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(B) Using Bernoulli's equation at the inside of the bottle (point $1$) and at the nozzle (point $2$):$p_{1} + \frac{1}{2} \rho v_{1}^{2} = p_{2} + \fr

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$10$

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(B) Using Bernoulli's equation at the inside of the bottle (point $1$) and at the nozzle (point $2$):$p_{1} + \frac{1}{2} ho v_{1}^{2} = p_{2} + \frac{1}{2} ho v_{2}^{2}$Here,$p_{1} = p_{	ext{atm}} + \frac{F}{A}$ and $p_{2} = p_{	ext{atm}}$. Since the squeezing is slow,we assume $v_{1} \approx 0$.Therefore,$p_{	ext{atm}} + \frac{F}{A} = p_{	ext{atm}} + \frac{1}{2} ho v_{2}^{2} \Rightarrow v_{2}^{2} = \frac{2F}{ho A}$ .......... $(i)$For a horizontal projectile motion from height $h$,the range $R$ is given by $R = v_{2} \sqrt{\frac{2h}{g}}$.Squaring both sides,$R^{2} = v_{2}^{2} \left(\frac{2h}{g}ight) \Rightarrow v_{2}^{2} = \frac{R^{2}g}{2h}$ .......... $(ii)$Equating $(i)$ and $(ii)$:$\frac{2F}{ho A} = \frac{R^{2}g}{2h} \Rightarrow F = \frac{R^{2}g ho A}{4h}$Given $R = 2 \,m$,$g = 10 \,m/s^{2}$,$ho = 1000 \,kg/m^{3}$,$A = 10 \,cm^{2} = 10 	imes 10^{-4} \,m^{2} = 10^{-3} \,m^{2}$,and $h = 1 \,m$:$F = \frac{(2)^{2} 	imes 10 	imes 1000 	imes 10^{-3}}{4 	imes 1} = \frac{4 	imes 10 	imes 1}{4} = 10 \,N$.

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