A solid steel ball of diameter 3.6~mm$3.6\mathrm{~mm}$ acquired terminal velocity 2.45×10⁻²~m/s$2.45\times10^{-2}\mathrm{~m/s}$ while falling under gravity through an oil of density 925~kg~m⁻³$925\mathrm{~kg~m}^{-3}$. Take density of steel as 7825~kg~m⁻³$7825\mathrm{~kg~m}^{-3}$ and g as 9.8~m/s²$9.8\mathrm{~m/s}^{2}$. The viscosity of the oil in SI unit is :

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