## 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) $\frac{L}{2\pi}$

Ans: (C)

Solution: 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. Ans: 1

Solution: $v_{efflux} = \sqrt{2 g h}$

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

Ans: (a,d)

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

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

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

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

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 ,

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

## A spherical bubble inside water has radius R . Take the pressure inside the bubble and the water pressure to be po ….

Q: A spherical bubble inside water has radius R . Take the pressure inside the bubble and the water pressure to be Po . The bubble now gets compressed radially in an adiabatic manner so that its radius becomes (R-a) . For a << R the magnitude of the work done in the process is given by (4πPoR2 a )X , Where X is a constant and γ = Cp/Cv = 41/30 . The value of X is ……

Ans: (2.05)

Solution: $\displaystyle P_1 V_1^{gamma} = P_2 V_2^{gamma}$

$\displaystyle P_0 (\frac{4}{3}\pi R^3)^{gamma} = P_2 (\frac{4}{3}\pi (R-a)^3)^{gamma}$

$\displaystyle P_2 = P_0 (\frac{R}{R-a})^{3\gamma}$

$\displaystyle W = \frac{P_1 V_1 -P_2 V_2}{\gamma – 1}$

$\displaystyle W = \frac{P_0 (\frac{4}{3}\pi R^3) – P_0 (\frac{R}{R-a})^{3\gamma} (\frac{4}{3}\pi (R-a)^3)}{\frac{41}{30} – 1}$

$\displaystyle W = \frac{P_0 \frac{4}{3}\pi}{\frac{11}{30}} (R^3 – (R)^{41/10} (R-a)^{-11/10})$

$\displaystyle W = \frac{P_0 \frac{4}{3}\pi}{\frac{11}{30}} (R^3 – (R)^{41/10}R^{-11/10} (1-a/R)^{-11/10})$

$\displaystyle W = \frac{P_0 \frac{4}{3}\pi}{\frac{11}{30}} (R^3 – (R)^3 (1-a/R)^{-11/10})$

$\displaystyle W = 4\pi P_0 R^2 a$

Work done is given by (4πPoR2 a )X

On comparing X = 1

## A hot air balloon is carrying some passengers , and a few sandbags of mass 1 kg so each so that its total mass is 480 kg …..

Q: A hot air balloon is carrying some passengers , and a few sandbags of mass 1 kg each so that its total mass is 480 kg . Its effective volume giving the balloon its buoyancy is V . The balloon is floating at an equilibrium height of 100 m . When N number of sandbags are thrown out , the balloon rises to a new equilibrium height close to 150 m with its volume V remaining unchanged . If the variation of the density of air with height h from the ground $p h_o (h) = \rho_o e^{\frac{-h}{h_o}}$ ; where ρo = 1.25 kg/m^3 and ho = 6000 m , the value of N is —–

Ans: (4)

Solution : Weight = Upthrust

m g = Fu

480 × 10 = ρ V g

$\displaystyle 480 \times 10 = \rho_o e^{\frac{-h}{h_o}} V g$

$\displaystyle 480 \times 10 = \rho_o e^{\frac{-100}{6000}} V g$ …(i)

$\displaystyle ( 480 – N \times 1) 10 = \rho’ V g$

$\displaystyle (480 -N ) 10 = \rho_o e^{\frac{-150}{6000}} V g$ …(ii)

Dividing (i) by (ii)

$\displaystyle \frac{480}{480-N} = e^{\frac{150 – 100}{6000}}$

$\displaystyle \frac{480}{480-N} = e^{\frac{50}{6000}}$

$\displaystyle \frac{480}{480-N} = e^{\frac{1}{120}}$

N = 4

## A train with cross-sectional area St is moving with speed vo inside a long tunnel of cross-sectional area S0 (S0 = 4 St) …….

