UPSC NDA-1 Archives

41. Which one of the following is the best shape of a solid metal rod to f

Which one of the following is the best shape of a solid metal rod to form the top end of a lightning conductor ?

Figure (a)

Figure (b)

Figure (c)

Figure (d)

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2024

Download PDF Attempt Online

The best shape for the top end of a lightning conductor is a sharp point (spike).

A sharp point concentrates the electric field lines from a charged cloud. This high electric field at the tip promotes corona discharge, which is a continuous leakage of charge from the conductor into the air. This discharge helps to neutralize the surrounding air and reduce the potential difference between the cloud and the ground, thereby reducing the likelihood of a direct lightning strike to the protected structure.

Blunt or rounded ends (like options b, c, and d) are less effective at concentrating the electric field and facilitating corona discharge compared to a sharp point. Modern lightning protection systems sometimes use other designs, but traditionally, the sharp point has been considered superior for this purpose.

Join Our Telegram Channel

42. If the current through an electrical machine running on direct current

If the current through an electrical machine running on direct current is 15 A and the machine runs for 10 minutes, the charge that passes through the machine during this time is :

Join Our Telegram Channel

1.50 C

150 C

900 C

9000 C

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2024

Download PDF Attempt Online

The charge (Q) that passes through a machine is given by the product of the current (I) and the time (t) for which the current flows: Q = I * t.

The current is given in Amperes (A), and the time must be in seconds (s) to get the charge in Coulombs (C). The given time is 10 minutes, which is 10 * 60 = 600 seconds. The current is 15 A.

Calculating the charge: Q = 15 A * 600 s = 9000 Coulombs. This represents the total amount of electric charge that flows through the electrical machine over the 10-minute period.

43. An electric circuit is given below. V₁ = 1 V and Resistance R = 1000 Ω

An electric circuit is given below. V₁ = 1 V and Resistance R = 1000 Ω.
The current through the resistance R is very close to 1 mA and the voltage across point A and B, VAB = 1 V. Now the circuit is changed to :
where value of V₂ = 5 V. The internal resistances of both the batteries are 0.1 Ω. The current through the resistance R is about :

1.0 mA

1.2 mA

3.0 mA

5.0 mA

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2024

Download PDF Attempt Online

The first part of the question describes a circuit with a battery V1=1V, resistor R=1000Ω, where current is ~1mA and VAB=1V. This indicates R is connected across points A and B, and V1 is the source across A and B. The current is V/R = 1V/1000Ω = 1mA, consistent with the description.
The second part describes a changed circuit with V2=5V and internal resistances of both batteries (V1 and V2) are 0.1Ω. Assuming A and B are still the points across which R is connected, and both batteries are connected between A and B, the most plausible configuration is that the batteries V1 (1V, 0.1Ω) and V2 (5V, 0.1Ω) are connected in parallel across A and B, and the resistor R (1000Ω) is also connected between A and B.
In a parallel circuit with multiple voltage sources, we can use nodal analysis or source transformation. Using nodal analysis with B as the reference node (0V), the voltage at A is V_A.
Current through R: I_R = V_A / R
Current from V1 source: I_1 = (V1 – V_A) / r1 = (1 – V_A) / 0.1
Current from V2 source: I_2 = (V2 – V_A) / r2 = (5 – V_A) / 0.1
Applying KCL at node A (sum of currents leaving A is zero):
I_R + I_1 + I_2 = 0 is incorrect. The sources are supplying current towards A, so currents entering A should be summed to currents leaving A. Let’s assume currents leave A towards R, and into the positive terminals of V1 and V2. This means V_A is higher than the battery terminals connected to A. The positive terminals are connected to A, and negative terminals to B. So current flows from A through R to B, from A through V1 (r1) to B, and from A through V2 (r2) to B.
Currents leaving A: I_R = V_A / R, I_1 = V_A – V1 / r1? No. This implies V_A is higher than the source terminal.
Let’s redraw with proper node labels. A and B are the terminals. R is between A and B. Battery 1 (V1, r1) is between A and B (positive to A, negative to B). Battery 2 (V2, r2) is between A and B (positive to A, negative to B).
Using superposition or combining parallel sources:
Equivalent voltage source V_eq = (V1/r1 + V2/r2) / (1/r1 + 1/r2)
Equivalent internal resistance r_eq = 1 / (1/r1 + 1/r2)
V_eq = (1V/0.1Ω + 5V/0.1Ω) / (1/0.1Ω + 1/0.1Ω) = (10 A + 50 A) / (10 S + 10 S) = 60 A / 20 S = 3 V.
r_eq = 1 / (10 S + 10 S) = 1 / 20 S = 0.05 Ω.
Now the circuit is a single equivalent source V_eq=3V with internal resistance r_eq=0.05Ω connected across R=1000Ω.
The voltage across R is V_AB = V_eq * R / (R + r_eq) = 3V * 1000Ω / (1000Ω + 0.05Ω) = 3V * 1000 / 1000.05.
V_AB ≈ 3V.
The current through R is I_R = V_AB / R = (3V * 1000 / 1000.05) / 1000Ω = 3V / 1000.05Ω.
I_R ≈ 3 / 1000 = 0.003 A = 3 mA.
Calculating precisely: 3 / 1000.05 ≈ 0.00299985 A ≈ 2.99985 mA.
This is very close to 3.0 mA.

