## A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole …

Q: A large open tank has two small holes in its vertical wall as shown in figure. One is a square hole of side ‘L’ at a depth ‘4y’ from the top and the other is a circular hole of radius ‘R’ at a depth y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then, ‘R’ is equal to :

(A) $\frac{L}{\sqrt{2\pi}}$

(B) $2 \pi L$

(C) $\sqrt{\frac{2}{\pi}} L$

(D) $\frac{L}{2\pi}$

Click to See Solution :
Ans: (C)

Sol: Let v1 and v2 be the velocity of efflux from square and circular hole respectively. A1 and A2 be cross-section areas of square and circular holes.

$v_1 = \sqrt{8 g y}$ ; $v_2 = \sqrt{2 g y}$

The volume of water coming out of square and circular hole per second is

$Q_1 = v_1 A_1 = \sqrt{8 g y} L^2$

$Q_2 = v_2 A_2 = \sqrt{2 g y} \pi R^2$

$Q_1 = Q_2$

$R = \sqrt{\frac{2}{\pi}} L$

## A water tank stands on the roof of a building as shown. Find the value of h ….

Q: A water tank stands on the roof of a building as shown. Find the value of h (in m) for which the horizontal
distance ‘x’ covered by the water is greatest.

Click to See Solution :
Ans: 1
Sol: $v_{efflux} = \sqrt{2 g h}$

Time of fall , $t = \sqrt{\frac{2(4-h)}{g}}$

$x = v_{efflux} t = 2\sqrt{h(4-h)}$

the roots of x are (0,4) and the maximum of x is at h = 2. The permitted value of h is 0 to 1 clearly h = 1 will
give the maximum value of x in this interval

Aliter Solution:
If the column of water itself were from ground upto a
height of 4m, h = 2m would give the maximum range
x. Farther the hole is from this midpoint, lower the
range. Here the nearest point possible to this midpoint
is the base of the container. Hence

## A large tank of cross-section area A contains liquid of density ρ . A cylinder of density ρ / 4 and length l …

Q: COMPREHENSION :
A large tank of cross-section area A contains liquid of density ρ . A cylinder of density ρ / 4 and length l , and cross- section area a (a <<A) is kept in equilibrium by applying an external vertically downward force as shown. The cylinder is just submerged in liquid. At t = 0 the external force is removed instantaneously. Assume that water level in the tank remains constant.

1 . The acceleration of cylinder immediately after the external force is removed is

(A) g

(B) 2 g

(C) 3 g

(D) zero

2 . The speed of the cylinder when it reaches its equilibrium position is

(A) $\frac{1}{2}\sqrt{g l}$

(B) $\frac{3}{2}\sqrt{g l}$

(C) $\sqrt{2 g l}$

(D) $2 \sqrt{g l}$

3 . After its release at t = 0, the time taken by cylinder to reach its equilibrium position for the first time is

(A) $\frac{\pi}{8}\sqrt{\frac{l}{g}}$

(B) $\frac{\pi}{3}\sqrt{\frac{l}{g}}$

(C) $\frac{\pi}{4}\sqrt{\frac{l}{g}}$

(D) $\frac{\pi}{2}\sqrt{\frac{l}{g}}$

Click to See Solution :

Ans: 1. (C) ; 2.(B) ; 3.(C)

Sol:1 .

2 . The density of liquid is four times that of cylinder, hence in equilibrium postion one fourth of the cylinder is submerged. So as the cylinder is released from initial postion, it moves by 3l/4 to reach its equlibrium position. The upward motion in this time is SHM. Therefore require velocity is vmax = ω A ; $\omega = \sqrt{\frac{4 g}{l}}$ and $A = \frac{3 l}{4}$ . Therefore $v_{max} = \frac{3}{2} \sqrt{g l}$

3 . The require time is one fourth of time period of SHM , therefore $t = \frac{\pi}{2 \omega} = \frac{\pi}{4}\sqrt{\frac{l}{g}}$

## A beaker of radius r is filled with water (Refractive index = 4/3) up to height H as shown in the figure on the left …..

Q: A beaker of radius r is filled with water (Refractive index = 4/3) up to height H as shown in the figure on the left . The beaker is kept on a horizontal table rotating with angular speed  ω . This makes water surface curved so that the difference in height of water level at the center and at the circumference  of the beaker is h (h<< H , h<< r) as shown in the figure on the right . Take this surface to be approximately spherical with a radius of curvature R . Which of the following is/are correct ? (g is the acceleration due to gravity )

(a) $\displaystyle R = \frac{h^2 + r^2}{2h}$

(b) $\displaystyle R = \frac{3 r^2}{2h}$

(c) Apparent depth of the bottom of beaker is close to $\displaystyle v = [\frac{3H}{2}(1 + \frac{\omega^2 H}{2g})^{-1}]$

(d) Apparent depth of the bottom of beaker is close to $\displaystyle v = [\frac{3H}{4}(1 + \frac{\omega^2 H}{4g})^{-1}]$

Click to See Solution :
Ans: (a,d)

Sol:

$\displaystyle (R-h)^2 + r^2 = R^2$

$\displaystyle R^2 + h^2 – 2 R h + r^2 = R^2$

$\displaystyle h^2 + r^2 = 2Rh$

$\displaystyle R = \frac{h^2 + r^2}{2h}$

Since h << r

$\displaystyle R = \frac{r^2}{2h}$

$\displaystyle h = \frac{\omega^2 r^2}{2 g}$

$\displaystyle R = \frac{r^2 2 g}{2\omega^2 r^2}$

$\displaystyle R = \frac{g}{\omega^2}$

$\displaystyle \frac{\mu_1}{v} – \frac{\mu_2}{u} = \frac{\mu_1 – \mu_2}{R}$

$\displaystyle \frac{1}{v} – \frac{4}{3u} = \frac{1 – 4/3}{R}$

$\displaystyle \frac{1}{v} = – (\frac{1}{3R} + \frac{4}{3(H-h)})$

$\displaystyle \frac{1}{v} = – (\frac{1}{3R} + \frac{4}{3H})$ (Since h << H)

Putting the value of R we get ,

$\displaystyle v = – [\frac{3H}{4}(1 + \frac{\omega^2 H}{4 g})^{-1}]$