File No.25/07/19/12/2014
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
ANDHRA PRADESH - TELANGANA
2014-2015 PROGRAM M E
4TH - CHAPTER- SOLUTION
ELECTRICITY
1Sol. An energy which makes our life comfortable by doing a large number of jobs is electric energy
2Sol. A device in an electric circuit which can turn on/off the electric circuit is called an electric switch
3Sol. Electric generator is a big source of electric current.
4Sol. The overhead electric cables used for the transmission of electric current are made of Aluminium.
5Sol. Polarisation in simple voltaic cell can be covercome by using chemical means.
6Sol. Dry cells is an improvement on wet le-clanche cell.
7Sol. The negative terminal of a bichromate cell is amalgamated zinc rod
8Sol. The material used to make an element of electric heater is nichrome
9Sol. Transformer is an source of electric current.
10Sol. When electric current is passed through acidulated water, it splits into hydrogen and oxgen. This effect
of electric current is called chemical effect.
11Sol. In a metallic conductor, electric current thought to be due to the movement of electrons.
12Sol. In a simple voltaic cell, the positive plate consists of copper.
13Sol. The major defects of the simple voltaic cell are called local action & polarisation.
14Sol. The chemical substances used for eliminating polarisation in simple voltaic cell are called oxidising agents
& reducing agents
15Sol. When zinc plate is rubbed with mercury2014
the pure
- zinc
2015
metal floats up on the surface of mercury and
impurities in zinc settle in the layer of mercury.
16Sol. The positive terminal of simple le-clanche cell consists of Carbon.
17Sol. The liquid used in bichromate cell is potassium bichromate solution acidified with sulphuric acid.
18Sol. Electric generator source of electric current.
19Sol. The overhead electric cables used for the transmission of electric current are made of aluminium.
20Sol. Human body is a conductor of electricity .
1Sol. Here, Q = 0.8C, t = 8 s.
Q
t
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Using, I =
we get, I =
0.8C
8s = 0.1 A
2Sol. Here, I = 0.2 A, t = 30 s
Using, I =
Q
, we get
t
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PHYSICS - CHAPTER SOLUTIONS - 4
File No.25/07/19/12/2014
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Q = I x t = 0.2 x 30 = 6 C
If n electrons flow and e is the charge on an electron, then
Q = ne or n =
Q
6
 3.75x1019
=
19
e 1.6x10
3Sol. Here, I = 0.25 A, t = 2 hr = 2 x 60 x 60 s
Using
I=
Q
,
t
We get, Q = I x t = 0.25 x 2 x 60 x 60 = 1800 C
4Sol. Here, I = 0.4 A, t = 3 hr = 3 x 60 x 60 = 10800 s
Q
, we get
t
Using I =
Q = I x t = 0.4 x 10800 = 4320 C
5Sol. Here, n = 6.25 x 1018 electrons
Charge on each electron, e = 1.6 x 10-19 C
 Total charge flowing through the conductor, Q = ne = 6.25 x 1018 x 1.6 x 10-19 = 1 C
Time, t = 5 s
Q
, we get
t
Using I =
1C
= 0.2 A
5s
I=
6Sol. Here, I = 0.2 A; t = 1 hr = 60 x 60 = 3600 s
Q
, we get
t
Using I =
Q = It = 0.2 x 3600 = 720 C
7Sol. Here, I = 80  A = 80 x 10-6 A = 8 x 10-5 A
2014
(i) t = 1 second
Using, I =
- 2015
q
ne
=
, we get
t
t
5
It 8 x 10 A x 1s
n=
=
( e = 1.6 x 10-19 C)
1.6 x 1019 C 
e
 5 x 1014 electrons strike the screen every second.
(ii) I =
q
t
 q = It = 8 x 10-5 A x 2 minutes = 8 x 10-5 A x 2 x 60 seconds
= 960 x 10-5 C = 9600 x 10-6 C = 9600  C
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8Sol. Here, I = 0.1 A
(i) Using, I =
n=
PATHFINDER
ne
, we get
t
0.1A x 1s
It
=
= 6.25 x 1017 electron
1.6 x 1019 C
e
2
PHYSICS - CHAPTER SOLUTIONS - 4
File No.25/07/19/12/2014
(ii) Using I =
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
q
, we get
t
q = I t = 0.1 A x 1 hr = 0.1 A x 3600 s = 360 C.
9Sol. Given
W = 56.5 joules
Q = 50 coulombs
V=?
Potential difference V =
V=
Work done (W )
Ch arge transferred (Q )
56.5 J
50C = 1.13 J/C, V = 1.13 Volts
10Sol. Potential at the point A, V1=
Potential at the point B, V2 =
