The comprehensive approach in SCERT Class 10 Physics Solutions Chapter 5 Electric Energy: Consumption and Conservation Important Questions with Answers ensure conceptual clarity.
SSLC Physics Chapter 5 Important Questions Kerala Syllabus
Electric Energy: Consumption and Conservation Class 10 Important Questions
Question 1.
If intensity of electric current in a circuit is doubled the heat energy developed in this circuit increases _______________ times.
(2,12,4,14)
Answer:
4
Hint: Heat produced is directly proportional to the square of current. If current is doubled, heat energy will be increased to four times.
Question 2.
Which among the following is the energy transformation taking place when a storage battery is being charged?
a) chemical energy to electrical energy.
b) electrical energy to heat energy.
c) electrical energy to chemical energy.
d) magnetic energy to electrical energy.
Answer:
c) electrical energy to chemical energy
Question 3.
Some statements related to activities to be done to reduce carbon footprint is given
i. Increase domestic energy consumption.
ii Avoid wasting food.
iii. Use public transport.
iv. Reduce the use of reusable products.
Which are the correct statements related to it?
A. i and ii only
B. iv only
C. All are correct
D. ii and iii only
Answer:
D. ii and iii only
Question 4.
Some statements related to electrical power is given.
i. Electric power is measured in joules.
ii 1 kilowatt-hour is equal to 1000 joules.
iii Power is the quantity of work done by an electrical appliance per unit time.
iv. Power, P = \(\frac{V}{I}\)
Which are the correct statements related to it.
a) i and ii only
b) iii and iv only
c) iv only
d) iii only
Answer:
d) iii only
Question 5.
Statement: A heater A operating at 230 V draws 2 A current, its power obtained is 460 W.
Reason: The power can be calculated as the, product of voltage and current.
a) Statement and reason are correct; the reason explains the statement.
b) Statement and reason are correct; however, the reason does not explain the statement.
c) both the statement and the reason are incorrect.
d) Statement is incorrect; reason is correct
Answer:
a) Statement and reason are correct; the reason explains the statement.
Question 6.
The heat generated in a current carrying conductor can be explained by a famous law.
a) Write the name of this law.
b) Write the mathematical equation for this law, explain each letters used in the equation.
Answer:
a) Joule’s law
b) H = I2Rt or H ∝ I2 Rt
H – Heat generated in a conductor
I – Intensity of electric current
R – Resistance of the conductor
t – time for which current flows
Question 7.
Resistance of a 230 V heating device is 460 Ω. Calculate the heat energy produced by it in 10 minutes.
Answer:
H = \(\frac{V^2 t}{R}\)t, t = 10 × 60 s, R = 460 Ω
= \(\frac{230^2}{460}\) × 600 = 69000 J
H = 69000J
Question 8.
a) Name the part of a heating equipment in which the electric energy is converted into heat energy.
b) Name the substance used to make this part.
Answer:
a) Heating coil
b) Nichrome
Question 9.
In a house, 5 CF lamps each of 20 W work for 4 hours, and 4 fans each of 60 W work for 5 hours in a day. What will be the daily consumption of energy shown by the watt-hour meter?
Answer:
Electric energy (in kilowatt hour) = \(\frac{\text { Power in watt × time in hour }}{1000}\)
Energy consumed by 5 CF lamps = \(\frac{5 \times 20 \times 4}{1000}\) = 0.4 kWh
Energy consumed by 4 fans = \(\frac{4 \times 60 \times 5}{1000}\) = 1.2 kWh
Daily consumption of energy
= 0.4 + 1.2 = 1.6 kWh
= 1.6 units
Question 10.
How many units of energy will be consumed by a 100 W bulb in 30 days, if it works 10 hours everyday?
Answer:
Electric energy (in kilowatt hour) = \(\frac{\text { Power in watt × time in hour }}{1000}\)
Energy consumed by bulb in a day
= \(\frac{1 \times 100 \times 10}{1000}\)
= 1 unit
Energy consumed by bulb in a month
= 1 × 30 = 30 units
Question 11.
The government is promoting the installation of solar panels by giving incentives to boost electricity generation. Explain its significance in the context of the energy crisis.
Answer:
The over use of non-renewable fossil fules like coal, oil, and natural gas which contribute to pollution and global warming has resulted in an energy crisis for the entire world.
Because it is abundant, renewable, environmentally friendly, and accessible even in remote locations, solar energy is an excellent substitute. Solar thermal power plants use the sun’s heat to produce electricity, whereas solar cells directly turn sunlight into electrical power. In everyday life, gadgets like solar water heaters and stoves can help cut down on the amount of power and fossil fuels-used. We can lessen the effects of the energy crisis, conserve resources, and safeguard the environment by utilising more solar energy.
Question 12.
100W bulb works on 230V supply.
a) If voltage is halved, how much will be the power?
b) If voltage decreases to 1/4 times, how much will be the power?
c) If the voltage is doubled what happen to the bulb?
Answer:
a) R = \(\frac{\mathrm{V}^2}{\mathrm{P}}\) = \(\frac{230^2}{100}\) = 529
V = 115 V (voltage is halved)
P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) = \(\frac{115^2}{100}\) = 25 W
(When voltage is halved, power becomes 1/4 times the original power)
P = 25 W
b) V = 57.5 V (Voltage decreases to 1/4 times)
P = \(\frac{57.5^2}{529}\)
P = 6.25 W
c) Device will be damaged due to high voltage.
