UPSC NDA-2 Archives

1. Which one of the following is the major constituent of biogas ?

Which one of the following is the major constituent of biogas ?

[amp_mcq option1=”Carbon dioxide” option2=”Nitrous oxide” option3=”Methane” option4=”Oxygen” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2019

The correct option is C) Methane.

Biogas is a mixture of gases produced by the breakdown of organic matter in the absence of oxygen, primarily consisting of methane and carbon dioxide.

The typical composition of biogas is about 50-75% methane (CH4) and 25-50% carbon dioxide (CO2), with traces of other gases like hydrogen sulfide, nitrogen, and hydrogen. Methane is the primary combustible component that gives biogas its fuel value.

2. Which one of the following represents the correct order of electron re

Which one of the following represents the correct order of electron releasing tendency of metals ?

[amp_mcq option1=”Zn > Cu > Ag” option2=”Ag > Cu > Zn” option3=”Cu > Zn > Ag” option4=”Cu > Ag > Zn” correct=”option1″]

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UPSC NDA-2 – 2019

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The correct option is A) Zn > Cu > Ag.

The electron releasing tendency of metals is determined by their position in the electrochemical series or reactivity series. Metals higher in the series have a greater tendency to lose electrons and act as reducing agents, thus having higher electron releasing tendency.

The reactivity series orders metals based on their decreasing reactivity: K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu > Hg > Ag > Au. Zinc is higher in the series than copper, and copper is higher than silver. Therefore, the order of electron releasing tendency is Zn > Cu > Ag. This also corresponds to their relative standard electrode potentials (standard reduction potentials, where less positive or more negative potential indicates higher electron releasing tendency).

3. Which one of the following is not a monatomic element ?

Which one of the following is not a monatomic element ?

[amp_mcq option1=”Copper” option2=”Helium” option3=”Iodine” option4=”Barium” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2019

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The correct answer is C) Iodine.

A monatomic element is an element that exists stably as individual atoms rather than in molecules composed of two or more atoms. Helium (He) is a noble gas and exists as single atoms, making it monatomic. Copper (Cu) and Barium (Ba) are metals that exist as lattices of individual atoms in their solid state. Iodine (I) is a halogen that exists as a diatomic molecule ($I_2$) in its standard solid state and gaseous state (when it sublimes). Thus, Iodine is not a monatomic element; it is diatomic.

Noble gases (He, Ne, Ar, Kr, Xe, Rn) are the most common examples of naturally occurring monatomic elements. While metals exist as individual atoms in their solid structure, the term “monatomic” is most frequently used to contrast with diatomic or polyatomic molecular elements like $O_2$, $N_2$, $S_8$, $P_4$, and halogens ($F_2$, $Cl_2$, $Br_2$, $I_2$). Among the given options, Iodine is the only one that exists as a diatomic molecule in its elemental form.

4. A very large volume of hydrogen can be accommodated by making

A very large volume of hydrogen can be accommodated by making

[amp_mcq option1=”non-metallic hydrides.” option2=”hydrogen peroxide.” option3=”non-stoichiometric hydrides.” option4=”alkali metal hydrides.” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2019

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The correct answer is C) non-stoichiometric hydrides.

Certain transition metals (like palladium, titanium, vanadium, etc.) and their alloys can absorb large volumes of hydrogen, forming interstitial (non-stoichiometric) hydrides. In these hydrides, hydrogen atoms occupy interstitial sites within the metal lattice. This property is exploited for hydrogen storage, as these materials can store hydrogen at densities comparable to or even greater than liquid hydrogen, but at less extreme conditions.

Non-metallic hydrides are compounds like water (H2O), ammonia (NH3), etc. Hydrogen peroxide (H2O2) is a chemical compound. Alkali metal hydrides (like NaH, LiH) are ionic hydrides, which react vigorously with water and are not typically used for large-volume, reversible hydrogen storage applications in the same way as interstitial metal hydrides. The ability of metals like palladium to absorb hundreds of times their own volume of hydrogen is a classic example related to non-stoichiometric hydrides.

