Question

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 the other arm? (density of mercury=13.6*10^3kg/m^3 and density of water=1000kg/m^3).

Answer 1

Pascals Law = Pressure is equal at both arms
HDG=HDG
13.6x 1000=hx13.6x1000
H=1m

Mark as brainliest answer

Answer 2

The rise of mercury level in the other arm will be 1 cm.

Explanation:

Given Data :

In a simple 'U' tube,

=>  Two arms having equal amount of mercury

In one arm, the height of water poured (h) = 13.6 cm

$\text {Density of Mercury} = 13.6 \times 10^3 \frac{kg}{m^3}$

$\text {Density of water} = 1000 \frac{kg}{m^3}$

Let us consider h' be the raise of mercury level in another arm of the 'U' tube.

Apply the concepts of fluid mechanics,

The formula pressure of fluids is P = ρ × g× h

Where ρ = density of the fluid ; g = gravity ; h= height of the fluid

Pressure on one arm due to water = Pressure on another arm due to                                                                   Mercury

Density of water × g × height of the water = Density of mercury × g × height          of the mercury

(ρ × g× h) of water = (ρ × g× h) Mercury

 13.6 × 10³ × g = h' × 13.6 × 10³ × g

Gravity can be cancelled on both sides of the above equation

13.6 × 10³ = h' × 13.6 × 10³

$h' = \frac{13.6 \times 10^{3}}{13.6 \times 10^{3}} \text {cm}$

h' = 1 cm

Therefore When the water is poured to a height of 13.6 cm in an arm in a simple 'U' tube which contains mercury of equal amount in both the arm, the height of the mercury level rise in the other arm is 1 cm.

To Learn More ...

1) An open u tube is containing mercury kerosene of specific gravity 0.8 is poured into one of its limbs so that the length of column of kerosene is about 40 cm the level of mercury column in that limb is lowered approximately by how much?​

https://brainly.in/question/8105253