Class |
10th |

Subject |
Science (NCERT) |

Category |
Important Questions |

**Electricity Class 10 Science Chapter 11 Important Question Answer**

**Q1. The potential difference between the terminals of an electric heater is 80 V, when it draws a current of 5 A from the source. What current will the heater draw, if the potential difference is increased to 160 V?**

**Ans – **Using Ohm’s law, V = IR

In first instance, V = 80 volts, I = 5 A

Therefore, R = V/I = 80/5= 16 Ω

In second instance, V = 160 volts, R (calculated above) = 16 Ω

Therefore, I = V/R = 160/16 = 10 A.

Thus, the heater will draw a current of 10 Ampere, if the potential difference is increased to 160 V.

**Q2. An electric motor takes 4 A from a 220 V line. Find the power of the motor and energy consumed in 3 hours.**

**Ans – **Given, V = 220 V, I = 4 A,

t = 3 hour = 3×60×60 sec = 10800 sec

Power of motor P = VI

= 220V × 4 A

= 880 watt.

Energy consumed by motor in 3 hours will be

E = P × t

= 880 × 10800 joule

= 9.504 × 10^{6} j

**Q3. Why are the coils of electric irons and toasters are made of an alloy rather than a pure metal ? Most Important**

**Ans – **Because an alloy has a higher resistance than a pure metal. The alloys also do not melt easily at high temperature.

**Q4. 150 Joule of heat is produced each second in a 6Ω resistance. Find the potential difference across the resistor.**

**Ans – **Given, Heat energy H = 150 j, R = 6Ω, t = 1 sec

Using H = P × t

150 = P × 1

P = 150 watt

Now, using P =

150 =

V^{2} = 900

V = 30 V

Therefore, required potential difference is 30 V.

**Q5. A 6 Ω resistance emits heat energy at the rate of 125 J / s. Find the potential difference across the resistor.**

**Ans – **Given R = 6 Ω, P = 125 J/s, V = ?

Using, P =

125 =

V^{2} = 750

V = V = 27.39 V

**Q6. How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to give a total resistance of 1Ω ? Most Important**

**Ans – **If we connect all three resistors in parallel, we get

R_{e} = 1Ω

and we also need net resistance 1Ω

Therefore, we have to connect all given resistors in parallel.

**Q7. A lamp of resistance 20 Ω and a conductor of resistance 8 Ω are connected with a battery of 8 V in series. Calculate the : **

**(a) total resistance in the circuit**

**(b) current flowing through the circuit**

**Ans – **Given R_{1} = 20Ω , R_{2} = 8Ω, V = 8V

(a) Total resistance in series = R_{1 }+ R_{2} = 20Ω + 8Ω = 28Ω

(b) current flowing I = ?

Using V = IR

8 V = I × 28Ω

I = A

**Q8. An electric refrigerator of power 400 W is allowed to run 10 hrs. per day. What is the cost of energy to operate it for 30 days at Rs. 4.00 per kWh? Most Important**

**Ans – **The total energy consumed by the refrigerator in 30 days would be

400 W × 8.0 hour/day × 30 days = 120000 Wh = 120 kWh

Thus, the cost of energy to operate the refrigerator for 30 days is

120 kwh × Rs. 4.00 per kWh = Rs. 480

**Q9. On what factors does the resistance of a conductor depend? Most Important**

**Ans – **(i) The resistance of a conductor depends directly on its length

(ii) Depends inversely on its area of cross-section,

(iii) Also on the nature of material of the conductor

**Q10. What do you mean by earthing? Why should electrical appliances be earthed ? Most Important**

**Ans – **The third is the earth wire that has green insulation and this is connected to a metallic body deep inside earth. This process is called earthing.

It is used as a safety measure to ensure that any leakage of current to a metallic body does not give any severe shock to a user.

**Q11. Why is tungsten used almost exclusively for filament of electric lamps?**

**Ans – **Because it has a very high melting point due to which it does not melt even when it is heated to high temperatures due to the passage of electric current.

**Q12. An electric current of 0.5 A flows through the filament of an electric bulb for 5 min. What will be the electric charge flowing through that wire ?**

**Ans – **Given, I = 0.5 A, t = 5 min = 300s

Using, Q = It

= 0.5 A × 300s = 150 C

**Q13. An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?**

**Ans – **Given V = 220V, I = 0.50 A

Using, P = VI

P = 220 V × 0.50 A

P = 110 J/s = 110 W

**Q14. When a 12 V battery is connected across an unknown resistor there is a current of 2.5mA in the circuit. Find the value of resistance of the resistor.**

**Ans – **Given V = 12V, I = 2.5 mA = 2.5 × 10^{-3} A

Using V = IR

12 = 2.5 × 10^{-3} × R

R = 12/(2.5 × 10^{-3})

R = 4800 Ω

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