– The first part of the question establishes the context: R=1000Ω, V1=1V providing ~1mA and VAB=1V, suggesting R is connected across the 1V source.
– The second part adds V2=5V and internal resistances 0.1Ω for *both* batteries.
– Assuming a parallel connection of the two batteries and the resistor R between points A and B is the most likely configuration given the options and the structure implied by the first sentence.
– When voltage sources are in parallel, they can be combined into an equivalent source V_eq and r_eq.
– Formula for parallel sources: V_eq = (Σ Vi/ri) / (Σ 1/ri) and r_eq = 1 / (Σ 1/ri).
– Calculate V_eq and r_eq for V1 (1V, 0.1Ω) and V2 (5V, 0.1Ω).
– Calculate the current through R (1000Ω) when connected across the equivalent source.
– Calculation gives current ≈ 3 mA.

If the batteries were in series, the total voltage would be 6V (aiding) or 4V (opposing). Total internal resistance would be 0.2Ω. Current through 1000Ω would be approximately 6mA or 4mA. Neither of these is a very close match to the options (3.0 mA or 5.0 mA), whereas 3mA is a precise match for the parallel configuration calculation result of ~2.99985 mA. This strongly supports the parallel connection interpretation.

Join Our Telegram Channel

44. An infinite combination of resistors, each having resistance R = 4 Ω,

An infinite combination of resistors, each having resistance R = 4 Ω, is given below. What is the net resistance between the points A and B ? (Each resistance is of equal value, R = 4)

0 Ω

2 + 2√5 Ω

2 + √5 Ω

∞ Ω

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2024

Download PDF Attempt Online

This is an infinite ladder network of resistors. Let the equivalent resistance between points A and B be R_eq. Since the network is infinite and uniform, if we remove the first segment of the ladder (consisting of two horizontal resistors R and one vertical resistor R), the remaining infinite network to the right will have the same equivalent resistance R_eq. The circuit between A and B can be viewed as the first horizontal resistor R (on the top rail) in series with the parallel combination of the vertical resistor R and the equivalent resistance of the rest of the ladder (R_eq), and this combination is then in series with the first horizontal resistor R (on the bottom rail).
Wait, re-reading the visual interpretation and the standard ladder problem formulation: The equivalent resistance R_eq of the infinite ladder from point A to point B (across the first vertical rung) can be determined by considering that adding one more segment at the beginning does not change the total resistance. The segment added at the beginning consists of a horizontal R, a vertical R, and another horizontal R.
Let R_eq be the resistance of the infinite ladder starting from one junction pair (like A and B) to infinity. The structure is: A — R — J1 — …inf… ; B — R — K1 — …inf… ; J1 — R — K1.
The resistance between J1 and K1 onwards is also R_eq.
So, the resistance between A and B is R (A to J1) in series with the parallel combination of R (J1 to K1) and R_eq (J1 onwards), and this whole combination in series with R (K1 to B). This assumes A and B are the start points on the two rails *before* the first vertical resistor. This interpretation doesn’t match the diagram precisely, where A and B are connected to the points *before* the first horizontal resistors on the top and bottom rails respectively.

