UPSC NDA-2 Archives

21. Consider the following part of an electric circuit : [Image of circuit

Consider the following part of an electric circuit :
[Image of circuit diagram is implied here]
The total electrical resistance in the given part of the electric circuit is

$rac{15}{8}$ ohm

$rac{15}{7}$ ohm

15 ohm

$rac{17}{3}$ ohm

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While the circuit diagram is not provided in the text, assuming a common configuration that results in one of the given simple fractional values, let’s test a common parallel combination. If the circuit consists of two resistors with resistances 3 ohm and 5 ohm connected in parallel, the total electrical resistance ($R_{total}$) is given by the formula for parallel resistors:
$R_{total} = \frac{R_1 \times R_2}{R_1 + R_2}$
Let $R_1 = 3 \, \Omega$ and $R_2 = 5 \, \Omega$.
$R_{total} = \frac{3 \, \Omega \times 5 \, \Omega}{3 \, \Omega + 5 \, \Omega} = \frac{15 \, \Omega^2}{8 \, \Omega} = \frac{15}{8} \, \Omega$.
This result matches option A. This configuration is a plausible intended diagram for such a question structure.

– Resistors in parallel: The reciprocal of the total resistance is the sum of the reciprocals of individual resistances ($1/R_{total} = 1/R_1 + 1/R_2 + \dots$). For two resistors, this simplifies to $R_{total} = (R_1 \times R_2) / (R_1 + R_2)$.
– Assuming common resistor values (like 3 and 5 ohm) and standard circuit configurations (like parallel) often helps in solving such MCQs when the diagram is missing but options are specific.

Series combination of resistors ($R_{total} = R_1 + R_2 + \dots$) results in a resistance greater than any individual resistance. Parallel combination results in a resistance smaller than the smallest individual resistance. In this case, 15/8 = 1.875 is smaller than both 3 and 5 ohm. Other standard configurations like series-parallel combinations could also result in fractional resistances, but the 3||5 parallel arrangement directly yields 15/8.

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22. If the work done on the system or by the system is zero, which one of

If the work done on the system or by the system is zero, which one of the following statements for a gas kept at a certain temperature is correct ?

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Change in internal energy of the system is equal to flow of heat in or out of the system.

Change in internal energy of the system is less than heat transferred.

Change in internal energy of the system is more than the heat flow.

Cannot be determined.

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The first law of thermodynamics states that the change in internal energy of a system ($\Delta U$) is equal to the heat added to the system ($Q$) minus the work done by the system ($W$). This can be written as $\Delta U = Q – W$. The question states that the work done on the system or by the system is zero, meaning $W = 0$. Substituting this into the first law equation gives $\Delta U = Q – 0$, which simplifies to $\Delta U = Q$. Therefore, the change in internal energy of the system is equal to the flow of heat in or out of the system.

– First Law of Thermodynamics: $\Delta U = Q – W$.
– Work done is zero ($W=0$).
– The change in internal energy is directly equal to the heat transfer when no work is done.

This condition ($W=0$) occurs in processes where the volume of the system does not change (isochoric process) or when the system is rigid. Heat flowing into the system increases its internal energy ($Q > 0, \Delta U > 0$), and heat flowing out decreases its internal energy ($Q < 0, \Delta U < 0$).

23. A student measures certain lengths using a meter scale having least co

A student measures certain lengths using a meter scale having least count equal to 1 mm. Which one of the following measurements is more precise ?

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0$cdot$50 mm

29$cdot$07 cm

0$cdot$925 m

910 mm

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Precision refers to the level of detail and resolution in a measurement. With a meter scale having a least count of 1 mm, measurements are typically read directly to the nearest 1 mm, and a careful estimation can often be made to the nearest tenth of the least count, i.e., 0.1 mm.
Let’s express all measurements in mm:
A) 0.50 mm – recorded to two decimal places. This suggests precision to 0.01 mm.
B) 29.07 cm = 290.7 mm – recorded to one decimal place in mm. This suggests precision to 0.1 mm.
C) 0.925 m = 925 mm – recorded to the nearest whole number in mm. This suggests precision to 1 mm.
D) 910 mm – recorded to the nearest whole number in mm (assuming the trailing zero is significant). This suggests precision to 1 mm.
Comparing the implied precisions: A (0.01 mm) < B (0.1 mm) < C and D (1 mm). Although measuring 0.50 mm or achieving 0.01 mm precision with a 1 mm least count scale is highly questionable or impossible in practice for a single measurement, the *way the number is recorded* implies a certain level of precision. Among the given options, 0.50 mm is recorded with the highest implied precision (to the hundredths of a millimeter). Therefore, as recorded, it is the most precise measurement listed.

