JEE Main 2014 · 11 Apr · Shift 1 · Physics Q10

← JEE Main 2014 · 11 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 10 of 90

An air bubble of radius 0.1 cm is in a liquid having surface tension 0.06 N/m and density $10^3$ kg/m $^3$ . The pressure inside the bubble is 1100 Nm $^{-2}$ greater than the atmospheric pressure. At what depth is the bubble below the surface of the liquid ? ( $g=9.8$ ms $^{-2}$ )

0.1 m

0.15 m

0.20 m

0.25 m

Answer and worked solution

Official answer: A

$P_{\text{in}}-P_{\text{atm}}=\rho gh+\dfrac{2T}r$ . Here $\dfrac{2T}r=\dfrac{0.12}{0.001}=120$ Pa, so
$\rho gh=1100-120=980\implies h=\frac{980}{1000\times9.8}=0.1\text{ m}.$

Topics:Surface tension Pressure and Pascal's law

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 10