UPSC Geoscientist Archives

41. Which of the following statements is/are not correct about the parli

Which of the following statements is/are not correct about the parliamentary practice of asking Questions in the Parliament of India ?

1. The first hour of every sitting in both the Houses of Parliament is devoted to asking questions, referred to as the ‘Question Hour’.
2. Questions are generally asked by the Members of Parliament from the Ministers.
3. ‘Half-an-Hour Discussions’ on matters arising out of questions already answered can be held only in the Lok Sabha.
4. Questions cannot be addressed to a private member.

Select the answer using the code given below :

1 and 2

2 and 3

3 and 4

2 and 4

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UPSC Geoscientist – 2024

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Statement 3 is incorrect because ‘Half-an-Hour Discussions’ on matters arising out of questions already answered can be held in *both* the Lok Sabha and the Rajya Sabha, not just the Lok Sabha. Statement 4 is incorrect because questions *can* be addressed to a private member if the question relates to a Bill, Resolution, or other matter for which that member is responsible in their capacity as a private member. Therefore, statements 3 and 4 are not correct.

– Statement 1 is correct: The first hour of every sitting in both Houses is the ‘Question Hour’.
– Statement 2 is correct: Questions are primarily directed at Ministers to hold the executive accountable.
– Statement 3 is incorrect: Half-an-Hour Discussions are held in both Houses.
– Statement 4 is incorrect: Questions can be addressed to private members under specific circumstances.

The rules regarding questions are detailed in the Rules of Procedure and Conduct of Business in the respective Houses of Parliament. Question Hour is an important mechanism for parliamentary oversight. While questions are mainly for Ministers, addressing them to private members ensures accountability for their legislative or parliamentary actions.

42. Which of the following statements is not correct with respect to the

Which of the following statements is not correct with respect to the types of questions asked in the Lok Sabha?

A ‘Starred Question’ is the one to which only a written answer is given.

A ‘Short Notice Question’ is the one which relates to a matter of urgent public importance.

An ‘Unstarred Question’ is the one to which an oral answer is not given.

No supplementary question can be asked on an ‘Unstarred Question’.

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Statement A is incorrect because a ‘Starred Question’ is one for which an oral answer is required from the Minister, and supplementary questions can be asked based on the answer. A written answer is given for an ‘Unstarred Question’.

– Starred Question: Requires oral answer, supplementary questions allowed. Distinguished by an asterisk (*).
– Unstarred Question: Requires written answer, no supplementary questions allowed.
– Short Notice Question: Relates to a matter of urgent public importance, requires an oral answer and consent of the Minister and Speaker. Can be asked with a notice period shorter than the usual 10 days.

Questions are an important tool for Members of Parliament to seek information from the government, hold Ministers accountable, and raise matters of public interest during the ‘Question Hour’, which is the first hour of every parliamentary sitting.

43. A small metallic sphere is held in a non-uniform electrostatic field a

A small metallic sphere is held in a non-uniform electrostatic field as shown in the figure. Also shown are two points P and Q on the surface of the sphere. Which one of the following statements is true for the sphere?

Sphere does not experience a force in the field.

Surface charge density at point P is positive while at point Q is negative.

Sphere experiences a force towards the left.

Sphere experiences a force towards the right.

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When a conductor is placed in a non-uniform electrostatic field, charges are induced on its surface. Due to the non-uniformity of the field, the forces on the induced positive and negative charges do not cancel out, resulting in a net force on the conductor.

A conductor in an electrostatic field is an equipotential surface, and the electric field lines are perpendicular to its surface. In a non-uniform field (indicated by varying density of field lines), the side facing the stronger field region (denser lines, like point P) will accumulate induced charge opposite to the field source, and the opposite side (sparser lines, like point Q) will accumulate the same type of charge as the source. The forces on these induced charges are F = qE. Since the field strength E is non-uniform, the force on the induced charge in the stronger field region will be greater.

The diagram shows field lines entering the sphere predominantly from the left, indicating a stronger field on the left (around P) and a weaker field on the right (around Q). Induced negative charge will accumulate at P, and induced positive charge at Q (assuming the field source is effectively positive and to the left, causing field lines to point rightwards). The force on the negative charge at P (F_P) will be towards the left (opposite to the field direction). The force on the positive charge at Q (F_Q) will be towards the right (in the field direction). Since the field is stronger at P, |F_P| > |F_Q|. The net force F_net = F_P + F_Q is thus directed towards the left (towards the region of stronger field). Statement B is incorrect regarding charge signs.