5. Which one of the following statements regarding cathode rays is not

Which one of the following statements regarding cathode rays is not correct ?

[amp_mcq option1=”Cathode ray particles are electrons.” option2=”Cathode ray particles start from anode and move towards cathode.” option3=”In the absence of electrical and magnetic fields, cathode rays travel in straight lines.” option4=”Television picture tubes are cathode ray tubes.” correct=”option2″]

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UPSC NDA-2 – 2019

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The statement that is NOT correct regarding cathode rays is B) Cathode ray particles start from anode and move towards cathode.

Cathode rays are streams of negatively charged particles (electrons). In a discharge tube, these electrons are emitted from the cathode (the negative electrode) and are accelerated towards the anode (the positive electrode) by the electric field between them. Therefore, cathode rays move from the cathode towards the anode, not the other way around.

Statement A is correct: Cathode ray particles are indeed electrons. Statement C is correct: In the absence of external electric or magnetic fields, cathode rays travel in straight lines. Statement D is correct: Older style Television picture tubes (CRTs – Cathode Ray Tubes) and computer monitors utilized controlled beams of cathode rays to create images on a phosphorescent screen.

6. Light rays move in straight lines. But through an optical fibre, they

Light rays move in straight lines. But through an optical fibre, they can move in any type of zigzag path because

[amp_mcq option1=”the holes through the fibre are extremely fine.” option2=”light rays are absorbed at the entry end and relieved at the exit end of the fibre.” option3=”scattering of light occurs inside the fibre.” option4=”successive total internal reflections occur as a ray moves through the fibre.” correct=”option4″]

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UPSC NDA-2 – 2019

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The correct answer is D) successive total internal reflections occur as a ray moves through the fibre.

Optical fibres work on the principle of Total Internal Reflection (TIR). The fibre consists of a core with a higher refractive index surrounded by cladding with a lower refractive index. Light entering the core at suitable angles strikes the core-cladding interface at angles greater than the critical angle, causing the light to be reflected back into the core. This process repeats along the length of the fibre, guiding the light even if the fibre is bent.

Option A is irrelevant to the principle of light propagation. Option B is incorrect; light is transmitted, not absorbed and re-emitted in this way. Option C, scattering, generally leads to loss of signal, not efficient transmission through zigzag paths. TIR is the key phenomenon enabling light transmission through optical fibres over long distances with minimal loss.

7. Two bodies of mass M each are placed R distance apart. In another syst

Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed $\frac{R}{2}$ distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be

[amp_mcq option1=”16 F” option2=”1 F” option3=”4 F” option4=”None of the above” correct=”option1″]

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UPSC NDA-2 – 2019

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The correct answer is A) 16 F.

The gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton’s Law of Gravitation: $F = G \frac{m_1 m_2}{R^2}$, where G is the gravitational constant.
In the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G \frac{M \times M}{R^2} = G \frac{M^2}{R^2}$. This force is given as F. So, $F = G \frac{M^2}{R^2}$.
In the second system: $m_1′ = 2M$, $m_2′ = 2M$, $R_2 = R/2$. The force is $F_2 = G \frac{(2M) \times (2M)}{(R/2)^2}$.
Calculate $F_2$: $F_2 = G \frac{4M^2}{R^2/4} = G \frac{4M^2}{R^2} \times 4 = 16 G \frac{M^2}{R^2}$.
Substitute the expression for F: $F_2 = 16 F$.

The gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. Doubling the masses quadruples the product of masses ($2M \times 2M = 4M^2$). Halving the distance quarters the squared distance ($(R/2)^2 = R^2/4$), meaning the force is multiplied by 4 due to the inverse square law ($1 / (1/4) = 4$). The combined effect is a multiplication by $4 \times 4 = 16$.

8. The loudness of sound depends upon the

The loudness of sound depends upon the

[amp_mcq option1=”velocity of sound waves in the medium.” option2=”amplitude of the sound waves.” option3=”frequency of the sound waves.” option4=”frequency and velocity of the sound waves.” correct=”option2″]

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UPSC NDA-2 – 2019

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The correct answer is B) amplitude of the sound waves.