Let’s use the standard ladder network definition where A and B are the input points. The network structure repeats. Let R_eq be the equivalent resistance of the infinite network from points A and B.
Adding the first segment (R horizontal top, R vertical, R horizontal bottom) such that the rest of the infinite ladder connected after the first vertical resistor has resistance R_eq.
The total resistance R_eq is the resistance of the first horizontal top R, plus the parallel combination of the first vertical R and the resistance of the rest of the ladder. The rest of the ladder, starting from the points where the first vertical R connects to the rails, itself looks like the infinite ladder.
Let’s assume A and B are the input terminals, and the network extends to infinity. The resistance seen from A and B is R_AB. The first segment is horizontal R on top, horizontal R on bottom, and vertical R between them. This interpretation still doesn’t match the usual ladder structure formula derived above.

Let’s re-examine the formula derivation that matches option B. The formula R_eq = R * (1 + sqrt(5)) / 2 corresponds to an infinite ladder where *each segment* consists of a series resistance R on one rail, a series resistance R on the other rail, and a shunt resistance R between the rails. The resistance is calculated between corresponding points on the two rails at the beginning.
A — R — R — …
| |
R R
| |
B — R — R — …
Let R_eq be the resistance between A and B. If we add one more segment (R top, R bottom, R vertical) at the beginning, the resistance between the new input points should still be R_eq.
This structure is slightly different from the image. The image shows horizontal R, vertical R, horizontal R, vertical R…
A — R — J1
|
R
|
K1 — R — J2
|
R
|
K2 — …
B is connected to K1. This is not A and B being the input points of the ladder. A and B are connected as shown in the diagram.
Let the resistance of the infinite network starting from J1 and K1 onwards be R_inf.
The resistance between J1 and B is the series combination of R (J1 to K1) and R (K1 to B). So R_J1B = R + R = 2R.
The resistance between A and J1 is R.
The resistance from A to B is R (A to J1) in series with the parallel combination of R (vertical from J1) and the resistance of the rest of the network starting from K1. This is still confusing based on the diagram.

Let’s assume the question *implies* the standard infinite ladder network structure where A and B are the input terminals across the first vertical resistor, and the formula R_eq = R * (1 + sqrt(5)) / 2 applies to this structure.
In that case, with R = 4 Ω:
R_eq = 4 * (1 + sqrt(5)) / 2 = 2 * (1 + sqrt(5)) = 2 + 2√5 Ω.
This matches option B. Given this is a standard physics/engineering problem, it’s highly likely the question intends this standard structure and formula, despite the potentially ambiguous labeling of A and B in the diagram relative to the text description (“between the points A and B”). Let’s proceed with this interpretation as it’s the only one yielding a matching option.

Join Our Telegram Channel

– The network is an infinite ladder structure made of identical resistors R.
– The key to solving infinite networks is to recognize that adding or removing a finite number of repeating units does not change the overall equivalent resistance of the infinite part.
– Let the equivalent resistance of the infinite ladder from a typical junction pair onwards be R_eq.
– By considering the first segment and the remaining infinite part, a relationship can be set up to solve for R_eq.
– For a standard ladder with series resistors R on rails and shunt resistors R between rails, the equivalent resistance R_eq satisfies R_eq = R + (R * R_eq) / (R + R_eq). This leads to R_eq^2 – R * R_eq – R^2 = 0, with the positive solution R_eq = R * (1 + sqrt(5)) / 2.
– Applying R = 4 Ω, R_eq = 4 * (1 + sqrt(5)) / 2 = 2 * (1 + sqrt(5)) = 2 + 2√5 Ω.

The term (1 + sqrt(5)) / 2 is the golden ratio, often denoted by φ (phi). So, the equivalent resistance of such a ladder network is R * φ. Infinite networks are often solved using recurrence relations or by assuming the equivalent resistance of the infinite part is the same as the whole.

45. Which one of the following telescopes contains only mirrors ?

Which one of the following telescopes contains only mirrors ?

Join Our Telegram Channel

Galilean telescope

Keplerian telescope

Newtonian telescope

Schmidt telescope

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2023

Download PDF Attempt Online

A Newtonian telescope is a type of reflecting telescope that uses a concave primary mirror to collect and focus light, and a flat secondary mirror placed in the light path to redirect the light to an eyepiece positioned at the side of the telescope tube. The primary and secondary mirrors are the key optical components responsible for forming the image. It uses only mirrors for the main light path and image formation.

