JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 11)
The maximum and minimum distances of a comet from the Sun are 1.6 $$\times$$ 1012 m and 8.0 $$\times$$ 1010 m respectively. If the speed of the comet at the nearest point is 6 $$\times$$ 104 ms$$-$$1, the speed at the farthest point is :
3.0 $$\times$$ 103 m/s
6.0 $$\times$$ 103 m/s
1.5 $$\times$$ 103 m/s
4.5 $$\times$$ 103 m/s
Explanation
_16th_March_Morning_Shift_en_11_1.png)
v1 = 6 $$\times$$ 104 m/s
Let point 1 is nearest point,
and point 2 is farthest point.
Given, r1 = 8 $$\times$$ 1010 m & r2 = 1.6 $$\times$$ 1012 m
By angular momentum conservation
L1 = L2
mr1v1 = mr2v2
$$ \Rightarrow $$ v2 = $${{{r_1}{v_1}} \over {{r_2}}}$$
$$ \therefore $$ v2 = $${{8 \times {{10}^{10}} \times 6 \times {{10}^4}} \over {1.6 \times {{10}^{12}}}}$$
$$ \Rightarrow $$ v2 = 3.0 $$\times$$ 103 m/s
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