Part E Mock Test — Electricity & Magnetism

Answer: C

Explanation

Hydrogen bubbles collecting on the copper plate cause polarization, which increases the internal resistance and reduces the cell’s output.

Local action is different: it occurs mainly on the zinc plate due to impurities in zinc reacting with the electrolyte even when the circuit is not supplying current. Therefore, local action is not a consequence of hydrogen bubbles covering the copper plate.

Answer: A

Explanation:

As induced voltage is produced by a varying magnetic field. The induced voltage is proportional to the rate of change of magnetic flux linking the coil.

When an AC current flows through a wire, it generates a magnetic field that changes direction with the current. As the current alternates, the magnetic field expands and contracts, inducing an electromotive force (EMF) in any nearby conductor or coil of wire.

Answer: D

Explanation:

Given:

• Voltage, V = 260 V
• Power, P = 80 W
• Current, I =?

Recall that:

P = IV

80 = I × 260

I = \(\Large \frac{80}{260} \\ = \Large \frac{4}{13}\)

I = 0.31 A

Answer: D

Explanation:

Given:

• Capacitance, C = 0.5 µF
• Voltage, V = 10 V
• Charge, Q = ?

Recall that,

Q = CV;

Q = 10 × 0.5 × 10^-6 C

Q = 5 × 10^-6 C

Q = 5 µC

Answer: C

Explanation:

F =  BIL

F = \(\normalsize 0.4 \; \times \; 6 \; \times \; \Large \frac{15}{100}\)

F = \(\Large \frac{36}{100}\)

F = 0.36 N

Answer: C

Explanation:

For a coil (series R–L) on a.c, the impedance is

⇒ \(\normalsize Z=\frac{V}{I}\)

Given:

⇒ \(\normalsize V=100\text{ V (r.m.s)},\quad I=2\text{ A (r.m.s)},\quad R=30\,\Omega\)

1) Impedance: \(\normalsize Z= \Large \frac{100}{2} \\\normalsize = 50\,\Omega\)

2) For a series R–L circuit: \(\normalsize Z^2 = R^2 + X_L^2\)

So;

⇒ \(\normalsize X_L=\sqrt{Z^2-R^2}=\sqrt{50^2-30^2} \\ \normalsize =\sqrt{2500-900} \\ \normalsize =\sqrt{1600}\\ \normalsize =40\,\Omega\)

3) Inductive reactance: \(\normalsize X_L=\omega L\)

With the source frequency stated as \(\normalsize 50\pi\),

the matching option is obtained by using

⇒ \(\normalsize \omega = 100\,\text{rad s}^{-1}\)

equivalently \(\normalsize f= \Large \frac{50}{\pi}\normalsize \,\text{Hz}\) 

Hence,

⇒ \(\normalsize L= \Large \frac{X_L}{\omega} \\ = \Large \frac{40}{100}\\ = \normalsize 0.40\ \text{H}\)

Therefore, the inductance is \(\normalsize 0.40\,\text{H}\).

Answer: C

Solution

Total resistance in the circuit = external resistance + internal resistance

R_T = 2 Ω + 4 Ω = 6 Ω

Current:

⇒ \(\normalsize I= \Large \frac{E}{R_T}\\ = \Large\frac{12}{6}\)

⇒ \(\normalsize I=2\ \text{A}\)

Answer: B

Explanation:

In a step-up transformer, the electromotive force (EMF) in the secondary winding is indeed increasing compared to the primary winding, meaning the voltage on the secondary side is higher than the voltage on the primary side; this is the key characteristic of a step-up transformer.

Answer: B

Explanation:

Each cell has e.m.f. = 2.0 V. There are 8 cells connected in series.

If all were connected correctly:

• Total e.m.f. = 8 × 2.0 V = 16 V

But the question says two cells are wrongly connected (i.e., reversed), so each wrongly connected cell opposes the others. A reversed cell contributes −2.0 V instead of +2.0 V.

So the net e.m.f. is:

• Correctly connected cells = 6 cells → 6 × 2.0 V = 12 V
• Two reversed cells together contribute: (−2.0 V) + (−2.0 V) = −4 V
• Net e.m.f. = 12 V + (−4 V) = 8 V

Answer: 8 V

Answer: A

Explanation:

It is a step-up transformer because the number of turns in the secondary is higher than that of the primary.

Answer: B

Explanation: A dielectric increases capacitance \(C\). For the same applied voltage, energy stored \(U=\tfrac{1}{2}CV^{2}\) increases.

