Water coming out of a horizontal tube at a speed ? strikes normally a vertically wall close to the mouth of the tube and falls down vertically after impact. When the speed of water is increased to $2v$ .
Medium
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Thrust $=$ Rate of charge of linear momentum
Thrust $=\mathrm{mv}$ where $\mathrm{m}$ is mass striking per sec.
$=\rho a v \times v \Rightarrow \quad$ Thrust $=\rho a v^{2}$
Energy lost for second $=\frac{1}{2} m v^{2}=\frac{1}{2} \rho a v . v^{2}=\frac{1}{2} \rho a v^{3}$
so option $(\mathrm{B}, \mathrm{C})$ are correct.
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