Vidaara.orgClass 11 · Physics
CodeVID-P11-09-PPL-01
Pressure & Pascal's Law — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- 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. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Pressure is a:
- A.vector quantity
- B.scalar quantity
- C.tensor quantity
- D.dimensionless quantity
2.
1 atmosphere is approximately equal to:
- A.$10^3\ \text{Pa}$
- B.$1.013\times10^5\ \text{Pa}$
- C.$76\ \text{Pa}$
- D.$9.8\ \text{Pa}$
3.
Pressure inside a liquid at a given depth acts:
- A.only downward
- B.only sideways
- C.equally in all directions
- D.only upward
4.
A hydraulic lift works as a:
- A.force divider
- B.force multiplier
- C.speed multiplier
- D.pressure multiplier
5.
A mercury barometer column stands at about:
- A.$10\ \text{cm}$
- B.$76\ \text{cm}$
- C.$1\ \text{m}$
- D.$13.6\ \text{cm}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define pressure and write its SI unit.
7.
State Pascal's law.
8.
Find the pressure at a depth of 4 m in water. (Take $\rho=1000$, $g=10$, gauge only.)
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive the expression $P=P_0+\rho gh$ for the pressure at depth $h$ in a liquid.
10.
A hydraulic lift has piston areas $0.02\ \text{m}^2$ and $1\ \text{m}^2$. A 250 N force is applied to the small piston. Find the load lifted.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State Pascal's law and explain, with a labelled description, how a hydraulic lift uses it to multiply force. Why does the small input force lift a large load?
Answer Key
Section A — Multiple Choice Questions
- (B) scalar quantity
- (B) $1.013\times10^5\ \text{Pa}$
- (C) equally in all directions
- (B) force multiplier
- (B) $76\ \text{cm}$
Section B — Short Answer (2 marks)
- Pressure is the normal force per unit area, $P=\frac{F}{A}$; SI unit is the pascal (Pa).
- A pressure change applied to an enclosed incompressible fluid is transmitted undiminished to every part of the fluid and the container walls.
- $P=\rho gh=1000\times10\times4=4\times10^4\ \text{Pa}$.
Section C — Short Answer (3 marks)
- The weight of the liquid column of area $A$ and height $h$ is $\rho g h A$; dividing by $A$ and adding the surface pressure gives $P=P_0+\rho gh$.
- $F_2=F_1\frac{A_2}{A_1}=250\times\frac{1}{0.02}=12500\ \text{N}$.
Section D — Long Answer (5 marks)
- Pressure applied to an enclosed fluid is transmitted undiminished. In a lift, $P=\frac{F_1}{A_1}$ at the small piston is transmitted to the large piston giving $F_2=P A_2=F_1\frac{A_2}{A_1}$. Since $A_2>A_1$, $F_2>F_1$, so a small force lifts a large load (the small piston moves a larger distance, conserving work).
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