UPSC NDA-1 Archives

1. Why are the tyres of aircrafts made of conducting rubber? 1. So that

Why are the tyres of aircrafts made of conducting rubber?

1. So that the charge accumulated on the aircraft in flight, by rubbing the air, can easily be transferred to ground on landing.
2. So that the charge accumulated due to the operation of various electronic equipments in the aircraft in flight can easily be transferred to ground on landing.

Select the correct answer using the code given below.

1 only

2 only

Both 1 and 2

Neither 1 nor 2

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The correct option is C.

Aircraft accumulate static electric charge during flight due to friction with air (triboelectric effect) and the operation of onboard electronic equipment.

Statement 1 is correct: Friction with air (rubbing) during flight causes the aircraft to accumulate static charge, a phenomenon known as triboelectric charging. This is a significant source of charge buildup, especially in dry air.
Statement 2 is correct: The operation of various electronic systems within the aircraft can also contribute to the accumulation of static charge on the aircraft’s structure.
Conducting rubber tyres provide a path for this accumulated static charge to safely discharge to the ground upon landing, preventing the build-up of a large potential difference between the aircraft and the ground, which could otherwise lead to a spark. Such a spark could pose a fire hazard, especially during refueling, or cause damage to sensitive electronic components. Therefore, both reasons contribute to the necessity of conducting tyres.

2. A wooden box of mass 2 kg and dimensions (30 cm $\times$ 15 cm $\times

A wooden box of mass 2 kg and dimensions (30 cm $\times$ 15 cm $\times$ 10 cm) is placed on a table with sides 30 cm and 10 cm touching the tabletop. Which one of the following is the approximate pressure exerted on the table?

111.1 N/m$^2$

222.2 N/m$^2$

333.3 N/m$^2$

666.6 N/m$^2$

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The correct option is D.

Pressure is defined as force per unit area ($P = F/A$). The force exerted by the box on the table is its weight ($F = mg$). The area is the contact area between the box and the table.

Mass of the box $m = 2$ kg.
Assume acceleration due to gravity $g \approx 10$ m/s$^2$ for approximation, as is common in such problems.
Force (weight) $F = mg = 2 \text{ kg} \times 10 \text{ m/s}^2 = 20 \text{ N}$.
The dimensions of the box are 30 cm $\times$ 15 cm $\times$ 10 cm.
The box is placed with sides 30 cm and 10 cm touching the tabletop. The contact area is $A = (30 \text{ cm}) \times (10 \text{ cm})$.
Convert area to square meters: $A = (0.30 \text{ m}) \times (0.10 \text{ m}) = 0.03 \text{ m}^2$.
Pressure $P = F / A = 20 \text{ N} / 0.03 \text{ m}^2 = 20 / (3/100) \text{ N/m}^2 = (20 \times 100) / 3 \text{ N/m}^2 = 2000 / 3 \text{ N/m}^2$.
$2000 / 3 \approx 666.67 \text{ N/m}^2$.
Looking at the options, 666.6 N/m$^2$ is the closest value.
If $g = 9.8$ m/s$^2$ was used, $F = 2 \times 9.8 = 19.6$ N. $P = 19.6 / 0.03 = 1960 / 3 \approx 653.33$ N/m$^2$. 666.6 is still the closest option, suggesting $g=10$ m/s$^2$ was intended.

3. What is the magnification produced by a concave lens of focal length 1

What is the magnification produced by a concave lens of focal length 10 cm, when an image is formed at a distance of 5 cm from the lens?

2.0

1.0

0.5

0.33

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The correct option is C.

For a lens, the lens formula is $1/f = 1/v – 1/u$ and magnification is $m = v/u$. For a concave lens, the focal length $f$ is negative, and it forms virtual, upright, and diminished images, typically on the same side as the object (so image distance $v$ is negative).

Given focal length $f = -10$ cm (concave lens). The image is formed at a distance of 5 cm from the lens. Since it’s a concave lens, the image is virtual and formed on the same side as the object, so image distance $v = -5$ cm.
Using the lens formula: $1/f = 1/v – 1/u$
$1/(-10) = 1/(-5) – 1/u$
$-1/10 = -1/5 – 1/u$
$1/u = -1/5 + 1/10$
$1/u = -2/10 + 1/10$
$1/u = -1/10$
$u = -10$ cm (object distance is 10 cm in front of the lens).
Magnification $m = v/u = (-5 \text{ cm}) / (-10 \text{ cm}) = 0.5$.
The magnification is 0.5, indicating a diminished image.

4. Two resistors $R_1$ and $R_2$ arranged in parallel combination in an e

Two resistors $R_1$ and $R_2$ arranged in parallel combination in an electrical closed circuit are made of the same material and of same thickness. If the length of $R_2$ is twice the length of $R_1$, then the total resistance $R$ satisfies

$3R = 2R_1$

$3R = 2R_2$

$2R = 3R_1$

$2R = 3R_2$

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The correct option is A.

