Step 1: Count atoms from structure
Carbon (C) = 15 atoms
Hydrogen (H) = 11 atoms
Oxygen (O) = 4 atoms
Nitrogen (N) = 1 atom
Iodine (I) = 4 atoms
Step 2: Compute mass contribution
\[ \begin{aligned} \text{Mass of C} &= 15 \times 12 = 180 \, \text{g/mol} \\ \text{Mass of H} &= 11 \times 1 = 11 \, \text{g/mol} \\ \text{Mass of O} &= 4 \times 16 = 64 \, \text{g/mol} \\ \text{Mass of N} &= 1 \times 14 = 14 \, \text{g/mol} \\ \text{Mass of I} &= 4 \times 127 = 508 \, \text{g/mol} \end{aligned} \]
Step 3: Total molar mass of thyroxine
\[ \text{Molar mass} = 180 + 11 + 64 + 14 + 508 = 777 \, \text{g/mol} \]
Step 4: Percentage of Iodine
\[ \text{Percentage of Iodine} = \frac{508}{777} \times 100 \approx 65.37\% \]
Correct Answer: 65%
Step 1:
Thyroxine contains the elements carbon (C), hydrogen (H), oxygen (O), nitrogen (N), and iodine (I). To calculate the percentage of iodine, we first need to determine the total molar mass of thyroxine and the mass of iodine present in the compound.
Let's assume the chemical formula of thyroxine is C₁₄H₁₈I₄N₂O₄ (based on the provided structure in the image).
Step 2:
To find the molar mass of thyroxine, we add the molar masses of each element based on its quantity in the formula:
- Carbon (C): 12 × 14 = 168 g/mol
- Hydrogen (H): 1 × 18 = 18 g/mol
- Iodine (I): 127 × 4 = 508 g/mol
- Nitrogen (N): 14 × 2 = 28 g/mol
- Oxygen (O): 16 × 4 = 64 g/mol
Now, sum these values:
\[
\text{Molar mass of thyroxine} = 168 + 18 + 508 + 28 + 64 = 786 \, \text{g/mol}
\]
Step 3:
The mass of iodine in thyroxine is 508 g/mol (from 4 iodine atoms). Now, calculate the percentage of iodine in the compound:
\[
\text{Percentage of iodine} = \left( \frac{508}{786} \right) \times 100 = 64.6\%
\]
Step 4:
Rounding this value to the nearest integer gives the percentage of iodine in thyroxine as 65%.
Final Answer:
The percentage of iodine in thyroxine is 65%.

A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 