W
12
=
=4V
Q
3
W
16
=
= 5.3 V
Q
3
 Potential difference between the two points A and B.
=(V2 - V1) = (5.3 V - 4 V) = 1.3 V.
11Sol. Here, Q = 96000 C, t = 1 h = 60 x 60 = 3600 s, V = 50 V
Heat produced, W = QV = 96000 C x 50 V = 48 x 105 J.
12Sol :t = 0.8s, Q = 24 Coulomb, I = ?
We have,
Q=Ixt
I=
Q
24
=
= 30 amp
t
0.8
13Sol. Q = 1.5 Coulomb, W = 9 J, V =?
We have,
W = QV
V=
2014 - 2015
W
9
90
=
=
=6V
Q
1.5
15
14Sol. V = 9 volt, Q = 4 Coulomb, W = ? W
We have,
V=
W
Q
W = QV = 9 x 4 = 36 Joule
15Sol. V = 250 volt, W = 125 joule, Q = ?
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We have,
V=
W
Q
Q=
W
V
Q=
125
250
125 Q = 250 = 0.5 Coulomb
16Sol. W = 48 joule, Q = 6 Coulomb, V = ?
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PHYSICS - CHAPTER SOLUTIONS - 4
File No.25/07/19/12/2014
We have
We have i =
19.
V=
48
6
Q
t
Q=ixt
Q=2x3
Q = 6 Coulomb
i = 20 amp, t = 1/2 minute, Q = ?
(a) We have, . t = 1/2 minute = 30 s
Q=ixt
Q = 600 Coulomb
(b) e = 1.6 x 10-19 Coulomb, Q = 600 Coulomb, n = ?
We have, Q = ne
n=
20.
W
Q
V = 8 volt
i = 2 amp, t = 3 s, Q = ?
17.
18.
V=
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Q
e
n=
600
1.6 X 1019
n = 375 X 1019
n = 3.75 X 1021 electrons
e = 1.6 X 10-19C, Q = 1 Coulomb, n = ?
We have
n=
Q
e
n=
1.0
1.6 X 1019
n = 0.625 X 1019
2014 n = 6.25 X 1018 electrons Ans.
Number of electrons = number of protons
Number of protons = 6.25 X 1018
2015
BRAIN TWISTERS
1.
2.
3.
4.
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5.
The defects in voltaic cell are local action polarisation.
A dry cell consists of zncl2, NH4Cl, Manganese dioxide and Carbon.
A leclanche cell consists of carbon rod, Porous pot, powdered carbon and anganese dioxide
The Italian scientist voltair , invented the first electric cell
The dry cell is a modified form of a lechlanche cell
In an Bichromate cell the zine rod acts as a negative pole
The path along which electric current flows is called electric circuit
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PHYSICS - CHAPTER SOLUTIONS - 4
File No.25/07/19/12/2014
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
Coating of a conductor with a non-conductor is called insulator
The materials which allow the electric current to pass through them are called conductors
6.
7.
8.
9.
A) Potential difference
B) Metals
C) Semiconductor
D) Superconductor
a) Current
b) Charge
Volt
Conductors
Silicon
Volt
Ampere
Coulomb
c) Potential
Volt
d) Work
Joule
a) Electrolyte in voltaic cell
b) Electrolyte in lechlanche cell
c) Electrolyte in Bichromate cell
d) Electrolyte in a cell made of
a) Bulbs in series
b) Bulbs in parallel
Dilute H2SO4
Ammonium chloride solution
Dilute H2SO4 + Potassium Bi - chromate
Lime juice 4) Dilute H2SO4
E=E1 + E2 + E3, Failure of any bulb leads to a break in the circuit
Brightness of each bulb remains the same,
even when one bulb is removed
Volt
Volt
c) Unit of emf
d) Unit of potential difference
10.
11.
12.
13.
14.
15.
The metal to be purified by eletrolysis is connected to the anode of the voltameter, and the metallic ions
discharge at the cathode of the voltaic cell.
Both are correct but Reason doesnot explain Assertion.
MNO2 is a depolariser in le-clanche cell, because MNO2 is goud reducing gent.
Both are correct and Reason explains Assertion.
The positive charge is considered to be at higher potential and a negative charge at a lower potential
The earth is considered to be at zero potential.
2014 - 2015
Both are correct but, Reason doesnot explain Assertion.
A primary cell converts chemical energy into electrical energy
Assertion is correct but Reason is in correct.
If a voltage V is applied across the bulbs connected in series ,then the voltage across each bulb remains the
same
If a voltage V is applied across the bulbs connected in parallel ,then the voltage across each bulb remains
the same.
Assertion is in correct but Reason is correct.
q = it, q = 0.5C, t = 5 sec.
q 0.5

