Answer

Verified

33.6k+ views

**Hint:**In this question we have to do the analysis of pressure at the bottom of the flask. Doing so in the two given conditions will give us the new length of air columns.

**Complete step by step solution:**

Case 1: Before tilting;

Let the pressure of trapped air be ${P_1}$

This will be the pressure at the air-mercury interface.

Also atmospheric pressure ( ${P_0}$ ) = 76 cm of Hg.

Now at point on the air-mercury interface the pressure will be;

${P_0} + {P_{Hg}} = {P_1}$

${P_1} = 76 + 20 = 96$ (Equation: 1)

Now since air is an ideal gas it will have uniform pressure throughout it. So the pressure at the bottom of the flask will be ${P_1} = 96$

Let the area of the tube be A

So applying gas law we get;

${P_1}V = nRT$

Now we all know, $V = Al$

And $l = 43cm$

Thus $ \Rightarrow 96(A \times 43) = nRT$ (Equation: 2)

Now let us consider case 2 (after tilling);

The pressure at the air mercury interface ( ${P_2}$ ) is given by;

${P_2} = {P_0} + {P_{Hg}}\cos ({60^ \circ })$

Thus, $ \Rightarrow {P_2} = 79 + \dfrac{{20}}{2} = 86$

Now applying gas law we get;

${P_2}V = nRT$

$ \Rightarrow 86(Al') = nRT$ (Equation: 3) (here, $l'$ is the new length of air column)

From equation 2 and equation 3 we get;

$ \Rightarrow 96 \times 43A = 86Al'$

$\therefore l' = \dfrac{{96 \times 43}}{{86}} = 48cm$

**Therefore, the new length is 48 cm.**

**Note:**The volume is changing and hence the area is constant.

If the lid will be closed the volume won’t change and hence the area will change.

The pressure balancing should be done carefully.

Recently Updated Pages

To get a maximum current in an external resistance class 1 physics JEE_Main

f a body travels with constant acceleration which of class 1 physics JEE_Main

A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main

If the beams of electrons and protons move parallel class 1 physics JEE_Main

Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Other Pages

The force of interaction of two dipoles if the two class 12 physics JEE_Main

Normality of 03 M phosphorus acid H3PO3 is A 05 B 06 class 11 chemistry JEE_Main

How many grams of concentrated nitric acid solution class 11 chemistry JEE_Main

Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main

Dissolving 120g of urea molwt60 in 1000g of water gave class 11 chemistry JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main