UPSC NDA-1 Archives

41. Which one of the following is the correct arrangement of the given pla

Which one of the following is the correct arrangement of the given planets in descending order of their density (in gm/cm³)?

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”Earth

”Jupiter

”Earth

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The correct option is C.

Planets can be broadly classified into terrestrial (rocky) planets and Jovian (gas giant) planets. Terrestrial planets (like Earth and Venus) are generally much denser than gas giants (like Jupiter and Saturn) because they are primarily composed of rock and metal, while gas giants are mostly hydrogen and helium.

Approximate average densities in g/cm³:
– Earth: ~5.52
– Venus: ~5.24
– Jupiter: ~1.33
– Saturn: ~0.69 (Saturn is less dense than water!)
Arranging these in descending order: Earth > Venus > Jupiter > Saturn.

42. In terms of geological time scale, the quaternary period consists of t

In terms of geological time scale, the quaternary period consists of two epochs. They are :

Pleistocene and Pliocene

Holocene and Pleistocene

Pleistocene and Miocene

Holocene and Eocene

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The Quaternary period, the most recent geological period in the Cenozoic Era, is divided into two epochs: the Pleistocene and the Holocene.

The Pleistocene epoch is characterized by repeated glaciation events and the evolution of modern humans. The Holocene epoch is the current epoch, beginning after the last glacial period ended, approximately 11,700 years ago, and is characterized by the rise of human civilization.

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43. Which of the following statements is/are correct ? 1. Hypocenter is

Which of the following statements is/are correct ?

1. Hypocenter is the point on the surface of the Earth, nearest to the focus.
2. Velocity of earthquake waves is higher in denser materials.
3. P waves move faster and are the first to arrive at the surface of the Earth.

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1 and 2

2 and 3

1 and 3

3 only

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The correct option is B.

Statement 1 is incorrect because the hypocenter is the point of origin of the earthquake *within* the Earth, while the epicenter is the point *on the surface* directly above the hypocenter. Statement 2 is generally correct as seismic wave velocity increases with the rigidity and density of the medium. Statement 3 is correct; P waves (primary or compressional waves) are faster than S waves (secondary or shear waves) and arrive first.

1. Hypocenter (or focus) is the point below the Earth’s surface where the earthquake originates. The epicenter is the point on the Earth’s surface directly above the hypocenter.
2. Seismic wave speeds depend on the elastic properties and density of the material. Wave speed is proportional to the square root of the ratio of an elastic modulus (like bulk modulus or shear modulus) to density. While density increases with depth, the elastic moduli generally increase faster, leading to an overall increase in wave speed with depth in the Earth’s mantle and core.
3. P waves are longitudinal waves that can travel through solids, liquids, and gases. S waves are transverse waves that can only travel through solids. Because P waves involve compression and expansion, they travel faster than S waves, which involve shearing motion.

44. Which one of the following is not an igneous rock ?

Which one of the following is not an igneous rock ?

Granite

Slate

Basalt

Gabbro

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The correct option is B.

Igneous rocks are formed from the cooling and solidification of molten rock (magma or lava). Metamorphic rocks are formed from existing rocks (igneous, sedimentary, or other metamorphic rocks) that have been changed by heat, pressure, or chemical reactions. Slate is a metamorphic rock.

Granite is a coarse-grained intrusive igneous rock. Basalt is a fine-grained extrusive igneous rock. Gabbro is a coarse-grained intrusive igneous rock, compositionally similar to basalt. Slate is a fine-grained, foliated metamorphic rock formed by the low-grade regional metamorphism of shale or mudstone.

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45. Two forces of 5.0 N each are acting on a point mass. If the angle betw

Two forces of 5.0 N each are acting on a point mass. If the angle between the forces is 60°, then the net force acting on the point mass has magnitude close to :

8.6 N

4.3 N

5.0 N

6.7 N

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The correct option is A.

The magnitude of the resultant of two vectors (forces) can be found using the formula derived from the law of cosines: $|R| = \sqrt{|F_1|^2 + |F_2|^2 + 2|F_1||F_2|\cos\theta}$, where $\theta$ is the angle between the forces.

