UPSC NDA-2 Archives

21. To help deep-sea divers breathe, they carry cylinders of oxygen mixed

To help deep-sea divers breathe, they carry cylinders of oxygen mixed with

[amp_mcq option1=”chlorine” option2=”helium” option3=”nitrogen” option4=”ozone” correct=”option2″]

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To help deep-sea divers breathe, they carry cylinders of oxygen mixed with helium.

At high pressures experienced during deep-sea dives, the nitrogen in air can cause nitrogen narcosis, which impairs judgment and coordination. Oxygen at high partial pressures can also become toxic. To avoid these issues, the breathing gas mixture often replaces nitrogen with helium, which is less narcotic and also less dense than nitrogen, making breathing easier at depth. This mixture of oxygen and helium is called Heliox.

Other gas mixtures like Trimix (oxygen, helium, and nitrogen) are also used depending on the depth and duration of the dive. Chlorine and ozone are toxic gases and would not be used in breathing mixtures.

22. Which one of the following solutions is not capable of conducting

Which one of the following solutions is not capable of conducting electricity?

[amp_mcq option1=”Copper sulphate” option2=”Sodium chloride” option3=”Sugar” option4=”Sodium hydroxide” correct=”option3″]

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Sugar solution is not capable of conducting electricity.

For a solution to conduct electricity, it must contain free-moving ions. Solutions of ionic compounds (like Copper sulphate, Sodium chloride, Sodium hydroxide) dissociate into ions when dissolved in water, allowing them to conduct electricity. Sugar (sucrose) is a covalent compound; when it dissolves in water, it remains as neutral molecules and does not produce ions.

Electrolytes are substances that produce ions when dissolved in a solvent, making the solution electrically conductive. Non-electrolytes, like sugar, dissolve but do not ionize, and their solutions do not conduct electricity.

23. Which one of the following metals floats in cold water?

Which one of the following metals floats in cold water?

[amp_mcq option1=”Magnesium” option2=”Calcium” option3=”Potassium” option4=”Copper” correct=”option3″]

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Potassium is a metal that floats in cold water.

Metals less dense than water (density < 1 g/cm³) and which react with water tend to float. Potassium (density ~0.89 g/cm³) and Sodium (density ~0.97 g/cm³) are alkali metals that are less dense than water and react vigorously with cold water, producing hydrogen gas which aids flotation. Calcium (density ~1.55 g/cm³) is denser than water but reacts with cold water, and the hydrogen bubbles produced stick to its surface, causing it to float temporarily. Magnesium and Copper do not react with cold water in a way that causes them to float.

Highly reactive metals like Potassium and Sodium react exothermically with cold water to form hydroxides and hydrogen gas. This reaction can be quite vigorous. While Calcium also reacts with cold water and floats, Potassium and Sodium are well-known examples of metals that float and move rapidly on the surface during the reaction. Among the options provided, Potassium is the most definitive answer for a metal that floats in cold water due to its low density and vigorous reaction.

24. Which one of the following is the correct reactivity series with

Which one of the following is the correct reactivity series with water?

[amp_mcq option1=”Zinc > Iron > Lead > Copper” option2=”Copper > Lead > Zinc > Iron” option3=”Copper > Zinc > Iron > Lead” option4=”Zinc > Copper > Iron > Lead” correct=”option1″]

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The correct reactivity series with water among the given metals is Zinc > Iron > Lead > Copper.

The reactivity series lists metals in order of decreasing reactivity. Metals higher in the series are more reactive. In the given options, Zinc is more reactive than Iron, which is more reactive than Lead, which is more reactive than Copper.

Highly reactive metals like Potassium and Sodium react vigorously with cold water. Moderately reactive metals like Magnesium and Zinc react with hot water or steam. Less reactive metals like Iron, Lead, Copper, Silver, and Gold do not react with water or steam under normal conditions, or react very slowly (like Iron rusting in the presence of water and oxygen). The general order of reactivity involving water reaction is K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu > Ag > Au. Comparing the given metals (Zn, Fe, Pb, Cu) shows the order Zn > Fe > Pb > Cu.

