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Circular Motion question

2021 · 27 Aug · Shift 1 · Q47
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  5. /2021 · 27 Aug · Shift 1 · Q47

Circular Motion question

2021 · 27 Aug · Shift 1 · Q47

JEE MainPhysicsCircular MotionMCQ+4 / −1
A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second? Given : 1 ly = 9.46 ×\times× 1015 m 1 AU = 1.5 ×\times× 1011 m
  1. A
    4.1 ×\times× 108 s
  2. B
    4.5 ×\times× 1010 s
  3. C
    3.5 ×\times× 106 s
  4. D
    7.2 ×\times× 108 s
View written solutionFree

Correct answer: NONE OF THE GIVEN OPTIONS; COMPUTED TIME = 2.2 × 10^5 S

  1. Use arc length relation

For a circle, s=rθs = r\thetas=rθ where sss is arc length and θ\thetaθ is in radians.

Given:

  • arc length =4.4 ly= 4.4\ \text{ly}=4.4 ly
  • angle subtended =4 rad= 4\ \text{rad}=4 rad

So radius is r=sθ=4.44 ly=1.1 lyr = \frac{s}{\theta} = \frac{4.4}{4}\text{ ly} = 1.1\text{ ly}r=θs​=44.4​ ly=1.1 ly

  1. Find circumference of the circle

Circumference: C=2πr=2π(1.1)=2.2π lyC = 2\pi r = 2\pi(1.1) = 2.2\pi\ \text{ly}C=2πr=2π(1.1)=2.2π ly

For 444 revolutions, total distance traveled is D=4C=4(2.2π)=8.8π lyD = 4C = 4(2.2\pi) = 8.8\pi\ \text{ly}D=4C=4(2.2π)=8.8π ly

Using π≈3.14\pi \approx 3.14π≈3.14, D≈8.8×3.14=27.632 lyD \approx 8.8 \times 3.14 = 27.632\ \text{ly}D≈8.8×3.14=27.632 ly

  1. Convert distance into meters

Given 1 ly=9.46×1015 m1\ \text{ly} = 9.46 \times 10^{15}\ \text{m}1 ly=9.46×1015 m

Hence, D=27.632×9.46×1015D = 27.632 \times 9.46 \times 10^{15}D=27.632×9.46×1015 D≈261.0×1015=2.61×1017 mD \approx 261.0 \times 10^{15} = 2.61 \times 10^{17}\ \text{m}D≈261.0×1015=2.61×1017 m

  1. Convert speed into m/s

Given speed: v=8 AU/sv = 8\ \text{AU/s}v=8 AU/s

Also, 1 AU=1.5×1011 m1\ \text{AU} = 1.5 \times 10^{11}\ \text{m}1 AU=1.5×1011 m

So, v=8×1.5×1011=1.2×1012 m/sv = 8 \times 1.5 \times 10^{11} = 1.2 \times 10^{12}\ \text{m/s}v=8×1.5×1011=1.2×1012 m/s

  1. Compute time

t=Dv=2.61×10171.2×1012t = \frac{D}{v} = \frac{2.61 \times 10^{17}}{1.2 \times 10^{12}}t=vD​=1.2×10122.61×1017​ t=2.175×105 st = 2.175 \times 10^5\ \text{s}t=2.175×105 s

So the time required is approximately 2.2×105 s\boxed{2.2 \times 10^5\ \text{s}}2.2×105 s​

  1. Check options

The computed answer 2.2×105 s2.2 \times 10^5\,\text{s}2.2×105s does not match any of the given options.

Therefore, none of the listed options is correct.

It appears the question or options may contain a typo. If the angle were interpreted differently or the speed unit were different, one might get an option closer to the list, but with the given data the correct value is: 2.2×105 s\boxed{2.2 \times 10^5\ \text{s}}2.2×105 s​

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