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:
- The dimension of \( v \) is \( [L T^{-1}] \).
- The dimension of \( \frac{At^2 + Bt}{C + t} \) should match the dimension of \( v \), i.e., \( [L T^{-1}] \).
- For the numerator \( At^2 + Bt \), dimensions of both terms must be consistent.
- \( A \) has dimensions of \( [ML^2T^{-3}] \), as \( A t^2 \) gives \( [ML^2T^{-1}] \), which balances the \( [L T^{-1}] \) dimension of velocity.
- \( B \) has dimensions of \( [MLT^{-3}] \), as it has to balance the dimension of velocity when multiplied by \( t \).
- The denominator \( C + t \) has dimensions of \( [T] \), so \( C \) must have dimensions \( [L T^{-2}] \).
Thus, the dimension of \( A \), \( B \), and \( C \) is \( [ML^2T^{-3}] \).
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:
Match List-I with List-II.
Choose the correct answer from the options given below :


The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?