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Physics · Gravitation

JEE Main 2024 — 8 April, Shift 1 — Question 41

Two planets AA and BB having masses m1\mathrm{m}_{1} and m2\mathrm{m}_{2} move around the sun in circular orbits of r1r_{1} and r2r_{2} radii respectively. If angular momentum of A is L and that of BB is 3L3 L, the ratio of time period (TATB)\left(\frac{T_{A}}{T_{B}}\right) is:

  1. Option A:

    (r2r1)32\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{\frac{3}{2}}

  2. Option B:

    (r1r2)3\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3}

  3. Option C:

    127(m2m1)3\frac{1}{27}\left(\frac{m_{2}}{m_{1}}\right)^{3}

    Correct
  4. Option D:

    27(m1m2)327\left(\frac{m_{1}}{m_{2}}\right)^{3}

Answer: C

Step-by-step solution

πr12 TA=L2 m1……\frac{\pi \mathrm{r}_{1}^{2}}{\mathrm{~T}_{\mathrm{A}}}=\frac{\mathrm{L}}{2 \mathrm{~m}_{1}} \ldots \ldots.

πr22 TB=3 L2 m2\frac{\pi \mathrm{r}_{2}^{2}}{\mathrm{~T}_{\mathrm{B}}}=\frac{3 \mathrm{~L}}{2 \mathrm{~m}_{2}}

⇒TATB=3⋅ m1 m2⋅(r1r2)2\Rightarrow \frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}=3 \cdot \frac{\mathrm{~m}_{1}}{\mathrm{~m}_{2}} \cdot\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2} (TATB)2=(r1r2)3⇒(r1r2)2=(TATB)43\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{2}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3} \Rightarrow\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{2}=\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{\frac{4}{3}}

⇒127⋅( m2 m1)3=(TATB)\Rightarrow \frac{1}{27} \cdot\left(\frac{\mathrm{~m}_{2}}{\mathrm{~m}_{1}}\right)^{3}=\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 1
Subject
Physics
Chapter
Gravitation
Topic
Planetary Motion & Binary Star System (Kepler's Law)