## A paper disc of radius R from which a hole of radius r is cut out is floating in a liquid of the surface tension S…

Q: A paper disc of radius R from which a hole of radius r is cut out is floating in a liquid of the surface tension S. The force on the disc due to the surface tension is

(a) S × 2π R

(b) S × 2πr

(c) S × 2π (R – r)

(d) S × 2π (R + r)

Ans: (d)

Sol: Surface tension S = F/l  ⇒ F = S l

F = S (2 π R + 2 π r )

F = S × 2 π(R + r)

## Two parallel wires each of length 10 cm are 0.5 cm apart. A film of water is formed between them….

Q: Two parallel wires each of length 10 cm are 0.5 cm apart. A film of water is formed between them. If surface tension of water is 0.072 N/m, then the work done in increasing the distance between the wires by 1 mm is

(a) 1.44 × 10-5 J

(b) 1.72 × 10-5 J

(c) 1.44 × 10-4 J

(d) 1.72 × 10-4 J

Ans: (a)

Sol : $\displaystyle \Delta A = 2[10\times 10^{-2} \times 1 \times 10^{-3}]$

= 2 × 10-4 m2

W = S × ΔA

W = 0.072 × 2 × 10-4

= 0.144 × 10-4

= 1.44 × 10-5 J

## The length of a needle floating on water is 2.5 cm. The minimum force in addition to its weight…

Q: The length of a needle floating on water is 2.5 cm. The minimum force in addition to its weight needed to lift the needle above the surface of water will be (surface tension of water is 0.072 N/m)

(a) 3.6× 10-3 N

(b) 10-2 N

(c) 9 × 10-4 N

(d) 6 × 10-4 N

Ans: (a)

Sol: F = S × 2 l

= 0.072 × 2 × 2.5 × 10-2

= 0.072 × 5 × 10-2

= 0.36 × 10-2

= 3.6 × 10-3 N

## A solid sphere falls with a terminal velocity of 20 ms-1 in air. If it is allowed to fall in vacuum.

Q: A solid sphere falls with a terminal velocity of 20 ms-1 in air. If it is allowed to fall in vacuum.

(a) terminal velocity will be 20 ms-1

(b) terminal velocity will be less than 20 ms-1

(c) terminal velocity will be greater than 20 ms-1

(d) no terminal velocity will be attained

Ans: (d)

Sol:

Because there is no resistive force

## Water rises to a height h in a capillary tube of cross-sectional area A. the height to which water will rise…

Q: Water rises to a height h in a capillary tube of cross-sectional area A. the height to which water will rise in a capillary tube of cross-sectional area 4A will be

(a) h

(b) h/2

(c) h/4

(d) 4h

Ans: (b)

Sol: As cross-sectional area becomes 4 times , the radius becomes 2 times

$\displaystyle h \propto \frac{1}{r}$

h r = constant

Hence , h becomes h/2