Question 13.
Fill in the blanks of the table suitably.
| Device | Energy change | Effect of electricity |
| Electric bulb | (a) | Lighting effect |
| (b) | Electrical energy is converted to heat energy | (c) |
| Mixie | (d) | Mechanical effect |
| Electric kettle | (e) | (f) |
Answer:
(a) – Electrical energy,
(b) – Electric Iron,
(c) – Heating effect,
(d) – Electric energy to mechanical energy,
(e) – electrical energy to heat energy,
(f) – heating effect.
Question 14.
A heating coil with 60 Ω resistance is connected to a 240V supply.
a) Calculate the power of the appliance.
b) Calculate the amount of heat generated by this heating coil in 5 minutes.
c) If this appliance continuously work for 10 hours, calculate the energy consumed in commercial units.
Answer:
a) Power of the appliance
P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) = \(\frac{240^2}{60}\) = 960 W
b) t = 5 min = 5 × 60 = 300 s
Heat generated = \(\frac{\mathrm{V}^2}{\mathrm{t}}\)
H = 960 × 300 = 288000 J
c) Electric energy (in kilowatt hour)
= \(\frac{\text { Power in watt } \mathrm{x} \text { time in hour }}{1000}\)
Electric energy = \(\frac{1 \times 960 \times 10}{1000}\) = 9.6 units
Question 15.
Energy conservation is very important for a better tomorrow.
a) What do you mean by energy crisis?
b) How do people contribute to energy crisis (any two)
c) Write two actions to reduce your daily electrical energy consumption, (any two)
Answer:
a) Energy crisis is the increase in demand for energy and the decrease in its availability.
b)
- Careless use of electricity at homes, transport, and industries
- Over dependence on fossil fuels like coal, oil, and natural gas
- Not making proper use of natural energy resources.
- Slow action and lack of awareness about saving energy
c)
- Make a transition to energy-efficient appliances.
- Use LED Bulbs-LEDs use less power and last longer than normal bulbs.
- Turn off electric appliances when not in use- Always switch off fans, lights, and appliances when you leave a room.
- Use natural light during the day
Question 16.
An electric heater has a resistance of 690 Ω and is designed to operate at 230 V.
(a) What are the characteristics of a good heating coil?
(b) Calculate the heat energy produced by this heater when it works for 30 minutes.
Answer:
Resistance, R = 690 Ω,
time, t = 30 minute = 30 × 60 = 1800 s
Voltage, V = 230 V
(a) Characteristics of a good heating coil are
- High oxidation resistance
- Ability to remain in red hot condition for a long time.
- High melting point
- High resistivity
(b) Heat generated,
H = \(\frac{V^2 t}{R}\) = \(\frac{230^2 \times 1800}{690}\) = 138000 J
Question 17.
An electric heater of 1000 Ω works on a 230 V supply
a) Write down the energy change taking place in the electric heater.
b) State the law to find the quantity of heat generated in an electric heater?
c) Calculate the electrical energy consumed when the heater works for three hours.
Answer:
a) Electrical energy is converted to heat energy
b) The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I2), the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law.
c) V = 230V, R = 1000 Ω, t = 3 hour = 3 × 60 × 60 = 10800 s
H = \(\frac{V^2 t}{R}\) = \(\frac{230^2 \times 10800}{1000}\) = 571320J
Question 18.
Observe the figure.
a) Write down the name of the device.
b) What is it used for?
c) In a house 5 LED lamps each of 20 W works for 5 hours and one laptop of 50 W works for 2 hours daily.. Calculate electrical energy consumption in one month in commercial units.
Answer:
a) Watt-hour meter
b) The quantity of electric energy consumed in houses can be measured directly by a watt hour meter.
c) Electric energy (in kilowatt hour)
= \(\frac{\text { Power in watt × time in hour }}{1000}\)
Electric energy consumed by lamps
= \(\frac{5 \times 20 \times 5}{1000}\) = 0.5 units
Electric energy consumed by laptop
= \(\frac{1 \times 50 \times 2}{1000}\) = 0.1 units
Total energy consumption in a day
= 0.5 + 0.1 = 0.6 units
Total energy consumption in a month
= 0.6 × 30 = 18 units
Question 19.
The figure shows two heating coils connected in an electric circuit.
a) Find the current through each coil.
b) If current flows through the circuit for 5 minutes, which coil gets heated more? Find the heat
developed in that coil.
Answer:
a) V = 230 V, R = 115 Ω
Current through 115 Ω coil
I = \(\frac{V}{R}\) = \(\frac{230}{115}\) = 2 A
V = 230 V, R = 46 Ω
Current through 46 Ω coil
I = \(\frac{V}{R}\) = \(\frac{230}{46}\) = 5 A
b) The coil having 46 Ω. resistance gets heated up more (Even though V (voltage), t (time of current flow) is same in both coils, 46 Ω coil has less resistance, So more current flows through it and more heat energy is produced)
Heat energy produced in 46 Ω coil.
Question 20.
2 A current is drawn by a heating coil when 230 V potential difference is applied.
a) Which material is commonly used to make a heating coil?
b) What is the quantity of charge that flows through this coil in 5 minutes?
c) What is the resistance of the coil?
a) Find the current through each coil.
Answer:
a) Nichrome is used to the bating coil.
b) Charge (Q) = I × t = 2 × 300 = 600 C
c) Resistance (R) = \(\frac{V}{R}\) = \(\frac{230}{2}\) = 115 Ω