 0.1Amp
t
5
i = 1.5 amp
t = 2 sec.
q = it = 1.5  2 = 3 coloumb
i = 1 amp,
t = 1 sec, n = ?
i
16.
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17.
it
11
 6.25 1018
q = it = ne  n = 
e 1.6 10 19
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PHYSICS - CHAPTER SOLUTIONS - 4
File No.25/07/19/12/2014
18.
q = 60,
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
t = 4 sec.
q 60

 15 Amp
t 4
Four electgronic cells each of emf 1 V connected in parallel, then the combined emf is 1 V
There are 2 kinds of electric charge.
i=
19.
20.
1.
2.
Positive electrification of a body is due to the deficiency of electrons
q = 5C
d = 0.5
w = 10 J
p. d = ?
w 10
p.d = q  5  2 Volts
3.
i = 4 amp
n=?
t = 1 sec.
q  ne  it  n 
w it
4 1
 
 2.5 1019
q e 1.6 10 19
= 25 × 1018
4.
q = 25 Coulombs
w = 75 joule
w 75
p.d = q  25  3 J / coloumb
5.
2014 - 2015
= 3 Volts
In an Bichromate cell the glass vessel containing a mixture of solutions are K2Cr2o7 + H2SO4
3V
4V
6.
5V
7.
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8.
3V, 4V, 5V are connected in parallel, then the total emf of the cobination is equal to maximum value of emf
present in circuit So, 5V
1V , 1.5V , and 2V are connected in series.
then total emf = 1 + 1.5 + 2 = 4.5 Volt.
IV, 1.5V, 2V are connected in parallel then
Total emf = maximum voltage = 2V.
Given q = 5C of charge moves from A to B
WA = 20J, WB = 32J
WA
20
then, Potential at A = q  5  4 Volts
Potential B =
PATHFINDER
WB 32

 6.4 Volts
q
5
6
PHYSICS - CHAPTER SOLUTIONS - 4
File No.25/07/19/12/2014
9.
Current =
=
10.
no.of electrons  charge of electron
time
109 1.6 1019  4
 1.6  4  6.4 Amp to wards left
1
i = 3 mA = 3  10–3 Amp
t = 1 min = 60 sec.
q = ne = it  n = n 
11.
12.
13.
VII CLASS - IIT/N.T.S.E FOUNDATION - OLYMPIAD
p.d = VB – VA = 6.4 - 4 = 2.4 Volts.
Given = 1019  particles
1019 protons, and
1019 electrons moves right per second t = 1 sec.
310 3  60
it
 11.251018
=
1.6 1019
e
When ther is an electric current through a conducting wire along its length, then an electic field must exist inside
the wire but parallel to it
When a motor car is started its light becomes slightly dim because of starter draws high current
i = 1 mA
n=?
it 1103 1

 6.25 1019
e 1.6 10 19
14.
15.
16.
17.
Over loading of an electric circuit implies drawing of large current
The material for electric fuse Tin-lead alloy.
The heating elkement of an electric heater should be made with a material, which should have high specific
resistance anf high melting point
The electric bulbs have tunsten filaments of same lenght. If one of them given 60 watts and the other 100 watts
then 100 watts bulbs has thicker filament
5
18.
1
bulb 1 glows brightly than remaining bulbs.
3
4
2014 - 2015
2
A
B
19.
C
Brightness of A increases but that of B decreases
S
20.
If 60 columb of charge passes through a cross section of a conductor in 4 sec,
q
60

 15 Amp .
t
4
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i
PATHFINDER
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PHYSICS - CHAPTER SOLUTIONS - 4