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The dimensions of surface tension are

$[{MLT}^{- 1}]$

$[{ML}^{2} T^{- 2}]$

$[{ML}^{0} T^{- 2}]$

$[{ML}^{1} T^{- 2}]$

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Explanation

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A wooden stick 2m long is floating on the surface of water. The surface tension of water 0.07 N/m. By putting soap solution on one side of the sticks the surface tension is reduced to 0.06 N/m. The net force on the stick will be

0.07 N

0.06 N

0.01 N

0.02 N

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Explanation

(d) Net force on stick $= F_{1} - F_{2} = (T_{1} - T_{2}) \times l$

=(0.07-0.06) $\times$ l = 0.01x2 = 0.02 N

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The force required to take away a flat circular plate of radius 2 cm from the surface of water, will be (the surface tension of water is 70 dyne/cm)

$280 π dyne$

$250 π dyne$

$140 π dyne$

$210 π dyne$

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Explanation

(a) Force

$F = 2 πrT = 2 π \times 2 \times 70 = 280 π dyne$

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Surface tension may be defined as

The work done per unit area in increasing the surface area of a liquid under isothermal condition

The work done per unit area in increasing the surface area of a liquid under adiabatic condition

The work done per unit area in increasing the surface area of a liquid under both isothermal and adiabatic conditions

Free surface energy per unit volume

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Explanation

(a)

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Energy needed in breaking a drop of radius R into n drops of radii r is given by

$4 πT ({nr}^{2} - R^{2})$

$\frac{4}{3} π (r^{3} n - R^{2})$

$4 πT (R^{2} - {nr}^{2})$

$4 πT (r^{2} n + R^{2})$

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Explanation

(a) Energy needed = Increment in surface energy
= (surface energy of n small drops) – (surface energy of one big drop)

$= n 4 {πr}^{2} T - 4 {πR}^{2} T = 4 πT ({nr}^{2} - R^{2})$

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The potential energy of a molecule on the surface of liquid compared to one inside the liquid is -

Zero

Smaller

The same

Greater

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Explanation

(d)

When the surface area of the liquid is increased molecules from the interior of the liquid rise to the surface. This work is stored in the molecules in the form of ofpotential energy. Thus, the potential energy of the molecules lying on the surface is greater than that of the molecules in the interior of the liquid.

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Two droplets merge with each other and forms a large droplet. In this process

Energy is liberated

Energy is absorbed

Neither liberated nor absorbed

Some mass is converted into energy

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Explanation

(a) When two droplets merge with each other, their surface energy decreases.

$W = T (∆ A)$ =(negative) i.e. energy is released

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Radius of a soap bubble is 'r', surface tension of soap solution is T. Then without increasing the temperature, how much energy will be needed to double its radius

$4 {πr}^{2} T$

$2 {πr}^{2} T$

$12 {πr}^{2} T$

$24 {πr}^{2} T$

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Explanation

(d) $W = 8 πT (R_{2}^{2} - R_{1}^{2}) = 8 π [(2 r)^{2} - (r)^{2}] = 24 {πr}^{2} T$

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Work done in splitting a drop of water of 1 mm radius into $10^{6}$ droplets is (Surface tension of water $= 72 \times 10^{- 3} J / m^{2}$ )

$9.58 \times 10^{- 5} J$

$8.95 \times 10^{- 5} J$

$5.89 \times 10^{- 5} J$

$5.98 \times 10^{- 5} J$

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Explanation

(b) Work done in splitting a water drop of radius R into n drops of equal size

$= 4 {πR}^{2} T (n^{1 / 3} - 1)$

$= 4 π \times (10^{- 3})^{2} \times 72 \times 10^{- 3} \times (10^{6 / 3} - 1) = 4 π \times 10^{- 6} \times 72 \times 10^{- 3} \times 99 = 8.95 \times 10^{- 5} J$

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The amount of work done in blowing a soap bubble such that its diameter increases from d to D is (T= surface tension of the solution)

$4 π (D^{2} - d^{2}) T$

$8 π (D^{2} - d^{2}) T$

$π (D^{2} - d^{2}) T$

$2 π (D^{2} - d^{2}) T$

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Explanation

(d)

$W = T \times 8 π (r_{2}^{2} - r_{1}^{2}) = T \times 8 π (\frac{D^{2}}{4} - \frac{d^{2}}{4}) = 2 π (D^{2} - d^{2}) T$

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