Given below are two statements: Statement I: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate from the options given below:

Related Formula

At terminal velocity, downward force equals upward drag (assuming negligible buoyancy):

Core Logic

Since the density of the medium is negligible, we ignore buoyant forces. The mass M$M$ of the ball is specified to remain the same in both cases, despite the change in radius (implying the material density of the second ball is lower).

Step 1: Setup Proportionality

Since M$M$, g$g$, and η$\eta$ are all constants:

Step 2: Evaluating the Ratio

For the second ball, r' = 2r$r' = 2r$. Therefore, the new terminal velocity v'$v'$ is:

Pattern Recognition

Read the constraints carefully. Usually, questions keep material density uniform (v ∝ r²$v \propto r^2$). However, this specifically says "same mass". This shifts the formula dependency from v ∝ r²$v \propto r^2$ entirely to v ∝ 1/r$v \propto 1/r$ because M$M$ acts as a constant numerator.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Related Formula

Core Logic

When dropped, three forces act on the ball: Gravity downwards, Buoyant force upwards, and Viscous drag upwards.

Let (4π r³ g(ρ - ρL))/(3m) = K₁$\frac{4\pi r^3 g(\rho - \rho_L)}{3m} = K_1$ and (6πη r)/(m) = K₂$\frac{6\pi\eta r}{m} = K_2$.

Integrating from t=0, v=0$t=0, v=0$:

This is an exponential curve starting from the origin and asymptotically approaching the terminal velocity VT = K₁ / K₂$V_T = K_1 / K_2$.

Chapter Mix

Class 11 Physics: Mechanical Properties Of Fluids