25. When the pitch of sound increases, which one of the following

When the pitch of sound increases, which one of the following increases?

[amp_mcq option1=”Intensity” option2=”Loudness” option3=”Wavelength” option4=”Frequency” correct=”option4″]

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The correct answer is Frequency.

The pitch of a sound is determined by its frequency. A higher frequency corresponds to a higher pitch, and a lower frequency corresponds to a lower pitch.

Intensity is related to the amplitude of the sound wave and is perceived as loudness. Wavelength is inversely proportional to frequency for a constant speed of sound. Loudness is the subjective perception of sound intensity.

26. An electric bulb is connected to 220 V generator. The current drawn is

An electric bulb is connected to 220 V generator. The current drawn is 600 mA. What is the power of the bulb?

[amp_mcq option1=”132 W” option2=”13.2 W” option3=”1320 W” option4=”13200 W” correct=”option1″]

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UPSC NDA-2 – 2022

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The power (P) of an electrical device connected to a circuit is given by the formula:
P = V × I
Where V is the voltage across the device and I is the current flowing through it.
Given:
Voltage (V) = 220 V
Current (I) = 600 mA
First, convert the current from milliamperes (mA) to amperes (A):
1 A = 1000 mA
So, 600 mA = 600 / 1000 A = 0.6 A.
Now, calculate the power:
P = 220 V × 0.6 A
P = 220 × (6/10) W
P = 22 × 6 W
P = 132 W.

Power is calculated as the product of voltage and current (P=VI). Ensure units are in Volts and Amperes to get Power in Watts.

This calculation is a fundamental application of Ohm’s law and the power formula in basic electricity. The resistance of the bulb could also be calculated using Ohm’s law (V=IR), but it is not needed to find the power. The power rating (in Watts) indicates the rate at which the bulb consumes electrical energy.

27. Which one of the following wavelengths corresponds to the wavelength o

Which one of the following wavelengths corresponds to the wavelength of X-rays?

[amp_mcq option1=”500 nm” option2=”5000 nm” option3=”100 nm” option4=”1 nm” correct=”option4″]

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Electromagnetic radiation is classified by wavelength or frequency. The electromagnetic spectrum includes (in order of decreasing wavelength): Radio waves, Microwaves, Infrared radiation, Visible light, Ultraviolet radiation, X-rays, and Gamma rays.
Visible light wavelengths are typically in the range of 400 nm to 700 nm.
Ultraviolet (UV) radiation has wavelengths shorter than visible light, usually in the range of 10 nm to 400 nm.
X-rays have even shorter wavelengths, generally ranging from about 0.01 nm to 10 nm.
Gamma rays have the shortest wavelengths, typically less than 0.1 nm.

Comparing the given options to the typical range of X-ray wavelengths (0.01 nm – 10 nm):
A) 500 nm falls in the visible light spectrum.
B) 5000 nm (5 µm) falls in the infrared spectrum.
C) 100 nm falls in the ultraviolet spectrum.
D) 1 nm falls within the typical range of X-ray wavelengths.

X-rays are high-energy photons commonly used in medical imaging (radiography, CT scans) and material analysis (X-ray crystallography, fluorescence). They are produced when high-speed electrons strike a metal target or by electron transitions in atoms.

28. A pressure cooker cooks food faster by

A pressure cooker cooks food faster by

[amp_mcq option1=”increasing the boiling point of water” option2=”decreasing the boiling point of water” option3=”increasing the melting point of water” option4=”decreasing the melting point of water” correct=”option1″]

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A pressure cooker works by sealing the cooking pot tightly, which prevents steam from escaping. As water heats up, it turns into steam, increasing the pressure inside the cooker. According to the relationship between pressure and boiling point, increasing the pressure on a liquid increases its boiling point.