– Telescopes are classified as refractors (using lenses), reflectors (using mirrors), or catadioptric (using both lenses and mirrors).
– Galilean and Keplerian telescopes are types of refracting telescopes, using lenses.
– A Newtonian telescope is a type of reflecting telescope, using mirrors (a primary parabolic mirror and a secondary flat mirror).
– A Schmidt telescope (e.g., Schmidt-Cassegrain) uses both a corrector plate (lens) and mirrors.
– The question asks for a telescope containing *only* mirrors in its main optical path for image formation.

Reflecting telescopes, like the Newtonian, have advantages over refracting telescopes, such as the absence of chromatic aberration (color fringing) which occurs in lenses, and they can be made much larger since mirrors are supported from the back.

46. Which one among the following is true for the speed of sound in a give

Which one among the following is true for the speed of sound in a given medium ?

Join Our Telegram Channel

Speed of sound remains same at all frequencies

Speed of sound is faster at higher frequencies

Speed of sound is slower at higher frequencies

Speed of sound is slower at higher wavelengths

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2023

Download PDF Attempt Online

The speed of sound in a given medium is primarily determined by the properties of the medium itself, such as its elasticity (or compressibility) and density. In a specific medium under uniform conditions (like constant temperature and pressure for air), the speed of sound is generally constant and independent of the frequency or wavelength of the sound wave. The relationship v = fλ (speed = frequency × wavelength) holds true, meaning that if the frequency changes, the wavelength changes accordingly to keep the speed constant.

– Speed of sound depends on the properties of the medium (elasticity and density).
– For a given medium under specific conditions, the speed of sound is constant.
– The relationship v = fλ implies an inverse relationship between frequency (f) and wavelength (λ) for a constant speed (v).
– In contrast, for light waves in a *dispersive* medium (like glass), speed *does* depend on frequency/wavelength. However, for sound waves in typical media like air, water, or solids, dispersion effects related to frequency are usually negligible unless considering very specific materials or high frequencies outside the audible range.

Factors that affect the speed of sound in air include temperature (speed increases with temperature), humidity (speed increases slightly with humidity), and pressure (for an ideal gas, speed is independent of pressure at constant temperature and density). The speed of sound is generally fastest in solids, slower in liquids, and slowest in gases.

47. In the dispersion of white light by a common glass prism, which one am

In the dispersion of white light by a common glass prism, which one among the following is correct ?

Red light deviates the most because red light has highest speed in prism

Blue light deviates the most because blue light has highest speed in prism

Red light deviates the most because red light has lowest speed in prism

Blue light deviates the most because blue light has lowest speed in prism

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2023

Download PDF Attempt Online

When white light passes through a prism, it disperses into its constituent colors (VIBGYOR) because the refractive index of the prism material is different for different wavelengths of light. This phenomenon is called dispersion. The amount of deviation experienced by a color depends on the refractive index for that color; greater refractive index leads to greater deviation. For most transparent materials like glass, the refractive index is higher for shorter wavelengths (blue, violet) and lower for longer wavelengths (red). Since the speed of light in a medium is given by v = c/n (where c is the speed of light in vacuum and n is the refractive index), a higher refractive index means a lower speed. Therefore, blue light (shorter wavelength, higher refractive index) deviates the most and has the lowest speed in the prism, while red light (longer wavelength, lower refractive index) deviates the least and has the highest speed.

– Dispersion occurs because the refractive index (n) of the medium depends on the wavelength (λ).
– In most transparent materials (normal dispersion), refractive index decreases as wavelength increases (n_violet > n_blue > … > n_red).
– Deviation (δ) in a prism depends on the refractive index (for a small angle prism, δ ≈ (n-1)A). Higher n means greater deviation.
– Speed of light in the medium (v) is inversely proportional to the refractive index (v = c/n). Higher n means lower speed.
– Thus, blue light has a higher refractive index, deviates the most, and has the lowest speed in the prism.

The order of colors in the dispersed spectrum is typically remembered by the acronym VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). Violet light deviates the most, followed by Indigo, Blue, etc., with Red deviating the least.

