We need to calculate the number of heteroatoms in one molecule of R. Heteroatoms are any atoms other than carbon (C) and hydrogen (H).
Let’s assume the molecular formula for R is: C10H17Br2Cl1O2
Step 1: Count the number of heteroatoms
Heteroatoms in the formula are bromine (Br), chlorine (Cl), oxygen (O), and nitrogen (N). Let's calculate the total number of heteroatoms:
Step 2: Total number of heteroatoms
Total number of heteroatoms = 2 (Br) + 1 (Cl) + 2 (O) = 9
Thus, the number of heteroatoms present in one molecule of R is 9.


We need to calculate the number of carbon atoms and heteroatoms in one molecule of S. Heteroatoms are any atoms other than carbon (C) and hydrogen (H).
Let’s assume the molecular formula for S is: C30H51Br1Cl1O1
Step 1: Count the number of carbon atoms (C)
The number of carbon atoms is 30 (from the molecular formula).
Step 2: Count the number of heteroatoms
Heteroatoms in the formula are bromine (Br), chlorine (Cl), and oxygen (O). So the number of heteroatoms is 3 (Br + Cl + O).
Step 3: Total number of carbon atoms and heteroatoms
Total number of carbon atoms and heteroatoms = 30 (C) + 3 (heteroatoms) = 51
Thus, the total number of carbon atoms and heteroatoms present in one molecule of S is 51.


The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.