Vidaara.orgClass 10 · Physics
CodeVID-P10-03-HEA-01
Heating Effect & Electric Power — 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.
The SI unit of electric power is the:
- A.joule
- B.watt
- C.volt
- D.ampere
2.
$1\ \text{kWh}$ equals:
- A.$3.6\times10^{6}\ \text{J}$
- B.$3600\ \text{J}$
- C.$1000\ \text{J}$
- D.$10^{3}\ \text{kJ}$
3.
Heat produced is proportional to:
- A.$I$
- B.$I^2$
- C.$\frac{1}{I}$
- D.$\frac{1}{I^2}$
4.
A bulb filament is usually made of:
- A.copper
- B.nichrome
- C.tungsten
- D.iron
5.
A 100 W bulb on 200 V draws a current of:
- A.$0.5\ \text{A}$
- B.$2\ \text{A}$
- C.$5\ \text{A}$
- D.$20\ \text{A}$
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Joule's law of heating.
7.
A 220 V, 1100 W heater — find the current it draws.
8.
Why is the fuse wire of low melting point?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A $10\ \Omega$ resistor carries 2 A for 5 minutes. Find the heat produced.
10.
Find the energy (in kWh) used by a 500 W TV running 4 hours daily for 30 days.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive the three forms of electric power $P=VI=I^2R=\frac{V^2}{R}$ using Ohm's law, and use them to find the resistance of a 60 W, 120 V lamp.
Answer Key
Section A — Multiple Choice Questions
- (B) watt
- (A) $3.6\times10^{6}\ \text{J}$
- (B) $I^2$
- (C) tungsten
- (A) $0.5\ \text{A}$
Section B — Short Answer (2 marks)
- $H=I^2Rt$; heat depends on $I^2$, $R$ and $t$.
- $5\ \text{A}$.
- So it melts quickly on overload and breaks the circuit.
Section C — Short Answer (3 marks)
- $H=I^2Rt=4\times10\times300=12000\ \text{J}$.
- $0.5\times4\times30=60\ \text{kWh}$.
Section D — Long Answer (5 marks)
- From $P=VI$ and $V=IR$ get $P=I^2R=\frac{V^2}{R}$; $R=\frac{V^2}{P}=\frac{14400}{60}=240\ \Omega$.
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