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1151.

Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer»

We know that,

I=V/R

Since the resistance remains constant, so the current is directly proportional to the potential difference. If the potential difference is halved, the current also gets halved.

1152.

Which among iron and mercury is a better conductor of electricity ?

Answer»

Solution :

Iron is the best condoctor between iron and mercury

1153.

Which of the following is a good conductor of electricity.A. impure waterB. glassC. eboniteD. wood

Answer» Correct Answer - A
Among the given options impure water is the better conductor of electricity.
1154.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer»

Coils of electric toasters and electric irons are made of an alloy rather than a pure metal because
(i) the resistivity of an alloy is much higher than that of pure metal and
(ii) an alloy does not undergo oxidation easily even at high temperature.

1155.

Which material is the best conductor of electricity at room temperature ?

Answer» Silver is the best conducting material at room temperature as its resistivity is the lowest of all the materials.
1156.

Use the data in Table 12.2 to answer the following -(a) Which among iron and mercury is better conductor? (b) Which material is the best conductor?

Answer» (a) Resistivity of iron `= 10.0xx10^(-8)Omega`
Resistivity of mercury `=94.0xx10^(-8)Omega`
Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.
(b) It can observed form Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is best conductor.
1157.

Name the material which is the best conductor of electricity. 

Answer»

Solution : 

Silver metal.

1158.

Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer»

a) Resistivity of iron = 10.0 × 10-8

Resistivity of Mercury = 94.0 × 10-8

Resistivity of an alloy is greater than iron. By this we conclude that, Iron is good conductor of heat comparing to Mercury.

b) Resistivity of silver is less, hence it is a good conductor of heat.

1159.

Which material is the best conductor?

Answer»

We know that a good conductor of electricity should have a low resistivity and a poor conductor of electricity will have a high resistivity. Silver has the lowest resistivity of 1.60 x 10-8 Ω m, which means that silver offers the least resistance to the flow of current through it. So, silver is the best conductor of electricity.

1160.

Judge the equivalent resistance when the following are connected in parallel (i) 1Ω and 106 Ω), (ii) 1Ω and 103 Ω and 106 Ω.

Answer»

(i) When a number of  resistance are connected in parallel, then their combined resistance is less than the smallest individual resistance. Therefore, equivalent resistance will be less than 1Ω
(ii) In this case, also the equivalent resistance will be less than 1Ω

1161.

When a `4Omega` resistor is connected across the terminals of a 12V battery, the number of coulombs passing through the resistor per second is:A. 0.3B. 3C. 4D. 12

Answer» Correct Answer - B
1162.

A resistance of 20 ohms has a current of 2 amperes flowing in it. What potential difference is there between its ends ?

Answer»

Solution : 

R = 20ohms 

I = 2amp 

We know that 

V = IR 

Thus, V = 2 x 20 

V = 40V

1163.

A constant voltage is applied between two ends of a uniform conducting wire. If both the length and radius of the wire is doubled then what happens to the heat produced in the wire?

Answer»

We know that resistance of a conducting wire is 

R = \(\frac{pl}{A}\) = \(\frac{Pl}{πr^2}\)

If length l and radius r are doubled, then resistance will become half. But heat produced H = \(\frac{V^2}{R}.\)

Hence, heat produced per second will become thrice.

1164.

A potential difference of 20 volts is applied across the ends of a resistance of 5 ohms. What current will flow in the resistance ?

Answer»

Solution :

Potential difference, V = 20V 

Resistance, R = 5ohms Current, I = ? 

We know that 

V=IR 20 = I x 5 I = 20/5 = 4 A 

1165.

When a potential difference of `2 V` is applied across the ends of a wire of `5 m` length, a current of `1 A` is found to flow through it. Calculate (i) the resistance per unit length of the wire, (ii) the resistance of `2 m` length of the wire. (iii) the resistance across the ends of the wire if it doubled on itself.

Answer» Correct Answer - (i) `0.4 Omega//m`
(ii) `0.8 Omega`
(iii) `0.5 Omega`.
(i) `R = (V)/(I) = (2 V)/(1 A) = (2 V)/(1 A) = 2 Omega`. Resistance//unit length `= (2 Omega)/(5 m) = 0.4 Omega//m`
(ii) Resistance of `2 m` length of the wire `= 0.4 xx2 = 0.8 Omega`
(iii) When the wire is doubled on itself, it will be equivalent to two resistances (each of value `1 Omega`) parallel to each other. The resultant resistance across the ends of this folded wire `= (1 Omega)/(2) = 0.5 Omega`.
1166.

Important Class 10 Science MCQ Questions of Electricity with Answers?

Answer»

Class 10 Science MCQ questions of Electricity with Answers were arranged based on the most recent pattern of exams. We have Given Electricity Class 10 Science MCQ Questions with Answers to assist students with understanding the idea well overall.

