A cylinder containing water up to a height of h cm has a hole of cross-section area - cm2 at its bottom- It is counterpoised in a balance- When water just begins to flow out- the initial change in weight is -

The correct option is B 2ghα Upward thrust= d(mv)dt =vdmdt =v(α v)ρ=αρ v^2 =α× 2gh ( ρ=1g cm^ 3 ) =2ghα ...

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Standard IX

Physics

Pascal's Law

A cylinder co...

Question

A cylinder containing water up to a height of $h c m$ has a hole of cross-section area $α c m^{2}$ at its bottom. It is counterpoised in a balance. When water just begins to flow out, the initial change in weight is :

2 g h α

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\sqrt{2 g h α}

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\sqrt{g h α}

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None of these

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Solution

The correct option is B $2 g h α$
Upward thrust= $\frac{d (m v)}{d t}$
$= v \frac{d m}{d t}$
$= v (α v) ρ = α ρ v^{2}$
$= α \times 2 g h$ ( $ρ = 1 g c m^{- 3}$ )
$= 2 g h α$

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A cylinder containing water up to a height of h cm has a hole of cross-section area α cm2 at its bottom. It is counterpoised in a balance. When water just begins to flow out, the initial change in weight is :

The correct option is B 2ghαUpward thrust=d(mv)dt=vdmdt=v(αv)ρ=αρv2=α×2gh(ρ=1gcm−3)=2ghα

A cylinder containing water up to a height of $h c m$ has a hole of cross-section area $α c m^{2}$ at its bottom. It is counterpoised in a balance. When water just begins to flow out, the initial change in weight is :

The correct option is B $2 g h α$
Upward thrust= $\frac{d (m v)}{d t}$
$= v \frac{d m}{d t}$
$= v (α v) ρ = α ρ v^{2}$
$= α \times 2 g h$ ( $ρ = 1 g c m^{- 3}$ )
$= 2 g h α$