Vidaara.orgClass 11 · Physics
CodeVID-P11-09-CH-01
Mechanical Properties of Fluids — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks. Show all working for Sections B, C and D.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
Pressure at depth $h$ in a liquid is:
- A.$P_0-\rho gh$
- B.$P_0+\rho gh$
- C.$\rho g$
- D.$\frac{P_0}{\rho gh}$
2.
A hydraulic lift is based on:
- A.Archimedes' principle
- B.Pascal's law
- C.Bernoulli's principle
- D.Stokes' law
3.
The buoyant force equals the weight of:
- A.the body
- B.fluid displaced
- C.the container
- D.air above
4.
Terminal velocity of a sphere is proportional to:
- A.$r$
- B.$r^2$
- C.$\frac{1}{r}$
- D.$r^3$
5.
The equation of continuity is:
- A.$A_1v_1=A_2v_2$
- B.$P_1v_1=P_2v_2$
- C.$A_1v_2=A_2v_1$
- D.$\frac{A_1}{v_1}=\frac{A_2}{v_2}$
6.
The excess pressure inside a soap bubble of radius $R$ is:
- A.$\frac{2T}{R}$
- B.$\frac{4T}{R}$
- C.$\frac{T}{R}$
- D.$\frac{T}{2R}$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
State Pascal's law.
8.
State Archimedes' principle.
9.
Write Bernoulli's equation and name the three energy terms.
10.
Define surface tension and give its SI unit.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
Derive the expression for the terminal velocity of a sphere falling through a viscous fluid.
12.
Water flows through a horizontal pipe whose area falls from $5\ \text{cm}^2$ to $2\ \text{cm}^2$. If the speed in the wide part is $4\ \text{m/s}$, find the speed in the narrow part.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
State and explain Bernoulli's principle and derive it from the work–energy theorem for streamline flow. Use it to explain the lift on an aeroplane wing.
14.
Define surface tension and capillary rise. Derive $h=\frac{2T\cos\theta}{\rho g r}$ and explain why water rises while mercury is depressed in a glass capillary.
Answer Key
Section A — Multiple Choice Questions
- (B) $P_0+\rho gh$
- (B) Pascal's law
- (B) fluid displaced
- (B) $r^2$
- (A) $A_1v_1=A_2v_2$
- (B) $\frac{4T}{R}$
Section B — Short Answer (2 marks)
- A pressure change applied to an enclosed incompressible fluid is transmitted undiminished throughout the fluid and to the walls of the container.
- A body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces.
- $P+\frac{1}{2}\rho v^2+\rho gh=\text{const}$: pressure energy, kinetic energy and potential energy per unit volume.
- Force per unit length along a liquid surface, $T=\frac{F}{L}$; SI unit N/m.
Section C — Short Answer (3 marks)
- At terminal velocity, weight $=$ buoyancy $+$ viscous drag: $\frac{4}{3}\pi r^3\rho g=\frac{4}{3}\pi r^3\sigma g+6\pi\eta r v_t$, giving $v_t=\frac{2r^2(\rho-\sigma)g}{9\eta}$.
- $v_2=\frac{A_1v_1}{A_2}=\frac{5\times4}{2}=10\ \text{m/s}$.
Section D — Long Answer (5 marks)
- For steady, incompressible, non-viscous flow the work done by pressure differences equals the change in kinetic plus potential energy, giving $P+\frac{1}{2}\rho v^2+\rho gh=\text{const}$ along a streamline. Faster air over the curved top of a wing has lower pressure than the slower air below, and this pressure difference produces an upward lift force.
- Surface tension is force per unit length on a liquid surface. In capillary rise, the upward surface-tension force $2\pi r T\cos\theta$ balances the column weight $\pi r^2 h\rho g$, giving $h=\frac{2T\cos\theta}{\rho g r}$. Water on glass has an acute angle of contact ($\cos\theta>0$) so it rises; mercury has an obtuse angle ($\cos\theta<0$) so it is depressed.
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