Vidaara.orgClass 9 · Physics
CodeVID-P9-04-PWR-01
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.
- Take $g=10\,\text{m/s}^2$ unless stated otherwise. Show all working for Sections B, C and D.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Power is the rate of doing
- A.force
- B.work
- C.mass
- D.distance
2.
$1\,\text{W}$ equals
- A.$1\,\text{J}\times1\,\text{s}$
- B.$1\,\text{J/s}$
- C.$1\,\text{N/s}$
- D.$1\,\text{J/m}$
3.
The commercial unit of electrical energy is the
- A.joule
- B.watt
- C.kilowatt-hour
- D.newton
4.
In free fall, as a body descends its potential energy
- A.increases
- B.decreases
- C.stays same
- D.becomes zero at once
5.
$1\,\text{kWh}$ in joules is
- A.$3600$
- B.$3.6\times10^5$
- C.$3.6\times10^6$
- D.$1000$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Define one watt.
7.
A motor does 1200 J of work in 4 s. Find its power.
8.
State the law of conservation of energy.
9.
Convert $1\,\text{kWh}$ into joules.
Section C — Short Answer (3 marks)
3 × 3 = 9 marks
10.
A 100 kg load is raised 5 m in 10 s. Find the power used. Take $g=10\,\text{m/s}^2$.
11.
A 2 kg ball is dropped from 5 m. Find its speed at the ground using energy conservation.
12.
A 1.5 kW heater runs for 4 hours. Find the energy used in kWh.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
13.
Show how energy is conserved for a 1 kg ball dropped from 20 m, by finding PE and KE at the top, at 10 m, and at the ground. Take $g=10\,\text{m/s}^2$.
Answer Key
Section A — Multiple Choice Questions
- (B) work
- (B) $1\,\text{J/s}$
- (C) kilowatt-hour
- (B) decreases
- (C) $3.6\times10^6$
Section B — Short Answer (2 marks)
- One watt is the power when one joule of work is done in one second, $1\,\text{W}=1\,\text{J/s}$.
- $P=\frac{W}{t}=\frac{1200}{4}=300\,\text{W}$.
- Energy can neither be created nor destroyed; it can only be transformed from one form to another, so total energy is constant.
- $1\,\text{kWh}=1000\times3600=3.6\times10^6\,\text{J}$.
Section C — Short Answer (3 marks)
- $W=mgh=100\times10\times5=5000\,\text{J}$; $P=\frac{5000}{10}=500\,\text{W}$.
- $mgh=\frac{1}{2}mv^2\Rightarrow v^2=2gh=2\times10\times5=100\Rightarrow v=10\,\text{m/s}$.
- Energy $=P\times t=1.5\times4=6\,\text{kWh}$.
Section D — Long Answer (5 marks)
- Top (20 m): PE $=1\times10\times20=200\,\text{J}$, KE $=0$, total 200 J. At 10 m: PE $=100\,\text{J}$, KE $=100\,\text{J}$, total 200 J. Ground: PE $=0$, KE $=200\,\text{J}$, total 200 J. Total mechanical energy is constant at 200 J.
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