The force exerted by mercury on the bottom of the tube can be calculated as:
\[ F = P_0A + \rho_m ghA \]
where:
- \( P_0 = 10^5 \, \text{Nm}^{-2} \) (atmospheric pressure),
- \( A = \pi r^2 = \frac{22}{7} \times (2 \times 10^{-2})^2 \) (area of the base),
- \( \rho_m = 1.36 \times 10^4 \, \text{kg m}^{-3} \) (density of mercury),
- \( g = 10 \, \text{ms}^{-2} \),
- \( h = 30 \times 10^{-2} \, \text{m} \) (height of mercury column).
Calculating \( A \):
\[ A = \frac{22}{7} \times (2 \times 10^{-2})^2 \]
Substitute into the force equation:
\[ F = 10^5 \times \frac{22}{7} \times (2 \times 10^{-2})^2 + 1.36 \times 10^4 \times 10 \times (30 \times 10^{-2}) \times \frac{22}{7} \times (2 \times 10^{-2})^2 \]
Solving this:
\[ F = 51.29 + 125.71 = 177 \, \text{N}. \]
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 \( 4.606 \times 10^{-2} \text{ s}^{-1} \).