Q: A train with cross-sectional area St is moving with speed vo inside a long tunnel of cross-sectional area S0 (S0 = 4 St) . Assume that almost all the air (density ρ) in front of the train flows back between its sides and the walls of the tunnel . Also , the air flow with respect to train is steady and laminar . Take the ambient pressure and that inside the train to be p0 . If the pressure in the region between the sides of the train and the tunnel walls is p then $p_0 – p = \frac{7}{2N}\rho v_t^2$ . The value of N is ———

Ans: (9)

Solution: (with respect to train)
Applying Bernoulli’s equation

$\displaystyle p_0 + \frac{1}{2}\rho v_t^2 = p + \frac{1}{2}\rho v^2$

$\displaystyle p_0 – p = \frac{1}{2}\rho ( v^2 -v_t^2)$ ….(i)

From equation of continuity ,

$\displaystyle 4 S_t v_t = v \times 3 S_t$

$\displaystyle v = \frac{4}{3} v_t$ ….(ii)

From equation (i) & (ii)

$\displaystyle p_0 – p = \frac{1}{2}\rho (\frac{16}{9} v_t^2 -v_t^2)$

$\displaystyle p_0 – p = \frac{7}{18}\rho v_t^2$

N = 9

## When water is filled carefully in a glass , one can fill it to a height h above the rim of the glass due to the surface tension of water ….

Q: When water is filled carefully in a glass , one can fill it to a height h above the rim of the glass due to the surface tension of water . To calculate h just before water starts flowing , model the shape of water above the rim as a disc of thickness h having semicircular edges as shown schematically in the figure . When pressure of water at the bottom of this disc exceeds what can be withstood due to due to surface tension , the water surface breaks near the rim and water starts flowing from there . If the density of water , its surface tension and the acceleration due to gravity are 103 kg/m^3 , 0.07 N/m and 10 m/s^2 respectively , the value of h (in mm) is ……… Ans: (3.74)

Sol: Pressure at the bottom of disc = Pressure due to surface tension

$\displaystyle \rho g h = T (\frac{1}{R_1} + \frac{1}{R_2})$

R1 >> R2

$\displaystyle \frac{1}{R_1} << \frac{1}{R_2}$ and R2 = h/2

$\displaystyle \rho g h = T (\frac{1}{R_1} + \frac{1}{R_2})$

$\displaystyle \rho g h = T (0 + \frac{1}{h/2})$

$\displaystyle h^2 = \frac{2 T}{\rho g}$

$\displaystyle h = \sqrt{\frac{2 T}{\rho g}}$

$\displaystyle h = \sqrt{\frac{2 \times 0.07}{10^3 \times 10}}$

$h = n\sqrt{14} mm$ = 3.741

## An Open – ended U-tube of uniform cross-sectional area contains water (density 10^3 kg/m^3) . Initially the water level stands

Q: An Open – ended U-tube of uniform cross-sectional area contains water (density 103 kg/m3) . Initially the water level stands at 0.29 m from the bottom in each arm . Kerosene oil (a water immiscible liquid) of density 800 kg/m3 is added to the left arm until its length is 0.1 m as shown in figure . The ratio (h1/h2) of the height of the liquid in the two arms is (a) 15/14

(b) 35/33

(c) 7/6

(d) 5/4

Ans: (b)

Sol: As h1 + h2 = 0.29 x 2 + 0.1

h1 + h2 = 0.68 …(i)

P0 + ρk g (0.1) + ρw g (h1 -0.1) – ρw g h2 = P0

ρk g (0.1) + ρw g h1 – ρw g x 0.1 = ρw g h2

800 x 10 x 0.1 + 1000 x 10 x h1 – 1000 x 10 x 0.1 = 1000 x 10 x h2

10000 (h1 -h2) = 200

h1 – h2 = 0.02 …(ii)

Solving (i) & (ii) we get

h1 = 0.35 m and h2 = 0.33 m

h1/h2 = 35/33

## When a long glass capillary tube of radius 0.015 cm is dipped in a liquid , the liquid rises to a height of 15 cm within it …..

Q: When a long glass capillary tube of radius 0.015 cm is dipped in a liquid , the liquid rises to a height of 15 cm within it . If the contact angle between the liquid and glass to close to 0° , the surface tension of the liquid in millinewton m-1 is (ρliquid = 900 kg/m-3 , g = 10 m/s2) (give Answer in closest integer) ….

Click to See Solution :
Ans: (101)

Sol: Here r = 0.015 cm , ρliquid = 900 kg/m-3 , h = 15 cm g = 10 m/s^2 , θ = 0°

Height to which liquid will rise , $\displaystyle h = \frac{2 S cos\theta}{\rho g r }$

$\displaystyle S = \frac{h \rho g r}{2 cos\theta}$

$\displaystyle S = \frac{0.15 \times 900 \times 10 \times 0.015 \times 10^{-2}}{2 cos0^o}$

S = 0.10125 N/m = 101.25 mN/m