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  • Please help! - Properties of Solids and Liquids - JEE Main-2

When an air bubble of radius r rises from the bottom to the surface of a lake, its radius becomes\frac{5r}{4} . Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature) :

  • Option 1)

    11.2 m

  • Option 2)

    8.7 m

  • Option 3)

    9.5 m

  • Option 4)

    10.5 m

 

Answers (3)

best_answer

As we learnt

Change in Pressure of bubble in air -

\Delta P=\left ( \frac{2T}{R} \right )\times 2= \frac{4T}{R}

- wherein

T- Temperature

R- Radius

 

 At bottom surface

\rho _{1}=\rho _{a}+\rho gh+\frac{4s}{r}

p_{2}-p _{a}=\frac{4s}{\left ( \frac{5r}{4} \right )}=\left ( \frac{16s}{5r} \right )

Also,P_{1}v_{1}=P_{2}v_{2}

P_{1}.\frac{4\Pi }{3}r^{3}=P_{2}.\frac{4\Pi }{3}*\frac{125}{64}.r^{3}

P_{1}= \frac{125}{64}P_{2}\: \: or\: \: \frac{P_{1}}{P_{2}}=\frac{125}{64}

\therefore \frac{P_{a}+\rho gh+\frac{4s}{r}}{P_{a}+\frac{16s}{5r}}= \frac{125}{64}

Ignoring Surface Tension

\therefore \frac{P_{a}+\rho gh}{P_{a}}=\frac{125}{64} \: \: or\: \: 1+\frac{\rho gh}{P_{a}}=\frac{125}{64}

\therefore \frac{\rho gh}{P_{a}}=\frac{61}{64} \: \: or\: \: \rho gh=\left ( \frac{61}{64} \right )\rho g*10

h=9.5m


Option 1)

11.2 m

Option 2)

8.7 m

Option 3)

9.5 m

Option 4)

10.5 m

Posted by

Avinash

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

p2v2=p1v1---------(1) {ideal gas equation}{inside air bubble}

p2=patm=10*d*g.............{d= density of water}

v2=43*pi*(5r/4)^3

v1=4//3*pi*r^3

p1= p2 + H*d*g= 10*d*g + H*d*g

(1)-------> (10*d*g + H*d*g)4/3*pi*r^3= 10*d*g4/3*pi*(5r/4)^3

==>(10 + H)= 10(5/4)^3  

==> H = 10*(125/64) - 10

H=(1250-640)/64= 610/64= 9.5m

 

Posted by

devtantra

View full answer

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