Question

A cylindrical container of radius ${}^{\prime }{R}^{\prime }$ and height ${}^{\prime }{h}^{\prime }$ is completely filled with a liquid. Two horizontal L – shaped pipes of small cross-sectional area ${}^{\prime }{a}^{\prime }$ are connected to the cylinder as shown in the figure. Now the two pipes are opened and fluid starts coming out of the pipes horizontally in opposite directions. Then the torque due to ejected liquid on the system is

• A. $4agh\rho R$
• B. $8agh\rho R$
• C. $2agh\rho R$
• D. None of these

Answer 1

The correct option is A $4agh\rho R$

As we know, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$-----(1)
Where, $r=$distance between two limbs i.e. $4R$
$F=\frac{d\left(mv\right)}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$
$F=v\frac{dm}{dt}$ ($\because$ velocity is constant)
$F=v\times \rho \times a\times v$ ($\because \frac{dm}{dt}=\rho av$)
$F=\rho a{v}^2$
$\tau =4R\times \rho a{v}^2$
$v=\sqrt{2gH}$
From diagram we get $H=\frac{h}{2}$
so, $v^2=2g\frac{h}{2}=gh$
put $v^2$ in equation (1)
$\tau =4R\rho agh$