The maximum and minimum distance of a comet from the sun are 8 x 10¹² m and 1.6 x10¹² m. If its velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest

The correct option is 12.

When a comet orbits the Sun under the influence of a central gravitational force, its angular momentum is conserved. The angular momentum $L$ of the comet is given by the formula:
$L=mvr\sin \left(\theta \right)$
where $m$ is the mass of the comet, $v$ is its velocity, $r$ is its distance from the Sun, and $\theta$ is the angle between the velocity vector and the position vector.

At the closest point (nearest to the Sun) and the farthest point (farthest from the Sun) of the elliptical orbit, the velocity vector is perpendicular to the position vector. This means $\theta =90°$ and $\sin (90°)=1$. Therefore, at these two positions, the conservation of angular momentum simplifies to:
$v_{\text{nearest}}{r}_{\text{nearest}}=v_{\text{farthest}}{r}_{\text{farthest}}$

We are given the following values from the problem statement:
- Maximum distance (farthest point), $r_{\text{farthest}}=8\times {10}^{12}\text{m}$
- Minimum distance (nearest point), $r_{\text{nearest}}=1.6\times {10}^{12}\text{m}$
- Velocity at the nearest point, $v_{\text{nearest}}=60\text{m/s}$

Substitute these values into the conservation of angular momentum equation:
$60\times (1.6\times {10}^{12})=v_{\text{farthest}}\times (8\times {10}^{12})$
To solve for the velocity at the farthest point, $v_{\text{farthest}}$:
$v_{\text{farthest}}=\frac{60\times 1.6\times {10}^{12}}{8\times {10}^{12}}$
Since ${10}^{12}$ cancels out from both the numerator and the denominator, we get:
$v_{\text{farthest}}=\frac{60\times 1.6}{8}$
$v_{\text{farthest}}=60\times 0.2=12\text{m/s}$
Thus, the velocity of the comet when it is farthest from the Sun is 12 m/s.