A vertical column of liquid $h_{0}$ long at $0^{o} C$ balances another column of same liquid at $t^{o} C$ . If r is the absolute expansion of that liquid, height of second liquid column is :

h_{0} (1 - r t)

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\frac{h_{0}}{(1 - r t)}

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h_{0} (1 + r t)

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\frac{h_{0}}{1 + r t}

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Solution

The correct option is A $h_{0} (1 + r t)$
$h_{t} = h_{0} (1 + γ Δ t)$
$γ = r$ as given
$∴ h_{t} = h_{0} (1 + r t)$

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A vertical column of liquid h0 long at 0oC balances another column of same liquid at toC . If r is the absolute expansion of that liquid, height of second liquid column is :

The correct option is A h0(1+rt)ht=h0(1+γΔt)γ=r as given∴ht=h0(1+rt)

A vertical column of liquid $h_{0}$ long at $0^{o} C$ balances another column of same liquid at $t^{o} C$ . If r is the absolute expansion of that liquid, height of second liquid column is :

The correct option is A $h_{0} (1 + r t)$
$h_{t} = h_{0} (1 + γ Δ t)$
$γ = r$ as given
$∴ h_{t} = h_{0} (1 + r t)$