UPSC NDA-1 Archives

11. Which one of the following is not a main greenhouse gas ?

Which one of the following is not a main greenhouse gas ?

[amp_mcq option1=”Water vapour” option2=”Oxygen” option3=”Carbon dioxide” option4=”Methane” correct=”option2″]

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Oxygen (O₂) is not considered a main greenhouse gas. Greenhouse gases are atmospheric gases that absorb and emit infrared radiation, trapping heat in the atmosphere.

Main greenhouse gases include water vapour (H₂O), carbon dioxide (CO₂), methane (CH₄), nitrous oxide (N₂O), and ozone (O₃). Oxygen and Nitrogen, which make up the bulk of the atmosphere, do not absorb infrared radiation effectively and are not greenhouse gases.

Water vapour is the most abundant greenhouse gas in the atmosphere and contributes significantly to the natural greenhouse effect. Carbon dioxide and methane are major contributors to anthropogenic (human-caused) climate change due to their increased concentrations from activities like burning fossil fuels and agriculture.

12. A positive charge is moving towards south in a space where magnetic fi

A positive charge is moving towards south in a space where magnetic field is pointing in the north direction. The moving charge will experience :

[amp_mcq option1=”a deflecting force towards north direction.” option2=”a deflecting force towards east direction.” option3=”a deflecting force towards west direction.” option4=”no deflecting force.” correct=”option4″]

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The force experienced by a positive charge (q > 0) moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the Lorentz force formula: $\vec{F} = q (\vec{v} \times \vec{B})$.

The direction of the velocity $\vec{v}$ is towards south, and the direction of the magnetic field $\vec{B}$ is towards north. These two directions are opposite to each other. The angle between $\vec{v}$ and $\vec{B}$ is 180 degrees.

The magnitude of the cross product $\vec{v} \times \vec{B}$ is $|\vec{v}| |\vec{B}| \sin(\theta)$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Since $\theta = 180^{\circ}$, $\sin(180^{\circ}) = 0$. Therefore, the magnitude of the force $|\vec{F}| = q |\vec{v}| |\vec{B}| \sin(180^{\circ}) = 0$. The moving charge will experience no deflecting force.

13. Which one of the following statements is not true for a flute, a mus

Which one of the following statements is not true for a flute, a musical instrument ?

[amp_mcq option1=”Momentum of waves on the blowing jet determines the loudness of the produced note.” option2=”Arrival time of the waves on the blowing jet determines the pitch of the produced note.” option3=”Sound comes from a vibrating column of air inside the flute.” option4=”Sound comes from a vibrating column of air inside as well as outside the flute.” correct=”option4″]

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In a flute, the sound primarily originates from the vibration of the column of air *inside* the tube. The tube acts as a resonant cavity that reinforces specific frequencies of vibration initiated by the air jet blown across the lip plate.

While there is complex airflow and some turbulence outside the flute, the characteristic musical tone and pitch are produced by the standing waves formed within the air column confined by the tube. The tube length (modified by finger holes) determines the resonant frequencies (pitches).

Statement A is plausible as the force/momentum of the air jet relates to the energy input and amplitude of vibration, affecting loudness. Statement C is true, describing the essential mechanism. Statement B is also arguably true in the sense that the timing of feedback from the air column resonance influences the jet’s oscillation rate, thus affecting the pitch. However, stating that sound comes from a vibrating column of air *outside* as a *main* source is incorrect; the primary resonant vibration is internal.

14. A railway wagon (open at the top) of mass M₁ is moving with speed v₁ a

A railway wagon (open at the top) of mass M₁ is moving with speed v₁ along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M₂ and speed becomes v₂. Taking the rain to be falling vertically and water stationary inside the wagon, the relation between the two speeds v₁ and v₂ is :

[amp_mcq option1=”v₁ = v₂” option2=”1/2 M₁v₁² < 1/2 M₂v₂²" option3="M₁v₁ = M₂v₂" option4="M₁v₁ < M₂v₂" correct="option3"]

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As the rain falls vertically, it has no horizontal momentum. The horizontal momentum of the system (wagon + water) is conserved because there are no external horizontal forces acting on it (assuming negligible friction and air resistance).

