KCET · Physics · Units and Dimensions
Dimensional formula for the universal gravitational constant is
- A \(\left[\mathrm{M}^{-1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]\)
- B \(\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]\)
- C \(\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
- D \(\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-1}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}=\frac{\mathrm{GM}_{1} \mathrm{M}_{2}}{\mathrm{R}^{2}} \)
\( \Rightarrow \mathrm{G}=\frac{\mathrm{FR}^{2}}{\mathrm{M}_{1} \mathrm{M}_{2}} \)
\( \therefore[\mathrm{G}]=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{\left[\mathrm{M}^{2}\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
\( \Rightarrow \mathrm{G}=\frac{\mathrm{FR}^{2}}{\mathrm{M}_{1} \mathrm{M}_{2}} \)
\( \therefore[\mathrm{G}]=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}{\left[\mathrm{M}^{2}\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\)
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