We need to find the dimensional formula of angular impulse.
Angular impulse is defined as the product of torque and time. Torque is the rotational analogue of force, given by \( \vec{\tau} = \vec{r} \times \vec{F} \). The dimensional formula can be derived from this relationship.
Step 1: Write the definition of angular impulse.
\[ \text{Angular Impulse} = \text{Torque} \times \text{Time} \]Step 2: Find the dimensional formula of torque.
Torque (\( \tau \)) = Force × Perpendicular Distance
\[ [\text{Torque}] = [\text{Force}] \times [\text{Distance}] \]The dimensional formula of force is \( [M L T^{-2}] \), and distance is \( [L] \).
\[ [\text{Torque}] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}] \]Step 3: Find the dimensional formula of time.
\[ [\text{Time}] = [T] \]Step 4: Combine the dimensions to get the dimensional formula for angular impulse.
\[ [\text{Angular Impulse}] = [\text{Torque}] \times [\text{Time}] = [M L^2 T^{-2}] \times [T] = [M L^2 T^{-1}] \]Thus, the dimensional formula of angular impulse is [M L² T⁻¹].
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,

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 : \(A_2 \rightleftharpoons 2A\)