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 quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of:
The expression given below shows the variation of velocity \( v \) with time \( t \): \[ v = \frac{At^2 + Bt}{C + t} \] The dimension of \( A \), \( B \), and \( C \) is:
Match List-I with List-II.
Choose the correct answer from the options given below :

Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.