The initial horizontal momentum of the wagon is $P_1 = M_1 v_1$. After the rain has collected, the total mass is $M_2$ and the speed is $v_2$. The final horizontal momentum is $P_2 = M_2 v_2$. By conservation of horizontal momentum, $P_1 = P_2$, so $M_1 v_1 = M_2 v_2$.

This is an example of an inelastic collision/process where mass is added to the system. Kinetic energy is not conserved in this case; the kinetic energy of the system decreases because the incoming water had zero horizontal kinetic energy relative to the ground.

15. Shown in the figure are two plane mirrors XY and YZ (XY ⊥ YZ) joined a

Shown in the figure are two plane mirrors XY and YZ (XY ⊥ YZ) joined at their edge. Also shown is a light ray falling on one of the mirrors and reflected back parallel to its original path as a result of this arrangement. The two mirrors are now rotated by an angle θ to their new position X’Y’Z’, as shown. As a result the new reflected ray is at an angle α from the original reflected ray. Then :

[amp_mcq option1=”α = 0″ option2=”α = θ” option3=”α = 2θ” option4=”α = 4θ” correct=”option1″]

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For a ray of light reflecting successively from two plane mirrors placed perpendicular to each other, the final reflected ray is always anti-parallel to the incident ray, regardless of the initial angle of incidence.

This property is characteristic of a corner reflector made of two perpendicular mirrors. The total deviation after two reflections is 180 degrees. When the mirrors are rotated together by an angle θ, the angle between them remains 90 degrees. Therefore, the property holds true, and the final reflected ray remains anti-parallel to the incident ray.

Since the original reflected ray is anti-parallel to the incident ray and the new reflected ray is also anti-parallel to the *same* incident ray, the two reflected rays are parallel to each other and in the same direction (anti-parallel to the incident ray). The angle between them, α, is 0.

16. A rectangle ABCD is kept in front of a concave mirror of focal length

A rectangle ABCD is kept in front of a concave mirror of focal length f with its corners A and B being, respectively, at distances 2f and 3f from the mirror with AB along the principal axis as shown in the figure. It forms an image A’B’C’D’ in front of the mirror. What is the ratio of B’C’ to A’D’ ?

[amp_mcq option1=”1″ option2=”2″ option3=”1/2″ option4=”2/3″ correct=”option3″]

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The ratio of B’C’ to A’D’ is 1/2.

For a concave mirror, the magnification (m) is given by m = -v/u, where v is the image distance and u is the object distance. The size of the image perpendicular to the axis (like B’C’ and A’D’) is related to the object size (BC and AD) by the magnification: Image size = |m| * Object size. Since ABCD is a rectangle, AD = BC.

Using the mirror formula 1/f = 1/u + 1/v:
For point A at u_A = -2f (assuming standard sign convention where f is negative for concave mirror, but distances are given as 2f and 3f. Let’s assume f > 0 is magnitude, and u is negative): u_A = -2f. 1/f = 1/(-2f) + 1/v_A => 1/v_A = 1/f + 1/(2f) = 3/(2f) => v_A = 2f/3. No, if A is at 2f from the mirror, and f is the focal length, then 2f is the radius of curvature C. Object at C forms image at C. So u_A = -2f, then v_A = -2f. Magnification m_A = -v_A/u_A = -(-2f)/(-2f) = -1. A’D’ = |m_A| * AD = AD.
For point B at u_B = -3f: 1/f = 1/(-3f) + 1/v_B => 1/v_B = 1/f + 1/(3f) = 4/(3f) => v_B = 3f/4. Wait, the standard formula for concave mirror with positive f is 1/f = 1/u + 1/v. Let’s use this and assume distances are positive. u_A = 2f, u_B = 3f, f=f. 1/v_A = 1/f – 1/u_A = 1/f – 1/(2f) = (2-1)/(2f) = 1/(2f) => v_A = 2f. Magnification m_A = -v_A/u_A = -(2f)/(2f) = -1. A’D’ = |-1| * AD = AD.
1/v_B = 1/f – 1/u_B = 1/f – 1/(3f) = (3-1)/(3f) = 2/(3f) => v_B = 3f/2. Magnification m_B = -v_B/u_B = -(3f/2)/(3f) = -1/2. B’C’ = |-1/2| * BC = 1/2 * BC.
Since AD = BC, the ratio B’C’ / A’D’ = (1/2 * BC) / AD = 1/2 * (BC/AD) = 1/2 * 1 = 1/2.