44. Shown in the circuit is a 20 V battery connected to an arrangement of

Shown in the circuit is a 20 V battery connected to an arrangement of four resistors. If the potential difference between points P and Q in the circuit is zero then :

R = 2 Ω

R = 8 Ω

R = 18 Ω

R = 4 Ω

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The circuit can be analysed as a parallel combination of two series branches. The potential difference between points P and Q is zero when the ratio of resistances in the two branches satisfies the condition for a balanced Wheatstone bridge.

Let the branches be from A to B. Branch 1 has resistors 10Ω and R in series, with P between them. Branch 2 has 5Ω and 10Ω in series, with Q between them. For the potential difference between P and Q to be zero (V_P = V_Q), the potential division must be the same in both branches relative to the terminals A and B. This occurs when the bridge is balanced, i.e., the ratio of resistances in the arms is equal: 10Ω / R = 5Ω / 10Ω.

From the balanced bridge condition, 10 / R = 5 / 10. Cross-multiplying gives 10 * 10 = 5 * R, so 100 = 5R, which means R = 20 Ω. However, 20 Ω is not among the options. This suggests a potential error in the question or provided options. Assuming the intended answer is one of the options provided, and based on external information about this specific question from past exams, option B (R=8 Ω) is indicated as correct in some sources, despite standard calculation leading to R=20 Ω. There is likely an inconsistency in the problem statement as presented or its intended solution.

45. 100 grams of ice at 0°C is put in 240 grams of water at 5°C. The mixtu

100 grams of ice at 0°C is put in 240 grams of water at 5°C. The mixture finally comes to equilibrium at 0°C with m grams of ice melted in the process. Value of m is close to : (latent heat of ice = 3.33 × 10⁵ J/kg and specific heat of water = 4.2 J g^-1°C)

0 g

10 g

5 g

15 g

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When the ice at 0°C is put into water at 5°C, the water will lose heat as it cools down to 0°C. This heat is then used to melt the ice at 0°C.

The heat lost by the water is Q_water = m_water * c_water * ΔT_water. Given m_water = 240 g, c_water = 4.2 J g⁻¹°C⁻¹, and ΔT_water = 5°C – 0°C = 5°C. Q_water = (240 g) * (4.2 J g⁻¹°C⁻¹) * (5°C) = 5040 J. This heat melts a mass m of ice. The heat required to melt ice is Q_melt = m * L_f. Given L_f = 3.33 × 10⁵ J/kg = 3.33 × 10⁵ J / 1000 g = 333 J/g.

Setting the heat lost by water equal to the heat gained by the melting ice: 5040 J = m (g) * 333 J/g. m = 5040 / 333 g ≈ 15.135 g. The value of m is closest to 15 g among the given options.

46. Ions in a solution are subject to a uniform electric field. If each io

Ions in a solution are subject to a uniform electric field. If each ion carries charge q and has radius R, then the ionic current I due to the applied field will depend on q and R as :

I ∝ qR2

I ∝ q2R

I ∝ qR0

I ∝ q2R-1

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Ions in a solution move under the influence of an electric field (drift velocity), and this motion constitutes an ionic current. The drift velocity is proportional to the electric force and inversely proportional to the viscous drag force.

The electric force on an ion is F_e = qE. Assuming Stokes’ law for the viscous drag force in a solution, F_d = 6πηRv, where η is viscosity, R is the ion’s radius, and v is its drift velocity. In steady state, F_e = F_d, so qE = 6πηRv. The drift velocity is v = qE / (6πηR). The ionic current I is proportional to the charge of each ion (q), the number density of ions (n), their drift velocity (v), and the cross-sectional area (A): I = nqvA.

Substituting the expression for drift velocity, I ∝ q * (qE/R) * A (assuming n, E, η, A are constant). Thus, I ∝ q²/R = q²R⁻¹.

47. Magnetic dipole moment of a square current loop of side 1 cm and carry

Magnetic dipole moment of a square current loop of side 1 cm and carrying a current of 0.10 A is :

3.14 × 10-5 A m2

1.00 × 10-5 A m2

6.28 × 10-3 A m2

4.00 × 10-3 A m2

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The magnitude of the magnetic dipole moment (m) of a planar current loop is given by the product of the current (I) flowing through the loop and the area (A) of the loop: m = IA.

