Question

A cylindrical vessel is full of water (density $\rho$ = 1 gm/cc) is placed on horizontal ground as shown. A small hole of area (= 1 cm$^2$) is made on a curved wall at a height of 2m from the bottom of container (container does not move) choose CORRECT statement(s) (g = 10 m/s$^2$) :-

(A) Horizontal range of water jet is maximum when water level is at height of 8 m from ground

(B) Maximum horizontal force on container by table is 1 Newton

(C) Maximum horizontal range of water jet on ground is $4\sqrt{5}$ meter

(D) Horizontal range of water jet will decrease then increase with decrease of water level.

PHYSICS / Assginment # 09 (Fluid Dynamics)

• A. Horizontal range of water jet is maximum when water level is at height of 8 m from ground
• B. Maximum horizontal force on container by table is 1 Newton
• C. Maximum horizontal range of water jet on ground is $4\sqrt{5}$ meter
• D. Horizontal range of water jet will decrease then increase with decrease of water level.

Answer 1

Answer: (C)

C

Answer 2

The velocity of water efflux from a hole at a depth $d$ below the free surface is given by Torricelli's law: $v = \sqrt{2gd}$. The horizontal range $R$ of the water jet is given by $R = v \times t$. The time of flight $t$ is determined by the vertical motion: $h = \frac{1}{2}gt^2$, so $t = \sqrt{\frac{2h}{g}}$. Thus, $R = \sqrt{2gd} \times \sqrt{\frac{2h}{g}} = \sqrt{4hd}$, where $d$ is the depth of the hole from the free surface and $h$ is the height of the hole from the ground. If $H$ is the height of the water level from the ground, then $d = H-h$. So, $R = \sqrt{4h(H-h)}$.

(A) For a fixed hole height $h=2$ m, the range is $R(H) = \sqrt{4 \times 2 \times (H-2)} = \sqrt{8(H-2)}$. This is an increasing function of $H$. Thus, the range is not maximum at $H=8$ m.

(B) The horizontal force exerted by the water jet is $F = \rho A v^2 = \rho A (2g(H-h))$. With $\rho = 1000$ kg/m$^3$, $A = 10^{-4}$ m$^2$, $g = 10$ m/s$^2$, and $H=7$ m, $h=2$ m, we get $d=5$ m, $v=10$ m/s. $F = 1000 \times 10^{-4} \times (10)^2 = 10$ N. This force increases with $H$.

(C) The range $R = \sqrt{4h(H-h)}$. If $h=2$ m, $R = \sqrt{8(H-2)}$. If the maximum water level is $H=12$ m, then $R = \sqrt{8(12-2)} = \sqrt{80} = 4\sqrt{5}$ m. This is stated as the maximum horizontal range.

(D) For a fixed hole height $h=2$ m, the range $R(H) = \sqrt{8(H-2)}$ decreases as the water level $H$ decreases (for $H>2$).