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On a smooth inclined plane (with angle of inclination $α = 37 °$ ), a vessel containing liquid is being pushed upwards with an acceleration $5 m s^{- 2}$ . If $g = 10 m s^{- 2}$ then angle ( $θ$ ) that the liquid surface makes with the horizontal is (Given, $\sin 37 ° = \frac{3}{5}$ )

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$37 °$

$\tan^{- 1} (\frac{5}{13})$

$\tan^{- 1} (\frac{4}{13})$

$53 °$

answer is C.

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Detailed Solution

$a_{x} {=a}_{0} {cosα=5cos37}^{0} {=4ms}^{-2}$

$a_{y} {=a}_{0} {sinα=5sin37}^{0} {=3ms}^{-2}$

Now, ${θ=tan}^{-1} (\frac{a_{x}}{a_{y} +g}) {=tan}^{-1} (\frac{4}{3+10}) {=tan}^{-1} (\frac{4}{13})$

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