CBSE X Science Effects of Current - EducareHut (Next Gen Knowledge Point)

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#1. Which of the following gases are filled in electric bulbs?

Helium and Neon

Neon and Argon

Argon and Hydrogen

Argon and Nitrogen

#2. A boy records that 4000 joule of work is required to transfer 10 coulomb of charge between two points of a resistor of 50 Ω. The current passing through it is

2 A

4 A

8 A

16 A

Explanation: (c) Work done in transferring the charge

W= qV = qlR …….. (V = IR)

#3. The least resistance obtained by using 2 Ω, 4 Ω, 1 Ω and 100 Ω is

< 100 Ω

< 4 Ω

< 1 Ω

> 2 Ω

Explanation: (c) In parallel combination, the equivalent resistance is smaller than the least resistance used in the circuit.

#4. Two resistors of resistance 2 and 4 when connected to a battery will have

same current flowing through them when connected in parallel

same current flowing through them when connected in series

same potential difference across them when connected in series

different potential difference across them when connected in parallel

Explanation: In series combination of resistor, the current through both the resistor are same but potential difference across each will be different.

In parallel combination, current across each resistor will be different but the potential difference will be same.

#5. A coil in the heater consume power P on passing current. If it is cut into halves and joined in parallel, it will consume power

P/2

#6. 1 mV is equal to:

10 volt

1000 volt

10^-3 volt

10^-6 volt

#7. Electrical resistivity of any given metallic wire depends upon

its thickness

its shape

nature of the material

its length

#8. A wire of length /, made of material resistivity ρ is cut into two equal parts. The resistivity of the two parts are equal to,

ρ2

2 ρ

4 ρ

Explanation: (a) Resistivity of the material depends only on the nature of material not dimensions

#9. A fuse wire repeatedly gets burnt when used with a good heater. It is advised to use a fuse wire of

more length

less radius

less length

more radius

Explanation: (d) In order to get the working of heater properly, fused wire of higher rating must be used.

#10. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?

1/5 Ω

10 Ω

5 Ω

1 Ω

Explanation: The maximum resistance is obtained when resistors are connected in series combination.

Thus equivalent resistance obtained by connecting five resistors of resistance 1/5 Ω each, in series = (1/5 1/5 1/5 1/5) = 1 Ω

#11. The heating element of an electric iron is made up of:

copper

nichrome

aluminium

iron

#12. A cell, a resistor, a key, and an ammeter are arranged as shown in the circuit diagrams. The current recorded in the ammeter will be

maximum in (i)

maximum in (ii)

maximum in (iii)

same in all the cases

Explanation :(d) Ammeter is always connected in series with in the circuit. The reading is independent from its location.

#13. The resistance whose V-I graph is given below is

Explanation: (b) Resistance = slope line of V-I graph =

(9-0)/(15-0)=9/15=3/5

#14. Electric potential is a:

scalar quantity

vector quantity

neither scalar nor vector

sometimes scalar and sometimes vector

#15. The resistance of hot filament of the bulb is about 10 times the cold resistance. What will be the resistance of 100 W-220 V lamp, when not in use?

48 Ω

400 Ω

484 Ω

48.4 Ω

R= V^2/P

#16. To get 2 Ω resistance using only 6 Ω resistors, the number of them required is

Explanation: (b) Three resistors of 2 Ω is required to get 6 Ω because resultant is more than individual so they all must be connected in series.

#17. Unit of electric power may also be expressed as

volt ampere

kilowatt hour

watt second

joule second

Explanation:

Electric power = voltage × current

SI Unit of voltage = Volt

SI Unit of current = Ampere

#18. In an electrical circuit, two resistors of 2 and 4 respectively are connected in series to a 6V battery. The heat dissipated by the 4 resistor in 5s will be

10J

20J

30J

Explanation:

Here, first resistor, R1 = 2Ω

And second resistor, R2 = 4Ω

Voltage of cell, V = 6V

Time taken = t = 5s

Total resistance of the circuit = R = R1 R2 = 2 4 = 6 Ω

Current, I = V/R = 6/6 = 1A

Heat dissipated by the 4Ω resistor in 5s is given as,

H = I²Rt

⟹ H = 1 x 1 × 4 × 5 = 20J

#19. Which of the following represents voltage?

Work done /(Current x Time)

Word done x Charge

(Work done x Time) / Current

Work done x Charge x Time

#20. In an electrical circuit three incandescent bulbs. A, B and C of rating 40W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

Brightness of all the bulbs will be the same

Brightness of bulb A will be the maximum

Brightness of bulb B will be more than that of A

Brightness of bulb C will be less than that of B

Explanation: We know that power is defined as rate of doing work. A bulb consumes electric energy and produces heat and light. Now, bulb with more power rating will produce more heat and light or we can say that

Power rating of bulb is directly proportional to the brightness produced by bulb.

Therefore, brightness of bulb B with power rating 60 W will be more than the brightness of bulb A having power rating as 40W.

#21. Two devices are connected between two points say A and B in parallel. The physical quantity that will remain the same between the two points is

Current

Voltage

Resistance

None of these

Explanation:(b) In parallel combination, voltage remains same across two points.

#22. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly

10^20

10^16

10^18

10^23

Explanation: (a) Q = ne and Q = It

∴ ne = It

n=IT/e= (1×16)/(1.6×10^-19)

10²⁰

#23. An Electric kettle Consumes 1 KW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

1 A

2 A

4 A

5 A

Explanation: Here, power = P = 1 KW = 1000 W

Voltage = V = 220 V

Current = I = ?

Now, I = P/V = 1000/220 = 4.5 A

Now rating of fuse wire must be slightly greater than 4.5 A i.e., 5 A.

#24. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

100%

200%

300%

400%

If I is current and R is resistance then,

Power, P = I²R

Power in first case, P₁ = I²R

100% increase in current means that current becomes 2I

Power in second case, P₂ = (2I)²R = 4I²R

Now, increase in dissipated power = P₂– P₁ = 4I²R – I²R = 3I²R

Percentage increase in dissipated power = 3P₁/ P₁ × 100 = 300%

#25. Electrical resistivity of a given metallic wire depends upon

Its length

Its thickness

Its shape

Nature of the material

Explanation: The resistivity of a material is constant for a particular temperature at a constant temperature.
Resistivity of material does not depend on length, thickness and shape of the material. It only depends on the temperature.

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