Question 55: Gravitation

Question

$ 	ext { Match the LIST-I with LIST-II } $
List - I
List - II
A.
$
	ext { Gravitational constant }
$
I.
$
\left[\mathrm{LT}^{-2}ight]
$
B.
$
	ext { Gravitational potential energy }
$
II.
$
\left[\mathrm{L}^2 \mathrm{~T}^{-2}ight]
$
C.
$
	ext { Gravitational potential }
$
III.
$
\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]
$
D.
$
	ext { Acceleration due to gravity }
$
IV.
$
\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}ight]
$
Choose the correct answer from the options given below:

Accepted Answer

HereÃ¢ÂÂs how each quantity lines up with its dimensional formula:
Gravitational constant, $G$
Ã¢ÂÂ¢ From NewtonÃ¢ÂÂs law $F = G\frac{m_1m_2}{r^2}$
Ã¢ÂÂ¢ $[G] = \frac{[F]\,[r]^2}{[m]^2} = \frac{(MLT^{-2})\;L^2}{M^2} = M^{-1}L^3T^{-2}$
Ã¢ÂÂ IV
Gravitational potential energy, $U$
Ã¢ÂÂ¢ As work or $U = mgh$
Ã¢ÂÂ¢ $[U] = [F]\,[h] = (MLT^{-2})\;L = ML^2T^{-2}$
Ã¢ÂÂ III
Gravitational potential, $\phi$
Ã¢ÂÂ¢ Energy per unit mass: $\phi = \frac{U}{m}$
Ã¢ÂÂ¢ $[\phi] = \frac{ML^2T^{-2}}{M} = L^2T^{-2}$
Ã¢ÂÂ II
Acceleration due to gravity, $g$
Ã¢ÂÂ¢ Just an acceleration
Ã¢ÂÂ¢ $[g] = LT^{-2}$
Ã¢ÂÂ I
Matching them gives
AÃ¢ÂÂIV,Ã¢ÂÂBÃ¢ÂÂIII,Ã¢ÂÂCÃ¢ÂÂII,Ã¢ÂÂDÃ¢ÂÂI
That corresponds to
Option A
.

Answer

A-IV, B-III, C-II, D-I

Answer

A-II, B-IV, C-III, D-I

Answer

A-I, B-III, C-IV, D-II

Answer

A-III, B-II, C-I, D-IV

Gravitation

Choose the correct answer: