According to the principle of homogeneity of dimensions, the dimensions on the left-hand side (LHS) must match those on the right-hand side (RHS).
1. Check Dimensions of Each Term in Option (3):
Consider:
\[ T^2 = \frac{4\pi^2 r^3}{GM}. \] - The dimensions of \( T^2 \) are \([T^2]\).
- The dimensions of \( G \) (gravitational constant) are \([M^{-1}L^3T^{-2}]\).
- The dimensions of \( M \) are \([M]\).
- The dimensions of \( r \) (radius) are \([L]\).
2. Dimensional Analysis:
Substitute the dimensions into RHS:
\[ \left[\frac{L^3}{M \times M^{-1}L^3T^{-2}}\right] = [T^2]. \] Since both sides have the dimension of \([T^2]\), option (3) is dimensionally correct.
Answer: \( \frac{4\pi^2 r^3}{GM} \)
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32