The ratio of powers of two motors is frac{3 s | vedclass

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(D) Average Power is defined as the total work done divided by the total time taken.$P = \frac{W}{t} = \frac{mgh}{t}$Given the ratio of powers of two

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$16$

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(D) Average Power is defined as the total work done divided by the total time taken.$P = \frac{W}{t} = \frac{mgh}{t}$Given the ratio of powers of two motors is $\frac{P_1}{P_2} = \frac{3 \sqrt{x}}{\sqrt{x}+1}$.For the first motor: $m_1 = 300 \ kg$,$t_1 = 5 \ minutes = 300 \ s$,$h = 100 \ m$.For the second motor: $m_2 = 50 \ kg$,$t_2 = 2 \ minutes = 120 \ s$,$h = 100 \ m$.Calculating the ratio of powers:$\frac{P_1}{P_2} = \frac{m_1 g h / t_1}{m_2 g h / t_2} = \frac{m_1}{t_1} 	imes \frac{t_2}{m_2}$$\frac{P_1}{P_2} = \frac{300}{5} 	imes \frac{2}{50} = 60 	imes 0.04 = 2.4$Equating the two expressions:$\frac{3 \sqrt{x}}{\sqrt{x}+1} = 2.4$$3 \sqrt{x} = 2.4 \sqrt{x} + 2.4$$0.6 \sqrt{x} = 2.4$$\sqrt{x} = 4$$x = 16$

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

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