At the surface:
\[mg = 300 \, \text{N}\]
\[m = \frac{300}{g_s}\]
At depth \( \frac{R}{4} \):
\[g_d = g_s \left( 1 - \frac{d}{R} \right)\]
where \( d = \frac{R}{4} \).
\[g_d = g_s \left( 1 - \frac{R}{4R} \right) = g_s \cdot \frac{3}{4}\]
The weight at depth \( \frac{R}{4} \) is:
\[\text{Weight} = m \times g_d = m \times \frac{3 g_s}{4}\]
\[= \frac{3}{4} \times 300 = 225 \, \text{N}\]
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)
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