A liquid will not wet the surface of a solid, if the angle of contact is

1. 0$°$

2. 45$°$

3. 60$°$

4. >90$°$

Drops of a liquid are usually more spherical in shape than larger drops of same liquid because

1. force of surface tension is equal and opposite to force of gravity

2. force of surface tension predominates the force of gravity

3. force of gravity predominates the force of surface tension

4. force of gravity and surface tension are equal and in same direction

A vessel whose bottom has a round hole with diameter 0.1 mm, is filled with water. The maximum height upto which water can be filled without leakage is

(Surface tension of water =75 dyne/cm and g = 1000 cm s^-2)

1. 100 cm

2. 75 cm

3. 50 cm

4. 30 cm

A spherical liquid drop of radius R is divided into eight equal droplets. If surface tension is T, then the work done in this process will be

1. $2\pi R^{2}T$

2. $3\pi R^{2}T$

3. $4\pi R^{2}T$

4. $5\pi R^{2}T$

There are two hole $O_1 and O_2$ in a tank of height H . The water emerging from $O_1 and O_2$ strikes the ground at the same points, as shown in figure. Then

1. $H=h_1+h_2$

2. $H=h_2-h_1$

3. $H=\sqrt{h_1{h}_2}$

4. None of these

With the increase in temperature, the surface tension of a liquid

(1) increases

(2) decreases

(3) remains the same

(4) changes erratically

Water flows in a streamlined manner through a capiliary tube of radius a, the pressure difference being p and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2p, the rate of flow becomes

(1) 4Q

(2) Q

(3) $\frac{Q}{4}$

(4) $\frac{Q}{8}$

An object of w and density $\rho$ is submerged in liquid of density $\sigma$, its apparent weight will be

(1) $\left(\rho -\sigma \right)$

(2) $\left(\rho -\sigma \right)/w$

(3) $w\left(1-\frac{\sigma }{\rho }\right)$

(4) $w\left(1-\frac{\rho }{\sigma }\right)$

A person swimming in a fresh water pool swims, obeying

(1) Newton's second law

(2) gravitational law

(3) Newton's third law

(4) Newton's first law 

A container has a small hole at its bottom. Area of cross-section of the hole is $A_1$ and that of the container is $A_2$. Liquid is poured in the containor at a constant rate Q $m^3/s$
The maximum level of liquid in the container will be

(1) $\frac{Q^2}{2g A_1{A}_2}$

(2) $\frac{Q^2}{2g {A_1}^2}$

(3) $\frac{Q}{2g A_1{A}_2}$

(4) $\frac{Q^2}{2g {A_2}^2}$