Given $|F_1| = 5.0$ N, $|F_2| = 5.0$ N, and $\theta = 60°$. The magnitude of the resultant force is $|R| = \sqrt{5.0^2 + 5.0^2 + 2 \cdot 5.0 \cdot 5.0 \cdot \cos(60°)}$.
$|R| = \sqrt{25 + 25 + 2 \cdot 25 \cdot (1/2)} = \sqrt{50 + 25} = \sqrt{75}$.
$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$.
Using the approximation $\sqrt{3} \approx 1.732$, $|R| \approx 5 \times 1.732 = 8.66$.
The closest value among the options is 8.6 N.

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46. A particle is moving in a circle of radius R with a constant speed v.

A particle is moving in a circle of radius R with a constant speed v. Its average acceleration over the time when it moves over half the circle is :

$rac{v^2}{R}$

$rac{pi v^2}{2R}$

$rac{2v^2}{pi R}$

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The correct option is C.

Average acceleration is defined as the change in velocity divided by the time taken. For a particle moving in a circle with constant speed, the velocity vector changes direction continuously. Over half a circle, the initial and final velocity vectors are in opposite directions.

Let the particle start at $\theta=0$ (position vector $R\hat{i}$) and move counter-clockwise. The initial velocity is $\vec{v}_i = v\hat{j}$. After moving half a circle ($\pi$ radians), it is at $\theta=\pi$ (position vector $-R\hat{i}$). The final velocity is $\vec{v}_f = -v\hat{j}$. The change in velocity is $\Delta\vec{v} = \vec{v}_f – \vec{v}_i = -v\hat{j} – v\hat{j} = -2v\hat{j}$. The magnitude of the change in velocity is $|\Delta\vec{v}| = 2v$. The distance covered is half the circumference, $\pi R$. The time taken is $t = (\pi R) / v$. The average acceleration is $\vec{a}_{avg} = \Delta\vec{v}/t = (-2v\hat{j}) / (\pi R / v) = -\frac{2v^2}{\pi R}\hat{j}$. The magnitude of the average acceleration is $|\vec{a}_{avg}| = \frac{2v^2}{\pi R}$.

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47. Two identical containers X and Y are connected at the bottom by a thin

Two identical containers X and Y are connected at the bottom by a thin tube of negligible volume. The tube has a valve in it, as shown in the figure. Initially container X has a liquid filled up to height h in it and container Y is empty. When the valve is opened, both containers have equal amount of liquid in equilibrium. If the initial (before the valve is opened) potential energy of the liquid is Pᵢ and the final potential energy is Pғ, then:

Pᵢ=Pғ

Pғ = 1/4 * Pᵢ

Pᵢ=2Pғ

Pғ = 1/8 * Pᵢ

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The correct option is C.

Potential energy of a continuous mass distribution like liquid is the integral of $y \cdot dm \cdot g$. For a liquid column of uniform density in a cylindrical container, the center of mass is at half the height. The potential energy is proportional to the mass and the height of the center of mass. When the liquid redistributes to reach equilibrium in connected identical containers, the total volume is conserved, and the liquid levels become equal.

Let A be the cross-sectional area of the identical containers. Initially, all liquid is in X up to height h. Volume $V = Ah$. Mass $m = \rho V = \rho Ah$. The initial potential energy is $P_i = mg(h/2) = (\rho Ah)g(h/2) = \frac{1}{2}\rho Ag h^2$. When the valve is opened, the liquid distributes equally by volume between X and Y, and the levels are equal. Let the final height in both be $h_f$. Total volume $Ah = A h_f + A h_f = 2Ah_f$, so $h_f = h/2$. The liquid in X has mass $m_X = \rho A (h/2)$, and its CM is at $(h/2)/2 = h/4$. PE in X is $P_{fX} = m_X g (h/4) = (\rho Ah/2)g(h/4) = \frac{1}{8}\rho Ag h^2$. Similarly, PE in Y is $P_{fY} = \frac{1}{8}\rho Ag h^2$. The total final potential energy is $P_f = P_{fX} + P_{fY} = \frac{1}{8}\rho Ag h^2 + \frac{1}{8}\rho Ag h^2 = \frac{1}{4}\rho Ag h^2$. Comparing $P_i$ and $P_f$: $P_i = \frac{1}{2}\rho Ag h^2$ and $P_f = \frac{1}{4}\rho Ag h^2$. Thus, $P_i = 2P_f$.

