| Updated On - Nov 19, 2024
JEE Main 27 Jan Shift 2 2024 Question Paper with Solutions and Answer Key PDF is available here. NTA conducted JEE Main 2024 Jan 27 Shift 2 exam from 3 PM to 6 PM. The question paper for JEE Main 2024 Jan 27 Shift 2 includes 90 questions equally divided into Physics, Chemistry and Maths. Candidates must attempt 75 questions in a 3-hour time duration. The official JEE Main 2024 paper 1 question paper for Jan 27 Shift 1 provided in the article below.
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JEE Main 27 Jan Shift 2 2024 Questions with Solution
| Question No. | Question | Options | Correct Answer | Solution |
|---|---|---|---|---|
| 1 | Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying arctan(x) + arctan(2x) = π/4 is: | More than 2, 1, 2, 0 | 1 | Given tan⁻¹(x) + tan⁻¹(2x) = π/4, we simplify to find 2x = (1-x)/(1+x). Solving this quadratic equation, only one positive solution x = (-3 + sqrt(17))/4 exists. |
| 2 | Consider the function f: (0,2) -> R defined by f(x) = x^2 + 2/x, and the function g(x) by min{f(t)} for 0 < t ≤ x, 0 < x ≤ 1 and g(x) = (3/2 + x) for 1 < x <2. Which statement about g is correct? | g is continuous but not differentiable at x = 1, g is not continuous for all x in (0,2), g is neither continuous nor differentiable at x = 1, g is continuous and differentiable for all x in (0,2) | g is continuous but not differentiable at x = 1 | The left-hand and right-hand limits equal g(1) for continuity at x = 1, but differentiability fails as derivatives from each side do not match at x = 1. |
| 3 | Let R be the region between lines 3x - y + 1 = 0 and x + 2y - 5 = 0 containing the origin. Find values of a for which (a^2, a + 1) lie in R: | (−3,−1) ∪ (-1/3,1), (−3,0) ∪ (-1/3,1), (−3,0) ∪ (2/3,1), (−3,−1) ∪ (-1/3,1) | (−3,0) ∪ (-1/3,1) | By analyzing inequalities from line equations, we find a in (−3,0) ∪ (-1/3,1). |
| 4 | The 20th term from the end of the progression 20, 19.25, 18.5, 17.75, ..., -129.25 is: | -118, -110, -115, -100 | -115 | The sequence is reversed, and the 20th term from the end is found by treating it as a standard arithmetic progression (A.P.) problem. |
| 5 | Let f: R - {-1/2} -> R and g: R - {-5/2} -> R be defined as f(x) = (2x + 3) / (2x + 1) and g(x) = |x| + 1 / (2x + 5). What is the domain of f ◦ g? | R - {-5/2}, R, R - {-7/4}, R - {-5/2, -7/4} | R - {-5/2} | Since f(g(x)) is undefined at x = -5/2, we exclude x = -5/2 from the domain of f ◦ g. |
| 6 | For 0 < a < 1, find the value of the integral ∫[0,π] dx / (1 - 2a cos(x) + a^2): | (π / (1 - a^2)), (π / (1 + a^2)), (π2 / (π + a^2)), (π2 / (π - a^2)) | π / (1 - a^2) | To solve the integral, complete the square in the denominator and use a trigonometric substitution to simplify. |
| 7 | Let g(x) = 3f(x/3) + f(3 - x) where f''(x) > 0 for all x in (0,3). If g is decreasing in (0, α) and increasing in (α,3), what is 8α? | 24, 0, 18, 20 | 18 | Using f''(x) > 0, g(x) is analyzed for intervals where it is decreasing and increasing to find α = 9/4, leading to 8α = 18. |
| 8 | If lim(x→0) (3 + α sin(x) + β cos(x) + log(1 - x)) / (3 tan^2(x)) = 1/3, then 2α - β is equal to: | 2, 7, 5, 1 | 5 | Using Taylor series expansions around x = 0, solve for α and β to get 2α - β = 5. |
| 9 | If α and β are the roots of x² - x - 1 = 0, and Sn = 2023α^n + 2024β^n, then: | 2S₁₂ = S₁₁ + S₁₀, S₁₂ = S₁₁ + S₁₀, 2S₁₁ = S₁₂ + S₁₀, S₁₁ = S₁₀ + S₁₂ | S₁₂ = S₁₁ + S₁₀ | Using the recurrence relation for roots of quadratic equations, we find S₁₂ = S₁₁ + S₁₀. |
