tp = $\frac{d}{v} = \frac{10^{5}}{2\times 10^{8}} = 0.5 \times 10^{-3}$
No of bits = tp x bandwidth = $0.5 \times 10^{-3} \times 154 \times 10 ^{6} = 77000$
tp= 105 / 2*108 = 0.5 *103
??
Consider reading d link given below. The ques is exactly similar. I hope it solves ur query. :)
https://gateoverflow.in/9562/how-many-bits-fit-in-a-cable
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