A car of mass 2 metric tons is moving up a smooth inclined plane with a velocity of 72KMPH. If the power of the engine is 8 KW, the angle of inclination of the inclined plane is

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$\cos^{- 1} (\frac{48}{49})$

$\cos^{- 1} (\frac{1}{49})$

$\sin^{- 1} (\frac{1}{49})$

${Tan}^{- 1} (\frac{1}{49})$

answer is D.

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Detailed Solution

We know that,

$F = m g \sin θ F = 2000 \times 10 \sin θ F = 20000 S i n θ$

but,

$P = F \times V 8000 = 20000 \sin θ \times \frac{72000}{3600} \sin θ = \frac{1}{49}$

Hence the correct answer is $θ = \sin^{- 1} (\frac{1}{49})$ .

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