The work done in blowing a soap bubble of \(20\text{ cm}\) radius is (surface tension of soap solution is \(0.03\text{ N/m}\))
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:
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 \(L = \frac{B}{\rho g}\).
The velocity of a small ball of mass \(M\) and density \(d\), when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be
When the ball reaches terminal velocity, net force is zero: \(F_v + F_B = Mg\). The buoyant force is \(F_B = V \rho_{\text{glycerine}} g = V \left(\frac{d}{2}\right) g = \frac{Mg}{2}\). Thus, the viscous force is \(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\).
A steel rod has a radius of 10 mm and a length of 1 m. A 100 kN force stretches it along its length. The elongation in rod is (Assume young’s modulus \(Y = 2 times 10^{11}\text{ N/m}^2\))
Formula: \(Delta L = frac{FL}{AY}\). Area \(A = pi r^2 = 3.14 times 10^{-4}\text{ m}^2\. Substituting the given values yields \(Delta L = frac{10^5 times 1}{3.14 times 10^{-4} times 2 times 10^{11}} approx 1.59\text{ mm}\.
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:
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:
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
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
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 \(F_v = mg - \frac{mg}{2} = \frac{mg}{2}\).
The increase in pressure required to decrease the 400 litre volume of a liquid by 0.001% is (Bulk modulus of the liquid is \(2.1 \times 10^9 \text{ N/m}^2\))
Bulk modulus \(B = -\frac{\Delta P}{\Delta V/V}\). Ignoring the negative sign for magnitude, \(Delta P = B \frac{\Delta V}{V} = (2.1 \times 10^9) \times \left(\frac{0.001}{100}\right) = 2.1 \times 10^4 \text{ Pa} = 21 \text{ kPa}\).
The Young’s modulus of brass and steel are \(1 \times 10^{11}\text{ N/m}^2\) and \(2 \times 10^{11}\text{ N/m}^2\) respectively. If wires of both materials, having same length, are loaded with same weight, then they both extend by 4 mm. Ratio of the radii of two wires \(R_B : R_S\) is
Since length, load, and extension are the same: \(Y = \frac{FL}{\pi R^2 \Delta L} ⇒ R^2 \propto \frac{1}{Y} ⇒ \frac{R_B}{R_S} = \sqrt{\frac{Y_S}{Y_B}} = \sqrt{\frac{2 \times 10^{11}}{1 \times 10^{11}}} = \sqrt{2} : 1\).