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 \(W \approx 16 \times 10^{-3}\text{ J}\).
Terminal velocity of iron ball of radius \(1\text{ mm}\) in glycerine is \(v_1\) and \(v_2\) is the terminal velocity of lead ball of radius \(2\text{ mm}\) in glycerine then \(v_1 : v_2\) is : [\(\rho_{\text{glycerine}} = 13.6\text{ g/cc}\), \(\sigma_{\text{Fe}} = 7.6\text{ g/cc}\), \(\sigma_{\text{Pb}} = 11.6\text{ g/cc}\)]
1. 0.47
2. 0.58
3. 0.75
4. 0.21
View Answer
Terminal velocity \(v_T \propto r^2(\sigma - \rho)\). For iron, \(v_1 \propto 1^2(7.6 - 13.6) = -6\). For lead, \(v_2 \propto 2^2(11.6 - 13.6) = -8\). Thus, the ratio is \(v_1 : v_2 = -6 / -8 = 0.75\).
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:
1. P
2. \(\frac{3P}{4}\)
3. \(\frac{P}{2}\)
4. \(\frac{P}{4}\)
View Answer
Rate of flow \(Q = \frac{\pi P r^4}{8 \eta l}\). Under new conditions, \(Q' = 2Q\), \(r' = 2r\), and \(l' = 2l\), giving \(Q' = \frac{\pi P' (2r)^4}{8 \eta (2l)} = 8 \left(\frac{\pi P' r^4}{8 \eta l}\). Thus, \(8 P' = 2 P\), yielding \(P' = P/4\).
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 water tank resting on the floor has two small holes vertically one above the other. The holes are \(h_1\) \(text{cm}\) and \(h_2\) \(text{cm}\) above the floor. How high does water stand in the tank if the jets from the holes hits the floor at the same point ?
1. \(h_1 + h_2\)
2. \(h_2 - h_1\)
3. \(\frac{h_1^2 + h_2^2}{2}\)
4. \(\frac{h_2^2 - h_1^2}{2}\)
View Answer
For equal horizontal range, the height \(H\) of the water level in the tank must satisfy \(h_1(H - h_1) = h_2(H - h_2)\). Solving for \(H\) gives \(H(h_2 - h_1) = h_2^2 - h_1^2\), which simplifies to \(H = h_1 + h_2\).
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 \(v_2 = 20\text{ m/s}\).
The atmospheric pressure at a place is \(10^5\text{ Pa}\). If liquid of specific gravity equal to 2, be employed as the barometric liquid, the barometric height will be (\(g = 10\text{ m/s}^2\))
1. 5 m
2. 3.2 m
3. 7 m
4. 4.5 m
View Answer
Using the relation \(P = \rho g h\), where density \(\rho = 2 \times 10^3\text{ kg/m}^3\) (specific gravity is 2). Substituting the values: \(10^5 = 2 \times 10^3 \times 10 \times h\). Solving for \(h\) gives \(h = 5\text{ m}\).