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Properties of Matter question

2023 · Shift 2 · Q45
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Properties of Matter question

2023 · Shift 2 · Q45

JEE AdvancedPhysicsProperties of MatterNumerical+4 / −1
An incompressible liquid is kept in a container having a weightless piston with a hole. A capillary tube of inner radius 0.1 mm0.1 \mathrm{~mm}0.1 mm is dipped vertically into the liquid through the airtight piston hole, as shown in the figure. The air in the container is isothermally compressed from its original volume V0V_0V0​ to 100101V0\frac{100}{101} V_0101100​V0​ with the movable piston. Considering air as an ideal gas, the height (h)(h)(h) of the liquid column in the capillary above the liquid level in cm\mathrm{cm}cm is ‾\underline{\hspace{2cm}}​. [Given: Surface tension of the liquid is 0.075 N m−10.075 \mathrm{~N} \mathrm{~m}^{-1}0.075 N m−1, atmospheric pressure is 105 N m−210^5 \mathrm{~N} \mathrm{~m}^{-2}105 N m−2, acceleration due to gravity (g)(\mathrm{g})(g) is 10 m s−210 \mathrm{~m} \mathrm{~s}^{-2}10 m s−2, density of the liquid is 103 kg m−310^3 \mathrm{~kg} \mathrm{~m}^{-3}103 kg m−3 and contact angle of capillary surface with the liquid is zero] JEE Advanced 2023 Paper 2 Online Physics - Properties of Matter Question 3 English
Numerical answer
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Correct answer: 25

  1. Pressure increase due to isothermal compression of air

Since the air is compressed isothermally from volume V0V_0V0​ to 100101V0\dfrac{100}{101}V_0101100​V0​, we use Boyle’s law:

P1V1=P2V2P_1V_1=P_2V_2P1​V1​=P2​V2​

Initially, the air pressure is atmospheric:

P1=Patm=105 PaP_1=P_{\text{atm}}=10^5\ \text{Pa}P1​=Patm​=105 Pa

So,

P2=P1V1V2=105×V0(100/101)V0P_2= P_1\frac{V_1}{V_2}=10^5\times \frac{V_0}{(100/101)V_0}P2​=P1​V2​V1​​=105×(100/101)V0​V0​​

P2=105×101100=1.01×105 PaP_2=10^5\times \frac{101}{100}=1.01\times 10^5\ \text{Pa}P2​=105×100101​=1.01×105 Pa

Hence the excess pressure inside the container is

ΔP=P2−Patm=1.01×105−105=103 Pa\Delta P=P_2-P_{\text{atm}}=1.01\times 10^5-10^5=10^3\ \text{Pa}ΔP=P2​−Patm​=1.01×105−105=103 Pa

  1. Pressure balance at the meniscus in the capillary

The liquid rises in the capillary because of:

  • excess air pressure inside container: ΔP=1000 Pa\Delta P=1000\ \text{Pa}ΔP=1000 Pa
  • capillary pressure due to surface tension:

2Tcos⁡θr\frac{2T\cos\theta}{r}r2Tcosθ​

Given:

  • T=0.075 N/mT=0.075\ \text{N/m}T=0.075 N/m
  • θ=0∘⇒cos⁡θ=1\theta=0^\circ \Rightarrow \cos\theta=1θ=0∘⇒cosθ=1
  • r=0.1 mm=10−4 mr=0.1\ \text{mm}=10^{-4}\ \text{m}r=0.1 mm=10−4 m

Therefore,

2Tr=2×0.07510−4=1500 Pa\frac{2T}{r}=\frac{2\times 0.075}{10^{-4}}=1500\ \text{Pa}r2T​=10−42×0.075​=1500 Pa

This total upward pressure supports a liquid column of height hhh:

ρgh=ΔP+2Tr\rho gh=\Delta P+\frac{2T}{r}ρgh=ΔP+r2T​

Substitute values:

1000×10×h=1000+15001000\times 10\times h = 1000+15001000×10×h=1000+1500

104h=250010^4 h=2500104h=2500

h=2500104=0.25 mh=\frac{2500}{10^4}=0.25\ \text{m}h=1042500​=0.25 m

  1. Convert into cm

0.25 m=25 cm0.25\ \text{m}=25\ \text{cm}0.25 m=25 cm

  1. Final answer

25\boxed{25}25​

The derived answer matches the stored correct answer.

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