By increasing the pressure inside, the boiling point of water is raised from 100°C (at standard atmospheric pressure) to a higher temperature, typically around 120-125°C. Food cooks faster at these higher temperatures.

Conversely, at high altitudes where atmospheric pressure is lower, water boils at temperatures below 100°C, which means food takes longer to cook compared to cooking at sea level. Pressure cookers counteract this effect or speed up cooking at any altitude by artificially increasing the pressure and thus the boiling point.

29. An object is made of two equal parts by volume; one part has density $

An object is made of two equal parts by volume; one part has density $\rho_0$ and the other part has density $2\rho_0$. What is the average density of the object?

[amp_mcq option1=”$3\rho_0$” option2=”$\frac{3}{2}\rho_0$” option3=”$\rho_0$” option4=”$\frac{1}{2}\rho_0$” correct=”option2″]

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Let V be the total volume of the object. The object is made of two equal parts by volume, so the volume of each part is $V_1 = V_2 = V/2$.
Let $\rho_1$ be the density of the first part and $\rho_2$ be the density of the second part.
Given: $\rho_1 = \rho_0$ and $\rho_2 = 2\rho_0$.
The mass of the first part is $m_1 = \rho_1 \times V_1 = \rho_0 \times (V/2)$.
The mass of the second part is $m_2 = \rho_2 \times V_2 = 2\rho_0 \times (V/2) = \rho_0 V$.
The total mass of the object is $M = m_1 + m_2 = \rho_0 (V/2) + \rho_0 V = \rho_0 V (\frac{1}{2} + 1) = \rho_0 V (\frac{3}{2})$.
The total volume of the object is $V_{\text{total}} = V_1 + V_2 = V/2 + V/2 = V$.
The average density of the object is $\rho_{\text{avg}} = \frac{M}{V_{\text{total}}} = \frac{\rho_0 V (3/2)}{V} = \frac{3}{2}\rho_0$.

When calculating average density for parts of equal volume, the average density is the simple arithmetic mean of the densities. However, in this case, the masses are different. The calculation involves finding the total mass and dividing by the total volume.

If the parts were of equal mass instead of equal volume, the calculation would be different, involving the reciprocal of the average of reciprocals (harmonic mean of densities).

30. Which one of the following statements regarding a current-carrying sol

Which one of the following statements regarding a current-carrying solenoid is not correct?

[amp_mcq option1=”The magnetic field inside the solenoid is uniform.” option2=”The current-carrying solenoid behaves like a bar magnet.” option3=”The magnetic field inside the solenoid increases with increase in current.” option4=”If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.” correct=”option4″]

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Let’s analyze each statement about a current-carrying solenoid:
A) The magnetic field inside a *long* solenoid is approximately uniform and directed along the axis of the solenoid, except near the ends. This statement is correct for an ideal or long solenoid.
B) A current-carrying solenoid creates a magnetic field pattern similar to that of a bar magnet, with magnetic poles at its ends. This statement is correct.
C) The magnitude of the magnetic field inside a solenoid is given by $B = \mu n I$, where $\mu$ is the permeability of the core material, $n$ is the number of turns per unit length, and $I$ is the current. The field is directly proportional to the current ($I$). So, increasing the current increases the magnetic field. This statement is correct.
D) If a soft iron bar (a ferromagnetic material with high permeability) is inserted inside the solenoid, the magnetic field inside increases significantly. This happens because the soft iron gets strongly magnetized in the direction of the solenoid’s field, and its own magnetic field adds to the field produced by the current. The permeability of soft iron is much greater than the permeability of air or vacuum ($\mu >> \mu_0$). The magnetic field does *not* remain the same; it increases. This statement is incorrect.

Ferromagnetic materials like soft iron dramatically increase the magnetic field strength when placed inside a solenoid or coil because they become strongly magnetized, effectively increasing the magnetic permeability of the core.

This property is used to make electromagnets, where a soft iron core is inserted into a solenoid to produce a very strong magnetic field when current flows.