[3 Mark Question Answer]

Question 13 Marks

Explain three electrical hazards.

Answer

The three electrical hazards are :
$1.$ Overloading : If too many devices connected to the same circuit are operated at the same time, the wires draw a much larger current from the mains than they can handle. This is called overloading. This causes heating in the circuit that can lead to sparking and electric fires.
$2.$ Short circuiting : Due to defective or loose connections or damaged wiring, the live and the neutral wires may sometimes touch each other. This results in a sudden large current flowing through the circuit. This is called short circuiting. The enormous heat produced due to a large current can damage electrical devices such as sound systems, bulbs, mixer grinders, televisions and so on, or even cause a fire.
$3.$ Electric shocks : Due to negligence if a person touches an exposed live wire in a circuit, he will receive a severe shock. Electric shocks can often be fatal.

View full question & answer→

Question 23 Marks

Describe the role of switches and electric meters in houses.

Answer

A switch$-$is a small device used to open or close a circuit. It is always connected to the live wire. The front cover of a switch is made of an insulated substance such as bakelite or ebonite, whereas the working portion inside the cover is made of highly conducting material such as brass or, copper. When the switch is in the on mode, the metal strip inside the switch connects to the live wire and the current flows through the circuit.
When the switch is in the off mode, the connection with the live wire is disrupted and no current flows through the circuit. This is how a switch helps in making or breaking a circuit. An electric meter records the consumption of electric energy directly in kilowatt$-$hour $\text{(kWh)}$ units.

View full question & answer→

Question 33 Marks

Describe the working of an electric fuse.

Answer

A fuse is essentially a thin wire made of an alloy $(60 \%$ copper, $40 \%$ lead$)$ with a melting point slightly below the melting point of the connecting wires. It is fixed between two clamps of copper mounted on a porcelain grip. When the grip is inserted in the casing, the fuse wire completes the circuit.
As soon as there is overheating caused due to any of the reasons such as overloading, fluctuations in current, or short circuiting, the fuse wire melts. This breaks the circuit and disrupts the flow of current in it, thus saving a device from getting damaged or preventing a fire. The fuse wire should be replaced to reuse the fuse.

View full question & answer→

Question 43 Marks

Five $\text{LED}$ bulbs of $8\ W$ each, two fans of $40 W$ each, an electric motor of $3\ kW$, and a geyser of $5\ kW$ are used for $8$ hours per day by a family. Calculate the total energy consumed in $30$ days and the cost of electricity used in one day at the rate of $₹.5$ per unit.

Answer

Total power $= 5 \times 80/1000\ kW + 2 \times 40/1000\ kW + 1 \times 3\ kW +1\times 5\ kW = 8.48\ kW$
Electrical energy consumed in $1$ day $= 8.48\ kW \times 8 h = 67.84\ kWh$
Electrical energy consumed in $30$ days $= 67.84\ kWh \times 30 = 2035.2\ kWh$
Cost of $1\ kWh$ of energy $= ₹.5$
Cost of energy used in $1$ day $= ₹.5 \times 67.84 = ₹. 339.2$

View full question & answer→

Question 53 Marks

A heater is rated at $5\ kW$. Calculate the cost of energy consumed by it in $30$ days if it is run for $2$ hours daily. The cost of $1\ kWh$ is priced at $₹.4.20.$

Answer

Power of a heater $= 5\ kW$
Electrical energy consumed by the heater in $1$ day $= 5\ kW \times 2 h = 10\ kWh$
Electrical energy consumed by the heater in $30$ days $= 10\ kWh \times 30 = 300\ kWh$
Cost of $1\ kWh$ of electrical energy $= ₹. 4.20$
Cost of $300\ kWh$ of electrical energy $= ₹. 4.20 \times 300 = ₹ .1260$
Hence, the cost of energy consumed by the heater in $30$ days is $₹ .1260.$

View full question & answer→

Question 63 Marks

An electric kettle rated at $2.2 \ kW$ works for $4$ hours a day. Calculate the number of units consumed by it in $20$ days.

Answer

Power of an electric kettle $= 2.2\ kW$
Electrical energy consumed by the kettle in $1$ day
$= 2.2\ kW \times 4 h = 8.8\ kWh$
Electrical energy consumed by the kettle in $20$ days
$= 8.8\ kWh \times 20 = 176\ kWh$

View full question & answer→

Question 73 Marks

A fan of $60 W$ runs for $75$ hours in a month. How much electrical energy is consumed?

Answer

Power of the fan $= 60W = 60/1000\ kW = 0.06\ kW$
Energy consumed by the fan in $75$ hours $= 0.06 W \times 75 h = 4.5\ kWh$
So, electrical energy consumed by a fan of $60 W$ in a month is $4.5\ kWh.$

View full question & answer→

PHYSICS [3 Mark Question Answer]

Test yourself on this topic