In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used.
The variation of the terminal velocity (\(v\)) with the ratio of density of spherical ball (\(\sigma\)) to density of the liquid (\(\rho\)), is best represented by:
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Always start with Stokes' law when solving terminal velocity questions.
Terminal velocity depends on the density difference \((\sigma-\rho)\).
Express the equation in the form \(y=mx+c\) to identify the graph.
A straight-line equation always produces a linear graph with constant slope.
Straight line having positive slope and non-zero intercept
Parabolic curve
Hyperbolic curve
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The Correct Option isB
Solution and Explanation
Step 1: Recall Stokes' terminal velocity. A sphere falling through a viscous liquid reaches $v = \dfrac{2 r^2 g}{9\eta}(\sigma - \rho)$, where $\sigma$ is the ball density and $\rho$ the liquid density. Step 2: Note what is fixed. Here radius $r$, $g$, $\eta$ and $\rho$ are all constant; only $\sigma$ changes from ball to ball. We plot $v$ against $\sigma/\rho$. Step 3: Factor out the liquid density. Write $\sigma - \rho = \rho\left(\dfrac{\sigma}{\rho} - 1\right)$ to bring in the ratio $\sigma/\rho$. \[ v = \frac{2 r^2 g \rho}{9\eta}\left(\frac{\sigma}{\rho} - 1\right) \] Step 4: Expand into straight-line form. Let $m = \dfrac{2 r^2 g \rho}{9\eta}$. Then $v = m\left(\dfrac{\sigma}{\rho}\right) - m$. Step 5: Compare with $y = mx + c$. Here $x = \sigma/\rho$, the slope is $m > 0$, and the intercept is $-m$, which is non-zero. Step 6: Identify the graph. The plot is a straight line with positive slope and a non-zero (negative) intercept, not a parabola or hyperbola. This is option (B). \[ \boxed{\text{Straight line having positive slope and non-zero intercept}} \]
