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For each natural number k, let Ck denotes the circle with radius k units and centre at the origin. On the Ck–, a particle moves k units in the counter clockwise direction. After completing its motion on Ck, the particle moves to Ck+l in some well defined manner, where l > 0.The motion of the particle continues in this manner.

On the basis of above information, answer the following questions:

1. Let l = 1, the particle starts at (1, 0). If the particle crossing the positive direction of the x-axis for the first time on the circle Cn, then n is equal to

a. 3

b. 5

c. 7

d. 8

2. If k ∈ N and l = 1, the particle starts (–1, 0) the particle cross x-axis again at

a. (3, 0)

b. (1, 0)

c. (4, 0)

d. (2, 0)

3. If k ∈ N and l = 1, the particle moves in the radial direction from circle Ck to Ck +1. If particle starts form the point (–1, 0), then

a. it will cross the +ve y-axis at (0, 4)

b. it will cross the –ve y-axis at (0, –4)

c. it will cross the +ve y-axis at (0, 5)

d. it will cross the –ve y-axis at (0, –5)

4. If k ∈ N and l = R , particle moves tangentially from the circle Ck to Ck+1, such that the length of tangent is equal to k units itself. If particle starts form the point (1, 0), then

a. the particle will cross x-axis again at x = 3

b. the particle will cross x-axis again at x = 4

c. the particle will cross +ve x-axis again at x = 2√2

d. the particle will cross +ve x-axis again at x ∈ (2√2, 4)

5. Let the particle starts from the point (2, 0) and moves π/2 units, on circle C2 in the counterclockwise direction, then itn moves on the circle C3 along the tangential path, let this straight line (tangential path traced by particle) intersect the circle C3 at the points. A and B tangents drawn at A and B intersect at

a. \(\left(\frac 9{2\sqrt 2};\frac 9{2\sqrt2}\right)\)

b. (9√2, 9√2)

c. (9, 9)

d. (√2, √2)

1 Answer

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by (51.1k points)
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Best answer

1. c. 7

2. c. (4, 0)

3. c. it will cross the +ve y-axis at (0, 5)

4. d. the particle will cross +ve x-axis again at x ∈ (2√2, 4)

5. a. \(\left(\frac 9{2\sqrt 2};\frac 9{2\sqrt2}\right)\)

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