Two objects A & B of equal density and radius rA = 1 mm and rB = 2 mm are moving in same medium then find the ratio of their terminal velocity \( \frac{v_{B}}{v_{A}}\) in the medium.
View Answer
The terminal velocity v of a spherical object moving through a viscous medium is directly proportional to the square of its radius, expressed as v
\(r^2 \), when the density and the medium remain constant.
\( \frac{v_B}{v_A} = \left(\frac{r_B}{r_A}\right)^2 = \left(\frac{2}{1}\right)^2 = 4 \)
Thus, the required ratio \( \frac{v_B}{v_A} \) is 4 (Option 3).
The viscous drag acting on a metal sphere of diameter \(1\text{ mm}\), falling through a fluid of viscosity \(0.8\text{ Pa s}\) with a velocity of \(2\text{ m s}^{-1}\) is equal to
1. \(1.5 \times 10^{-3} \text{ N}\)
2. \(20 \times 10^{-3} \text{ N}\)
3. \(15 \times 10^{-3} \text{ N}\)
4. \(30 \times 10^{-3} \text{ N}\)
View Answer
According to Stokes' law, \(F = 6\pi \eta r v\). Here \(r = 0.5 \times 10^{-3}\text{ m}\), \(\eta = 0.8\text{ Pa s}\), and \(v = 2\text{ m s}^{-1}\). Calculating, we get \( F = 6 \times 3.14 \times 0.8 \times 0.5 \times 10^{-3} \times 2 \approx 15 \times 10^{-3}\text{ N}\).
Given below are two statements:
Assertion (A): A hydrogen-filled balloon stops rising after it has attained a certain height in the sky.
Reason (R): The atmospheric pressure decreases with height and becomes zero when maximum height is attained by balloon.
1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. Both (A) and (R) are false.
View Answer
As the balloon rises, the density of air decreases, leading to a decrease in buoyant force until it equals the weight of the balloon, so it stops rising. Thus, Assertion is true. However, atmospheric pressure does not become zero at this height, making Reason false.
Given below are two statements:
Assertion (A): When two soap bubbles of different radii are brought into contact, the common interface of contact bulges into the bubble of larger radii.
Reason (R): Pressure inside a soap bubble of a lesser radius is more than pressure inside the soap bubble of a larger radius.
1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. Both (A) and (R) are false.
View Answer
The excess pressure inside a soap bubble is \(P = \frac{4T}{R}\). Thus, pressure is greater inside the smaller bubble (lesser radius). When in contact, the common interface bulges towards the larger bubble due to this pressure difference. Both are true and Reason explains Assertion.
Given below are two statements:
Assertion (A): Young’s modulus for a perfectly plastic body is zero.
Reason (R): For a perfectly plastic body, restoring force is zero.
1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. Both (A) and (R) are false.
View Answer
For a perfectly plastic body, there is no tendency to regain its shape, meaning the restoring force (and thus stress) is zero. Since Young's modulus \(Y = \frac{\text{stress}}{\text{strain}}\), it is also zero. Both statements are true and Reason is the correct explanation.
A small ball of density \(rho\) is immersed in a liquid of density \(sigma (\sigma > \rho)\) to a depth h and released. The height above the surface of water upto which the ball jumps is:
1. \(\left(\frac{\sigma}{\rho} - 1\right)h\)
2. \(\left(\frac{\sigma}{\rho} + 1\right)h\)
3. \(\left(\frac{\rho}{\sigma} - 1\right)h\)
4. \(\left(\frac{\rho}{\sigma} + 1\right)h\)
View Answer
Applying the work-energy theorem, work done by the buoyant force over depth \(h\) equals total work done against gravity: \(V \sigma g h = V \rho g (h + H)\). Solving for height \(H\) gives \(H = \left(\frac{\sigma}{\rho} - 1\right)h\).
An object of mass m and density \(\rho\) is falling in a viscous liquid of density \(\frac{\rho}{4}\). When it attains terminal velocity, then viscous force acting on it will be:
1. \(\frac{mg}{4}\)
2. \(\frac{3mg}{4}\)
3. \(\frac{mg}{3}\)
4. 3mg
View Answer
At terminal velocity, the upward forces balance the downward force: \(F_v + F_b = mg\). Since buoyant force is \(F_b = V \rho_l g = \left(\frac{m}{\rho}\right)\left(\frac{\rho}{4}\right)g = \frac{mg}{4}\), we get \(F_v = mg - \frac{mg}{4} = \frac{3mg}{4}\).