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(B) Given: Internal diameter of the hose pipe, $d_1 = 4 \,cm$. Radius $r_1 = \frac{d_1}{2} = 2 \,cm = 2 	imes 10^{-2} \,m$.Speed of water through the

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(B) Given: Internal diameter of the hose pipe, $d_1 = 4 \,cm$. Radius $r_1 = \frac{d_1}{2} = 2 \,cm = 2 	imes 10^{-2} \,m$.Speed of water through the hose pipe, $v_1 = 1 \,ms^{-1}$.Speed of water through the nozzle, $v_2 = 4 \,ms^{-1}$.Let the diameter of the nozzle be $d_2$ and its radius be $r_2$.According to the equation of continuity for an incompressible fluid, $A_1 v_1 = A_2 v_2$.Substituting the area $A = \pi r^2$, we get $\pi r_1^2 v_1 = \pi r_2^2 v_2$.$r_2^2 = \frac{r_1^2 v_1}{v_2} = \frac{(2 	imes 10^{-2} \,m)^2 	imes 1 \,ms^{-1}}{4 \,ms^{-1}}$.$r_2^2 = \frac{4 	imes 10^{-4}}{4} \,m^2 = 10^{-4} \,m^2$.Taking the square root, $r_2 = 10^{-2} \,m = 1 \,cm$.The diameter of the nozzle is $d_2 = 2 r_2 = 2 	imes 1 \,cm = 2 \,cm$.

Water flows through a hose pipe whose internal diameter is $4 \,cm$ at a speed of $1 \,ms^{-1}$. If water has to emerge at a speed of $4 \,ms^{-1}$, then the diameter of the nozzle should be (in $\,cm$)

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