UPSC NDA-1

31. The displacement-time graph of a particle acted upon by a constant for

The displacement-time graph of a particle acted upon by a constant force is

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a straight line
a circle
a parabola
any curve depending upon initial conditions
This question was previously asked in
UPSC NDA-1 – 2015
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The correct option is C. The displacement-time graph of a particle acted upon by a constant force is a parabola.
According to Newton’s second law, a constant force (F) acting on a particle of constant mass (m) results in constant acceleration (a), where F = ma.
For motion under constant acceleration, the displacement (s) of a particle from its initial position as a function of time (t) is given by the kinematic equation: s = ut + (1/2)at², where u is the initial velocity and a is the constant acceleration.
This equation is a quadratic function of time (t).
The general form of the displacement-time graph for constant acceleration is a parabola. If the initial velocity is zero (u=0), the equation simplifies to s = (1/2)at², which is clearly a parabola passing through the origin. If there is an initial velocity, the parabola is shifted and possibly rotated, but remains a parabola. A straight line graph represents constant velocity (zero acceleration, thus zero net force).

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32. If radius of the earth were to shrink by 1%, its mass remaining the sa

If radius of the earth were to shrink by 1%, its mass remaining the same, g would decrease by nearly

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1%
2%
3%
4%
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UPSC NDA-1 – 2015
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The correct option is B. g would increase by nearly 2%. (Note: The question contains a likely typo stating “decrease” instead of “increase” or “change”).
The acceleration due to gravity (g) on the surface of a sphere of mass M and radius R is given by g = GM/R², where G is the gravitational constant.
If the radius R shrinks by 1%, the new radius R’ = R – 0.01R = 0.99R. The mass M remains the same.
The new gravity g’ on the surface would be g’ = GM/(R’)² = GM/(0.99R)² = GM/(0.9801R²) = (1/0.9801) * (GM/R²) ≈ 1.01928 * g.
The change in gravity is g’ – g = 1.01928g – g = 0.01928g.
The percentage change in gravity is ((g’ – g)/g) * 100% = (0.01928g / g) * 100% ≈ 1.93%.
Using the approximation for small changes, if R changes by a small fraction ε (R’ = R(1-ε)), then g changes by approximately 2ε (g’ ≈ g(1+2ε)). Here ε = 0.01 (1% decrease in R). So g changes by approximately 2 * 0.01 = 0.02, which is a 2% increase.
A decrease in Earth’s radius while keeping the mass constant would result in a stronger gravitational pull at the surface, hence an *increase* in ‘g’. The question states “g would decrease by nearly”, which contradicts the physics. Assuming the question intends to ask about the magnitude of the percentage change calculated from the given change in radius, the change is approximately 2% (an increase). Option B (2%) is the closest value to the calculated change. It is highly probable that “decrease” is a typo.

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33. Ultrasonic waves of frequency 3 × 10⁵ Hz are passed through a medium w

Ultrasonic waves of frequency 3 × 10⁵ Hz are passed through a medium where speed of sound is 10 times that in air (Speed of sound in air is 300 m/s). The wavelength of this wave in that medium will be of the order of

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1 cm
10 cm
100 cm
0·1 cm
This question was previously asked in
UPSC NDA-1 – 2015
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The correct option is A. The wavelength will be of the order of 1 cm.
The relationship between wave speed (v), frequency (f), and wavelength (λ) is given by v = fλ.
Given frequency f = 3 × 10⁵ Hz.
Speed of sound in air v_air = 300 m/s.
Speed of sound in the medium v_medium = 10 × v_air = 10 × 300 m/s = 3000 m/s.
We need to find the wavelength λ in the medium. Using the formula λ = v_medium / f:
λ = 3000 m/s / (3 × 10⁵ Hz) = 3000 / 300000 m = 1/100 m = 0.01 m.
To convert meters to centimeters, we multiply by 100:
0.01 m × 100 cm/m = 1 cm.
The wavelength of the ultrasonic wave in the given medium is 1 cm.

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34. Statement-I: Due to diffused or irregular reflection of light, a close

Statement-I: Due to diffused or irregular reflection of light, a closed room gets light even if no direct sunlight falls inside the room.
Statement-II: Irregular reflection, where the reflected rays are not parallel, does not follow the laws of reflection.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
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The correct option is C. Statement I is true, but Statement II is false.
Statement I is true. Light entering a room, whether direct sunlight or light from outside reflected by various surfaces, undergoes diffused reflection when it strikes the walls, furniture, and other objects inside the room. Diffused reflection scatters light in multiple directions, illuminating the entire room even areas not directly hit by the initial light rays.
Statement II is false. Irregular reflection (diffused reflection) occurs when light strikes a rough or uneven surface. While the reflected rays scatter in various directions, *each individual ray* still obeys the laws of reflection: the angle of incidence equals the angle of reflection, and the incident ray, the normal to the surface at the point of incidence, and the reflected ray all lie in the same plane. The scattering happens because the normals to the surface at different points are oriented in different directions.
The difference between regular (specular) reflection and irregular (diffused) reflection lies in the nature of the reflecting surface, not in the adherence to the laws of reflection. A smooth surface causes regular reflection (like a mirror), while a rough surface causes diffused reflection (like a wall).

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35. Statement-I: While putting clothes for drying up, we spread them out.

