Arranging the letters in alphabetical order: NAGPUR
Starting with \( A \): \( 5! = 120 \) positions
Starting with \( G \): \( 5! = 120 \) positions, cumulative: 240
Starting with \( N \) and \( A \): \( 4! = 24 \) positions, cumulative: 264
Starting with \( N \) and \( G \): \( 4! = 24 \) positions, cumulative: 288
Starting with \( N \) and \( P \): \( 4! = 24 \) positions, cumulative: 312
Now, starting with \( N \), \( R \), and \( A \):
\[ \text{NRAGUP} = 1, \text{ cumulative: 313} \]
\[ \text{NRAGPU} = 1, \text{ cumulative: 314} \]
\[ \text{NRAPGU} = 1, \text{ cumulative: 315} \]
Thus, the word at the \( 315^{\text{th}} \) position is NRAPGU.
To determine the 315th word in the dictionary arrangement of the letters of the word "NAGPUR", we follow these steps:
We calculate the total number of permutations starting with each letter until we reach the desired position:
Now, calculate from NP to find position 315:
Continuing with NRA (R can be followed by A, G, P, U forming 4! each):
Thus, the word at the 315th position is NRAPGU.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)