– The precision of a measurement is indicated by the number of significant figures and the position of the last significant digit relative to the least count of the instrument.
– With a scale of 1 mm least count, readings are usually taken to the nearest 1 mm or estimated to the nearest 0.1 mm.
– The number of decimal places in the unit of the least count often suggests the level of precision in the recorded value.

– Recording 0.50 mm suggests an uncertainty in the hundredths place, typically interpreted as $\pm 0.005$ mm, which is much smaller than the least count of 1 mm.
– Recording 290.7 mm suggests uncertainty in the tenths place, typically interpreted as $\pm 0.05$ mm, which is a reasonable estimation precision with a 1 mm scale.
– Recording 925 mm or 910 mm suggests uncertainty in the units place, typically interpreted as $\pm 0.5$ mm, which is standard for direct readings.
– Despite the practical limitations, the recording 0.50 mm implies the highest level of precision among the given options.

24. “The sum of emfs and potential differences around a closed loop equals

“The sum of emfs and potential differences around a closed loop equals zero” is a consequence of

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Ohm's law.

Conservation of charge.

Conservation of momentum.

Conservation of energy.

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Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of the potential differences (voltages) around any closed loop in a circuit is zero. This law is a direct consequence of the principle of conservation of energy. As a unit charge traverses a closed loop and returns to its starting point, the net work done on it by the electric field (and thus the net change in potential energy) must be zero if no energy is gained or lost within the loop from non-electric sources (which is accounted for by EMFs).

– Kirchhoff’s Voltage Law (KVL) is mathematically expressed as $\sum V = 0$ around any closed loop.
– KVL is based on the conservative nature of the electric field and the conservation of energy.
– Potential difference is defined as the change in potential energy per unit charge. Traversing a closed loop means returning to the initial potential, so the total change in potential (and potential energy) is zero.

– Ohm’s law ($V=IR$) is a specific relationship for a resistive component and is used *within* a loop analysis, but KVL is a fundamental principle for the loop itself.
– Conservation of charge is the basis for Kirchhoff’s Current Law (KCL), which deals with currents at a junction.
– Conservation of momentum is a principle applied in mechanics, not directly related to circuit laws in this manner.

25. A car starts from Bengaluru, goes 50 km in a straight line towards sou

A car starts from Bengaluru, goes 50 km in a straight line towards south, immediately turns around and returns to Bengaluru. The time taken for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip

is 0.

is 50 km/hr.

is 25 km/hr.

cannot be calculated without knowing acceleration.

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UPSC NDA-2 – 2019

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Average velocity is defined as the total displacement divided by the total time taken. Displacement is the change in position from the starting point to the ending point. The car starts from Bengaluru and returns to Bengaluru. Therefore, the initial position and the final position are the same. This means the total displacement is zero.
Total displacement = Final position – Initial position = Bengaluru – Bengaluru = 0 km.
Total time taken = 2 hours.
Average velocity = Total displacement / Total time = 0 km / 2 hours = 0 km/hr.
The magnitude of the average velocity is the absolute value of the average velocity, which is $|0| = 0$.

– Average velocity is a vector quantity and depends on displacement, not total distance traveled.
– Displacement is the shortest straight-line distance from the starting point to the ending point, along with direction. If the start and end points are the same, displacement is zero.
– Average speed is a scalar quantity and depends on the total distance traveled divided by the total time. In this case, the average speed is 100 km / 2 hr = 50 km/hr.

– The path taken (straight line towards south and back) is relevant for calculating distance but not displacement when the trip is a round trip back to the origin.
– The acceleration is not constant during the trip (it changes direction when turning around), but this information is not needed to calculate average velocity, which only depends on total displacement and total time.

