An element of p-block forms a species of type EH$_4^{+}$, which when passed through a basic solution of K$_2$[HgI$_4$], forms a brown ppt. Select the correct option:
Show Hint
In p-block elements, the formation of brown precipitates with mercury salts is a characteristic reaction of ammonia (NH$_3$).
Element E has maximum electron affinity in its group.
EH$_3$ is phosphine.
Show Solution
The Correct Option isB
Solution and Explanation
Concept:
The total moment of inertia is the sum of the moments of inertia of the individual spheres. Since the axis does not pass through the centers (it is tangential), we use the Parallel Axis Theorem: \(I = I_{cm} + md^2\), where \(I_{cm} = \frac{2}{5}mR^2\) for a solid sphere and \(d=R\).
Step 1: Calculate \(I_1\).
\(m_1 = 10\), \(R_1 = 0.2\) m.
\[ I_1 = \frac{2}{5}m_1 R_1^2 + m_1 R_1^2 = \frac{7}{5}m_1 R_1^2 \]
\[ I_1 = 1.4 \times 10 \times (0.2)^2 = 14 \times 0.04 = 0.56\,\text{kg m}^2 \]
Step 2: Calculate \(I_2\).
\(m_2 = 5\), \(R_2 = 0.1\) m.
\[ I_2 = \frac{7}{5}m_2 R_2^2 \]
\[ I_2 = 1.4 \times 5 \times (0.1)^2 = 7 \times 0.01 = 0.07\,\text{kg m}^2 \]
Step 3: Total Moment of Inertia.
\[ I_{total} = I_1 + I_2 = 0.56 + 0.07 = 0.63\,\text{kg m}^2 \]
\[
\boxed{I = 0.63\,\text{kg m}^2}
\]
