padikam.in - AI-Powered Learning for Kerala Syllabus

Q1.

What is the SI unit of heat energy produced in an electrical conductor?

[1 Mark]

(A)

Watt (W)

(B)

Joule (J)

✓ Correct

(C)

Ampere (A)

(D)

Volt (V)

Step-by-Step Solution & KaTeX Derivation:

Heat energy $H$ is measured in Joules (J) according to SI standards. Power is measured in Watts (W).

Q2.

According to Joule's Law, which equation represents heat energy $H$ produced by current $I$ through resistance $R$ in time $t$ ?

[1 Mark]

(A)

$H = V \times I$

(B)

$H = I^2 R t$

✓ Correct

(C)

$H = \frac{V}{R} t$

(D)

$H = P t^2$

Step-by-Step Solution & KaTeX Derivation:

Joule's Law of heating states $H = I^2 R t$ , where heat is directly proportional to square of current, resistance, and time.

Q3.

If the electric current flowing through a fixed resistor is doubled, how does the heat energy generated change?

[1 Mark]

(A)

Increases by 2 times

(B)

Increases by 4 times

✓ Correct

(C)

Decreases to half

(D)

Remains unchanged

Step-by-Step Solution & KaTeX Derivation:

Since $H \propto I^2$ , doubling current ( $2^2 = 4$ ) increases heat generation by 4 times.

Q4.

Which of the following is the correct formula for electric power $P$ in terms of voltage $V$ and resistance $R$ ?

[1 Mark]

(A)

$P = V \times R$

(B)

$P = \frac{V^2}{R}$

✓ Correct

(C)

$P = I^2 V$

(D)

$P = V R t$

Step-by-Step Solution & KaTeX Derivation:

Electric power $P = V I$ . Substituting $I = V / R$ gives $P = \frac{V^2}{R}$ .

Q5.

Why is a safety fuse wire constructed using a tin-lead alloy?

[1 Mark]

(A)

High melting point & zero resistance

(B)

Low melting point & appropriate resistance

✓ Correct

(C)

High boiling point & high conductivity

(D)

Low ductility

Step-by-Step Solution & KaTeX Derivation:

Fuse wire has a low melting point so that when excess current flows ( $I^2R$ heat builds up), it melts quickly to break the circuit.

Q6.

What is the commercial unit of electrical energy consumed in households?

[1 Mark]

(A)

Watt-second

(B)

Kilowatt-hour (kWh)

✓ Correct

(C)

Volt-Ampere

(D)

Ohm-metre

Step-by-Step Solution & KaTeX Derivation:

1 Kilowatt-hour (kWh) $= 3.6 \times 10^6\text{ Joules} = 1\text{ Unit}$ of electricity.

Q7.

An electrical appliance rated $230\text{ V}$ consumes $920\text{ W}$ power. What is the current drawn by the appliance?

[1 Mark]

(A)

$2\text{ A}$

(B)

$4\text{ A}$

✓ Correct

(C)

$6\text{ A}$

(D)

$10\text{ A}$

Step-by-Step Solution & KaTeX Derivation:

Using $P = V \times I \Rightarrow I = \frac{P}{V} = \frac{920}{230} = 4\text{ Amperes (A)}$ .

Q8.

How are high-power electrical appliances connected in domestic distribution wiring?

[1 Mark]

(A)

Series connection

(B)

Parallel connection

✓ Correct

(C)

Single line loop

(D)

Open circuit

Step-by-Step Solution & KaTeX Derivation:

Appliances are connected in parallel so each appliance receives full supply voltage ( $230\text{ V}$ ) independently.

Q9.

What is the primary function of earthing in household electrical wiring?

[1 Mark]

(A)

Increase electrical voltage

(B)

Prevent electric shock during insulation leakage

✓ Correct

(C)

Reduce electricity bill

(D)

Boost current frequency

Step-by-Step Solution & KaTeX Derivation:

Earthing connects the metal body of appliances to the ground, providing a safe low-resistance path for leakage currents.

Q10.

An electric heater of $100\ \Omega$ resistance operates on a $200\text{ V}$ line for $10\text{ s}$ . Calculate heat $H$ generated.

[1 Mark]

(A)

$2,000\text{ J}$

(B)

$4,000\text{ J}$

✓ Correct

(C)

$20,000\text{ J}$

(D)

$40,000\text{ J}$

Step-by-Step Solution & KaTeX Derivation:

Using $H = \frac{V^2}{R} t = \frac{200^2}{100} \times 10 = \frac{40,000}{100} \times 10 = 400 \times 10 = 4,000\text{ Joules (J)}$ .

Effects of Electric Current — Joule's Law & Power