Physics numerical in Hindi:- chapter - Electricity(Electric Current)
Q.1. The resistance of a wire is 40Ω. If the length of the wire is doubled, find the resistance of the new wire.
A) 140
B) 60
C) 80
D) 160Ω
ANSWER= D) 160Ω
Explain:- 1. Trick solution :-
(n)² × R
=(2)²× 40
= 160Ω
Q. 2. The resistance of a wire is 40Ω. The length of that wire has been increased to 5 times without any change in volume. Find the resistance of the new wire.
A) 1000Ω
B) 600
C) 800
D) 1600Ω
ANSWER= A) 1000Ω
Explain:- 1. Trick solution :-
(n)² × R
= (5)² × 40
= 1000Ω
Q. 3. Resistance of a wire is RΩ. If the length of this wire is stretched n times. The resistance of the new wire will be :-
A) nR
B) n²R
C) R/ n
D) n /R
ANSWER= B) n²R
Explain:-
Q.4. The resistance of a wire is 4Ω whose length is L. What will be the resistance of 2L length of this wire?
A) 16Ω
B) 8Ω
C) 32Ω
D) 6Ω
ANSWER= B) 8Ω
Explain:- R∝ L
R∝ 1/a
This means that the more times the length is increased, the more times the resistance will increase.
Hence ans = 2×4
= 8Ω
Q.5. The resistance of a wire is RΩ. This wire is sliced in the middle. What will be the resistance of each piece of wire?
A) R/4
B) 2R
C) R/2
D) 4R
ANSWER= C) R/2
Explain:-
Q. 6. The length of a wire of resistance R is doubled by stretching. What will be the resistance of the new wire?
A) 8R
B) 2R
C) R/2
D) 4R
ANSWER= D) 4R
Explain:- 1. Trick solution :-
(n)²×R
=(2)²×R
=4R
Q. 7. The length of a wire of resistance 6Ω is stretched 5 times. The resistance of the new wire will be :-
A) 169
B) 200
C) 150
D) 120
ANSWER= C) 150
Explain:- Trick solution :-
(5)²×6 = 25×6
= 150
Q.8. An electric bulb is marked with 100 W and 200 V. Find the resistance of the bulb.
A) 200Ω
B) 300Ω
C) 150Ω
D) 400Ω
ANSWER= D) 400Ω
Explain:- (200)²÷ 100= 40000÷ 100
= 400Ω
{Formula : R = V²/P}
Q.9. An electric bulb is marked 50W and 100V. Find the resistance of the bulb.
A) 200Ω
B) 30Ω
C) 150Ω
D) 40Ω
ANSWER= A) 200Ω
Explain:- (100)² 50 = 10000÷ 50
= 200Ω
{Formula :- R = V²/P}
Q. 10. What will be the ratio of the resistance of an electric bulb of 40 W and 60 W?
A) 1/2
B) 1/3
C) 2/3
D) 3/2
ANSWER= D) 3/2
Explain:- Resistance of 40W bulb = V²/40
Resistance of 60 W bulb = V²/60
Required ratio = (V²/40)÷ (V²/60)
= 60÷40
=3/2
Q.11. Electric bulbs of 50W and 100W have been used in homes. Which one will have more resistance?
A) of 50 W.
B) of 100 W.
C) Equal of both.
D) cannot be determined.
ANSWER= A) of 50W.
