To solve this problem, we need to understand Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution (T) of a planet is directly proportional to the cube of the semi-major axis of its orbit (r). Mathematically, it is written as:
T^2 \propto r^3
This implies:
\left(\frac{T_1}{T_2}\right)^2 = \left(\frac{r_1}{r_2}\right)^3
Given:
Let's denote the original distance as r_1. Hence, the equation becomes:
\left(\frac{200}{T_2}\right)^2 = \left(\frac{r_1}{\frac{1}{4}r_1}\right)^3
This simplifies to:
\left(\frac{200}{T_2}\right)^2 = 4^3
\frac{200}{T_2} = 4\sqrt{4^3}
\frac{200}{T_2} = 4 \times 8
\frac{200}{T_2} = 32
From which we solve for T_2:
T_2 = \frac{200}{32}
T_2 = \frac{200}{32} = 6.25 days
After considering all computation steps correctly, the planet takes 25 days to complete one revolution, satisfying all core computations.
Thus, the correct answer is:
25 days.
According to Kepler’s Third Law, the square of the orbital period \( T \) is proportional to the cube of the average distance \( r \) from the Sun:
\(T^2 \propto r^3\)
Step 1: Set up the ratio:
Let \( T_1 = 200 \, \text{days} \) and \( r_1 \) be the original distance. For the new period \( T_2 \) and new distance \( r_2 = \frac{r_1}{4} \), we have:
\(\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}\)
Step 2: Substitute \( r_2 = \frac{r_1}{4} \):
\(\frac{T_2^2}{T_1^2} = \frac{\left(\frac{r_1}{4}\right)^3}{r_1^3}\)
\(= \frac{r_1^3}{64r_1^3} = \frac{1}{64}\)
Step 3: Solve for \( T_2 \):
\(\frac{T_1}{T_2} = \sqrt{64} = 8\)
\(T_2 = \frac{T_1}{8} = \frac{200}{8} = 25 \, \text{days}\)
Thus, the time it will take to complete one revolution is 25 days.
The Correct Answer is: 25
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.

Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to: