1. Given current equation

   $i(t)=5\sin(49\pi t-30^\circ)$

   For a pure inductor,

   $v=L\frac{di}{dt}$

2. Differentiate the current

   $\frac{di}{dt}=5\cdot 49\pi \cos(49\pi t-30^\circ)$

   Therefore,

   $v= L\cdot 5\cdot 49\pi \cos(49\pi t-30^\circ)$

3. Substitute $L=30\text{ mH}=30\times 10^{-3}\text{ H}$

   $v=30\times 10^{-3}\times 5\times 49\pi \cos(49\pi t-30^\circ)$

4. Compute the amplitude

   Using $\pi=\dfrac{22}{7}$,

   $49\pi=49\cdot \frac{22}{7}=154$

   So,

   $V_0=30\times 10^{-3}\times 5\times 154$

   $V_0=0.03\times 770=23.1$

5. Convert cosine to sine form

   Since

   $\cos\theta=\sin(\theta+90^\circ)$

   we get

   $v=23.1\sin\big((49\pi t-30^\circ)+90^\circ\big)$

   $v=23.1\sin(49\pi t+60^\circ)$

6. Match with the options

   This corresponds to:

   Option D

Final Answer

$\boxed{23.1\sin(49\pi t+60^\circ)}$