Join Our Telegram Channel

48. An electric bulb is rated as 220 V and 80 W. When it is operated on 11

An electric bulb is rated as 220 V and 80 W. When it is operated on 110 V, the power rating would be :

80 W

60 W

40 W

20 W

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2023

Download PDF Attempt Online

The power rating of an electric bulb is given by the formula P = V^2 / R, where P is power, V is voltage, and R is resistance. The resistance of the bulb remains constant (assuming its temperature change is negligible for this calculation). First, calculate the resistance using the rated values: R = V_rated^2 / P_rated = (220 V)^2 / 80 W = 48400 / 80 = 605 Ω. When operated on 110 V, the power consumed is P_new = V_new^2 / R = (110 V)^2 / 605 Ω = 12100 / 605 = 20 W.

– The resistance of the bulb is a fixed property.
– Power is proportional to the square of the voltage when resistance is constant (P ∝ V^2).
– Since the voltage is halved (110V is half of 220V), the power will be reduced by a factor of (1/2)^2 = 1/4.
– New Power = Rated Power * (New Voltage / Rated Voltage)^2 = 80 W * (110 V / 220 V)^2 = 80 W * (1/2)^2 = 80 W * (1/4) = 20 W.

This calculation assumes the resistance R is constant, which is a reasonable approximation for this type of problem, although in reality, the resistance of a filament bulb changes with temperature.

Join Our Telegram Channel

49. Which one of the following statements is true ?

Which one of the following statements is true ?

Join Our Telegram Channel

The force of gravity of the Earth on the Moon is greater than the force of gravity of the Moon on the Earth.

The force of gravity of the Moon on the Earth is greater than the force of gravity of the Earth on the Moon.

The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude and are in the same direction.

The force of gravity of the Earth on the Moon and of the Moon on the Earth are equal in magnitude but are in opposite directions.

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2023

Download PDF Attempt Online

According to Newton’s Law of Universal Gravitation, the force of gravity exerted by the Earth on the Moon is equal in magnitude to the force of gravity exerted by the Moon on the Earth. This is an application of Newton’s Third Law of Motion (for every action, there is an equal and opposite reaction). These forces are always attractive and act along the line joining the centers of the two bodies, thus they are in opposite directions.

– Newton’s Law of Universal Gravitation states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them (F = G * m1 * m2 / r^2). This formula is symmetrical with respect to m1 and m2, showing the force is the same magnitude for both.
– Newton’s Third Law of Motion states that for every action, there is an equal and opposite reaction. The force Earth exerts on the Moon is the action, and the force Moon exerts on the Earth is the reaction.
– The forces are attractive, meaning Earth pulls the Moon towards it, and the Moon pulls the Earth towards it. Therefore, the forces are in opposite directions.

Although the forces are equal in magnitude, their effects are different due to the difference in mass. The Earth’s mass is much greater than the Moon’s mass. Therefore, the acceleration of the Moon towards the Earth (due to the Earth’s pull) is much greater than the acceleration of the Earth towards the Moon (due to the Moon’s pull, which causes tides).

50. By which one of the following amendments, was Article 51A, relating to

By which one of the following amendments, was Article 51A, relating to the Fundamental Duties, inserted into the Constitution of India ?

Join Our Telegram Channel

The Constitution (42nd Amendment) Act

The Constitution (44th Amendment) Act

The Constitution (85th Amendment) Act

The Constitution (92nd Amendment) Act

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2023

Download PDF Attempt Online

Part IVA (Article 51A) relating to Fundamental Duties was inserted into the Constitution of India by the Constitution (42nd Amendment) Act, 1976. This amendment is also known as the ‘Mini-Constitution’ due to the significant number of changes it introduced.

– Fundamental Duties were not part of the original Constitution.
– They were added based on the recommendations of the Swaran Singh Committee.
– The 42nd Amendment Act, 1976, added ten Fundamental Duties.
– An eleventh Fundamental Duty was added later by the 86th Amendment Act, 2002.

The 44th Amendment Act, 1978, primarily aimed to reverse some of the changes made by the 42nd Amendment, particularly concerning emergency provisions and property rights. The 85th Amendment Act relates to reservations in promotion. The 92nd Amendment Act, 2003, related to the Eighth Schedule of the Constitution, adding Bodo, Dogri, Maithili, and Santali languages.