Loudness is the perceptual quality of sound that is most closely associated with intensity. Intensity is proportional to the square of the amplitude of the sound wave. A larger amplitude corresponds to a higher intensity and thus a louder sound.

The frequency of sound waves determines the pitch of the sound. The velocity of sound waves in a medium depends on the properties of the medium (like density and elasticity) and temperature, but it does not directly determine the loudness for a given source and medium. While velocity and frequency are related to wavelength ($\text{velocity} = \text{frequency} \times \text{wavelength}$), loudness is fundamentally related to the energy or power carried by the wave, which is proportional to the square of its amplitude.

9. A pendulum clock is lifted to a height where the gravitational acceler

A pendulum clock is lifted to a height where the gravitational acceleration has a certain value g. Another pendulum clock of same length but of double the mass of the bob is lifted to another height where the gravitational acceleration is g/2. The time period of the second pendulum would be :
(in terms of period T of the first pendulum)

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

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UPSC NDA-2 – 2019

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The correct answer is A) $\sqrt{2}$ T.

The time period (T) of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$, where L is the length of the pendulum and g is the acceleration due to gravity. The mass of the bob does not affect the time period of a simple pendulum.
For the first pendulum: Length $L_1=L$, gravity $g_1=g$. Time period $T_1 = 2\pi \sqrt{\frac{L}{g}} = T$.
For the second pendulum: Length $L_2=L$ (stated as same length), mass $M_2=2M_1$ (mass does not affect T), gravity $g_2=g/2$. Time period $T_2 = 2\pi \sqrt{\frac{L_2}{g_2}} = 2\pi \sqrt{\frac{L}{g/2}} = 2\pi \sqrt{\frac{2L}{g}}$.
We can rewrite $T_2$ in terms of $T_1$: $T_2 = \sqrt{2} \times (2\pi \sqrt{\frac{L}{g}}) = \sqrt{2} T_1 = \sqrt{2} T$.

The time period of a simple pendulum is independent of the mass and amplitude (for small oscillations) of the bob. It depends only on the length of the string and the local acceleration due to gravity. The question tests the understanding that mass does not influence the time period and how the time period scales with changes in gravitational acceleration.

10. Water is heated with a coil of resistance R connected to domestic supp

Water is heated with a coil of resistance R connected to domestic supply. The rise of temperature of water will depend on
1. Supply voltage.
2. Current passing through the coil.
3. Time for which voltage is supplied.
Select the correct answer from among the following :

[amp_mcq option1=”1, 2 and 3″ option2=”1 and 2 only” option3=”1 only” option4=”2 and 3 only” correct=”option1″]

This question was previously asked in

UPSC NDA-2 – 2019

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The correct answer is A) 1, 2 and 3.

When a resistance coil is connected to a domestic supply, it generates heat according to Joule’s law of heating. The heat produced (H) is given by $H = I^2 R t$, where I is the current, R is the resistance, and t is the time. Using Ohm’s law ($V = IR$), this can also be expressed as $H = \frac{V^2}{R} t$ or $H = V I t$. The rise in temperature of the water is proportional to the heat supplied to it. Therefore, the rise in temperature depends on the supply voltage (V), the current passing through the coil (I), and the time for which the voltage is supplied (t). All three factors influence the amount of heat generated and transferred to the water.

The specific heat capacity and mass of the water also influence the final temperature rise, but the question asks what the heat *production* (and thus potential temperature rise) depends on from the given options related to the electrical supply and coil. The supply voltage and the resistance of the coil determine the current ($I = V/R$). So, if voltage and time are given, the current is implicitly determined by the resistance, and heat depends on V, R, and t. If voltage and current are given, resistance is implicitly determined ($R=V/I$), and heat depends on V, I, and t. If current, resistance, and time are given, heat is directly $I^2 R t$. Since the options list voltage, current, and time separately, it means that the heat produced (and temperature rise) is a function of all these parameters in one form of the heat equation or another.