5Pa$ between two points separated by $1km$. The change in the kinetic energy per kg of the oil flowing at these points is :(Density of oil $=800kg{m}^{-3}$)

• A. $6.25x{10}^{-3}Jk{g}^{-1}$
• B. $5.25x{10}^{-4}Jk{g}^{-1}$
• C. $3.25x{10}^{-5}Jk{g}^{-1}$
• D. $4.25x{10}^{-2}Jk{g}^{-1}$

Answer 1

The correct option is A $6.25x{10}^{-3}Jk{g}^{-1}$

from Bernoulli's therom,
$P_1+\frac{1}{2}p{V}_1^2=P_2+\frac{1}{2}p{V}_2^2$
$\Rightarrow 5Pa-\frac{1}{2}\times 800\times (V_2^2-V_1^2)$
$\Rightarrow \frac{100}{800}=V_2^2-V_1^2\Rightarrow 0.0125=V_2^2-V_1^2$
$\Rightarrow \frac{1}{2}m(V_2^2-V_1^2)=\frac{1}{2}\times 1\times 0.0125$[where m=1kgs]
$\Rightarrow$ Change in K.E per kg $=6.24\times {10}^{-3}/k{h}^{-1}$