\(\frac{2}{9}F\)
\(\frac{16}{9}F\)
\(\frac{8}{9}F\)
\(F\)
The correct answer is (C) : \(\frac{8}{9}F\)
Let the masses are m and distance between them is l, then
\(F=\frac{Gm^2}{I^2}\)
When 1/3rd mass is transferred to the other then masses will be 4m/3 and 2m/3. So new force will be
\(F^′=\frac{G\frac{4m}{3}×\frac{2m}{3}}{I^2}\)
\(=\frac{8}{9}\frac{Gm^2}{I^2}=\frac{8}{9}F\)
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is:
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].