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Properties of Matter question

2023 · 29 Jan · Shift 1 · Q52
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  5. /2023 · 29 Jan · Shift 1 · Q52

Properties of Matter question

2023 · 29 Jan · Shift 1 · Q52

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Surface tension of a soap bubble is 2.0×10−2Nm−12.0 \times 10^{-2} \mathrm{Nm}^{-1}2.0×10−2Nm−1. Work done to increase the radius of soap bubble from 3.5 cm3.5 \mathrm{~cm}3.5 cm to 7 cm7 \mathrm{~cm}7 cm will be: Take [π=227]\left[\pi=\frac{22}{7}\right][π=722​]
  1. A
    18.48×10−4 J18 .48 \times 10^{-4} \mathrm{~J}18.48×10−4 J
  2. B
    5.76×10−4 J5.76 \times 10^{-4} \mathrm{~J}5.76×10−4 J
  3. C
    0.72×10−4 J0.72 \times 10^{-4} \mathrm{~J}0.72×10−4 J
  4. D
    9.24×10−4 J9.24 \times 10^{-4} \mathrm{~J}9.24×10−4 J
View written solutionFree

Correct answer: A

  1. Key concept

For a soap bubble, there are two surfaces, so the excess surface energy is E=2T⋅4πr2=8πTr2E = 2T \cdot 4\pi r^2 = 8\pi T r^2E=2T⋅4πr2=8πTr2 Hence, the work done in increasing the radius from r1r_1r1​ to r2r_2r2​ is the increase in surface energy: W=8πT(r22−r12)W = 8\pi T (r_2^2 - r_1^2)W=8πT(r22​−r12​)

  1. Given data

T=2.0×10−2 N m−1T = 2.0 \times 10^{-2}\, \text{N m}^{-1}T=2.0×10−2N m−1 r1=3.5 cm=3.5×10−2 mr_1 = 3.5\, \text{cm} = 3.5 \times 10^{-2}\, \text{m}r1​=3.5cm=3.5×10−2m r2=7 cm=7×10−2 mr_2 = 7\, \text{cm} = 7 \times 10^{-2}\, \text{m}r2​=7cm=7×10−2m

  1. Compute r22−r12r_2^2 - r_1^2r22​−r12​

r22−r12=(7×10−2)2−(3.5×10−2)2r_2^2 - r_1^2 = (7\times 10^{-2})^2 - (3.5\times 10^{-2})^2r22​−r12​=(7×10−2)2−(3.5×10−2)2 =49×10−4−12.25×10−4= 49\times 10^{-4} - 12.25\times 10^{-4}=49×10−4−12.25×10−4 =36.75×10−4= 36.75\times 10^{-4}=36.75×10−4 =3.675×10−3 m2= 3.675\times 10^{-3}\, \text{m}^2=3.675×10−3m2

  1. Substitute into formula

W=8πT(r22−r12)W = 8\pi T (r_2^2-r_1^2)W=8πT(r22​−r12​) Using π=227\pi = \dfrac{22}{7}π=722​, W=8⋅227⋅2×10−2⋅3.675×10−3W = 8 \cdot \frac{22}{7} \cdot 2\times 10^{-2} \cdot 3.675\times 10^{-3}W=8⋅722​⋅2×10−2⋅3.675×10−3

Now, 8⋅2×10−2=16×10−2=0.168 \cdot 2\times 10^{-2} = 16\times 10^{-2} = 0.168⋅2×10−2=16×10−2=0.16 So, W=227⋅0.16⋅3.675×10−3W = \frac{22}{7} \cdot 0.16 \cdot 3.675\times 10^{-3}W=722​⋅0.16⋅3.675×10−3 0.16⋅3.675=0.5880.16 \cdot 3.675 = 0.5880.16⋅3.675=0.588 Thus, W=227⋅0.588×10−3W = \frac{22}{7} \cdot 0.588\times 10^{-3}W=722​⋅0.588×10−3 =1.848×10−3 J= 1.848\times 10^{-3}\, \text{J}=1.848×10−3J

  1. Write in the given form

1.848×10−3=18.48×10−4 J1.848\times 10^{-3} = 18.48\times 10^{-4}\, \text{J}1.848×10−3=18.48×10−4J

  1. Option check

This matches Option A.

Final Answer: 18.48×10−4 J\boxed{18.48\times 10^{-4}\, \text{J}}18.48×10−4J​

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