Explain:- Resistance of 50W bulb = V²÷50
Resistance of 100W bulb = V²÷ 100
V²÷50 > V² 100
Q. 12. An electric fan is marked 880 W 220 V. What is the resistance of the electric fan?
A) 50Ω
B) 60Ω
C) 55
D) 40Ω
ANSWER= C) 55Ω
Explain:- ( 220 ×220)÷ 880 = 220÷ 4
= 55Ω
Q. 13. A resistance of 3Ω is connected across a 9V battery. How much current will flow through the resistance? :-
A) 5 A
B) 3 A
C) 9 A
D) 4 A
ANSWER= B) 3A
Explain:- V = iR
→i = V÷R
= 9÷3
= 3A
Q.14. 0.5 coulomb of charge per second flows through an electric filament. Find the number of electrons flowing per second. :-
A) 2.35 × 10¹²
B) 13.17 × 10¹⁹
C) 6.25 × 10¹⁸
D) 3.125 × 10¹⁸
ANSWER= D) 3.125 × 10¹⁸
Explain:- Number of electrons in 1 coulomb = 6.25 × 10¹⁸
Hence number of electrons in 0.5 coulomb charge = 0.5 × 6.25 × 10¹⁸
= 3.125 × 10¹⁸
Q.15. If the amount of electric current is 300 mA, then find the total number of electrons that will flow in 1 minute:-
A) 2.35 × 10¹²
B) 13.17 × 10¹⁹
C) 1.125 × 10²⁰
D) 3.125 × 10¹⁸
ANSWER= C) 1.125 × 10²⁰
Explain:- Current = 300mA
= 300÷1000A
= .30 A
time = 1 minute
= 60 seconds
charge = current × time
= .30 × 60
= 18 coulombs
Now number of electrons in 1 coulomb = 6.25 × 10¹⁸
Number of electrons in 18 coulombs = 18 ×6.25 ×10¹⁸
= 112.5 × 10¹⁸
= 1.125 × 10²⁰
Q.16. If 1.25A current flows in an electric lamp. Find the number of electrons that flow through that lamp every day. :-
A) 6.75 × 10²³
B) 7.92 × 10³²
C) 5.43 × 10¹²⁰
D) 3.21 × 10²⁰
ANSWER= A) 6.75 × 10²³
Explain:- Current = 1.25A
Time = 24 × 3600 seconds
=86400 seconds
charge = current × time
= 1.25 × 86400
= 108000 coulombs
Number of electrons in 1 coulomb = 6.25 × 10¹⁸
Number of electrons in 10800 coulombs
= 108000 × 6.25 ×10¹⁸
= 6.75 × 10²³
Q.17. The starter motor of a car draws a current of 300A from a 12 volt battery. If the car starts in 2 minutes, then how much energy is consumed by the motor in 2 minutes of the battery:-
A) 400 kj
B) 375 kj
C) 500 kj
D) 432 kj
ANSWER= D) 432 kj
Explain:- {Formula:- Electric energy = v×i×t}
time = 2 minutes
= 2 × 60 seconds
= 120 seconds
v×i×t = 12 ×300×120
= 432000 joules
= 432 kilojoules
Q.18. A 5A current electric fuse can bear a maximum of 1 watt of electrical power. Find the resistance of the fuse. :-
A) 0.06Ω
B) 0.04Ω
C) 0.05Ω
D) 0.03Ω
ANSWER= B) 0.04Ω
Explain:- {Formula :- P =V×i}
V= P÷i
= 1÷5
= .20 volts
Now, R = v/I
= .20/5
= 0.04Ω
Q.19. The resistance of an electric fan is 10 . If the amount of current is 10 amperes, then how much energy will be consumed in 10 minutes?:-
A) 6 × 10³ joules
B) 5 × 10³ joules
C) 3.5 × 10⁶ joules
D) 6 × 10⁵ joules
ANSWER= D) 6 × 10⁵ joules
Explain:- {Formula:- Electric energy = i²×R×t}
t = 10 minutes
= 10 × 60
= 600 seconds
Electric energy = 10² × 10 × 600
= 6 × 10⁵ joules
Q.20. An electric lamp is marked 60 W - 230 V. If the cost of 1 KW electric energy is Rs 1.25, then how much will it cost to use this lamp for 8 hours?:-
A) Rs 0.20
B) Rs 1.0
C) Rs.0.60
D) Rs 1.50
ANSWER= C) Rs.0.60
Explain:- Electric energy = Electric power × Time
= 60 × 8 × 3600 joules
Now, 3.6 × 10⁵ Joule = 1 KW
1 joule = {1÷(3.6 ×10⁵)} KW
60×8×3600 joules = {(60×8×3600)÷(3.6×10⁵)} KW
= .48 KW
Now 1KW = Rs.1.25
0.48 KW = 0.48×1.25
= Rs.0.60
Q.21. A boiler with a capacity of 2 KW is used for 1 hour each day. How many units of electricity will it consume in 30 days? :-
A) 60 units
B) 50 units
C) 30 units
D) 120 units
ANSWER= A) 60 units
Explain:- 1 KW h = 1 unit
2KWh is being used per day i.e. 2 units are being used.
60 units of electricity will be consumed in 30 days.
Q.22. A 2 ampere current flowing in a conductor produces 80 joules of heat in 10 seconds. Find the amount of resistance. :-
A) 2.5
B) 1.5
C) 2
D) 0.75
ANSWER= C) 2
Explain:- H = i²×R×t
R = H÷ i²t
= 80 2²× 10
= 2Ω
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