JEE Main 2021 18th March Morning Shift Question Paper with Solutions

Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:

$T^2 \propto r^3$

where:

• $T$ is the orbital period
• $r$ is the radius of the circular orbit (which serves as the semi-major axis in this case)

Considering two satellites, one with period $T$ and radius $R$, and another with unknown period $T'$ and radius $9R$, we can form an equation:

$\frac{{T'}^2}{T^2} = \frac{(9R)^3}{R^3}$

This simplifies to:

$\frac{{T'}^2}{T^2} = 729$

Taking the square root of both sides, we get:

$T' = T \times \sqrt{729} = T \times 27$

So, the period of another satellite in a circular orbit of radius $9R$ is $27T$.

In the double-slit experiment, the position of a bright fringe is given by the formula :

$y = \frac{{m\lambda L}}{{d}}$,

where :

• m is the order of the fringe (1 for the first bright fringe, 2 for the second, etc.)
• $\lambda$ is the wavelength of the light (in meters)
• L is the distance from the slits to the screen (in meters)
• d is the distance between the slits (in meters)

We need to find the difference in position between the first and third bright fringes. So, we find the position of both and subtract the position of the first from the position of the third :

$y_3 = \frac{{3\lambda L}}{{d}}$

$y_1 = \frac{{\lambda L}}{{d}}$

Then, the difference between the third and the first bright fringes is :

$y_3 - y_1 = 2\frac{{\lambda L}}{{d}}$

Now, let's plug the given values: $\lambda = 5890 \,Å = 5890 \times 10^{-10} \, m$ (since 1 Å = $10^{-10}$ meters), L = 0.5 m, and d = 0.5 mm = 0.5 $\times 10^{-3}$ m :

$y_3 - y_1 = 2 \times \frac{5890 \times 10^{-10} \, \text{m} \times 0.5 \, \text{m}}{0.5 \times 10^{-3} \, \text{m}} = 1178 \times 10^{-6} \, \text{m}$

So, the correct answer is 1178 $\times 10^{-6}$ m, which corresponds to Option B.