| 10 | Let A and B be two finite sets with m and n elements, respectively. If the total subsets of set A are 56 more than B's subsets, then the distance of the point P(m,n) from Q(-2,-3) is: | 10, 6, 4, 8 | 10 | Solving 2^m = 2^n + 56, we find m = 6, n = 3. The distance between points P(6,3) and Q(-2,-3) is calculated as 10. |
| 11 | The values of α for which the determinant |1 3 2; α+3 2 1; 1 1 3| = 0 lie in the interval: | (-2,1), (-3,0), (-3/2,3/2), (0,3) | (-3,0) | Expanding the determinant and solving for α yields the interval (-3,0). |
| 12 | An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and the second draw gives all black balls is: | 5/256, 5/715, 3/715, 3/256 | 3/715 | The probability of drawing 4 white balls in the first draw is 15/1365. After removing 4 white balls, there are 9 black balls left. The probability of drawing 4 black balls in the second draw is 126/330. The required probability is the product of these probabilities, giving 3/715. |
| 13 | The integral ∫(x^8 - x^2) dx / ((x^12 + 3x^6 + 1) arctan(x^3 + 1/x^3)) is equal to: | log(arctan(x^3 + 1/x^3)^(1/3) + C), log(arctan(x^3 + 1/x^3)^(1/2) + C), log(arctan(x^3 + 1/x^3) + C), log(arctan(x^3 + 1/x^3)^3 + C) | log(arctan(x^3 + 1/x^3)^(1/3) + C) | Using the substitution u = arctan(x^3 + 1/x^3), we simplify the integral to du/u, which evaluates to log(u) + C. The final answer is in the given form. |
| 14 | If 2tan^2(θ) - 5sec(θ) = 1 has exactly 7 solutions in the interval [0, nπ/2] for the least value of n ∈ N, then Σ(k=1 to n) k / 2k is equal to: | 1/215(214 - 14), 1/214(215 - 15), 1 - 15/213, 1/213(214 - 15) | 1/213(214 - 15) | Setting sec(θ) = t, we solve the quadratic equation to find that n = 13. Using the formula for the sum, Σ = (1/213) * (214 - 15). |
| 15 | The position vectors of the vertices A, B, and C of a triangle are A = 2i - 3j + 3k, B = 2i + 2j + 3k, and C = -i + j + 3k respectively. If ℓ is the length of the angle bisector AD of ∠BAC, then 2ℓ^2 equals: | 49, 42, 50, 45 | 45 | Using the position vectors and midpoint formula, we find the length of AD and calculate 2ℓ^2 as 45. |
| 16 | If y = y(x) is the solution curve of the differential equation (x^2 - 4) dy - (y^2 - 3y) dx = 0, x > 2, y(4) = 3/2, and the slope of the curve is never zero, then y(10) equals: | 3/(1 + 8^(1/4)), 3/(1 + 2√2), 3/(1 - 2√2), 3/(1 - 8^(1/4)) | 3/(1 + 8^(1/4)) | Separating variables and solving the differential equation, we integrate both sides and substitute x = 10 to find y(10) = 3/(1 + 8^(1/4)). |
| 17 | If e₁ is the eccentricity of the hyperbola x²/16 - y²/9 = 1 and e₂ is the eccentricity of the ellipse x²/a² + y²/b² = 1, which passes through the foci of the hyperbola and satisfies e₁e₂ = 1, then the length of the chord of the ellipse parallel to the x-axis and passing through (0,2) is: | 4√5, 8√5/3, 10√5/3, 3√5 | 10√5/3 | For the ellipse x²/25 + y²/9 = 1, using the formula for the length of the chord parallel to the x-axis, we find the answer as 10√5/3. |
| 18 | Let α = (4!)! / (4!)^3! and β = (5!)! / (5!)^4!. Then: | α ∈ N and β ∉ N, α ∉ N and β ∈ N, α ∈ N and β ∈ N, α ∉ N and β ∉ N | α ∈ N and β ∈ N | Both α and β represent valid combinatorial expressions for arranging groups, making both natural numbers. |
| 19 | Let the position vectors of vertices A, B, and C of a triangle be 2i + 2j + k, i + 2j + 2k, and 2i + j + 2k respectively. Let l₁, l₂, and l₃ be lengths of perpendiculars from the orthocenter to sides AB, BC, and CA. Then l₁² + l₂² + l₃² equals: | 1/5, 1/2, 1/4, 1/3 | 1/2 | Using midpoint formula and perpendicular distance calculations, we find l₁² + l₂² + l₃² = 1/2. |