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26. Two substances of densities $\rho_1$ and $\rho_2$ are mixed in equal v

Two substances of densities $\rho_1$ and $\rho_2$ are mixed in equal volume and their relative density is 4. When they are mixed in equal masses, relative density is 3. The values of $\rho_1$ and $\rho_2$ respectively are

6, 2

3, 5

12, 4

9, 3

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Let the densities be $\rho_1$ and $\rho_2$. Relative density is numerically equal to density assuming water density is 1.
Case 1: Mixed in equal volume ($V$). Total mass $m = m_1 + m_2 = \rho_1 V + \rho_2 V = (\rho_1 + \rho_2)V$. Total volume $V_{total} = V + V = 2V$. Density $\rho_{mix, V} = \frac{(\rho_1 + \rho_2)V}{2V} = \frac{\rho_1 + \rho_2}{2}$. Given relative density is 4, so $\frac{\rho_1 + \rho_2}{2} = 4 \implies \rho_1 + \rho_2 = 8$.
Case 2: Mixed in equal masses ($m$). Volume $V_1 = \frac{m}{\rho_1}$, $V_2 = \frac{m}{\rho_2}$. Total mass $m_{total} = m + m = 2m$. Total volume $V_{total} = \frac{m}{\rho_1} + \frac{m}{\rho_2} = m\left(\frac{\rho_1 + \rho_2}{\rho_1 \rho_2}\right)$. Density $\rho_{mix, m} = \frac{2m}{m\left(\frac{\rho_1 + \rho_2}{\rho_1 \rho_2}\right)} = \frac{2\rho_1 \rho_2}{\rho_1 + \rho_2}$. Given relative density is 3, so $\frac{2\rho_1 \rho_2}{\rho_1 + \rho_2} = 3$.
Substitute $\rho_1 + \rho_2 = 8$ into the second equation: $\frac{2\rho_1 \rho_2}{8} = 3 \implies \frac{\rho_1 \rho_2}{4} = 3 \implies \rho_1 \rho_2 = 12$.
We need two numbers whose sum is 8 and product is 12. These are the roots of the quadratic equation $x^2 – 8x + 12 = 0$, which factors as $(x-2)(x-6) = 0$. The roots are 2 and 6. So, the densities are 2 and 6. Option A is (6, 2), which satisfies both conditions: (6+2)/2 = 4 and (2*6*2)/(6+2) = 24/8 = 3.

– Density of a mixture depends on the densities of the components and the proportions in which they are mixed (by mass or by volume).
– Average density when mixed by volume is the arithmetic mean of the densities.
– Average density when mixed by mass is the harmonic mean of the densities, weighted by mass proportion (or a form related to it). The formula derived $\frac{2 \rho_1 \rho_2}{\rho_1 + \rho_2}$ is 2 times the harmonic mean of $\rho_1$ and $\rho_2$.

– Relative density (or specific gravity) is the ratio of the density of a substance to the density of a reference substance (usually water).
– The problem effectively provides the arithmetic mean and a form of the harmonic mean of the two densities, allowing us to solve for the densities.

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27. A solid disc and a solid sphere have the same mass and same radius. Wh

A solid disc and a solid sphere have the same mass and same radius. Which one has the higher moment of inertia about its centre of mass ?

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The disc

The sphere

Both have the same moment of inertia

The information provided is not sufficient to answer the question

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The moment of inertia ($I$) for a solid disc about an axis through its center and perpendicular to its plane is $I_{disc} = \frac{1}{2}MR^2$. The moment of inertia for a solid sphere about an axis through its center is $I_{sphere} = \frac{2}{5}MR^2$. Given that both the disc and the sphere have the same mass ($M$) and radius ($R$), we compare the coefficients $\frac{1}{2}$ and $\frac{2}{5}$. Since $\frac{1}{2} = \frac{5}{10}$ and $\frac{2}{5} = \frac{4}{10}$, we have $\frac{1}{2} > \frac{2}{5}$. Therefore, $I_{disc} > I_{sphere}$. The solid disc has the higher moment of inertia.

– Moment of inertia depends on the mass distribution relative to the axis of rotation.
– Formulas for the moment of inertia of common shapes are standard results derived from integration.
– For objects of the same mass and radius, the object with more mass distributed further from the axis of rotation will have a higher moment of inertia. In the disc, all mass is at a distance up to R from the axis in a plane, whereas in the sphere, mass is distributed in a volume, including closer to the center.

