The ratio of powers of two motors is $\frac{3 \sqrt{x}}{\sqrt{x}+1}$, that are capable of raising $300 \mathrm{~kg}$ water in 5 minutes and $50 \mathrm{~kg}$ water in 2 minutes respectively from a well of $100 \mathrm{~m}$ deep. The value of $x$ will be

Let us first find the power required to lift the water using each motor. Let $P_1$ be the power of the first motor, and $P_2$ be the power of the second motor. <br/><br/> The work done in lifting the water is given by $W = mgh$, where $m$ is the mass of water lifted, $g$ is the acceleration due to gravity, and $h$ is the height through which the water is lifted. In this case, $m = 300\mathrm{~kg}$ and $h = 100\mathrm{~m}$ for the first motor, and $m = 50\mathrm{~kg}$ and $h = 100\mathrm{~m}$ for the second motor. <br/><br/> The work done in lifting the water in 5 minutes by the first motor is:<br/><br/> $$W_1 = mgh = (300\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 294000\mathrm{~J}$$ <br/><br/> The power required to do this work in 5 minutes is:<br/><br/> $$P_1 = \frac{W_1}{t_1} = \frac{294000\mathrm{~J}}{300\mathrm{~s}} = 980\mathrm{~W}$$ <br/><br/> The work done in lifting the water in 2 minutes by the second motor is:<br/><br/> $$W_2 = mgh = (50\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 49000\mathrm{~J}$$ <br/><br/> The power required to do this work in 2 minutes is:<br/><br/> $$P_2 = \frac{W_2}{t_2} = \frac{49000\mathrm{~J}}{120\mathrm{~s}} = 408.33\mathrm{~W}$$ <br/><br/> The ratio of the powers of the two motors is:<br/><br/> $\frac{P_1}{P_2} = \frac{980\mathrm{~W}}{408.33\mathrm{~W}} \approx 2.4$ <br/><br/> We are given that this ratio is equal to:<br/><br/> $\frac{3 \sqrt{x}}{\sqrt{x}+1}$ <br/><br/> We can solve for $x$ as follows:<br/><br/> $\frac{3 \sqrt{x}}{\sqrt{x}+1} = 2.4$<br/><br/> $3\sqrt{x} = 2.4(\sqrt{x}+1)$<br/><br/> $3\sqrt{x} = 2.4\sqrt{x} + 2.4$<br/><br/> $(3-2.4)\sqrt{x} = 2.4$<br/><br/> $0.6\sqrt{x} = 2.4$<br/><br/> $\sqrt{x} = 4$<br/><br/> $x = 16$<br/><br/> Therefore, the value of $x$ is 16.