| 20 | The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking, it was found that an observation was read as 10 instead of 12. If µ and σ² denote the mean and variance of the correct observations, then 15(µ + µ² + σ²) is equal to: | 2521 | 2521 | Correcting the observation and recalculating mean and variance, we find 15(µ + µ² + σ²) = 2521. |
| 21 | The values of α for which lines 2x - y + 3 = 0, 6x + 3y + 1 = 0, and ax + 2y - 2 = 0 do not form a triangle are: | (-2,1), (-3,0), (-3/2,3/2), (0,3) | (-3,0) | For lines not to form a triangle, they must be concurrent or parallel. Solving for α, we get (-3,0). |
| 22 | If the area of the region {(x, y) : 0 ≤ y ≤ min(2x, 6x - x²)} is A, then 12A is equal to: | 304 | 304 | We calculate the area by integrating the expression and find A. Then, 12A is calculated to be 304. |
| 23 | Let Λ be a 2×2 real matrix and I be the identity matrix of order 2. If the roots of the equation |Λ - xI| = 0 are -1 and 3, then the sum of the diagonal elements of the matrix Λ² is: | 10 | 10 | Using the trace and determinant properties of Λ and calculating Λ², we find the sum of the diagonal elements as 10. |
| 24 | If the sum of squares of all real values of α, for which the lines 2x - y + 3 = 0, 6x + 3y + 1 = 0, and ax + 2y - 2 = 0 do not form a triangle is p, then the greatest integer less than or equal to p is: | 32 | 32 | Solving for the concurrent or parallel condition of these lines, we find p and the integer part is 32. |
| 25 | The coefficient of x^2012 in the expansion of (1 - x)^2008(1 + x + x²)^2007 is: | 0 | 0 | Analyzing the terms in the expansion, we see there is no term for x^2012, hence the coefficient is 0. |
| 26 | If the solution curve of the differential equation dy/dx = (x + y - 2)/(x - y) passes through the point (2, 1), the value of y(10) equals: | 3/(1 + (8)^1/4) | 3/(1 + (8)^1/4) | Solving the differential equation and substituting x = 10, we find y(10) as 3/(1 + (8)^1/4). |
| 27 | Let f(x) = ∫₀ˣ g(t) log((1 - t)/(1 + t)) dt, where g is a continuous odd function. If ∫(-π/2 to π/2) (f(x) x² cos(x)/(1 + e^x)) dx = π/2 - α, then α is equal to: | 2 | 2 | Using the properties of odd functions and simplifying the integral, we find that α = 2. |
| 28 | Consider a circle (x - α)² + (y - β)² = 50, where α, β > 0. If the circle touches the line y + x = 0 at point P, whose distance from the origin is 4√2, then (α + β)² is equal to: | 100 | 100 | Using the distance and tangency conditions, we find that (α + β)² = 100. |
| 29 | The lines (x - 2)/1 = (y - 1)/(-1) = (z - 7)/8 and (x + 3)/4 = (y + 2)/3 = (z + 2)/1 intersect at the point P. If the distance of P from the line (x + 1)/2 = (y - 1)/3 = (z - 1)/1 is l, then 14l² is equal to: | 108 | 108 | Calculating the intersection point and using the distance formula, we find 14l² = 108. |
| 30 | Let the complex numbers α and 1/α lie on the circles |z - z₀| = 2 and |z - z₀| = 4 respectively, where z₀ = 1 + i. Then, the value of 100|α|² is: | 20 | 20 | Since α and 1/α lie on concentric circles with known radii, |α| = 2. Thus, 100|α|² = 20. |
| 31 | The equation of state of a real gas is given by P + (a/V²)(V - b) = RT, where P, V, and T are pressure, volume, and temperature. The dimensions of a/b² are similar to: | P, PV, RT, R | P | Using dimensional analysis on a/b², we find it has the same dimensions as pressure P. |
