Properties of a magnet Practice Questions - NDA Physics Question Bank

Question 1 of 10 Authentic UPSC NDA Question

Which of the following is NOT a property of a magnet?

Step-by-Step Logic & Solution

[CORE CONCEPT]: Magnets exhibit specific properties such as attraction to ferromagnetic materials (like iron) and the presence of a magnetic field. [STEP-BY-STEP SOLUTION]: While magnets attract iron and exhibit repulsion between like poles, they do not attract all metals (e.g., aluminum, copper). [SHORTCUT / EXAM TRICK]: Remember that only ferromagnetic materials are attracted to magnets. [COMMON MISTAKE TRAP]: Students may assume all metals are magnetic, leading to incorrect conclusions.

Question 2 of 10 Authentic UPSC NDA Question

For an electric circuit given below, the correctcombination of voltage (V) and current (I) is

Step-by-Step Logic & Solution

Solution

Option (b) is correct.

Explanation:

\[ V = I_1 R_1 \]

Given \( I_1 = 3A \):

\[ V = 3 \times 100 = 300V \]

Current through \( R_2 \):

\[ I_2 = \frac{V}{R_2} = \frac{300}{200} = 1.5 \, A \]

Current through \( R_3 \):

\[ I_3 = \frac{V}{R_3} = \frac{300}{300} = 1 \, A \]

Net current \( I \):

\[ I = I_1 + I_2 + I_3 = 3 + 1.5 + 1 = 5.5 \, A \]

📌 Hints / Properties Used:

Ohm's Law: \( V = IR \)
Current in parallel: \( I_{total} = I_1 + I_2 + I_3 \)

📊 Visual Diagram Suggestion:

A schematic showing the parallel connection of resistors with wires indicating current flows and labeled voltage sources to clarify voltage drops and current distribution.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 12 (Current Electricity).

Question 3 of 10 Authentic UPSC NDA Question

An incandescent electric bulb converts 20%of its power consumption into light and the remaining power is dissipated as heat. The bulb’s filament has a resistance of 200 W and 2 A current flows through it. If the bulb remains ON for 10 h and the rate of electricity chargeis ₹ 5/unit, then which among the followingis the correct amount of the money spent on producing light?

Step-by-Step Logic & Solution

Solution

Option (d) is correct.

Explanation:

\[ P = I^2R \]

Here, \( I = 2 \, \text{A} \) and \( R = 200 \, \Omega \).

\[ P = (2)^2 \times 200 = 800 \, \text{W} \]

Total power consumption in kW:

\[ P = 0.8 \, \text{kW} \]

Only 20% of this power is used for light:

\[ P_{\text{light}} = 0.8 \times \frac{20}{100} = 0.16 \, \text{kW} \]

Energy consumed over 10 hours:

\[ E = P_{\text{light}} \times t = 0.16 \, \text{kW} \times 10 \, \text{h} = 1.6 \, \text{kWh} = 1.6 \, \text{units} \]

Cost of electricity at ₹5/unit:

\[ \text{Cost} = 1.6 \times 5 = ₹8 \]

📌 Hints / Properties Used:

Power formula: \( P = I^2R \)
Energy calculation: \( E = P \times t \)
Cost: \( \text{Cost} = \text{Units} \times \text{Rate} \)

📊 Visual Diagram Suggestion:

A schematic circuit diagram representing the incandescent bulb, indicating the current and resistance values, along with power dissipation percentages would enhance comprehension.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 4 (Electromagnetic Induction).

Question 4 of 10 Authentic UPSC NDA Question

The AC mains domestic supply current in India changes direction in every:

Step-by-Step Logic & Solution

Correct Answer

✓ Option (d) is correct. AC current changes direction every 1/100 seconds.

📖 Governing Law / Core Concept

In alternating current (AC) systems, the frequency of the AC supply determines the time taken for the current to change direction. The formula for frequency (\( f \)) is given by:

\[ f = \frac{1}{T} \]

Where \( T \) is the time period in seconds. In India, the frequency of AC is 50 Hz.

🧪 Step-by-Step Breakdown

To calculate the time period \( T \) of the AC supply:

\[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \text{ s} \]

The AC current inverts its direction once during each half-cycle. Therefore, the time for a complete direction change is:

\[ T_{change} = \frac{T}{2} = \frac{0.02}{2} = 0.01 \text{ s} = \frac{1}{100} \text{ s} \]

🔍 Option Analysis

Option (a): 50 s - Incorrect; far too long for practical AC usage.
Option (b): 1/50 s - Incorrect; this would correspond to a higher frequency than typical AC.
Option (c): 100 s - Incorrect; outside the range of domestic supply.
Option (d): 1/100 s - Correct; matches the AC frequency in India.

