The ratio of powers of two motors is (3√(x))/(√(x) + 1), that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value of x will be (a) 16 (b) 4 (c) 2 (d) 2.4

Answer: (a). P = mgh/t. P₁ = (300 × 10 × 100)/(300) = 1000 W, P₂ = (50 × 10 × 100)/(120) = 1250/3 W (P₁)/(P₂) = 12/5 = (3√(x))/(√(x)+1) ⇒ 15√(x) = 12√(x) + 12 ⇒ √(x) = 4 So the answer is (a) x = 16.

The ratio of powers of two motors is (3√(x))/(√(x) + 1), that are… | JEE Main 2023 (13 Apr, Shift 1)

Q17JEE Main 2023 · 13 APR · SHIFT 1Medium

Power

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

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Answer: (a)

$P = \dfrac{mgh}{t}$ .
$P_1 = \dfrac{300 \times 10 \times 100}{300} = 1000\text{ W}$ , $P_2 = \dfrac{50 \times 10 \times 100}{120} = \dfrac{1250}{3}\text{ W}$
$\dfrac{P_1}{P_2} = \dfrac{12}{5} = \dfrac{3\sqrt{x}}{\sqrt{x}+1} \Rightarrow 15\sqrt{x} = 12\sqrt{x} + 12 \Rightarrow \sqrt{x} = 4$
So the answer is (a) $x = 16$ .

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