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48. A sphere of volume V is made of a material with lower density than wat

A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f₁V (f₁ < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f₂V. Then: [amp_mcq option1="f₂=f₁" option2="f₂ = (a/g) * f₁" option3="f₂ > f₁” option4=”f₂ = (g/(g-a)) * f₁” correct=”option1″]

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The correct option is A.

The fraction of volume submerged when an object floats in a fluid is determined by the ratio of the object’s density to the fluid’s density ($f = \rho_{object} / \rho_{fluid}$). This relationship arises from the condition that the buoyant force equals the weight of the object. Both the buoyant force (which depends on the effective acceleration) and the weight (which also depends on the effective acceleration) are directly proportional to the effective acceleration. Therefore, the ratio of densities, and hence the fraction of submerged volume, is independent of the magnitude of the uniform effective acceleration (like gravity) acting on the system.

On Earth, the buoyant force is $F_{B1} = \rho_w (f_1V) g$ and the weight is $W_1 = \rho_s V g$. Since it floats, $F_{B1} = W_1 \implies \rho_w f_1 V g = \rho_s V g \implies f_1 = \rho_s / \rho_w$. On the spaceship with effective acceleration $a$, the buoyant force is $F_{B2} = \rho_w (f_2V) a$ and the weight is $W_2 = \rho_s V a$. Since it floats, $F_{B2} = W_2 \implies \rho_w f_2 V a = \rho_s V a \implies f_2 = \rho_s / \rho_w$. Thus, $f_1 = f_2$. The magnitude of the effective acceleration affects the magnitude of the forces but not their ratio when determining the submerged fraction.

49. A stone is thrown horizontally from the top of a 20 m high building wi

A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s² and neglecting air resistance will give:

R=12m

R=18m

R=24m

R=30m

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The correct option is C.

For projectile motion, the horizontal and vertical components of motion are independent (neglecting air resistance). The time it takes for the stone to hit the ground is determined by the vertical distance and the acceleration due to gravity. The horizontal distance (range) is then calculated by multiplying the horizontal velocity by the time of flight.

The vertical motion is governed by $y = v_{y0}t + \frac{1}{2}gt^2$. Since the stone is thrown horizontally, $v_{y0} = 0$. The height is $y = 20$ m and $g = 10$ m/s². So, $20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \implies 20 = 5t^2 \implies t^2 = 4 \implies t = 2$ s (time of flight). The horizontal motion is uniform with velocity $v_x = 12$ m/s. The range $R$ is given by $R = v_x \cdot t = 12 \text{ m/s} \cdot 2 \text{ s} = 24$ m.

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50. Two identical spring balances S₁ and S₂ are connected one after the ot

Two identical spring balances S₁ and S₂ are connected one after the other and are held vertically as shown in the figure. A mass of 10 kg is hanging from S₂. If the readings on S₁ and S₂ are W₁ and W₂ respectively, then:

W₁=5kg and W₂=10kg

W₁=10kg and W₂=5kg

W₁=5kg and W₂=5kg

W₁=10kg and W₂=10kg

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The correct option is D.

When identical spring balances are connected in series and a mass is suspended from the lower one, the lower balance measures the weight of the suspended mass. The upper balance measures the total tension required to support everything hanging below it, which includes the lower balance (assuming negligible weight for the balance itself) and the suspended mass. Thus, both balances will read the weight of the 10 kg mass.

Spring balances measure force (tension). When used to measure weight under gravity, they are often calibrated to display mass (mass = weight / g). In this setup, the tension in the lower spring (S₂) is equal to the weight of the 10 kg mass. The tension in the upper spring (S₁) is equal to the sum of the tensions from S₂ and the weight of S₂ itself. Assuming the spring balances have negligible weight, S₁ measures the same tension as S₂, which is the weight of the 10 kg mass. If the mass is 10 kg, its weight is approximately 10g N. A spring balance calibrated in kg would read 10 kg for a 10 kg mass under standard gravity.

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