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(B) Let the mass of the (bucket $+$ water) be $M$. The mass of the weight is $M/2$.The system is connected by a rope over a pulley. The net force on t

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(B) Let the mass of the (bucket $+$ water) be $M$. The mass of the weight is $M/2$.The system is connected by a rope over a pulley. The net force on the system is $Mg - (M/2)g = Mg/2$.The total mass of the system is $M + M/2 = 3M/2$.The acceleration $a$ of the system is $a = \frac{	ext{Net Force}}{	ext{Total Mass}} = \frac{Mg/2}{3M/2} = g/3$.Since the bucket is moving downward with acceleration $a = g/3$,the effective acceleration due to gravity is $g_{eff} = g - a = g - g/3 = 2g/3$.The gauge pressure at the bottom is given by $P - P_0 = ho g_{eff} h$.Substituting the values: $ho = 1000 \ kg/m^3$,$g = 10 \ m/s^2$,$h = 0.15 \ m$.$P - P_0 = 1000 	imes (2 	imes 10 / 3) 	imes 0.15 = 1000 	imes (20/3) 	imes 0.15 = 1000 	imes 1 = 1000 \ Pa = 1 \ kPa$.

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