The maximum and minimum distances of a comet from the Sun are 1 . 6 × 10 ^ 12 m and 8 . 0 × 10 ^ 10 m respectively. If the speed of the comet at the nearest point is 6 × 10 ^ 4 ms ^ - 1 , the speed at the farthest point

Correct answer: C. 3 . 0 × 10 ^ 3 m s ^ - 1 By angular momentum conservation : m v _ 1 r _ 1 = m v _ 2 r _ 2 v _ 1 = 48 × 10 ^ 14 1 . 6 × 10 ^ 12 = 3000m s ^ - 1 = 3 × 10 ^ 3 m s ^ - 1 .

JEE Main 202116 MarchMorning Shift

Exam: JEE Main
Year: 2021
Date: 16 March
Shift: Morning Shift
Subject: Physics
Chapter: Rotational Motion

JEE Main 2021 · 16 March · Morning Shift · Physics · Rotational Motion

The maximum and minimum distances of a comet from the Sun are 1 . 6...

The maximum and minimum distances of a comet from the Sun are

1 . 6 \times {10}^{12}m

and

8 . 0 \times {10}^{10}m

respectively. If the speed of the comet at the nearest point is

6 \times {10}^{4}{ms}^{- 1}

, the speed at the farthest point is

Options

A $1 . 5 \times {10}^{3}m{s}^{- 1}$
B $6 . 6 \times {10}^{3}m{s}^{- 1}$
C $3 . 0 \times {10}^{3}m{s}^{- 1}$
D $4 . 5 \times {10}^{3}m{s}^{- 1}$

Correct answer

C. $3 . 0 \times {10}^{3}m{s}^{- 1}$

Step-by-step solution

By angular momentum conservation :

$m{v}_{1}{r}_{1} = m{v}_{2}{r}_{2}$

${v}_{1} = \dfrac{48 \times {10}^{14}}{1 . 6 \times {10}^{12}} = 3000m{s}^{- 1}$

$= 3 \times {10}^{3}m{s}^{- 1}$ .

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