A U-tube having identical limbs is partially, filled with water. An immiscible oil having density 0.8 g/cc is poured into one side until water rises by 25 cm on the other side. Find the difference in the levels of the free surfaces of the liquids

125 cm

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75 cm

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22.5 cm

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12.5cm

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Solution

The correct option is C 12.5cm
$it raise in height of water = 25$

$then diff b/w heights of water = 50$

$Since pressure of liq.s water$

$is same.$

$ρ_{w} g (50) = ρ_{0} g \times h$

$1 \times 50 = 0.8 \times 17$

$h = 62.5 c m$

$S o Δ h = 62.5 - 50$

$= 12.5 c m$

Answer:- $(D)$

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Similar questions

Q. An

U -

tube is partially filled with water. An immiscible oil having density

0.8 g / c c

, is poured into one side until water rises by

25 c m

, on the other side. The level of oil will stand higher than level of water by ?

Q. A

U -

tube is partially filled with water. Oil which does not mix with water is next poured into one side,until water rises by

2.5 c m

on the other side. If the density of oil is

0.8 g / c c

the oil level will stand higher than the water level by:

A U-tube is partially filled with water. Oil, which does not mix with water, is next poured into one side, until water rises by 25 cm on the other side. If the density of the oil is 0.8 g/cc, the oil level will stand higher than the water level by

Q. A U-tube is partially filled with water.Oil which does not mix with water is next poured into one side, until water rises by 25 cm on th other side.If the density of oil is 0.8, the oil level will stand higher than the water level by

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A U-tube having identical limbs is partially, filled with water. An immiscible oil having density 0.8 g/cc is poured into one side until water rises by 25 cm on the other side. Find the difference in the levels of the free surfaces of the liquids

The correct option is C 12.5cm it raise in height of water =25 then diff b/w heights of water =50 Since pressure of liq.s water is same. ρwg(50)=ρ0g×h1×50=0.8×17h=62.5 cmSoΔh=62.5−50=12.5 cmAnswer:-(D)

The correct option is C 12.5cm
$it raise in height of water = 25$

$then diff b/w heights of water = 50$

$Since pressure of liq.s water$

$is same.$

$ρ_{w} g (50) = ρ_{0} g \times h$

$1 \times 50 = 0.8 \times 17$

$h = 62.5 c m$

$S o Δ h = 62.5 - 50$

$= 12.5 c m$

Answer:- $(D)$