⚡ Mnemonic / Speed-Run

Remember: For AC systems, frequency and time period are inversely related. 50 Hz translates to a direction change of 1/100 seconds.

6. Visual Suggestion

Include a cycle diagram illustrating the AC waveform, highlighting the direction change points every 1/100 seconds.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class 12 Physics, Chapter on Alternating Current.

Question 5 of 10 Authentic UPSC NDA Question

If the length of a copper wire is increased by twice, then its resistivity will be

Step-by-Step Logic & Solution

✓

1. Correct Answer

Option C is correct. The resistivity of a material remains unchanged with length changes.

📖 Core Concept

Resistivity is a fundamental property of materials, independent of geometric attributes such as length or cross-sectional area.

💡 3. Explanation

Resistivity (\( \rho \)) is characterized by the formula:

\[ R = \rho \frac{L}{A} \]

where \( R \) is resistance, \( L \) is the length, and \( A \) is the cross-sectional area. Here, \( \rho \) is constant for a material under steady conditions and does not change regardless of the wire's length or shape. Therefore, increasing the length of the wire does not affect its resistivity.

❌ 4. Option Analysis

•Option A: Incorrect. Doubling the length influences resistance, not resistivity.
•Option B: Incorrect. Resistivity does not diminish with an increase in length.
•Option D: Incorrect. There is no correlation between the increase in length and resistivity.

⚡

NDA Speed-Run / Mnemonic Shortcut

Remember, resistivity (\( \rho \)) is solely a material characteristic and remains fixed regardless of length or shape variations.

🎯

5. Key Takeaways

Resistivity is unaffected by the wire's length or cross-sectional area.

Question 6 of 10 Authentic UPSC NDA Question

A current through a horizontal power line flow in east to west direction. What will be the direction of magnetic field at a point directly below when viewed from east end

Step-by-Step Logic & Solution

✓

1. Correct Answer

Option A is correct. The magnetic field direction is clockwise as observed from the east end of the wire.

📖 Core Concept

The direction of magnetic fields around current-carrying wires can be determined using the right-hand rule, which helps visualize their orientation in space.

💡 3. Explanation

The right-hand thumb rule states that if the thumb of the right hand points in the direction of the current (from east to west in this case), the direction in which the fingers curl gives the direction of the magnetic field lines.

\[ \text{Magnetic field direction} \rightarrow \text{Clockwise} \]

Therefore, at a point directly below the wire, when viewed from the east end, the magnetic field will indeed be in a clockwise direction in a plane perpendicular to the wire.

❌ 4. Option Analysis

• Option B: Anti-clockwise in a plane perpendicular to the wire - Incorrect because, based on the right-hand rule, the field is clockwise.
• Option C: Clockwise in a plane parallel to the wire - Incorrect; the observation is from below in a perpendicular plane.
• Option D: Anti-clockwise in a plane parallel to the wire - Incorrect since it misunderstands the trajectory of the magnetic field lines.

⚡

NDA Speed-Run / Mnemonic Shortcut

Use the right-hand rule by aligning the thumb with the current direction to quickly remember the magnetic field orientation.

🎯

5. Key Takeaways

Magnetic fields around current-carrying wires can be visualized using the right-hand rule; the field is clockwise when viewed from the east end.

Question 7 of 10 Authentic UPSC NDA Question

Which one of the following statements is NOT correct?

Step-by-Step Logic & Solution

✓

1. Correct Answer

Option D is correct. A voltmeter has low resistance, and an ammeter has high resistance is incorrect.

📖 Core Concept

Accurate measurement of electrical parameters requires ammeters and voltmeters to connect the circuit correctly based on their resistance characteristics.

💡 3. Explanation

An ammeter must be in series to measure current, having low resistance to avoid circuit disruption. A voltmeter requires parallel connection and has high resistance to prevent drawing significant current that would alter the voltage across the component.

❌ 4. Option Analysis

•Option A: Correctly states ammeters are connected in series.
•Option B: Correctly states voltmeters are connected in parallel.
•Option C: Correctly describes an ammeter's low resistance and a voltmeter's high resistance.

⚡

NDA Speed-Run / Mnemonic Shortcut

Remember: "Ammeters = Series + Low Resistance, Voltmeters = Parallel + High Resistance." This helps quickly recall their functions.

🎯

5. Key Takeaways

A voltmeter has high resistance and is used in parallel; an ammeter has low resistance and is used in series.

Question 8 of 10 Authentic UPSC NDA Question

If three resistors of 1 Ohm each, connect in parallel to each other, then resultant resistance is

Step-by-Step Logic & Solution

✓

1. Correct Answer

Option B is correct. The resultant resistance is 1/3 Ohm.

📖 Core Concept

In parallel circuits, the total resistance decreases as more resistors are added, calculated by summing the reciprocals of each individual resistor's resistance.