Statement-I: While putting clothes for drying up, we spread them out.
Statement-II: The rate of evaporation increases with an increase in surface area.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
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UPSC NDA-1 – 2015
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The correct option is A. Both statements are individually true, and Statement II is the correct explanation of Statement I.
Statement I is true. Spreading out clothes increases their surface area exposed to the air, which is a common practice to speed up drying.
Statement II is true. Evaporation is a surface phenomenon, where liquid molecules gain enough energy to escape into the gaseous phase from the surface of the liquid. The rate of evaporation is directly proportional to the surface area exposed to the surrounding environment (air). Other factors affecting evaporation rate include temperature, humidity, and wind speed.
By spreading clothes, we maximize the surface area from which water can evaporate, thus increasing the rate of drying. Statement II correctly explains why Statement I is an effective drying method.

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36. Statement-I: A body weighs less on a hill top than on earth’s surface

Statement-I: A body weighs less on a hill top than on earth’s surface even though its mass remains unchanged.
Statement-II: The acceleration due to gravity of the earth decreases with height.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
This question was previously asked in
UPSC NDA-1 – 2015
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The correct option is A. Both statements are individually true, and Statement II is the correct explanation of Statement I.
Statement I is true because weight (W) is defined as mass (m) multiplied by the acceleration due to gravity (g), W = mg. A body’s mass remains constant regardless of location.
Statement II is true because the acceleration due to gravity (g) decreases with increasing height above the Earth’s surface. The gravitational force, and hence ‘g’, depends on the distance from the center of the Earth. At a higher altitude like a hill top, this distance is greater than on the surface at sea level. The relationship is approximately g’ = g(R/(R+h))², where R is Earth’s radius and h is height. As h increases, g’ decreases.
Since weight is directly proportional to ‘g’ (with constant mass), a decrease in ‘g’ at a hill top compared to the Earth’s surface directly causes the body to weigh less. Therefore, Statement II provides the correct reason for Statement I.

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37. Statement-I: Oxygen gas is easily produced at a faster rate by heating

Statement-I: Oxygen gas is easily produced at a faster rate by heating a mixture of potassium chlorate and manganese dioxide than heating potassium chlorate alone.
Statement-II: Manganese dioxide acts as a negative catalyst.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
This question was previously asked in
UPSC NDA-1 – 2015
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The correct option is C. Statement I is true, but Statement II is false.
Statement I correctly describes the effect of adding manganese dioxide (MnO₂) to potassium chlorate (KClO₃) when heated. MnO₂ acts as a catalyst, speeding up the decomposition of KClO₃ into potassium chloride (KCl) and oxygen gas (O₂). The reaction is 2KClO₃ (s) → 2KCl (s) + 3O₂ (g).
Statement II claims that manganese dioxide acts as a negative catalyst. A negative catalyst (or inhibitor) slows down a reaction. Since MnO₂ increases the rate of decomposition of KClO₃, it is a positive catalyst.
Catalysts increase the rate of a chemical reaction without being consumed in the process. Positive catalysts increase the rate, while negative catalysts decrease it. Manganese dioxide is a common positive catalyst used in the laboratory preparation of oxygen from potassium chlorate.

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38. Statement-I: Diamond is very bright. Statement-II: Diamond has very lo

Statement-I: Diamond is very bright.
Statement-II: Diamond has very low refractive index.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
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UPSC NDA-1 – 2015
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Statement I is true, but Statement II is false.
Statement I is true because diamond is known for its exceptional brilliance and sparkle, which makes it a highly valued gemstone. Statement II is false; diamond actually has a very *high* refractive index (approximately 2.42), one of the highest among naturally occurring transparent minerals.
The brilliance and fire of a diamond are primarily due to its high refractive index and high dispersion (the ability to split white light into its constituent colours). The high refractive index causes significant bending of light entering the stone and allows for high internal reflection, leading to light bouncing around inside before exiting, creating sparkle. If diamond had a *low* refractive index, light would pass through with less bending and reflection, resulting in significantly less brilliance.

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39. Statement-I: Colour of nitrogen dioxide changes to colourless at low t

Statement-I: Colour of nitrogen dioxide changes to colourless at low temperature.
Statement-II: At low temperature Nitrogen tetroxide (N₂O₄) is formed which is colourless.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
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Both Statement I and Statement II are individually true, and Statement II is the correct explanation of Statement I.
Statement I is true; the reddish-brown colour of nitrogen dioxide gas disappears at low temperatures. Statement II is also true; at low temperatures, nitrogen dioxide dimerizes to form colourless nitrogen tetroxide (N₂O₄).
Nitrogen dioxide (NO₂) is an equilibrium mixture with its dimer, dinitrogen tetroxide (N₂O₄):
2NO₂(g) (reddish-brown) ⇌ N₂O₄(g) (colourless)
This dimerization reaction is exothermic (ΔH < 0). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium towards the exothermic reaction, favoring the formation of N₂O₄. As the temperature decreases, more NO₂ dimerizes into colourless N₂O₄, causing the colour of the gas mixture to fade to colourless. Statement II correctly identifies the formation of colourless N₂O₄ as the reason for the colour change.

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40. Statement-I: The granules of modern gunpowder (also called black powde

Statement-I: The granules of modern gunpowder (also called black powder) are typically coated with Graphite.
Statement-II: Graphite prevents the build-up of electrostatic charge.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
This question was previously asked in
UPSC NDA-1 – 2015
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Both Statement I and Statement II are individually true, and Statement II is the correct explanation of Statement I.
Statement I is true; graphite is commonly used as a coating for black powder granules. Statement II is also true; graphite is a good conductor and helps dissipate electrostatic charge.
Black powder (gunpowder) components, especially sulfur and potassium nitrate, can generate static electricity through friction during handling and transport. Static discharge can ignite the powder, leading to dangerous explosions. Coating the granules with a conductive material like graphite prevents static buildup, making the powder safer to handle and improving its flow properties. Therefore, Statement II provides the reason for the practice described in Statement I.

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