For resistors in parallel, the total resistance $R$ is given by $1/R = 1/R_1 + 1/R_2$. The resistance of a wire is $R = \rho L/A$, where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

Given that $R_1$ and $R_2$ are made of the same material ($\rho_1 = \rho_2 = \rho$) and have the same thickness (assuming same cross-sectional area $A_1 = A_2 = A$).
Let the length of $R_1$ be $L_1$. Then $R_1 = \rho L_1 / A$.
The length of $R_2$ is $L_2 = 2L_1$. Then $R_2 = \rho L_2 / A = \rho (2L_1) / A = 2 (\rho L_1 / A) = 2R_1$.
Now, the resistors are in parallel, so $1/R = 1/R_1 + 1/R_2$.
Substitute $R_2 = 2R_1$:
$1/R = 1/R_1 + 1/(2R_1)$
$1/R = (2 + 1) / (2R_1)$
$1/R = 3 / (2R_1)$
$R = (2R_1) / 3$
Multiplying both sides by 3 gives $3R = 2R_1$.

5. According to Fleming’s right-hand rule, if the forefinger indicates th

According to Fleming’s right-hand rule, if the forefinger indicates the direction of magnetic field and thumb shows the direction of motion of conductor, then the stretched middle finger will predict the direction of

force acting on the conductor

electric field

induced current

current

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The correct option is C.

Fleming’s Right-Hand Rule is used to determine the direction of the induced current in a conductor moving in a magnetic field.

According to Fleming’s Right-Hand Rule, if the thumb points in the direction of the motion of the conductor, the forefinger points in the direction of the magnetic field, then the middle finger points in the direction of the induced electric current. This rule is fundamental to understanding the operation of generators. Fleming’s Left-Hand Rule, in contrast, determines the direction of the force on a current-carrying conductor placed in a magnetic field.

6. All objects experience a buoyancy when they are immersed in a fluid. B

All objects experience a buoyancy when they are immersed in a fluid. Buoyancy is

a downward force

a downward pressure

an upward force

an upward pressure

This question was previously asked in

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The correct option is C.

Buoyancy is an upward force exerted by a fluid on an immersed object, opposing the object’s weight.

Archimedes’ principle states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. This force acts vertically upwards through the centroid of the displaced fluid volume. Buoyancy is a force, not a pressure, and its direction is always opposite to the direction of gravity on the fluid (hence, upward in the case of earth’s gravity).

7. A cell is unable to synthesize lipids. Which of its cell organelles mi

A cell is unable to synthesize lipids. Which of its cell organelles might be defective?

Smooth endoplasmic reticulum

Golgi bodies

Lysosomes

Mitochondria

This question was previously asked in

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The correct option is A.

The smooth endoplasmic reticulum (SER) is a major site for lipid synthesis in eukaryotic cells.

The smooth endoplasmic reticulum synthesizes various lipids, including phospholipids, steroids, and oils. It is also involved in the detoxification of drugs and poisons, and the storage of calcium ions. Golgi bodies are involved in protein and lipid modification and packaging. Lysosomes contain hydrolytic enzymes for intracellular digestion. Mitochondria are primarily involved in energy production through cellular respiration.

8. Which one of the following plants has unisexual flowers?

Which one of the following plants has unisexual flowers?

Papaya

Hibiscus

Mustard

Sunflower

This question was previously asked in

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The correct option is A.

Unisexual flowers contain either male reproductive organs (stamens) or female reproductive organs (pistil/carpel), but not both.

Plants with unisexual flowers can be monoecious (male and female flowers on the same plant, e.g., maize, cucumber) or dioecious (male and female flowers on separate plants, e.g., papaya, date palm). Papaya is a dioecious plant, having separate male and female trees with unisexual flowers. Hibiscus, Mustard, and Sunflower typically have bisexual flowers, containing both stamens and pistil/carpel within the same flower.

9. Which one of the following is the mechanism of action of oral contrace

Which one of the following is the mechanism of action of oral contraceptive pills?

They kill the egg.

They kill the sperm.

They kill the zygote.

They inhibit the release of egg.

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The correct option is D.

Oral contraceptive pills primarily work by inhibiting ovulation, which is the release of the egg from the ovary.

Combined oral contraceptive pills (containing estrogen and progestogen) and progestogen-only pills have multiple mechanisms of action, but the inhibition of ovulation is the most significant. They also cause changes in cervical mucus (making it thicker and less penetrable by sperm) and the uterine lining (making it less receptive to implantation), but preventing the release of the egg is the primary way they prevent pregnancy. They do not kill the egg, sperm, or a formed zygote.

10. Which one of the following reactions is an example of decomposition

Which one of the following reactions is an example of decomposition reaction?

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

2AgCl(s) Sunlight → 2Ag(s) + Cl₂(g)

CuO + H₂ Heat → Cu + H₂O

Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)

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The correct option is B.

A decomposition reaction is a reaction in which a single compound breaks down into two or more simpler substances.

Option A is a combustion reaction, where a substance reacts rapidly with oxygen to produce heat and light.
Option B, $2\text{AgCl(s)} \text{ Sunlight} \rightarrow 2\text{Ag(s)} + \text{Cl}_2\text{(g)}$, shows silver chloride (a single compound) breaking down into silver and chlorine (simpler substances) upon exposure to sunlight. This is a classic example of photolytic decomposition.
Option C is a reduction reaction, specifically the reduction of copper(II) oxide by hydrogen.
Option D is a displacement reaction, where iron displaces copper from copper sulphate solution.