The loop is a square with side length s = 1 cm. Convert the side length to meters: s = 1 cm = 0.01 m = 10⁻² m. The area of the square loop is A = s² = (10⁻² m)² = 10⁻⁴ m². The current is I = 0.10 A.

The magnitude of the magnetic dipole moment is m = I * A = (0.10 A) * (10⁻⁴ m²) = 0.1 * 10⁻⁴ A m² = 10⁻¹ * 10⁻⁴ A m² = 10⁻⁵ A m². This can also be written as 1.00 × 10⁻⁵ A m².

48. Consider metallic spheres of different radii r, all at the same electr

Consider metallic spheres of different radii r, all at the same electrostatic potential. Then the magnitude E of the electric field just outside these spheres depends on r as :

E ∝ r-2

E ∝ r-1

E ∝ r0

E ∝ r

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For a metallic sphere (conductor), the charge resides on its surface. The electrostatic potential V at the surface of a conducting sphere of radius r with charge Q is given by V = Q / (4πε₀r). The electric field E just outside the surface is given by E = Q / (4πε₀r²).

Given that all spheres are at the same electrostatic potential V, we can express the charge Q in terms of V and r: Q = V * (4πε₀r). Substitute this expression for Q into the formula for E.

E = Q / (4πε₀r²) = [V * (4πε₀r)] / (4πε₀r²) = V / r. Since V and 4πε₀ are constants for all spheres, the magnitude of the electric field just outside the sphere depends on the radius as E ∝ 1/r, which is E ∝ r⁻¹.

49. A particle moves in a circle of radius 4 m. Its linear speed is given

A particle moves in a circle of radius 4 m. Its linear speed is given by 4√3 t, where t is the time measured in seconds. At t = 1 s, the angle made by the resultant acceleration – r̂ with direction (+ r̂ is the radial direction) is given by φ. Which one among the following is the correct value of φ ?

tan⁻¹ (1/√3)

tan⁻¹ (1/√2)

tan⁻¹ (2/√3)

tan⁻¹ (√3)

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In circular motion, the resultant acceleration has two perpendicular components: tangential acceleration (a_t) and radial (centripetal) acceleration (a_r). The angle made by the resultant acceleration with the radial direction can be found using trigonometry.

The linear speed is v = 4√3 t. The tangential acceleration is a_t = dv/dt = d(4√3 t)/dt = 4√3 m/s². At t = 1 s, v = 4√3 m/s. The radial acceleration is a_r = v²/r = (4√3)² / 4 = (16*3) / 4 = 48/4 = 12 m/s². The radial direction +r̂ points towards the center. The total acceleration vector is the sum of a_t (tangential) and a_r (radial, inward). The angle φ between the resultant acceleration and the radial direction (+r̂, pointing inward) satisfies tan(φ) = a_t / a_r.

tan(φ) = (4√3) / 12 = √3 / 3 = 1/√3. Therefore, φ = tan⁻¹(1/√3).

50. A ball of mass 100 g falls freely from a height of h = 20 m and hits t

A ball of mass 100 g falls freely from a height of h = 20 m and hits the ground at a speed of 1·4 √gh. Take g = 10 m s⁻². Which one among the following is the correct value of the work done on the ball by air friction ?

0·4 J

0·5 J

0·3 J

1 J

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The work done on the ball by air friction can be determined using the Work-Energy Theorem, which states that the total work done on an object equals the change in its kinetic energy. The total work done is the sum of work done by gravity and work done by air friction.

Work done by gravity (a conservative force) is W_g = mgh = (0.1 kg)(10 m/s²)(20 m) = 20 J. The initial kinetic energy is KE_i = 0. The final kinetic energy is KE_f = ½ m v_f², where v_f = 1.4√gh. With g=10 m/s² and h=20m, v_f = 1.4√(10*20) = 1.4√200. So, v_f² = (1.4)² * 200 = 1.96 * 200 = 392 m²/s². KE_f = ½ (0.1 kg)(392 m²/s²) = 0.05 * 392 = 19.6 J.

According to the Work-Energy Theorem, W_total = W_g + W_air = ΔKE = KE_f – KE_i. 20 J + W_air = 19.6 J – 0 J. W_air = 19.6 J – 20 J = -0.4 J. The question asks for the value, and options are positive, implying magnitude. The magnitude of the work done by air friction is |-0.4 J| = 0.4 J.