QJEEM23Phc3f0: JEE Main 2023 Physics - Class 11 JEEM23Phc3f0 Solution: The ratio of powers of two motors is , that are capable of…

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.

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.

The work done in lifting the water in 5 minutes by the first motor is:

$W_1 = mgh = (300\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 294000\mathrm{~J}$

The power required to do this work in 5 minutes is:

$P_1 = \frac{W_1}{t_1} = \frac{294000\mathrm{~J}}{300\mathrm{~s}} = 980\mathrm{~W}$

The work done in lifting the water in 2 minutes by the second motor is:

$W_2 = mgh = (50\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 49000\mathrm{~J}$

The power required to do this work in 2 minutes is:

$P_2 = \frac{W_2}{t_2} = \frac{49000\mathrm{~J}}{120\mathrm{~s}} = 408.33\mathrm{~W}$

The ratio of the powers of the two motors is:

$\frac{P_1}{P_2} = \frac{980\mathrm{~W}}{408.33\mathrm{~W}} \approx 2.4$

We are given that this ratio is equal to:

$\frac{3 \sqrt{x}}{\sqrt{x}+1}$

We can solve for $x$ as follows:

$\frac{3 \sqrt{x}}{\sqrt{x}+1} = 2.4$

$3\sqrt{x} = 2.4(\sqrt{x}+1)$

$3\sqrt{x} = 2.4\sqrt{x} + 2.4$

$(3-2.4)\sqrt{x} = 2.4$

$0.6\sqrt{x} = 2.4$

$\sqrt{x} = 4$

$x = 16$

Therefore, the value of $x$ is 16.