A vertical glass tube, closed at the bottom contains mercury to a height of 100 cm at 300C.The height of mercury column at 50°C is.γ for Hg=1.8×10−4/∘C,αg=8×10−6/∘C

H1=H01+(γ−3α)t1H2=H01+(γ−3α)t2H1=H01+(γ−3α)t1H2=H01+(γ−3α)t2100H2=1+1.54×10−4×301+1.54×10−4×50H2=100.32 nearly

A vertical glass tube, closed at the bottom contains mercury to a height of 100 cm at 300C.The height of mercury column at 50°C is.γ for Hg=1.8×10−4/∘C,αg=8×10−6/∘C

H1=H01+(γ−3α)t1H2=H01+(γ−3α)t2H1=H01+(γ−3α)t1H2=H01+(γ−3α)t2100H2=1+1.54×10−4×301+1.54×10−4×50H2=100.32 nearly

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A vertical glass tube, closed at the bottom contains mercury to a height of 100 cm at 30⁰C.The height of mercury column at 50°C is.
$(γ for Hg = 1.8 \times 10^{- 4} /^{\circ} C, α_{g} = 8 \times 10^{- 6} /^{\circ} C)$

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10.0328 cm

100.328 mm

100.328 cm

1003.28 cm

answer is A.

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Detailed Solution

$\begin{array}{l} H_{1} = H_{0} (1 + (γ - 3 α) t_{1}) \\ H_{2} = H_{0} (1 + (γ - 3 α) t_{2}) \\ \frac{H_{1} = H_{0} (1 + (γ - 3 α) t_{1})}{H_{2} = H_{0} (1 + (γ - 3 α) t_{2})} \\ \frac{100}{H_{2}} = \frac{1 + (1.54 \times 10^{- 4} \times 30)}{1 + (1.54 \times 10^{- 4} \times 50)} \\ H_{2} = 100.32 nearly \end{array}$