Answer: A

Explanation

When two inductors are arranged in series, the total inductance is the sum of individual inductors.

4 H + 8 H = 12 H

Energy =  \(\Large \frac{L\:I^2}{2}\)

Energy =  \(\Large \frac{12\: \times \: 10^2}{2}\)

Energy =  \(\Large \frac{12\: \times \: 100}{2}\)

Energy =  \(\Large \frac{1200}{2}\)

= 600 J

Answer: D

Solution

Electric potential is work done in taking a unit mass or charge from infinity to that point

Electric potential = \(\Large \frac{electric\:force \: \times \: distance \: of \: separation}{charge}\)

But Electric force = \(\Large \frac{q^2}{4 \pi ε_o r^2}\)

∴ Electric potential = \(\Large \frac{\frac{q^2}{4 \pi ε_o r^2 } \normalsize \: \times \: r}{q}\)

= \(\Large\frac{ \frac{q^2}{4 \pi ε_o r}}{q}\)

= \(\Large \frac{q^2}{4 \pi ε_o r} \: \times \: \frac{1}{q}\)

= \(\Large \frac{q}{4 \pi ε_o r}\)

Answer: B

Explanation:

Capacitive reactance (X_C) measures the opposition to alternating current flow and is inversely proportional to both frequency and capacitance. The standard formula is:

X_C = \(\Large \frac{1}{2 \pi f C}\)

Where:

• f = 50 Hz
• C = 16 × 10^-6 F

∴ X_C = \(\Large \frac{1}{2 \: \times \: 3.142 \: \times 50 \: \times 16 \: \times \: 10^{-6}}\)

∴ X_C = \(\Large \frac{1}{0.0050265} \\ = \normalsize 198.9436 \: \Omega \\ \normalsize \approx 200 \: \Omega\)

Answer: A

Solution

First step: find z from data in the question

z = \(\Large \frac{M}{It} \\ = \Large \frac{1.8}{2.5 \: \times \: \left( 3\:\times \: 60\: \times \: 60\right)}\\ = \Large \frac{1.8}{27000} \\ = \normalsize 6.67 \: \times \: 10^{-5}\)

Therefore, the new M = zIt

M = \(\normalsize 6.67 \: \times \: 10^{-5} \: \times \: 4 \: \times \: \left ( 4.8\: \times \: 60 \: \times \: 60\right) \\ = \normalsize 6.67 \: \times \: 10^{-5} \: \times \: 4 \: \times \:17280\\ = \normalsize 4.56 \\ \normalsize \approx 4.6\: g\)

Answer: B

Explanation

Faraday's law of electromagnetic induction, also known as Faraday's law, is the basic law of electromagnetism which helps us predict how a magnetic field would interact with an electric circuit to produce an electromotive force (EMF). This phenomenon is known as electromagnetic induction.

Answer: B

Solution

In the case of a parallel configuration, each resistor has the same potential drop across it, and the currents through each resistor may be different

∴  \(\normalsize I = \Large \frac {V}{R}\)

⇒ \(\normalsize I = \Large \frac {12}{12}\)

= 1A

Answer: D

Explanation

Larger plates provide greater capacity to store electric charge. Therefore, as the area of the plates increase, capacitance increases.

Answer: A

Explanation:

The capacitors are connected in series

For capacitors in series, the effective capacitance is given as

⇒ \(\Large \frac{1}{C_T} = \frac{1}{C_1} \; + \; \frac{1}{C_2} \; + \; \frac{1}{C_3} \; + \; \frac{1}{C_4}......\frac{1}{C_n}\)

⇒ \(\Large \frac{1}{C_T} = \frac{1}{3} \; + \; \frac{1}{3} \; + \; \frac{1}{3}\)

⇒ \(\Large \frac{1}{C_T} = \frac{3}{3}\)

C_T = 1 µF

Answer: A

Explanation: A potentiometer measures e.m.f. without drawing current, making it suitable for accurate determination of internal resistance of a cell.

