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Solutions of Electricity (Page No- 67) - Physics By Lakhmir Singh, Class 10 | Extra Documents, Videos & Tests for Class 10 PDF Download

Lakhmir Singh Physics Class 10 Solutions Page No:67

Question 31:
The electrical resistivities of four materials P, Q, R and S are given below :

Solutions of Electricity (Page No- 67) - Physics By Lakhmir Singh, Class 10 | Extra Documents, Videos & Tests for Class 10

Which material will you use for making : (a) heating element of electric iron (b) connecting wires of electric iron (c) covering of connecting wires ? Give reason for your choice in each case.
Solution :
(a) S; because it has high resistivity of 11/10000000 ohm-m (it is actually nichrome).
(b) Q; because it has very low resistivity of 1.7/100000000 ohm-m (it is actually copper).
(c) R; because it has very very high resistivity of 1.0X100000000000000 ohm-m (it is actually rubber).

Question 32:
(a) How does the wire in the filament of a light bulb behave differently to the other wires in the circuit when the current flows ?
 (b) What property of the filament wire accounts for this difference ?

Solution :
(a) The filament wire becomes white hot where as other wires in the circuit do not get heated much.
(b) High resistance of filament wire accounts for this difference.

Question 33:
Two exactly similar heating resistances are connected (i) in series, and (ii) in parallel, in two different circuits, one by one. If the same current is passed through both the combinations, is more heat obtained per minute when they are connected in series or when they are connected in parallel ? Give reason for your answer.

Solution :
In series, because total resistance in series connection is more than that in parallel connection.

Question 34:
An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at ‘minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Calculate the current and resistance in each case.

Solution :

Given: V = 220V. Pmin=360W. Pmax = 840W 
For minimum heating case:
We know that
Pmin = VI
360 = 220X1
I = 1.63amp
R = V/I
R = 22011.63
R =134.96ohms 

For maximum heating case:
We know that
Pmax = VI
840 = 220X1
I = 3.81amp
R = V/I
R = 220/3.81
R = 57.74ohms 

Question 35:
Which electric heating devices in your home do you think have resistors which control the flow of electricity ?

Solution :
Electric iron , electric oven, water heater, room heater.

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FAQs on Solutions of Electricity (Page No- 67) - Physics By Lakhmir Singh, Class 10 - Extra Documents, Videos & Tests for Class 10

1. What are the different solutions of electricity?
Ans. The different solutions of electricity include using renewable energy sources such as solar power, wind power, hydroelectric power, and geothermal power. Additionally, energy conservation and efficiency measures can also be considered as solutions to reduce electricity consumption.
2. How can solar power be a solution to electricity?
Ans. Solar power is a solution to electricity as it harnesses the energy from the sun and converts it into electricity using solar panels. This renewable energy source is abundant, sustainable, and does not produce harmful emissions, making it an environmentally-friendly solution to meet our electricity needs.
3. What is the significance of wind power as a solution to electricity?
Ans. Wind power is a solution to electricity as it utilizes the kinetic energy from the wind to generate electricity through wind turbines. It is a clean and renewable energy source that can be harnessed in various locations, making it a reliable and sustainable solution for electricity generation.
4. How does hydroelectric power contribute to the solutions of electricity?
Ans. Hydroelectric power is a solution to electricity as it uses the force of flowing or falling water to generate electricity. It is a renewable energy source that produces electricity without emitting greenhouse gases. Hydroelectric power plants can be built on rivers or dams, providing a consistent and reliable source of electricity.
5. How can energy conservation and efficiency be considered as solutions to electricity?
Ans. Energy conservation and efficiency can be considered as solutions to electricity by reducing electricity consumption and optimizing energy usage. This can be achieved through practices such as using energy-efficient appliances, insulating homes, and promoting energy-saving habits. By conserving and using energy efficiently, we can reduce the demand for electricity and decrease our carbon footprint.
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