JEE Physics 2023 Question with Solution

Given: $T_1 = 1 \, \text{year}$, $R_1 = 1.5 \times 10^6 \, \text{km}$, $T_2 = 2.83 \, \text{years}$

Find: Distance of the imaginary planet from the Sun, $R_2$.

Using Kepler’s third law:

$$T^2 \propto R^3$$

or

$$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$$

Substitute the given values:

$$\frac{1}{(2.83)^2} = \frac{(1.5 \times 10^6)^3}{R_2^3}$$

So,

$$R_2^3 = \left[(2.83)^2 \cdot (1.5 \times 10^6)^3\right]$$

Taking the cube root:

$$R_2 = 3 \times 10^6 \, \text{km}$$

Therefore, the distance of the imaginary planet from the Sun is $3 \times 10^6 \, \text{km}$. The solution states the correct option is B, although this value matches option C in the listed options.

Given: $T_1 = 1 \, \text{year}$, $R_1 = 1.5 \times 10^6 \, \text{km}$, $T_2 = 2.83 \, \text{years}$

Find: $R_2$.

From Kepler’s third law:

$$\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3$$

Substitute:

$$\left(\frac{2.83}{1}\right)^2 = \left(\frac{R_2}{1.5 \times 10^6}\right)^3$$

Thus,

$$\left(\frac{R_2}{1.5 \times 10^6}\right)^3 = (2.83)^2$$ $$R_2 = \left[(2.83)^2 (1.5 \times 10^6)^3\right]^{1/3}$$

Using the working shown in the solution:

$$R_2 = 8^{1/3} \times 1.5 \times 10^6$$ $$R_2 = 3 \times 10^6 \, \text{km}$$

Hence, the numerical result is $3 \times 10^6 \, \text{km}$. This corresponds to option C, even though the solution labels the option as B.