Solutions of Electricity (Page No- 66) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

Class 10 Physics Solutions By Lakhmir Singh & Manjit Kaur

Class 10 : Solutions of Electricity (Page No- 66) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev

The document Solutions of Electricity (Page No- 66) - Physics By Lakhmir Singh, Class 10 Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Physics Solutions By Lakhmir Singh & Manjit Kaur.
All you need of Class 10 at this link: Class 10

Lakhmir Singh Physics Class 10 Solutions Page No:66

Question 1:
How does the heat H produced by a current passing through a fixed resistance wire depend on the magnitude of current I ?
Solution :
Heat produced is directly proportional to the square of current.

Question 2:
If the current passing through a conductor is doubled, what will be the change in heat produced ?

Solution :
Heat produced is directly proportional to the square of current.
If current I is doubled, heat H will be four times.

Question 3:
Name two effects produced by electric current.

Solution :
Two effects of produced by electric current are:
(a) Heating effect
(b) Magnectic effect

Question 4:
Which effect of current is utilised in an electric light bulb ?

Solution :
Heating effect

Question 5:
Which effect of current is utilised in the working of an electric fuse ?

Solution :
Heating effect

Question 6:
Name two devices which work on the heating effect of electric current.

Solution :
Electirc heater and electric fuse.

Question 7:
Name two gases which are filled in filament type electric light bulbs.

Solution :
Argon and nitrogen.

Question 8:
Explain why, filament type electric bulbs are not power efficient.

Solution :
Filament type electric bulbs are not power efficien because most of the electric power consumed by the filament of a bulb appears as heat and only a small amount of electric power is converted into light.

Question 9:
Why does the connecting cord of an electric heater not glow hot while the heating element does ?

Solution :
The connecting cord of the heater made of copper does not glow because negligible heat is produced in it by passing current (because of its extremely low resistance); but the heating element made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current (because of its high resistance).

Question 10:
(a) Write down the formula for the heat produced when a current I is passed through a resistor R for time
 (b) An electric iron of resistance 20 ohms draws a current of 5 amperes. Calculate the heat produced in 30 seconds.

Solution :

(a) Heat produced. H = I2Rt 
(b) Given: R = 20ohm, I = Samp, t = 30s 
We know that H = 12Rt 
H = 52x20x30 
H = 150001 

Question 11:
State three factors on which the heat produced by an electric current depends. How does it depend on these factors ?

Solution :
Heat produced by an electric current depends on the following factors:
(i) Heat produced is directly proportional to square of current.
(ii) Heat produced is directly proportional to resistance.
(iii) Heat produced is directly proportional to the time for which current flows.

Question 12:
(a) State and explain Joule’s law of heating.
 (b) A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination of resistances in half a minute.

Solution :

(a) Joules law of heating states that heat produced injou les when a current of I amperes flows in a wire of resistance R ohms for time t seconds is 
given by H = I2Rt. 
Thus the heat produced In a wire Is directly proportional to: 
(i) Square of current 
(ii) Resistance of wire 
(iii) Time for which current is passed 
(b) Given: R1= 40ohms. R= 60ohms lin seriesI.V = 220V. t = 30sec 
We know that 
Total resistance. R = 40+60 = 100ohms 
By Ohm's law, 
V = IR 
I = V/R 
I = 220/100 = 2.2amp 
Putting the values of I, R and t in eq. H = I2RT 
H = 2.22 X 100 X 30 
H =14520 J 

Question 13:
Why is an electric light bulb not filled with air ? Explain why argon or nitrogen is filled in an electric bulb.

Solution :
If air is filled in an electric bulb, then the extremely hot tungsten filament would burn up quickly in the oxygen of air. So, the electric bulb is filled with a chemically unreactive gas like argon or nitrogen. Thes gases do not react with the hot tungsten filament and hence prolong the life of the filament of the bulb.

Question 14:
Explain why, tungsten is used for making the filaments of electric bulbs.

Solution :
Tungsten is used for making the filaments of electric bulbs because it has a very high melting point. Due to its very high melting point, the tungsten filament can be kept white hot without melting away. Also, tungsten has high flexibility and low rate of evaporation at high temperature.

