125 small droplets each of radius r, combine to form a big drop. If surface tension of liquid is T. Then loss in surface potential energy during this process will be:
1. \(T \times 16\pi r^2\)
2. \(T \times 20\pi r^2\)
3. \(T \times 100\pi r^2\)
4. \(T \times 400\pi r^2\)
View Answer
Volume conservation gives \(R = 5r\). Initial surface area is \(A_i = 125 \times 4\pi r^2 = 500\pi r^2\) and final is \(A_f = 4\pi R^2 = 100\pi r^2\). The decrease in area is \(400\pi r^2\), so loss in surface energy is \(T \times 400\pi r^2\).
If the excess pressure inside a soap bubble is balanced by an oil column of height \(2\text{ mm}\), then the surface tension of soap solution will be : (\(r = 1\text{ cm}\) and density \(d = 0.8\text{ gm/cc}\))
1. \(4\text{ N/m}\)
2. \(4 \times 10^{-2}\text{ N/m}\)
3. \(4 \times 10^{-4}\text{ N/m}\)
4. 4 dyne/m
View Answer
Excess pressure in a soap bubble is \(\Delta P = \frac{4T}{r}\), and the pressure of the oil column is \(h d g\). Setting them equal, \(T = \frac{h d g r}{4}\). Substituting SI values gives \(T = \frac{2 \times 10^{-3} \times 800 \times 9.8 \times 10^{-2}}{4} \approx 4 \times 10^{-2}\text{ N/m}\).
A ring of radius \(1.5\text{ cm}\) is floating horizontally on the surface of water. If this ring has to be raised up then how much additional force has to be applied to lift ring : (Surface tension of water \(73 \times 10^{-3}\text{ Newton/metre}\))
1. \(1.37 \times 10^{-2}\text{ N}\)
2. \(2.3 \times 10^{-2}\text{ N}\)
3. \(5.1 \times 10^{-3}\text{ N}\)
4. \(4 \times 10^{-2}\text{ N}\)
View Answer
The additional force required to lift the ring is \(F = 2 \times (2\pi r T) = 4\pi r T\). Substituting \(r = 1.5 \times 10^{-2}\text{ m}\) and \(T = 73 \times 10^{-3}\text{ N/m}\) gives \(F = 4 \times 3.14 \times 1.5 \times 10^{-2} \times 73 \times 10^{-3} \approx 1.37 \times 10^{-2}\text{ N}\).
A liquid is flowing in a cylindrical pipe of internal diameter \(4\text{ cm}\) with a velocity of \(5\text{ m/s}\). If this tube is joined with another tube of internal diameter \(2\text{ cm}\) then the velocity of flow of liquid in the smaller tube will be (in \(\text{ms}^{-1}\))
View Answer
Using the equation of continuity, \(A_1 v_1 = A_2 v_2\), which ⇒ \(d_1^2 v_1 = d_2^2 v_2\). Substituting \(d_1 = 4\text{ cm}\), \(v_1 = 5\text{ m/s}\), and \(d_2 = 2\text{ cm}\) gives \(16 \times 5 = 4 \times v_2\), so \(v_2 = 20\text{ m/s}\).
Young’s modulus of rubber is \(10^4\text{ N/m}^2\) and area of cross-section is \(2\text{ cm}^2\). If force of \(2 \times 10^5\text{ dynes}\) is applied along its length, then length of wire becomes how much times of its initial length \(L\):- (Assume stress \(\propto\) strain)
1. three times
2. four times
3. two times
4. No change in length
View Answer
Converting Young's modulus to CGS units gives \(Y = 10^5\text{ dyne/cm}^2\). Stress is \(F/A = 2 \times 10^5 / 2 = 10^5\text{ dyne/cm}^2\). Since \(\text{Strain} = \text{Stress}/Y = 1\), we have \(\Delta L = L\). Thus, the final length becomes \(L + \Delta L = 2L\).
A metal block is experiencing an atmospheric pressure of \(1 \times 10^5\text{ N/m}^2\). When the same block is placed in a vacuum chamber, the fractional change in its volume is (the bulk modulus of metal is \(1.25 \times 10^{11}\text{ N/m}^2\))
1. \(4 \times 10^{-7}\)
2. \(2 \times 10^{-7}\)
3. \(8 \times 10^{-7}\)
4. \(1 \times 10^{-7}\)
View Answer
The bulk modulus is defined as \(B = \frac{\Delta P}{\Delta V/V}\). Moving to vacuum causes a pressure change of \(\Delta P = 10^5\text{ N/m}^2\). Thus, the fractional volume change is \(\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{10^5}{1.25 \times 10^{11}} = 8 \times 10^{-7}\).
When the load on a wire is increased from \(3\text{ kg wt}\) to \(5\text{ kg wt}\) the elongation increases from \(0.61\text{ mm}\) to \(1.02\text{ mm}\). The required work done during the extension of the wire is :
1. \(16 \times 10^{-3}\text{ J}\)
2. \(8 \times 10^{-2}\text{ J}\)
3. \(20 \times 10^{-2}\text{ J}\)
4. \(11 \times 10^{-3}\text{ J}\)
View Answer
The work done during the extension is \(W = \frac{1}{2} (F_2 x_2 - F_1 x_1)\). Converting values: \(F_1 = 3 \times 9.8\text{ N}\), \(F_2 = 5 \times 9.8\text{ N}\), \(x_1 = 0.61 \times 10^{-3}\text{ m}\), and \(x_2 = 1.02 \times 10^{-3}\text{ m}\) yields \(W \approx 16 \times 10^{-3}\text{ J}\).