Physics And Measurement Question 49
Question: The dimensions of universal gravitational constant are
[ CBSE PMT 1988, 92; 2004 ]
Options:
A) $ {{M}^{-2}}L^{2}{{T}^{-2}} $
B) $ {{M}^{-1}}L^{3}{{T}^{-2}} $
C) $ M{{L}^{-1}}{{T}^{-2}} $
D) $ ML^{2}{{T}^{-2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ F=\frac{Gm _{1}m _{2}}{d^{2}}\Rightarrow G=\frac{Fd^{2}}{m _{1}m _{2}} $
$ 1ly=9.46\times 10^{15}meter $
$ [G]=\frac{[ML{{T}^{-2}}][L^{2}]}{[M^{2}]}=[{{M}^{-1}}L^{3}{{T}^{-2}}] $
