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:
\[\text{Weight} = m \times g_d = m \times \frac{3 g_s}{4}\]
\[= \frac{3}{4} \times 300 = 225 \, \text{N}\]
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).