– Moment of inertia is a measure of an object’s resistance to changes in its rotational motion.
– A higher moment of inertia means it is harder to start or stop the rotation.
– The formulas used are for axes passing through the center of mass, as specified in the question.

28. If an object moves at a non-zero constant acceleration for a certain i

If an object moves at a non-zero constant acceleration for a certain interval of time, then the distance it covers in that time

depends on its initial velocity.

is independent of its initial velocity.

increases linearly with time.

depends on its initial displacement.

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UPSC NDA-2 – 2019

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The distance covered by an object moving with constant acceleration is given by the kinematic equation: $s = ut + \frac{1}{2}at^2$, where $s$ is the distance (or displacement magnitude if direction is constant), $u$ is the initial velocity, $a$ is the constant acceleration, and $t$ is the time interval. This equation clearly shows that the distance covered ($s$) depends on the initial velocity ($u$), assuming $a \neq 0$ and $t > 0$.

– For motion with constant acceleration, the relationship between distance, initial velocity, acceleration, and time is described by standard kinematic equations.
– The equation $s = ut + \frac{1}{2}at^2$ explicitly includes the initial velocity $u$.

– Option B is incorrect because the term $ut$ makes the distance dependent on initial velocity.
– Option C is incorrect because the presence of the $\frac{1}{2}at^2$ term (with $a \neq 0$) means the distance increases quadratically with time, not linearly.
– Option D is incorrect; the distance covered (change in position) is independent of the starting position (initial displacement). Initial displacement affects the final position, but not the distance traveled during the interval.

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29. A sample of ‘soft soap’ contains

A sample of ‘soft soap’ contains

Caesium

Potassium

Calcium

Magnesium

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Soaps are salts of fatty acids. The difference between hard soap and soft soap lies in the cation used. Hard soaps are typically the sodium salts of fatty acids (e.g., derived from NaOH reacting with fat/oil), while soft soaps are typically the potassium salts of fatty acids (e.g., derived from KOH reacting with fat/oil). Therefore, a sample of soft soap contains Potassium.

– Soap is produced by the saponification of fats or oils with a strong alkali.
– Hard soaps use sodium hydroxide (NaOH) as the alkali, resulting in sodium salts of fatty acids.
– Soft soaps use potassium hydroxide (KOH) as the alkali, resulting in potassium salts of fatty acids.

– Soft soaps are generally more soluble in water and produce a richer lather than hard soaps. They are often used in liquid soaps and shaving creams.
– Calcium and Magnesium ions are present in hard water and react with soap to form insoluble precipitates (“soap scum”), which is why soap is less effective in hard water. They are not typically added as components of soap itself.

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30. Which one of the following statements about water is not true ?

Which one of the following statements about water is not true ?

Hydrogen bonds are present in liquid water.

Water has a high boiling point.

Water has a high heat of fusion.

Water is a non-polar molecule.

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Statement D, “Water is a non-polar molecule”, is not true. Water (H2O) has a bent molecular geometry due to the presence of two lone pairs on the oxygen atom. The O-H covalent bonds are polar because oxygen is more electronegative than hydrogen. Because of the bent shape, the bond dipoles do not cancel out, resulting in a net dipole moment for the entire molecule. Therefore, water is a polar molecule.

– Water is a polar molecule due to its bent structure and polar covalent bonds.
– The polarity and the presence of hydrogen atoms bonded to a highly electronegative oxygen atom enable hydrogen bonding between water molecules.
– Hydrogen bonding is responsible for many of water’s unusual properties, such as high boiling point, high heat of fusion, high specific heat capacity, and high surface tension.

– Statement A is true; hydrogen bonds are extensive in liquid water, forming a dynamic network.
– Statement B is true; water has an exceptionally high boiling point (100°C at 1 atm) compared to other molecules of similar size and molar mass without hydrogen bonding (e.g., H2S, which boils at -60°C).
– Statement C is true; water requires a significant amount of energy to melt (heat of fusion) due to the need to break hydrogen bonds holding the ice lattice together.

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