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Properties of Matter question

2012 · Shift 0 · Q64
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Properties of Matter question

2012 · Shift 0 · Q64

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5×10−2  N1.5 \times {10^{ - 2}}\,\,N1.5×10−2N(see figure). The length of the slider is 30cm30cm30cm and its weight negligible. The surface tension of the liquid film is AIEEE 2012 Physics - Properties of Matter Question 260 English
  1. A
    0.0125  Nm−10.0125\,\,N{m^{ - 1}}0.0125Nm−1
  2. B
    0.1  Nm−10.1\,\,N{m^{ - 1}}0.1Nm−1
  3. C
    0.05  Nm−10.05\,\,N{m^{ - 1}}0.05Nm−1
  4. D
    0.025  Nm−10.025\,\,N{m^{ - 1}}0.025Nm−1
View written solutionFree

Correct answer: D

  1. Force due to surface tension on the slider

A liquid film has two free surfaces, so the total upward force due to surface tension on the slider is

F=2TlF = 2TlF=2Tl

where:

  • TTT = surface tension
  • lll = length of slider
  1. Given data
  • Supported weight: F=1.5×10−2 NF = 1.5 \times 10^{-2}\,\text{N}F=1.5×10−2N
  • Length of slider: l=30 cm=0.30 ml = 30\,\text{cm} = 0.30\,\text{m}l=30cm=0.30m
  1. Apply equilibrium condition

Since the film supports the weight,

2Tl=1.5×10−22Tl = 1.5 \times 10^{-2}2Tl=1.5×10−2

Substitute l=0.30l = 0.30l=0.30 m:

2T(0.30)=1.5×10−22T(0.30) = 1.5 \times 10^{-2}2T(0.30)=1.5×10−2

0.60T=1.5×10−20.60T = 1.5 \times 10^{-2}0.60T=1.5×10−2

T=1.5×10−20.60T = \frac{1.5 \times 10^{-2}}{0.60}T=0.601.5×10−2​

T=2.5×10−2 N m−1T = 2.5 \times 10^{-2}\,\text{N m}^{-1}T=2.5×10−2N m−1

T=0.025 N m−1T = 0.025\,\text{N m}^{-1}T=0.025N m−1

  1. Match with options

This corresponds to:

Option D: 0.025 N m−10.025\,\text{N m}^{-1}0.025N m−1

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