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 \(v_1 : v_2 = -6 / -8 = 0.75\).
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 \(h = 5\text{ m}\).
The work done in blowing a soap bubble of \(20\text{ cm}\) radius is (surface tension of soap solution is \(0.03\text{ N/m}\))
1. \(2.06 \times 10^{-2}\text{ J}\)
2. \(3.01 \times 10^{-2}\text{ J}\)
3. \(5.06 \times 10^{-2}\text{ J}\)
4. \(1.51 \times 10^{-2}\text{ J}\)
View Answer
The work done to create a soap bubble (which has two free surfaces) is given by \(W = 2T\Delta A = 8\pi R^2 T\). Substituting \(R = 0.2\text{ m}\) and \(T = 0.03\text{ N/m}\) gives \(W = 8 \times \pi \times (0.2)^2 \times 0.03 \approx 3.01 \times 10^{-2}\text{ J}\).
If \(\rho\) is the density of the material of a wire and \(B\) is the breaking stress, the greatest length of the wire that can hang freely without breaking is:
1. \(\frac{2B}{rho g}\)
2. \(\frac{rho}{Bg}\)
3. \(\frac{B}{rho g}\)
4. \(\frac{rho g}{2B}\)
View Answer
Breaking stress is \(B = \frac{\text{Maximum Tension}}{\text{Area}}\). For a wire of length \(L\) hanging freely, the maximum tension is at the support: \(T = mg = A L \rho g\). Hence, \(B = L \rho g\), which gives \(L = \frac{B}{\rho g}\).
Consider the following statements:
Statement A: When a capillary tube is dipped into a liquid, if the liquid neither rises nor falls in the capillary, then the angle of contact must be zero.
Statement B: The working of a Venturi-meter is based on Bernoulli’s theorem.
Based upon the above information, pick the correct option:
1. Both statement (A) and (B) are correct
2. Both statement (A) and (B) are incorrect
3. Statement (A) is correct while statement (B) is incorrect
4. Statement (A) is incorrect while statement (B) is correct
View Answer
If liquid neither rises nor falls, the net vertical surface tension force is zero, meaning the contact angle is \(90^\circ\). Venturi-meter works on Bernoulli's theorem, so only statement B is correct.
A small ball of mass \(m\) and density \(rho\) is dropped in a viscous liquid of density \(\rho_0\). After sometime, the ball falls with constant velocity. The net force acting on the ball after it attains constant velocity, will be:
1. \(mg\left(\frac{\rho_0}{\rho}-1\right)\)
2. \(mg(\rho - \rho_0)\)
3. \(mg\left(1-\frac{\rho}{\rho_0}\right)\)
4. Zero
View Answer
When a body falls with constant velocity (terminal velocity), its acceleration is zero. Thus, by Newton's second law, the net force on it is zero.
n identical small drops of water having radius \( r \) coalesce to form a bigger drop. If surface tension of water is \( T \) then excess pressure in bigger drop will be
1. \[ \frac{n^3 4T}{r} \]
2. \[ \frac{2T}{n^{\frac{1}{3}} r} \]
3. \[ \frac{4T}{nr} \]
4. \[ \frac{2T}{n^3 r} \]
View Answer
By conserving volume, \( \frac{4}{3} \pi R^3 = n \left(\frac{4}{3} \pi r^3\right) ⇒ R = n^{1/3} r \). The excess pressure in a single-surface liquid drop of radius \( R \) is given by \( \Delta P = \frac{2T}{R} = \frac{2T}{n^{1/3} r} \).
The velocity of a small ball of mass \(m\) and density \(d\), when dropped in a container filled with glycerine becomes constant after sometime. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be
1. \(2mg\)
2. \(\frac{mg}{2}\)
3. \(mg\)
4. \(\frac{3}{2}mg\)
View Answer
At terminal velocity, net force is zero: \(F_v + F_B = mg\). Here, buoyant force \(F_B = V\left(\frac{d}{2}\right)g = \frac{mg}{2}\). Thus, the viscous force is \(F_v = mg - \frac{mg}{2} = \frac{mg}{2}\).