| 32 | The total kinetic energy of 1 mole of oxygen at 27°C is: | 6845.5 J, 5942.0 J, 6232.5 J, 5670.5 J | 6232.5 J | Using the formula E = (5/2)nRT for a diatomic gas at 300 K, we find the total kinetic energy as 6232.5 J. |
| 33 | The primary side of a transformer is connected to a 230 V, 50 Hz supply. Turns ratio of primary to secondary winding is 10:1. Load resistance on the secondary side is 46 Ω. The power consumed in it is: | 12.5 W, 10.0 W, 11.5 W, 12.0 W | 11.5 W | Using the turns ratio to find V₂ and calculating power with P = V₂²/R, we find it to be 11.5 W. |
| 34 | During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of Cp/Cv for the gas is: | 5/3, 3/2, 7/5, 9/7 | 3/2 | Using the adiabatic relation and the given condition, we derive that the ratio of Cp/Cv (γ) is 3/2. |
| 35 | The threshold frequency of a metal with work function 6.63 eV is: | 16 ×10¹⁵ Hz, 16 ×10¹² Hz, 1.6 × 10¹² Hz, 1.6 × 10¹⁵ Hz | 1.6 × 10¹⁵ Hz | Using E = hν₀ with work function converted to joules, we find ν₀ as 1.6 × 10¹⁵ Hz. |
| 36 | A current of 200 µA deflects the coil of a moving coil galvanometer through 60°. The current to cause deflection through π/10 radians is: | 30 µA, 120 µA, 60 µA, 180 µA | 60 µA | Using proportionality of current and angle, we find the required current as 60 µA. |
| 37 | The atomic mass of 6C¹² is 12.000000 u and that of 6C¹³ is 13.003354 u. The required energy to remove a neutron from 6C¹³, if the mass of the neutron is 1.008665 u, will be: | 62.5 MeV, 6.25 MeV, 4.95 MeV, 49.5 MeV | 4.95 MeV | Using mass defect and E = ∆m x 931.5 MeV/u, we find the energy required as 4.95 MeV. |
| 38 | A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme and lowest position are equal. The angle (θ) of deflection in the extreme position is: | tan⁻¹(√2), 2tan⁻¹(1/2), tan⁻¹(1/2), 2tan⁻¹(1/√5) | 2tan⁻¹(1/2) | Using energy conservation and solving for θ, we find it as 2tan⁻¹(1/2). |
| 39 | Three voltmeters are joined as shown. When a potential difference is applied across A and B, their readings are V₁, V₂, and V₃. Choose the correct option. | V₁ = V₂, V₁ = V₃ - V₂, V₁ + V₂ > V₃, V₁ + V₂ = V₃ | V₁ + V₂ = V₃ | Applying Kirchhoff’s Voltage Law across the loop, we find that V₁ + V₂ = V₃. |
| 40 | The total kinetic energy of 1 mole of oxygen at 27°C is: | 6845.5 J, 5942.0 J, 6232.5 J, 5670.5 J | 6232.5 J | Using the formula E = (5/2)nRT for a diatomic gas at 300 K, we find the total kinetic energy as 6232.5 J. |
| 41 | Given that the angular speed of the moon in its orbit about the earth is greater than that of the earth around the sun, identify the reason. | Shorter period of orbit | Shorter period of orbit | Angular speed ω is inversely proportional to time period. Since the moon's period is shorter, ωmoon > ωearth. |
| 42 | The primary side of a transformer is connected to 230 V, 50 Hz supply with a turns ratio of 10:1. Load resistance is 46 Ω. The power consumed is: | 12.5 W, 10 W, 11.5 W, 12 W | 11.5 W | Using turns ratio and calculating V₂, then power with P = V₂²/R, we find it as 11.5 W. |
| 43 | During an adiabatic process, the pressure of a gas is proportional to the cube of its absolute temperature. The ratio of Cp/Cv is: | 5/3, 3/2, 7/5, 9/7 | 3/2 | Using the adiabatic relation and condition given, we derive γ = Cp/Cv as 3/2. |
| 44 | The threshold frequency of a metal with work function 6.63 eV is: | 16 ×10¹⁵ Hz, 16 ×10¹² Hz, 1.6 × 10¹² Hz, 1.6 × 10¹⁵ Hz | 1.6 × 10¹⁵ Hz | Using E = hν₀ with work function converted to joules, we find ν₀ as 1.6 × 10¹⁵ Hz. |