Practice Class 10 Science MCQ Questions of Electricity

1. 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

(a) current
(b) voltage
(c) resistance
(d) None of these

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

(a) 2
(b) 3
(c) 4
(d) 6

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

(a) more length
(b) less radius
(c) less length
(d) more radius

4. Keeping the potential difference constant, the resistance of the circuit is halved. The current will become:

(a) One-fourth
(b) Four times
(c) Half
(d) Double

5. When a current 'I' flows through a resistance 'R' for time 't' the electrical energy spent is given by

(a) \(IRt\)
(b) \(I^2Rt\)
(c) \(IR^2t\)
(d) \(\frac{I^2R}{t}\)

6. If the current flowing through a fixed resistor is halved, the heat produced in it will become:

(a) One-fourth
(b) One-half
(c) Double
(d) Four times

7. The resistivity does not change if

(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed

8. Unit of electric power may also be expressed as

(a) volt ampere
(b) kilowatt hour
(c) watt second
(d) Joule second

9. Electrical resistivity of a given metallic wire depends upon

(a) its length
(b) its thickness
(c) its shape
(d) nature of the material

10. The unit of e.m.f. of a cell is

(a) dyne
(b) volt
(c) ampere
(d) joule

11. Kilowatt hour is the unit of

(a) power
(b) energy
(c) impulse
(d) force

12. The current flow in conductor is due to:

(a) free electrons
(b) protons
(c) ions
(d) neutrons

13. Unit of resistance is:

(a) Ohm
(b) Ohm/cm
(c) Ohm-cm
(d) Volt

14. Ohm’s law is true for:

(a) only metallic conductors
(b) only non-metallic conductors
(c) only semi-conductors
(d) all of these.

15. Electric potential is a:

(a) scalar quantity
(b) vector quantity
(c) neither scalar nor vector
(d) sometimes scalar and sometimes vector

16. 1 mV is equal to:

(a) 10 volt
(b) 1000 volt
(c) \(10^{-3}\) volt
(d) \(10^{-6}\) volt

17. Coulomb is the SI unit of:

(a) charge
(b) current
(c) potential difference
(d) resistance

18. Materials which allow larger currents to flow through them are called

(a) insulators
(b) conductors
(c) semiconductors
(d) alloys

19. Conventionally, the direction of the current is taken as

(a) the direction of flow of negative charge
(b) the direction of flow of atoms
(c) the direction of flow of molecules
(d) the direction of flow of positive charge

20. The unit of specific resistance is

(a) ohm
(b) ohm
(c) ohm-metre
(d) ohm per metre

Answer key

1. (b) voltage

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

2. (b) 3

Explanation: 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.

3. (d) more radius

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

4. (d) Double

5. (b) \(I^2Rt\)

6. (a) One-fourth

7. (c) the shape of the resistor is changed

Explanation: The resistivity does not change if the shape of resistor is changed because nature of material will remain same.

8. (a) volt ampere

9. (d) nature of the material

10. (b) volt

11. (b) energy

12. (a) free electrons

13. (a) Ohm

14. (a) only metallic conductors

15. (a) scalar quantity

16. (c) \(10^{-3}\) volt

17. (a) charge

18. (b) conductors

19. (d) the direction of flow of positive charge

20. (c) ohm-metre

Click here to practice more MCQ Questions from Chapter Electricity Class 10 Science

1167.

Write true or false for the statementOur body can pass electricity through it.

Answer»

True Our body can pass electricity through it.

1168.

Write true or false for the statementAll metals are insulators of electricity.

Answer»

False All metals are insulators of electricity.

1169.

Write true or false for the statementThe earth wire protects us from an electric shock.

Answer»

True The earth wire protects us from an electric shock.

1170.

Three incandescent bulbs of 100W each are connected in series in an electrical circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.

Answer»

(a) No.
Reasons: Let R be the resistance of each bulb, then total resistance in series = 3R In series, current in each bulb is same.
So current drawn by each bulb connected in series is one third as compared to the current in each bulb in a parallel arrangement, so the bulbs connected in parallel combination glow more brightly.
(b) In a series arrangement, if one bulb is fused; then current in the bulbs connected in series will become zero, so bulbs will stop glowing. In a parallel arrangement, if one bulb is fused, the other two bulbs will continue to glow with same brightness.

1171.

A conducting sphere A of radius a, with charge Q, is placed concentrically inside a conducting shell B of radius b. B is earthed. C is the common centre of A and B.(a) The field at a distance r from C, where a ≤ r ≤ b, is k Q/r2.(b) The potential at a distance r from C, where a ≤ r ≤ b, is k Q/r.(c) The potential difference between A and B is kQ (1/a - 1/b).(d) The potential at a distance r from C, where a ≤ r ≤ b, is kQ (1/r - 1/b).

Answer»

Correct Answer is: (a, c, & d)

1172.

A spherical conductor A of radius r is placed concentrically inside a conducting shell B of radius R (R < r). A charge Q is given to A, and then A is joined to B by a metal wire. The charge flowing from A to B will be(a) Q(R/R + 1)(b) Q(R/ + r(c) Q(d) zero

Answer»

Correct Answer is: (c) Q

Method 1: When the two are joined by a metal wire, they become a single conductor. As charge can reside only on the outer surface of a conductor, the entire charge Q must flow to the outer sphere. 

Method 2: When charge of amount q has flown from A to B, the charge on A is (Q - q).

VA = k Q-q/r +k q/R

VB = k Q-q/R

∴ VA - VB = k(Q - q) (1/r - 1/R) > 0

∴ VA > VB for all values of q.

Charge will flow from A to B till q = Q.