Mechanical Properties of Fluids: NEET and JEE questions with solutions

Answer: (D) $400$

Continuity: $A_P v_P = A_Q v_Q$, so $v_Q = \dfrac{40}{20}v_P = 2v_P$.

Bernoulli (horizontal pipe):
$$P_P - P_Q = \frac{1}{2}\rho\left(v_Q^2 - v_P^2\right) = \frac{1}{2}\rho\left(3v_P^2\right)$$
$$15 = \frac{1}{2}(1000)(3v_P^2) \;\Rightarrow\; v_P^2 = 0.01 \;\Rightarrow\; v_P = 0.1\ \text{m s}^{-1} = 10\ \text{cm s}^{-1}$$

Rate of flow:
$$Q = A_P v_P = 40 \times 10 = 400\ \text{cm}^3\text{s}^{-1}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) see figure

Terminal velocity (Stokes' law):
$$v = \frac{2r^2(\sigma - \rho)g}{9\eta} = \frac{2r^2\rho g}{9\eta}\left(\frac{\sigma}{\rho} - 1\right)$$

For balls of the same radius, $v$ is a straight line in $\dfrac{\sigma}{\rho}$ that becomes zero at $\dfrac{\sigma}{\rho} = 1$ and increases linearly beyond it. (For $\sigma < \rho$ the ball does not sink, so there is no falling terminal velocity to measure.)

This matches option (1).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{d^2y}{dx^2} = \dfrac{\rho g}{S}y$

Take a thin strip of the liquid surface between $x$ and $x + dx$ (unit length along $z$). Measure $y$ from the level of the flat liquid surface far away, where the pressure just below the surface is atmospheric.

Surface tension pulls along the surface at both edges of the strip. Its net vertical component is
$$S\sin\theta(x + dx) - S\sin\theta(x) = S\frac{d\theta}{dx}dx = S\frac{d^2y}{dx^2}dx$$

Just below a surface at height $y$, the liquid pressure is lower than atmospheric by $\rho g y$. So the atmosphere pushes the strip down with a net force $\rho g y\,dx$.

For equilibrium:
$$S\frac{d^2y}{dx^2} = \rho g y \;\Rightarrow\; \frac{d^2y}{dx^2} = \frac{\rho g}{S}y$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $a = -\frac{1}{2},\ \alpha = -\frac{1}{2},\ \beta = -1,\ \gamma = \frac{1}{2},\ \delta = \frac{7}{2}$

Excess pressure inside the balloon: $\Delta P = \dfrac{kS}{r}$ ($k$ is a number, $2$ or $4$).

Bernoulli at the outlet: $\dfrac{1}{2}\rho v^2 = \dfrac{kS}{r}$, so
$$v = \sqrt{\frac{2kS}{\rho r}} \propto r^{-1/2} \;\Rightarrow\; a = -\frac{1}{2}$$

Rate of volume loss:
$$-\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = Av \;\Rightarrow\; -4\pi r^2\frac{dr}{dt} = A\sqrt{\frac{2kS}{\rho}}\,r^{-1/2}$$

$$dt = -\frac{4\pi}{A}\sqrt{\frac{\rho}{2kS}}\,r^{5/2}\,dr \;\Rightarrow\; T = \frac{4\pi}{A}\sqrt{\frac{\rho}{2kS}}\cdot\frac{2}{7}R^{7/2}$$

So $T \propto S^{-1/2}A^{-1}\rho^{1/2}R^{7/2}$: $\alpha = -\frac{1}{2}$, $\beta = -1$, $\gamma = \frac{1}{2}$, $\delta = \frac{7}{2}$.

Solution by Sreeraj P, M.Sc Physics