💡 3. Explanation

The formula for equivalent resistance \( R_{eq} \) for resistors in parallel is:

\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

For 3 resistors of 1 Ohm:

\[ R_1 = R_2 = R_3 = 1\, Ohm \]

Thus,

\[ \frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \]

Therefore,

\[ R_{eq} = \frac{1}{3}\, Ohm \]

💡 Hints & Visual Guide

📌 Hints / Properties Used:

Use \( R_{eq} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \) for parallel resistances.
Understand that the equivalent resistance reduces when resistors are added in parallel.

📊 Visual Diagram Suggestion:

[Illustration showing three 1 Ohm resistors connected in parallel, indicating the effective resistance computation as \( \frac{1}{3}\, Ohm \)]

Question 9 of 10 Authentic UPSC NDA Question

The work done in moving a charge of 2 coulombs( C) from point A to point B is 24 J. What is the potential difference between A and B?

Step-by-Step Logic & Solution

✓

1. Correct Answer

Option C is correct. [The potential difference is 12 V.]

📖 Core Concept

Potential difference (V) quantifies the work done per unit charge moved between two points in an electric field.

💡 3. Explanation

The formula for potential difference is given by:

V = W/Q

Given W = 24 J and Q = 2 C:

\[ V = \frac{24}{2} \]

Simplifying yields:

\[ V = 12 \, \text{V} \]

❌ 4. Option Analysis

•Option A: 48 V is incorrect; it suggests double the potential difference calculated.
•Option B: 6 V is incorrect due to miscalculation of W/Q.
•Option D: 0.08 V represents an unrealistically low potential difference for the work done.

⚡

NDA Speed-Run / Mnemonic Shortcut

Remember, V = W/Q helps quickly determine potential differences in electric problems.

🎯

5. Key Takeaways

V = W/Q defines the relationship between work and potential difference for electric charges.

Question 10 of 10 Authentic UPSC NDA Question

In which one of the following situations, the bulb would glow the most? (Consider allbatteries are the same.)

Step-by-Step Logic & Solution

Correct Answer

✓ Option (d) is correct. The two bulbs connected in parallel with two cells in series will glow brighter.

📖 Governing Law / Core Concept

The brightness of a bulb is directly proportional to the power it receives, represented by the formula:

\[ P = \frac{V^2}{R} \]

Where \( P \) is the power (in Watts), \( V \) is the voltage (in Volts), and \( R \) is the resistance (in Ohms).

🧪 Step-by-Step Breakdown

For bulbs connected in parallel, the voltage across each bulb remains constant. In this case, when two identical batteries are connected in series, the total voltage is:

\[ V_{\text{total}} = V_1 + V_2 \]

Given identical batteries, the total voltage becomes:

\[ V_{\text{total}} = 2V \]

In parallel, the power in each bulb is:

\[ P = \frac{(2V)^2}{R} = \frac{4V^2}{R} \]

Thus, each bulb will glow with maximum brightness, significantly brighter compared to series configurations.

🔍 Option Analysis

Options (a), (b), and (c) are series configurations, limiting voltage and power, resulting in dimmer bulbs.

⚡ Mnemonic / Speed-Run

For maximum brightness in circuits: Batteries in Series + Bulbs in Parallel = Highest Glow.

6. Visual Suggestion

Illustrate a circuit diagram showing two bulbs connected in parallel with two batteries in series, clearly indicating the equal voltage across each bulb for enhanced comprehension.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 9 (Electrical Circuits).

Properties of a magnet Questions

Solution

Solution

Correct Answer

📖 Governing Law / Core Concept

🧪 Step-by-Step Breakdown

🔍 Option Analysis

⚡ Mnemonic / Speed-Run

6. Visual Suggestion

📖 Factual Verification & Reference

1. Correct Answer

📖 Core Concept

💡 3. Explanation

❌ 4. Option Analysis

NDA Speed-Run / Mnemonic Shortcut

5. Key Takeaways

1. Correct Answer

📖 Core Concept

💡 3. Explanation

❌ 4. Option Analysis

NDA Speed-Run / Mnemonic Shortcut

5. Key Takeaways

1. Correct Answer

📖 Core Concept

💡 3. Explanation

❌ 4. Option Analysis

NDA Speed-Run / Mnemonic Shortcut

5. Key Takeaways

1. Correct Answer

📖 Core Concept

💡 3. Explanation

💡 Hints & Visual Guide

1. Correct Answer

📖 Core Concept

💡 3. Explanation

❌ 4. Option Analysis

NDA Speed-Run / Mnemonic Shortcut

5. Key Takeaways

Correct Answer

📖 Governing Law / Core Concept

🧪 Step-by-Step Breakdown

🔍 Option Analysis

⚡ Mnemonic / Speed-Run

6. Visual Suggestion