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Physics · Gravitation

JEE Main 2024 — 4 April, Shift 2 — Question 48

Applying the principle of homogeneity of dimensions, determine which one is correct. where T is time period, G is gravitational constant, MM is mass, rr is radius of orbit.

  1. Option A:

    T2=4π2rGM2\mathrm{T}^{2}=\frac{4 \pi^{2} r}{\mathrm{GM}^{2}}

  2. Option B:

    T2=4π2r3\mathrm{T}^{2}=4 \pi^{2} \mathrm{r}^{3}

  3. Option C:

    T2=4π2r3 GM \mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}^{3}}{\text { GM }}

    Correct
  4. Option D:

    T2=4π2r2GM\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}^{2}}{G M}

Answer: C

Step-by-step solution

According to principle of homogeneity dimension of LHS should be equal to dimensions of RHS so option (3) is correct.

T2=4π2r3GM\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}^{3}}{\mathrm{GM}}

[T2]=[L3][M−1 L3 T−2][M]\left[\mathrm{T}^{2}\right]=\frac{\left[\mathrm{L}^{3}\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right][\mathrm{M}]} (Dimension of GG is [M−1L3T−2]\left[M^{-1} L^{3} T^{-2}\right] ) [T2]=[L3][L3 T−2]=[T2]\left[\mathrm{T}^{2}\right]=\frac{\left[\mathrm{L}^{3}\right]}{\left[\mathrm{L}^{3} \mathrm{~T}^{-2}\right]}=\left[\mathrm{T}^{2}\right]

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Physics
Chapter
Gravitation
Topic
Planetary Motion & Binary Star System (Kepler's Law)

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