| 45 | A current of 200 µA deflects the coil of a moving coil galvanometer through 60°. The current to cause deflection through π/10 radians is: | 30 µA, 120 µA, 60 µA, 180 µA | 60 µA | Using proportionality of current and angle, we find the required current as 60 µA. |
| 46 | The atomic mass of 6C¹² is 12.000000 u and that of 6C¹³ is 13.003354 u. The required energy to remove a neutron from 6C¹³, if the mass of the neutron is 1.008665 u, will be: | 62.5 MeV, 6.25 MeV, 4.95 MeV, 49.5 MeV | 4.95 MeV | Using mass defect and E = ∆m x 931.5 MeV/u, we find the energy required as 4.95 MeV. |
| 47 | A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme and lowest position are equal. The angle (θ) of deflection in the extreme position is: | tan⁻¹(√2), 2tan⁻¹(1/2), tan⁻¹(1/2), 2tan⁻¹(1/√5) | 2tan⁻¹(1/2) | Using energy conservation and solving for θ, we find it as 2tan⁻¹(1/2). |
| 48 | Three voltmeters are joined as shown. When a potential difference is applied across A and B, their readings are V₁, V₂, and V₃. Choose the correct option. | V₁ = V₂, V₁ = V₃ - V₂, V₁ + V₂ > V₃, V₁ + V₂ = V₃ | V₁ + V₂ = V₃ | Applying Kirchhoff’s Voltage Law across the loop, we find that V₁ + V₂ = V₃. |
| 49 | The total kinetic energy of 1 mole of oxygen at 27°C is: | 6845.5 J, 5942.0 J, 6232.5 J, 5670.5 J | 6232.5 J | Using the formula E = (5/2)nRT for a diatomic gas at 300 K, we find the total kinetic energy as 6232.5 J. |
| 50 | The primary side of a transformer is connected to 230 V, 50 Hz supply with a turns ratio of 10:1. Load resistance is 46 Ω. The power consumed is: | 12.5 W, 10 W, 11.5 W, 12 W | 11.5 W | Using turns ratio and calculating V₂, then power with P = V₂²/R, we find it as 11.5 W. |
Also Check:
| JEE Main 2024 Paper Analysis | JEE Main 2024 Answer Key |
| JEE Main 2024 Cutoff | JEE Main 2024 Marks vs Rank |
JEE Main 2024 Jan 27 Shift 2 Question Paper by Coaching Institute
| Coaching Institutes | Question Paper with Solutions PDF |
|---|---|
| Aakash BYJUs | Download PDF |
| Reliable Institute | Physics Chemistry Maths |
| Resonance | Physics Chemistry Maths |
| Vedantu | Download PDF |
| Sri Chaitanya | Download PDF |
| FIIT JEE | To be updated |
JEE Main 27 Jan Shift 2 2024 Paper Analysis
JEE Main 2024 Jan 27 Shift 2 paper analysis for B.E./ B.Tech is updated here with details on the difficulty level of the exam, topics with the highest weightage in the exam, section-wise difficulty level, etc.
JEE Main 2024 Question Paper Pattern
| Feature | Question Paper Pattern |
|---|---|
| Examination Mode | Computer-based Test |
| Exam Language | 13 languages (English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu, and Urdu) |
| Number of Sections | Three- Physics, Chemistry, Mathematics |
| Exam Duration | 3 hours |
| Sectional Time Limit | None |
| Total Marks | 300 marks |
| Total Number of Questions Asked | 90 Questions |
| Total Number of Questions to be Answered | 75 questions |
| Type of Questions | MCQs and Numerical Answer Type Questions |
| Section-wise Number of Questions | Physics- 20 MCQs and 10 numerical type, Chemistry- 20 MCQs and 10 numerical type, Mathematics- 20 MCQs and 10 numerical type |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
Read More:
- JEE Main 2024 question paper pattern and marking scheme
- Most important chapters in JEE Mains 2024, Check chapter wise weightage here
JEE Main 2024 Question Paper Session 1 (January)
Those appearing for JEE Main 2024 can use the links below to practice and keep track of their exam preparation level by attempting the shift-wise JEE Main 2024 question paper provided below.