17. At which one of the following places did the Danes establish their set

At which one of the following places did the Danes establish their settlement in India ?

[amp_mcq option1=”Chinsura” option2=”Karaikal” option3=”Mahe” option4=”Tranquebar” correct=”option4″]

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The Danes established their settlements in India at Tranquebar (modern Tharangambadi) on the Coromandel Coast and Serampore (Frederiksnagore) in Bengal.

Tranquebar was established in 1620 and served as their primary base for over 200 years before being sold to the British in 1845. Serampore was another significant Danish possession.

Chinsura was a Dutch settlement. Karaikal and Mahe were French settlements.

18. Which one of the following kingdoms was founded by the two brothers Ha

Which one of the following kingdoms was founded by the two brothers Harihar and Bukka ?

[amp_mcq option1=”Bahmani” option2=”Vijayanagara” option3=”Malwa” option4=”Maratha” correct=”option2″]

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The Vijayanagara Empire was founded in 1336 CE by the two brothers, Harihar and Bukka, who were formerly associated with the Kakatiya kingdom of Warangal and later served the Delhi Sultanate.

Harihar I and Bukka Raya I established the Vijayanagara kingdom in the Deccan region in response to the instability and decline of the Delhi Sultanate’s control in the south.

The Bahmani kingdom was another major rival kingdom in the Deccan, founded by Ala-ud-Din Bahman Shah. Malwa was a kingdom in Central India. The Maratha kingdom rose later under Shivaji Maharaj in the 17th century.

19. Who among the following composed the ‘Prayag Prashasti’ of Samudragupt

Who among the following composed the ‘Prayag Prashasti’ of Samudragupta ?

[amp_mcq option1=”Harishena” option2=”Chand Bardai” option3=”Vishakhadatta” option4=”Kalidasa” correct=”option1″]

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The ‘Prayag Prashasti’, also known as the Allahabad Pillar Inscription, which describes the reign and military campaigns of the Gupta emperor Samudragupta, was composed by his court poet and minister, Harishena.

Harishena’s inscription is a significant historical source providing details about Samudragupta’s conquests and his policy towards different kingdoms.

Chand Bardai was a poet associated with Prithviraj Chauhan. Vishakhadatta was a playwright known for ‘Mudrarakshasa’ and ‘Devichandraguptam’. Kalidasa was a celebrated poet and dramatist of the Gupta period, often associated with Chandragupta II.

20. Which Governor General of Bengal underwent impeachment proceedings in

Which Governor General of Bengal underwent impeachment proceedings in the British Parliament ?

[amp_mcq option1=”Robert Clive” option2=”Henry Vansittart” option3=”Warren Hastings” option4=”Lord Cornwallis” correct=”option3″]

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Warren Hastings, the first Governor-General of Bengal, faced impeachment proceedings in the British Parliament.

The impeachment of Warren Hastings was initiated by Edmund Burke and others in the House of Commons in 1787, accusing him of misconduct and corruption during his tenure in India. The trial in the House of Lords lasted from 1788 to 1795, but he was eventually acquitted of all charges.

Robert Clive was a key figure in the early expansion of British power but did not face impeachment proceedings in Parliament in the same manner as Hastings, though his actions were investigated. Henry Vansittart was Governor of Bengal before Hastings. Lord Cornwallis succeeded Hastings and implemented significant reforms like the Permanent Settlement.