Answer: D

Solution

Peak value current, \(\normalsize I_o = \sqrt{2} A\)

Root-mean square value of current, \(\normalsize I_{rms} = \Large \frac{I_o}{\sqrt{2}}\)

⇒ \(\normalsize I_{rms} = \Large \frac{10}{\sqrt{2}}\)

⇒ \(\normalsize I_{rms} = 7.07\:A \\ \normalsize \approx 7.1\: A\)

Answer: C

Explanation:

Parameters given:

Inductance, L = 2 H

Frequency, f = \(\Large\frac{50}{\pi}\: \normalsize Hz\)

Reactance on the indctor

(Inductive reactance), X_L = Unknown

Recall; X_L = 2πfl

X_L= Inductive Reactance

X_L = \(\normalsize 2\pi \: \times \: \Large \frac{50}{\pi}\normalsize \: \times \: 2 \\ \normalsize = 200\: \Omega\)

Answer: C

Explanation

For resistor X:

⇒ \(\normalsize P_X = \Large \frac{V^2}{R_X}\)

Given:

• P_X = 10 W
• R_X = 4 Ω

⇒ \(\normalsize 10 = \Large \frac{V^2}{4}\)

⇒ \(\normalsize V^2 = 10 \times 4 \\ = \normalsize 40 \,V^2\)

For resistor Y (6 Ω):

⇒ \(\normalsize P_Y = \Large \frac{V^2}{R_Y} \\ = \Large \frac{40}{6} \\ = \normalsize 6.67 \: W \\ \normalsize \approx 6.7 \: W\)

Answer: C

Explanation:

Electrical work equals gain in kinetic energy:

⇒ \(\normalsize eV= \Large \frac{1}{2} \normalsize m v^{2}\).

Hence, \(\normalsize V= \Large \frac{m v^{2}}{2e} \\ =\Large\frac{(9.1\times 10^{-31})(4\times 10^{7})^{2}}{2(1.6\times 10^{-19})}\\ = \normalsize 4.55\times 10^{3} \: V\).

Answer: B

Explanation:

When a negatively charged rod is brought near the cap of a charged gold leaf electroscope, which has positive charges, the charges will neutralize one another and the leaf will collapse.

Answer: C

Explanation:

In a a Leclanché cell, also known as a zinc-carbon battery, is powered by chemical energy, as it generates electricity through a chemical reaction between the zinc anode and manganese dioxide cathode within the cell, converting stored chemical energy into electrical energy when used.

Answer: C

Explanation:

A current-carrying conductor placed in a magnetic field experiences a force whose direction is given by Fleming’s left-hand rule.

In the diagram:

• Magnetic field is from the north pole to the south pole (N → S), i.e. upward.
• The current in the conductor at X is perpendicular to the magnetic field

Applying Fleming’s left-hand rule:

⇒ \(\normalsize \text{Force} \perp \text{Current} \perp \text{Magnetic field}\)

The thumb (force direction) points horizontally toward P.

Hence, the force on the conductor is from M to P.

Therefore, the correct direction of force is MP.

Answer: D

Explanation:

⇒ \(\Large \frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2}\)

C_T = \(\Large \frac {C_1 \; \times \; C_2}{C_1 \; + \; C_2}\)

C_T = \(\Large \frac {3 \; \times \; 6}{3 \; + \; 6}\)

C_T = \(\Large \frac {18}{9}\) = 2 μF

Q = CV

⇒ 2 × 10⁻⁶ × 400

⇒ 800 × 10⁻⁶ C

Q = 8 × 10⁻⁴ C

Answer: B

Explanation:

Please note that in a series connection, the effective voltage across cells equals the simple addition of each cell voltage. So, for these 6 identical cells of 2 V each.

The effective e.m.f = 2 V + 2 V + 2 V + 2 V + 2 V + 2 V = 12 V

Answer: B

Explanation:

Recall: Q = CV

• Q = charge
• C = Capacitance
• V = Voltage/Potential difference

Q_e = C_e × V --------------------- *

Q_e = (5 + 10)20

Q_e = 300 μC

" I hope you know why C_e = (5 + 10)? " This is because the capacitors are in parallel.
Recall: If two or more resistors are arranged in parallel, they have the same voltage but different charges.

V = \(\Large \frac{Q_{e}}{C_{e}}\)

⇒ \(\normalsize Q_{e}=Q_{1}\: + \: Q_{2}\)

⇒ \(\normalsize V = V_{1}=V_{2}\)

For the 5 μf capacitor

20 = \(\Large \frac{Q_{1}}{5\: \times \: 10^{-6}}\)

⇒ \(\normalsize Q_{1} = 100 \: \times \: 10^{-6} = 10^{-4}C \;or\; 100\: \mu C\)

For the 10 μf capacitor

20 = \(\Large \frac{Q_{2}}{10\: \times \: 10^{-6}}\)

⇒ \(\normalsize Q_{2} = 200 \: \times \: 10^{-6} = 2 \: \times \: 10^{-4}C \;or\; 200\: \mu C\)

∴ the charges stored in P and Q, respectively, are 100 µC and 200 µC

Answer: B

Solution

Resistors are connected in parallel.

