An open U-tube contains mercury. When $13.6 c m$ of water is pourd into one of the arms of the tube then the mercury rise in the other arm from its initial level is:

1 c m

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0.5 c m

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10 c m

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5 c m

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Solution

The correct option is
B $0.5 c m$
$(2 x + h) P_{m} g = p_{m} h g + P_{w} g 13.6$
$2 P_{m} g x = P / w g 13.6 (\frac{P_{m}}{P_{w}} = 13.6)$
$x = 1 / 2 c m$

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Similar questions

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An open U-tube contains mercury. When 13.6cm of water is pourd into one of the arms of the tube then the mercury rise in the other arm from its initial level is:

The correct option is B 0.5cm(2x+h)Pmg=pmh g+Pw g 13.62Pm g x=P/w g 13.6(PmPw=13.6)x=1/2cm

An open U-tube contains mercury. When $13.6 c m$ of water is pourd into one of the arms of the tube then the mercury rise in the other arm from its initial level is:

The correct option is
B $0.5 c m$
$(2 x + h) P_{m} g = p_{m} h g + P_{w} g 13.6$
$2 P_{m} g x = P / w g 13.6 (\frac{P_{m}}{P_{w}} = 13.6)$
$x = 1 / 2 c m$