Question 15:
Explain why, the current that makes the heater element very hot, only slightly warms the connecting wires leading to the heater.

Solution :
The connecting wires of the heater get only slightly warm because they have extremely low resistance due to which negligible heat is produced in them by passing current.

Question 16:
When a current of 4.0 A passes through a certain resistor for 10 minutes, 2.88 x 104 J of heat are produced. Calculate :
 (a) the power of the resistor.
 (b) the voltage across the resistor.

Solution : Given: I = 4amp. t = 10min = 10x60 = 600sec. H = 2.88x104.J 

(a) We have

H = I2RT

28800=42xRx600

R-3ohms

We know that

P = 12xR

= 42x3

P = 48W

(b) V=?

We know that

V = IR

V = 4x3

V = 12V 

Question 17:
A heating coil has a resistance of 200 Ω. At what rate will heat be produced in it when a current of 2.5 A flows through it ?

Solution :

Given: R = 200ohms, I = 2.5amp.t = lsec
We know that
H = I2RT
H = 2.52X200X1
H = 1250 J/s  

Question 18: An electric heater of resistance 8Ω, takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater.

Solution :

Given: R=8ohms. I = 15amp. t = 1sec

We know that
H =12RT
H = 152X8X1
H = 1800J/s 

Question 19:
A resistance of 25 Ω is connected to a 12 V battery. Calculate the heat energy in joules generated per minute.

Solution : Given: R = 25ohms. V = 12V. H = ?. t = 60sec

V= I R
12=25X1
I = 0.48amp
We have
H = I2RT
H = 0.482X25X60
H = 345.6J 

Question 20:
100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor ?

Solution : Given:H = 100J, t = 1sec, R = 4ohms.

We know that

H=I2RT

100 = I2X4X1

100/4 = I2

I = 5amp 

V = IR

V = 5X4

= 20V 

Question 21:
(a) Derive the expression for the heat produced due to a current ‘I’ flowing for a time interval ‘t’ through a resistor ‘R’ having a potential difference ‘V’ across its ends. With which name is this relation known ?
 (b) How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V ?
 (c) The current passing through a room heater has been halved. What will happen to the heat produced by it ?
 (d) What is meant by the heating effect of current ? Give two applications of the heating effect of current.
 (e) Name the material which is used for making the filaments of an electric bulb.

Solution :

(a) When an electric charge Q moves against a p.d. V, the amount of work done is givenn by

W = Qx V ---------(1) 
We know, current, I = Q/t
Q = I x t -----(2)
By Ohm's law, — V/I = R
V = IxR  (3)
Putting eqs. (2) and (3) in eq, (1),
W = I x t x 1x R
W = I2Rt

Assuming that all the electrical work done is converted into heat energy, we get

Heat produced, H = I2Rt joules

This relation is known as Joule's law of heating. 

(b) Given:P = 12W. V = 12V. t = 60sec

Pr = VI

I = P/V = 12/12 = 1A

V = IR

R = V/I = 12/1 = 12ohm

H = 12Rt

H = 12x12x60

H = 720J

(c) The heat produced by the heater will become one-fourth because heat produced is cirectly proportional to the square of the current.

(d) When an electric current is passed through a high resistance wire. the wire becomes very hot and produces heat. This effect is knows as heating effect of current. This effect is used in room heaters and electric ovens.

(e) Tungsten is used for making the filaments of an electric bulb. 

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of Class 10

Dynamic Test

Content Category

Related Searches

Sample Paper

,

Solutions of Electricity (Page No- 66) - Physics By Lakhmir Singh

,

shortcuts and tricks

,

video lectures

,

Class 10 Class 10 Notes | EduRev

,

ppt

,

Extra Questions

,

MCQs

,

study material

,

past year papers

,

Exam

,

Viva Questions

,

Summary

,

Important questions

,

Previous Year Questions with Solutions

,

Class 10 Class 10 Notes | EduRev

,

mock tests for examination

,

Class 10 Class 10 Notes | EduRev

,

Free

,

Solutions of Electricity (Page No- 66) - Physics By Lakhmir Singh

,

Objective type Questions

,

practice quizzes

,

pdf

,

Solutions of Electricity (Page No- 66) - Physics By Lakhmir Singh

,

Semester Notes

;