$\text{Disk access time} =\text{Seek time}+\text{Rotational latency}+\text{Transfer time}$ (given that transfer time is neglected)
$\text{Seek time}=10 \text{ ms}$
$\text{Rotational speed}=6000 \text{ rpm}$
- $60\;s \to 6000 \text{ rotations}$
- $1 \text{ rotation}\to 60/6000 \;s$
- $\text{Rotational latency} =1/2 \times 60/6000 \;s =5 \text{ ms}$
Total time to transfer one library $=10+5=15 \text{ ms}$
$\therefore$ Total time to transfer $100$ libraries $=100 \times 15 \text{ ms}=1.5 s$
Correct Answer: $B$