\[(a)\;49\; cm/s \quad (b)\;98\; cm/s \quad (c)\;147\; cm/s \quad(d)\;196\; cm/s \]

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mass of disc=$ \large \frac{10}{1000}$ kg

Let speed of bullet $=v$

change in momentum per second $=\; 2 mvn$

where n is number of bullet, fired per second, m -mass of bullet

Therefore $Force$ applied by the bullets on the disk = $2mvn$

$\frac{10}{1000}$$ \times g=.2 \times 10 \times \large \frac{5}{1000}$$v$

$ v= 98 cm/s$

Hence b is the correct answer.

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