| Exam Date and Shift | Question Paper PDF |
|---|---|
| JEE Main 24 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 27 Jan Shift 1 2024 Question Paper | Check Here |
| JEE Main 29 Jan Shift 1 2024 Question Paper | Check Here |
| JEE Main 29 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 30 Jan Shift 1 2024 Question Paper | Check Here |
| JEE Main 30 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 31 Jan Shift 1 2024 Question Paper | Check Here |
| JEE Main 31 Jan Shift 2 2024 Question Paper | Check Here |
| JEE Main 1 Feb Shift 1 2024 Question Paper | Check Here |
| JEE Main 1 Feb Shift 2 2024 Question Paper | Check Here |
JEE Main Previous Year Question Paper
JEE Main Questions
1. An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is:
An electron rotates in a circle around a nucleus having positive charge Ze. Correct relation between total energy (E) of electron to its potential energy (U) is:
- E = 2U
- 2E = 3U
- E = U
- 2E = U
2. If G be the gravitational constant and u be thee nergy density then which of the following quantity have the dimension as that the \(\sqrt{UG}\)
If G be the gravitational constant and u be thee nergy density then which of the following quantity have the dimension as that the \(\sqrt{UG}\)
- Pressure gradient per unit mass
- Force per unit mass
- Gravitational potential
- Energy per unit mass
3. In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero.
In a co-axial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero.
- inside the outer conductor
- in between the two conductors
- outside the cable
- inside the inner conductor
4. Following gates section is connected in a complete suitable circuit.
For which of the following combination, bulb will glow (ON):
Following gates section is connected in a complete suitable circuit.
For which of the following combination, bulb will glow (ON):
For which of the following combination, bulb will glow (ON):
- A = 0, B = 1, C = 1, D = 1
- A = 1, B = 0, C = 0, D = 0
- A = 0, B = 0, C = 0, D = 1
- A = 1, B = 1, C = 1, D = 0
5. Match list-I with list-II:List-I List-II (A) Kinetic energy of planet \(- \frac{GMm}{a}\) (B) Gravitational Potential energy of Sun-planet system \(- \frac{GMm}{2a}\) (C) Total mechanical energy of planet \(\frac{GM}{r}\) (D) Escape energy at the surface of planet for unit mass object \(- \frac{GMm}{2a}\)
(Where a = radius of planet orbit, r = radius of planet, M = mass of Sun, m = mass of planet) Choose the correct answer from the options given below:
Match list-I with list-II:
| List-I | List-II |
|---|---|
| (A) Kinetic energy of planet | \(- \frac{GMm}{a}\) |
| (B) Gravitational Potential energy of Sun-planet system | \(- \frac{GMm}{2a}\) |
| (C) Total mechanical energy of planet | \(\frac{GM}{r}\) |
| (D) Escape energy at the surface of planet for unit mass object | \(- \frac{GMm}{2a}\) |
(Where a = radius of planet orbit, r = radius of planet, M = mass of Sun, m = mass of planet) Choose the correct answer from the options given below:
- (A) – II, (B) – I, (C) – IV, (D) – III
- (A) – III, (B) – IV, (C) – I, (D) – II
- (A) – I, (B) – IV, (C) – II, (D) – III
- (A) – I, (B) – II, (C) – III, (D) – IV
6. The angle between vector \( \vec{Q} \) and the resultant of \( (2\vec{Q} + 2\vec{P}) \) and \( (2\vec{Q} - 2\vec{P}) \) is:
The angle between vector \( \vec{Q} \) and the resultant of \( (2\vec{Q} + 2\vec{P}) \) and \( (2\vec{Q} - 2\vec{P}) \) is:
- \( 0^\circ \)
- \( \tan^{-1}\left(\frac{2\vec{Q} - 2\vec{P}}{2\vec{Q} + 2\vec{P}}\right) \)
- \( \tan^{-1}\left(\frac{P}{Q}\right) \)
- \( \tan^{-1}\left(\frac{2Q}{P}\right) \)
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