MHTCET 14 May 2023 Shift 2 Solved Paper

To determine how many bullets the man can fire per second without exceeding the maximum force he can exert, we need to compute the force applied due to the momentum change of the bullets being fired.The momentum $p$ of a single bullet is given by:$p=m \cdot v$where $m$ is the mass of the bullet and $v$ is the velocity of the bullet. Here, $m=30\text{ g}=0.03\text{ kg}$ (since $1\text{ g}=0.001\text{ kg}$ ) and $v=1000\text{ m}/\text{s}$.Now we calculate the momentum of a single bullet: $p=0.03\text{ kg} \cdot 1000\text{ m}/\text{s}=30\text{ kg} \cdot \text{m}/\text{s}$The rate of change of momentum gives us the force exerted by the bullets being fired. If $F$ is the force and $n$ is the number of bullets fired per second, the force is equal to the rate of change of momentum:$F = n \cdot \; \frac{\Delta p}{\Delta t}$Since $\Delta p$ is the change in momentum per bullet and $\Delta t = 1$ second (because $n$ is measured in bullets per second), the force exerted by $n$ bullets is:$F = n \cdot \; \frac{30\text{ kg} \cdot \text{m}/\text{s}}{1\text{ s}} = n \cdot 30\text{ N}$ We know the man can exert a maximum force of $F_{\max}=300\text{ N}$. Therefore, the maximum number of bullets ( $n_{\max}$ ) that can be fired per second without exceeding the maximum force is obtained by equating the two:$300\text{ N} = n_{\max} \cdot 30\text{ N}$Solving for $n_{\max}$ :$n_{\max} = \frac{300\text{ N}}{30\text{ N} / \text{bullet}} = 10\text{ bullets}/\text{s}$Thus, the man can fire at most 10 bullets per second. Hence, the correct answer is:Option C: 10