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.
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is:
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