⇒ \(\Large \frac{1}{R_{total}} = \frac{1}{R_1} \: + \: \frac{1}{R_2}\)

⇒ \(\Large \frac{1}{R_{total}} = \frac{1}{3}\: + \: \frac{1}{6}\)

⇒ \(\Large \frac{1}{R_{total}} = \frac{2 \: +\: 1}{6}\)

⇒ \(\Large \frac{1}{R_{total}} = \frac{3}{6}\)

⇒ \(\Large \frac{1}{R_{total}} = \frac{1}{2}\)

∴ \(\normalsize R_{total} = 2 \: \Omega\)

To calculate E.M.F use the formula:

⇒ \(\normalsize E = I(R\:+\:r)\)

r is negligible, therefore = 0

⇒ \(\normalsize E = 4(2\:+\:0)\)

⇒ \(\normalsize E = 4\: \times \: 2\)

⇒ \(\normalsize E = 8 \: V\)

Answer: B

Explanation:

The electroplating process uses an anode and a cathode. In electroplating, the metal dissolved from the anode can be plated onto the cathode. The anode is provided with direct current, oxidizing and dissolving its metal atoms in the electrolyte solution.

Answer: B

Explanation

r = resistance of the galvanometer

I = current through galvanometer = 10 mA or 0.01 A

V₁ = p.d across the galvanometer = I × r = 0.01 × r

V₁ = 0.01 r

V₂ = p.d across the multiplier = 5 - 0.01 r

R = resistance of the multiplier = 498 Ω

where R = \(\Large \frac{V}{I}\)

498 = \(\Large \frac{5\:-\:0.01r}{0.01}\)

cross multiply

498 × 0.01 = 5 - 0.01r

0.01r = 5 - 4.98

0.01r = 0.02

r = \(\Large \frac{0.02}{0.01}\)

r = 2 Ω

Answer: A

Explanation

A capacitor is used in the primary circuit of an induction coil. This is because when the circuit is broken, a high induced voltage is used to charge a capacitor to avoid sparks.

Answer: C

Explanation: A body is positively charged when it loses electrons (deficit). It is negatively charged when it gains electrons (excess). In metals, current is due to electron drift. Hence all are correct.

Answer: A

Explanation

In a home (domestic) wiring system, electrical appliances and lamps are connected in parallel so that:

• Each appliance receives the full mains voltage (its rated voltage), so it works efficiently.
• Each appliance can be controlled independently with its own switch.
• If one appliance fails or is switched off, the others continue to work, because each branch is separate.

Key idea: In a parallel circuit, the potential difference across each branch is the same as the supply voltage.

Therefore, the correct reason is: “Each appliance gets its rated voltage.”

Answer: A

Explanation:

I = 10sinωt

I = I₀sinωt

∴ I₀ = 10 A

⇒ \(\normalsize I_{r.m.s} = \Large \frac{I_o}{\sqrt{2}}\)

⇒ \(\normalsize I_{r.m.s} = \Large \frac{10}{\sqrt{2}}\)

⇒ \(\normalsize I_{r.m.s} = \Large \frac{10}{\sqrt{2}} \: \times \: \Large \frac{\sqrt{2}}{\sqrt{2}}\)

⇒ \(\normalsize I_{r.m.s} = \Large \frac{10\sqrt{2}}{\sqrt{2} \: \times \: \sqrt{2}}\)

⇒ \(\normalsize I_{r.m.s} = \Large \frac{10\sqrt{2}}{2}\)

⇒ \(\normalsize I_{r.m.s} = 5\sqrt{2}\: A\)

Answer: B

Explanation:

Given:

• Current, I = 4.6 A
• Resistance, R = 6.5 Ω
• Voltage, V = V

Recall that

V = IR

V = 6.5 × 4.6 = 29.9 V

Answer: D

Explanation:

The sensitivity of the galvanometer can be increased either by increasing the number of turns (N), using strong magnet or increasing the area of the coil

Answer: B

Solution

Number of tunes, n = 500

Area, A = 0.2 m × 0.2 m = 0.04 m²

change of flux density A B = 1.0 T - 0.4 T = 0.6 T

time, t = 5 s

Induced e.m.f, E = \(\Large\frac{NAB}{t}\)

Induced e.m.f, E = \(\Large \frac{500 \: \times \: 0.04 \: \times \: 0.6}{5}\)

E = 4 × 0.6 = 2.4 V