The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be :-

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