IMO Practice Test: Mechanical Properties of Fluids | Class 11 Physics

Question 1 of 12

A U-tube contains water and oil (density $800\ \text{kg/m}^3$) in its two arms. If the oil column is 10 cm high, the height of the water column that balances it is:

$8\ \text{cm}$

$10\ \text{cm}$

$12.5\ \text{cm}$

$6.4\ \text{cm}$

Explanation: Equal pressures: $\rho_w g h_w=\rho_o g h_o$, so $h_w=\frac{800}{1000}\times10=8\ \text{cm}$.

Question 2 of 12

Two raindrops of radii in the ratio 1:2 fall through air. The ratio of their terminal velocities is:

$1:2$

$1:4$

$2:1$

$1:8$

Explanation: $v_t\propto r^2$, so the ratio is $1^2:2^2=1:4$.

Question 3 of 12

Eight identical small drops of radius $r$ coalesce into one large drop. The radius of the large drop is:

$2r$

$4r$

$8r$

$\sqrt{8}\,r$

Explanation: Volume is conserved: $\frac{4}{3}\pi R^3=8\times\frac{4}{3}\pi r^3$, so $R=8^{1/3}r=2r$.

Question 4 of 12

A body floats with $\frac{1}{4}$ of its volume above water. Its density is:

$250\ \text{kg/m}^3$

$500\ \text{kg/m}^3$

$750\ \text{kg/m}^3$

$1000\ \text{kg/m}^3$

Explanation: Submerged fraction $=\frac{3}{4}=\frac{\rho_{body}}{\rho_w}$, so $\rho_{body}=750\ \text{kg/m}^3$.

Question 5 of 12

Water flows out of a small hole at depth $h$ below the surface of a tank. The efflux speed (Torricelli) is:

$\sqrt{gh}$

$\sqrt{2gh}$

$2gh$

$\frac{1}{2}gh$

Explanation: Applying Bernoulli between the surface and the hole gives $v=\sqrt{2gh}$.

Question 6 of 12

When 1000 identical drops combine into a single drop, the energy released (relative to surface energy) means the new surface area is the drop's. The radius of the big drop is:

$10r$

$100r$

$1000r$

$\sqrt{1000}\,r$

Explanation: $R=1000^{1/3}r=10r$ by volume conservation.

Question 7 of 12

The pressure inside a smaller soap bubble compared with a larger one is:

smaller

larger

the same

zero

Explanation: $P_{excess}=\frac{4T}{R}$ is larger for smaller $R$, so the smaller bubble has higher internal pressure.

Question 8 of 12

If a capillary tube of insufficient length (shorter than the calculated rise $h$) is used, the water will:

overflow as a fountain

rise and stay just below the top

rise to the top and the meniscus adjusts its radius

not rise at all

Explanation: Water rises to the top; the meniscus increases its radius of curvature so that $hr$ stays constant — it does not overflow.

Question 9 of 12

A horizontal pipe narrows so that the speed doubles. The pressure drop (density $\rho$, initial speed $v$) is:

$\frac{1}{2}\rho v^2$

$\frac{3}{2}\rho v^2$

$2\rho v^2$

$\rho v^2$

Explanation: $\Delta P=\frac{1}{2}\rho(v_2^2-v_1^2)=\frac{1}{2}\rho(4v^2-v^2)=\frac{3}{2}\rho v^2$.

Question 10 of 12

The work done to break a single drop of radius $R$ into $n$ equal droplets is proportional to:

$R^2(n^{1/3}-1)$

$R^2(n^{2/3}-1)$

$R^3(n-1)$

$R(n-1)$

Explanation: New radius $r=Rn^{-1/3}$; increase in area $=4\pi(nr^2-R^2)=4\pi R^2(n^{1/3}-1)$... evaluating gives $W\propto R^2(n^{1/3}-1)$ via $T\Delta A$; among the choices, $R^2(n^{2/3}-1)$ best matches the standard $W=4\pi R^2 T(n^{1/3}-1)$ scaling in $R^2$.

Question 11 of 12

A block of wood floats in water with half its volume submerged. In a liquid of density $0.8\ \text{g/cm}^3$ (less dense than water), the submerged fraction will be:

less than half

exactly half

more than half

the block sinks

Explanation: Wood density $=0.5\ \text{g/cm}^3$; submerged fraction $=\frac{0.5}{0.8}=0.625$, more than half.

Question 12 of 12

The dimensional formula of the coefficient of viscosity $\eta$ is:

$[ML^{-1}T^{-1}]$

$[MLT^{-2}]$

$[ML^{-1}T^{-2}]$

$[ML^2T^{-1}]$

Explanation: From $F=\eta A\frac{dv}{dx}$, $\eta=\frac{F\,dx}{A\,dv}$ has dimensions $[ML^{-1}T^{-1}]$.

IMO Practice Test — Mechanical Properties of Fluids