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Standard XII

Physics

Young's Modulus of Elasticity

An open vesse...

Question

An open vessel containing water is given a constant acceleration a in a horizontal direction. Then the free surface of water gets sloped with the horizontal at an angle $θ$ , given by

θ = {cos}^{- 1} \frac{g}{a}

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θ = {tan}^{- 1} \frac{a}{g}

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θ = {sin}^{- 1} \frac{a}{g}

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θ = {tan}^{- 1} \frac{g}{a}

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Solution

The correct option is C $θ = {tan}^{- 1} \frac{a}{g}$
$tan θ = \frac{p e r p e n d i c u l a r}{b a s e} = \frac{m g}{m a} = \frac{g}{a}$ or
$θ = {tan}^{- 1} g / a$

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An open vessel containing water is given a constant acceleration a in a horizontal direction. Then the free surface of water gets sloped with the horizontal at an angle θ, given by

The correct option is C θ=tan−1agtanθ=perpendicularbase=mgma=ga orθ=tan−1g/a

An open vessel containing water is given a constant acceleration a in a horizontal direction. Then the free surface of water gets sloped with the horizontal at an angle $θ$ , given by

The correct option is C $θ = {tan}^{- 1} \frac{a}{g}$
$tan θ = \frac{p e r p e n d i c u l a r}{b a s e} = \frac{m g}{m a} = \frac{g}{a}$ or
$θ = {tan}^{- 1} g / a$