Given:
- Lines \( L_{2n-1} \) (\( n = 1, 2, \dots, 10 \)) are parallel to each other.
- Lines \( L_{2n} \) (\( n = 1, 2, \dots, 10 \)) pass through a common point \( P \).
Step 1: Points of Intersection between \( L_{2n-1} \) and \( L_{2m} \)
Since all \( L_{2n-1} \) lines are parallel, they do not intersect among themselves. Similarly, all \( L_{2n} \) lines pass through the point \( P \), so they intersect at \( P \) and do not form additional intersection points among themselves.
However, each line \( L_{2n-1} \) intersects each line \( L_{2m} \) exactly once (since they are not parallel), leading to:
\[ 10 \times 10 = 100 \text{ intersection points} \]
Step 2: Points of Intersection among \( L_{2n} \) Lines
All \( L_{2n} \) lines pass through the common point \( P \). Therefore, there is exactly one intersection point among these lines at \( P \).
Step 3: Total Number of Points of Intersection
The total number of points of intersection is given by:
\[ 100 + 1 = 101 \]
Conclusion: The maximum number of points of intersection of pairs of lines from the set \( \{L_1, L_2, \dots, L_{20}\} \) is 101.
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 : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,