Mechanical Properties of Fluids: NEET previous year questions

Absolute pressure at depth $h$ is $P = P_0 + \rho g h$. So $100\times10^{5} = 1\times10^{5} + 1000\times10\times h$, giving $99\times10^{5} = 10^{4}\,h$, hence $h = 990\ \text{m}$.

Continuity gives $A_P v_P = A_Q v_Q$, so halving the area doubles the speed: $v_Q = 2v_P$. For a horizontal pipe Bernoulli's equation reduces to $\Delta P = \frac{1}{2}\rho(v_Q^2 - v_P^2) = \frac{3}{2}\rho v_P^2$. Substituting $15 = 1500\,v_P^2$ gives $v_P = 0.1\ \text{ms}^{-1} = 10\ \text{cm s}^{-1}$, and the volume flow rate is $A_P v_P = 40 \times 10 = 400\ \text{cm}^3\text{s}^{-1}$. NCERT Class 11, Mechanical Properties of Fluids.

Why the other options are wrong:

- (1) uses the narrow area $A_Q$ with the wide-section speed $v_P$, mismatching the pair
- (2) drops the factor 3 from $v_Q^2 - v_P^2$, effectively using $\Delta P = \frac{1}{2}\rho v_Q^2$ alone
- (3) mixes SI and centimetre-gram-second (CGS) units partway through, most often by leaving $\rho$ in $\text{kg m}^{-3}$ alongside areas in $\text{cm}^2$

Stokes' law gives $v = \frac{2r^2 g}{9\eta}(\sigma - \rho)$, which can be rewritten as $v = \frac{2r^2 g\rho}{9\eta}\left(\frac{\sigma}{\rho} - 1\right)$. With $r$ fixed this is a STRAIGHT LINE in $\sigma/\rho$ whose intercept on the horizontal axis is exactly at $\sigma/\rho = 1$. Below that value the ball is lighter than the liquid and does not sink at all, so the downward terminal velocity stays zero. NCERT Class 11, Mechanical Properties of Fluids (Stokes' law).

Why the other options are wrong:

- (2) takes $v \propto \sigma$ and lets the line pass through the origin, ignoring the buoyant term $-\rho$
- (3) places the zero-velocity crossing at some arbitrary density ratio rather than at the neutral-buoyancy point $\sigma = \rho$
- (4) treats terminal velocity as a fixed property of the ball once it sinks, missing its linear dependence on excess density

Capillary rise h is inversely proportional to r. The risen mass is ρ$\pi$r$^{2}$h, so it is proportional to r. Doubling the radius therefore doubles 5 g to 10 g. Chapter: Mechanical Properties of Fluids.