The area of cross-section of one limb of an U-tube is twice that of the other. Both the limbs contain mercury at the same level. Water is poured in the wider tube so that the mercury level in it goes down by $1 c m$ . the height of the water column is (density of water = $10^{3} k g m^{- 3}$ , density of mercury = $13.6 \times 10^{3} k g m^{- 3}$ ).

13.6 c m

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40.8 c m

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27.2 c m

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54.4 c m

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Solution

The correct option is B $40.8 c m$
$G i v e n d e n s i t y o f w a t e r = 10^{3} k g / m^{3} a n d d e n s i t y o f m e r c u r y = 13.6 \times 10^{3} k g / m^{3} N o w p r e s s u r e a t p o i n t A (P_{A}) = p r e s s u r e a t p o i n t B (P_{B}) A n d p r e s s u r e = ρ g h ρ_{A} g h_{A} = ρ_{B} g h_{B} 10^{3} g h = 13.6 \times 10^{3} \times 3 h = 40.8 c m$

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Similar questions

Q. A simple U tube contains mercury to the same level in both of its arms. If water is poured to a height of

13.6 c m

in one arm, The rise in mercury level in the other arm is x cm.Then x=?
Given : density of mercury =

13.6 \times 10^{3} k g m^{- 3}

and density of water =

10^{3} k g m^{- 3}

A simple U tube contains mercury to the same level in both of its arms. If water is poured to a height of 13.6 cm in one arm, how much will be the rise in mercury level in other arm? Density of mercury $13.6 \times 10^{3} k g / m^{3}$ and density of water = $1000 k g / m^{3}$ .