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

Results

#1. 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.

#2. 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

#3. Calculate the current flows through the 10 Ω resistor in the following circuit

1.2 A

0.6 A

0.2 A

2.0 A

Explanation: (b) In parallel, potential difference across each resistor will remain same. So, current through 10-Ω resistor

I = V/R=6/10 = 0.6 A

#4. 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.

#5. Identify the circuit in which the electrical components have been properly connected.

iii

Explanation: Essential conditions are necessary when electrical components are connected

Voltmeter should be connected in parallel.
Ammeter is always connected in series.
Positive terminals of voltmeter and ammeter should be connected to positive terminal of the cell and their negative terminals should be joined to the negative terminal of the cell.

Thus, the above conditions are satisfied in case (ii).

#6. 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

#7. 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.

#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. 1 kWh = ……….. J

3.6 × 10^-6 J

13.6 × 10^6 J

3.6 × 10^6 J

13.6 × 10^-6 J

#10. The proper representation of series combination of cells obtaining maximum potential is

iii

Explanation: Maximum potential is obtained when cells are connected in series such that, negative terminal of the cell is connected to the positive terminal of the second cell and so on, as shown in the following diagram.

#11. 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

#12. Which of the following is not correctly matched?

None

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

copper

nichrome

aluminium

iron

#15. The resistivity of insulators is of the order of

10^-8 Ω-m

10^1 Ω-m

10^-6 Ω-m

10^6 Ω-m

#16. Two wires of same length and area, made of two materials of resistivity ρ1 and ρ2 are connected in parallel V to a source of potential. The equivalent resistivity for the same length and area is

Explanation: (b) Equivalent resistance in parallel combination is

1/R_P =1/R₁ 1/R₂

For the same length and area of cross-section, R ∝ p (resistivity)

#17. 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

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

10^20

10^16

10^18

10^23

Current i= 1 A, time t=16s
Nuber of electrons=n
i=Q/t=ne/t
n=it/e=(1X16)/ (1.6*10^-19)=10^20

#19. Two wires of same length and area made of two materials of resistivity ρ1 and ρ2 are connected in series to a source of potential V. The equivalent resistivity for the same area is

Explanation: (a) For same length and area of cross-section R ∝ p.

For series combination, equivalent resistance is

Rs = R₁ R₂

⇒ Ps = ρ₁ ρ₂

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

Maximum in (i)

Maximum in (ii)

Maximum in (iii)

The same in all the cases

Explanation: In series connections the order of elements in the circuit will not affect the amount of current flowing in the circuit.

#21. The temperature of a conductor is increased. The graph best showing the variation of its resistance is

Explanation: (a) Resistance is directly proportional to temperature of the conductor.

#22. 1 mV is equal to:

10 volt

1000 volt

10^-3 volt

10^-6 volt

#23. 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.

#24. 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

#25. 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.

Finish