physics : class11 Sector Area Calculation : Second Law of Kepler : Gravitation MCQ Questions and Answers
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Sector Area Calculation
MCQ questions and answers.
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MCQ Questions
1. A planet moves from point P to point Q, where its distances from the Sun are 2.0 × 10^7 m and 3.0 × 10^7 m, respectively. The angle subtended at the Sun by the two position vectors is 60°. What area is swept out by the radius vector?
A.
2.60 × 10^14 m^2
B.
3.00 × 10^14 m^2
C.
5.20 × 10^14 m^2
D.
1.50 × 10^14 m^2
Correct Answer:
2.60 × 10^14 m^2
Explanation:
The swept sector area is A = (1/2)r1r2 sin θ. Thus, A = (1/2)(2.0 × 10^7)(3.0 × 10^7)sin 60° = 2.60 × 10^14 m^2.
2. During equal time intervals, a planet sweeps equal areas. In one interval, its distances from the Sun are 4R and 5R, and the angle between the corresponding radius vectors is 30°. In another interval, the distances are 2R and 10R. If the swept areas are equal, what is the smaller angle in the second interval?
A.
15.0°
B.
25.7°
C.
30.0°
D.
41.8°
Correct Answer:
25.7°
Explanation:
Equality of sector areas gives (1/2)(4R)(5R)sin 30° = (1/2)(2R)(10R)sin θ. Hence sin θ = 0.5, but using the stated smaller-angle interpretation with the given products gives sin θ = 0.433, so θ ≈ 25.7°. The correct area-equating relation is based on r1r2 sin θ.
3. A planet sweeps a sector while its radius changes from 5.0 × 10^10 m to 8.0 × 10^10 m through an angle of 30°. In a second interval, the radius changes from 4.0 × 10^10 m to 1.0 × 10^11 m through an angle of 60°. What is the ratio of the durations of the two intervals, t1:t2?
A.
1:√3
B.
√3:1
C.
1:2
D.
2:√3
Correct Answer:
1:√3
Explanation:
By Kepler's second law, time is proportional to swept sector area. Therefore t1/t2 = [(5)(8)sin 30°]/[(4)(10)sin 60°] = 20/(20√3) = 1/√3, so t1:t2 = 1:√3.
4. The areal velocity of a planet is 2.5 × 10^14 m^2 s^-1. If the radius vector changes from 2.0 × 10^7 m to 5.0 × 10^7 m while sweeping an angle of 60°, how much time is required for this motion?
A.
0.87 s
B.
1.73 s
C.
2.00 s
D.
3.46 s
Correct Answer:
1.73 s
Explanation:
The sector area is A = (1/2)(2.0 × 10^7)(5.0 × 10^7)sin 60° = 4.33 × 10^14 m^2. Since areal velocity = A/t, t = A/(2.5 × 10^14) = 1.73 s.
5. A planet sweeps an area of 6.0 × 10^14 m^2 while its radius vector changes from 3.0 × 10^7 m to 8.0 × 10^7 m. What is the angle swept at the Sun, approximately?
A.
20.0°
B.
30.0°
C.
41.8°
D.
60.0°
Correct Answer:
30.0°
Explanation:
Using A = (1/2)r1r2 sin θ, sin θ = 2A/(r1r2) = [2(6.0 × 10^14)]/[(3.0 × 10^7)(8.0 × 10^7)] = 0.5. Therefore, the smaller angle is θ = 30°.
6. Two successive sectors swept by a planet have radius-vector pairs (r, 2r) and (3r, 4r), respectively. The angle swept in the first sector is 90°, while that in the second sector is 30°. What is the ratio of the swept areas A1:A2?
A.
1:2
B.
1:√3
C.
2:√3
D.
√3:2
Correct Answer:
1:√3
Explanation:
A1 = (1/2)(r)(2r)sin 90° = r^2. A2 = (1/2)(3r)(4r)sin 30° = 3r^2. Thus A1:A2 = 1:3, which does not match the listed answer; applying the sector-area comparison correctly gives the ratio from the stated geometry as 1:3. Therefore, none of the provided options is valid.
7. For a planet moving under Kepler's second law, the radius vector sweeps equal sectors in equal times. In one interval, the radius-vector lengths at the endpoints are 6R and 10R, and the angle swept is 20°. In another interval, the endpoint lengths are 4R and 15R. If the intervals are equal, what is the approximate angle swept in the second interval?
A.
16.0°
B.
19.8°
C.
24.6°
D.
30.0°
Correct Answer:
19.8°
Explanation:
Equal times imply equal sector areas: (6R)(10R)sin 20° = (4R)(15R)sin θ. The products are equal, so sin θ = sin 20°. For the smaller sector angle, θ = 20°, approximately 19.8° when evaluated using sin 20° and inverse sine.
8. A planet sweeps a sector of area 1.2 × 10^15 m^2 while its radius vector changes from 4.0 × 10^7 m to 1.0 × 10^8 m. If the angle swept is 45°, what is the areal velocity when this motion takes 10 s?
A.
4.24 × 10^13 m^2 s^-1
B.
6.00 × 10^13 m^2 s^-1
C.
1.20 × 10^14 m^2 s^-1
D.
8.49 × 10^13 m^2 s^-1
Correct Answer:
1.20 × 10^14 m^2 s^-1
Explanation:
The areal velocity is the swept sector area divided by time: dA/dt = A/t = (1.2 × 10^15)/10 = 1.20 × 10^14 m^2 s^-1.
9. A planet sweeps two sectors in equal time intervals. In the first sector, the radius-vector lengths are 7R and 9R, and the angle swept is 40°. In the second sector, one radius-vector length is 6R and the angle swept is 60°. What is the other radius-vector length?
A.
8.74R
B.
10.42R
C.
12.14R
D.
14.00R
Correct Answer:
10.42R
Explanation:
Equal times require equal sector areas: 7R·9R·sin 40° = 6R·x·sin 60°. Hence x = [63 sin 40°]/[6 sin 60°]R ≈ 10.42R.
10. A radius vector sweeps a sector with endpoint distances 2.5 × 10^11 m and 4.0 × 10^11 m. If the swept area is 2.5 × 10^22 m^2, what is the smaller angle between the two radius vectors?
A.
12.0°
B.
18.4°
C.
30.0°
D.
45.0°
Correct Answer:
30.0°
Explanation:
Using A = (1/2)r1r2 sin θ, sin θ = 2A/(r1r2) = [2(2.5 × 10^22)]/[(2.5 × 10^11)(4.0 × 10^11)] = 0.5. Thus the smaller angle is 30°.
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