Given:
\[ A = \{1, 2, 3, \ldots, 100\} \]
Step 1: Define the relation \( R \)
\[ R \Rightarrow 2x = 3y \Rightarrow y = \frac{2x}{3} \]
Therefore, \[ R = \{(3, 2), (6, 4), (9, 6), \ldots, (99, 66)\} \]
Number of elements in \( R \): \[ n(R) = 33 \]
Since \( R \subset R_1 \)
Step 2: Define \( R_1 \)
\[ R_1 = \{(3, 2), (6, 4), (9, 6), \ldots, (99, 66), (2, 3), (4, 6), (6, 9), \ldots, (66, 99)\} \]
Step 3: Minimum number of elements
\[ \text{Minimum number of elements in } R_1 = 66 \]
Final Answer:
\[ \boxed{66} \]
The relation \( R \) consists of ordered pairs \( (x, y) \) such that \( 2x = 3y \). For \( x \) and \( y \) to satisfy this relation, \( x \) and \( y \) must form pairs with specific integer values that satisfy \( 2x = 3y \).
Thus, the pairs in \( R \) are:
\[ R = \{(3, 2), (6, 4), (9, 6), (12, 8), \ldots, (99, 66)\}. \]
There are 33 such pairs in \( R \), so:
\[ n(R) = 33. \]
To make \( R_1 \) symmetric, we include both \( (x, y) \) and \( (y, x) \) for each pair in \( R \). Thus, the pairs in \( R_1 \) are:
\[ R_1 = \{(3, 2), (2, 3), (6, 4), (4, 6), (9, 6), (6, 9), \ldots, (99, 66), (66, 99)\}. \]
This doubles the number of elements:
\[ n = 2 \times 33 = 66. \]
Therefore, the minimum value of \( n \) is: \[ 66 \]
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,