There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm ($1 \mathrm{atm}={10}^5 \mathrm{N}/{\mathrm{m}}^2$) then the velocity of water flowing from hole is 

(a) $\sqrt{400}$ m/s             (b) $\sqrt{600}$ m/s

(c) $\sqrt{60}$ m/s                (d) None of these

 

A cylindrical tank has a hole of $1 {\mathrm{cm}}^2$ in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of $70 {\mathrm{cm}}^3/\sec$. then the maximum height up to which water can rise in the tank is
(1) 2.5 cm                                             

(2) 5 cm

(3) 10 cm                                               

(4) 0.25 cm

A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, distance between the plates in m is

(1) 0.1                             

(2) 0.05

(3) 0.005                         

(4) 0.0005

Spherical balls of radius 'r' are falling in a viscous fluid of viscosity '$\mathrm{\eta }$' with a velocity 'v'. The retarding viscous force acting on the spherical ball is 

(1) inversely proportional to 'r' but directly proportional to velocity 'v'.

(2) directly proportional to both radius 'r' and velocity 'v'.

(3) inversely proportional to both radius 'r' and velocity 'v'.

(4) directly proportional to 'r' but inversely proportional to 'v'.

A small sphere of mass m is dropped from a great height. After it has fallen 100 m, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 m of fall is

(1) Greater than the work done by air friction in the second 100 m

(2) Less than the work done by air friction in the second 100 m

(3) Equal to 100 mg

(4) Greater than 100 mg

Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be 

(a) 10 cm per sec                                  (b) 2.5 cm per sec
(c) $5\times (4{)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ cm per sec                          (d) $5\times \sqrt{2}$ cm per sec

The rate of steady volume flow of water through a capillary tube of length 'l' and radius 'r' under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P)

(1) $\frac{\mathrm{V}}{16}$                                     

(2) $\frac{\mathrm{V}}{17}$

(3) $\frac{16\mathrm{V}}{17}$                                   

(4) $\frac{17\mathrm{V}}{16}$

A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is

(a) P                             (b) $\frac{3\mathrm{P}}{4}$

(c) $\frac{\mathrm{P}}{2}$                           (d) $\frac{\mathrm{P}}{4}$

We have two (narrow) capillary tubes T₁ and T₂. Their lengths are l₁ and l₂ and radii of cross-section are r₁ and r₂ respectively. The rate of flow of water under a pressure difference P through tube T₁ is 8cm³/sec. If l₁ = 2l₂ and r₁ =r₂, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)

(a) 4 cm³/sec                           (b) (16/3) cm³/sec

(c) (8/17) cm³/sec                    (d) None of these

The Reynolds number of a flow is the ratio of

(1) Gravity to viscous force

(2) Gravity force to pressure force

(3) Inertia forces to viscous force

(4) Viscous forces to pressure forces