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Properties of Matter question

2024 · Shift 2 · Q40
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Properties of Matter question

2024 · Shift 2 · Q40

JEE AdvancedPhysicsProperties of MatterMultiple correct+4 / −2
A table tennis ball has radius (3/2)×10−2 m(3 / 2) \times 10^{-2} \mathrm{~m}(3/2)×10−2 m and mass (22/7)×10−3 kg(22 / 7) \times 10^{-3} \mathrm{~kg}(22/7)×10−3 kg. It is slowly pushed down into a swimming pool to a depth of d=0.7 md=0.7 \mathrm{~m}d=0.7 m below the water surface and then released from rest. It emerges from the water surface at speed vvv, without getting wet, and rises up to a height HHH. Which of the following option(s) is(are) correct? [Given: π=22/7,g=10 m s−2\pi=22 / 7, g=10 \mathrm{~m} \mathrm{~s}^{-2}π=22/7,g=10 m s−2, density of water =1×103 kg m−3=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}=1×103 kg m−3, viscosity of water =1×10−3 Pa=1 \times 10^{-3} \mathrm{~Pa}=1×10−3 Pa-s.]
  1. A
    The work done in pushing the ball to the depth ddd is 0.077 J0.077 \mathrm{~J}0.077 J.
  2. B
    If we neglect the viscous force in water, then the speed v=7 m/sv=7 \mathrm{~m} / \mathrm{s}v=7 m/s.
  3. C
    If we neglect the viscous force in water, then the height H=1.4 mH=1.4 \mathrm{~m}H=1.4 m.
  4. D
    The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is 500/9500 / 9500/9.
View written solutionFree

Correct answer: A, B, D

  1. Given data

    Radius of ball: r=32×10−2 m=1.5×10−2 mr=\frac{3}{2}\times 10^{-2}\,\text{m}=1.5\times 10^{-2}\,\text{m}r=23​×10−2m=1.5×10−2m

    Mass of ball: m=227×10−3 kgm=\frac{22}{7}\times 10^{-3}\,\text{kg}m=722​×10−3kg

    Depth: d=0.7 md=0.7\,\text{m}d=0.7m

    Also, ρ=103 kg m−3,η=10−3 Pa⋅s,g=10 m s−2,π=227\rho=10^3\,\text{kg m}^{-3},\quad \eta=10^{-3}\,\text{Pa·s},\quad g=10\,\text{m s}^{-2},\quad \pi=\frac{22}{7}ρ=103kg m−3,η=10−3Pa⋅s,g=10m s−2,π=722​

  2. Volume of the ball

    V=43πr3V=\frac{4}{3}\pi r^3V=34​πr3

    Since r3=(1.5×10−2)3=3.375×10−6r^3=(1.5\times 10^{-2})^3=3.375\times 10^{-6}r3=(1.5×10−2)3=3.375×10−6

    Therefore, V=43⋅227⋅3.375×10−6V=\frac{4}{3}\cdot \frac{22}{7}\cdot 3.375\times 10^{-6}V=34​⋅722​⋅3.375×10−6

    V=4.5π×10−6=4.5⋅227×10−6V=4.5\pi\times 10^{-6}=4.5\cdot \frac{22}{7}\times 10^{-6}V=4.5π×10−6=4.5⋅722​×10−6

    V=997×10−6 m3V=\frac{99}{7}\times 10^{-6}\,\text{m}^3V=799​×10−6m3

  3. Buoyant force and weight

    Buoyant force when fully submerged: Fb=ρgVF_b=\rho gVFb​=ρgV

    Fb=103⋅10⋅997×10−6F_b=10^3\cdot 10\cdot \frac{99}{7}\times 10^{-6}Fb​=103⋅10⋅799​×10−6

    Fb=997×10−2 NF_b=\frac{99}{7}\times 10^{-2}\,\text{N}Fb​=799​×10−2N

    Weight: mg=(227×10−3)⋅10=227×10−2 Nmg=\left(\frac{22}{7}\times 10^{-3}\right)\cdot 10=\frac{22}{7}\times 10^{-2}\,\text{N}mg=(722​×10−3)⋅10=722​×10−2N

    Hence net upward force excluding viscosity: Fnet=Fb−mg=(997−227)×10−2F_{\text{net}}=F_b-mg=\left(\frac{99}{7}-\frac{22}{7}\right)\times 10^{-2}Fnet​=Fb​−mg=(799​−722​)×10−2

    Fnet=777×10−2=11×10−2=0.11 NF_{\text{net}}=\frac{77}{7}\times 10^{-2}=11\times 10^{-2}=0.11\,\text{N}Fnet​=777​×10−2=11×10−2=0.11N

  4. Option A: Work done in pushing ball down by depth ddd

    Since the ball remains fully submerged throughout the downward displacement, buoyant force and weight are constant.

    External force needed downward (quasi-statically): Fext=Fb−mg=0.11 NF_{\text{ext}}=F_b-mg=0.11\,\text{N}Fext​=Fb​−mg=0.11N

    Work done by external agent: W=Fextd=0.11×0.7=0.077 JW=F_{\text{ext}}d=0.11\times 0.7=0.077\,\text{J}W=Fext​d=0.11×0.7=0.077J

    So A is correct.

  5. Option B: Speed at emergence neglecting viscous force

    From release point to water surface, the net upward force is constant: Fnet=0.11 NF_{\text{net}}=0.11\,\text{N}Fnet​=0.11N

    Work-energy theorem: Fnetd=12mv2F_{\text{net}}d=\frac{1}{2}mv^2Fnet​d=21​mv2

    0.11×0.7=12(227×10−3)v20.11\times 0.7=\frac{1}{2}\left(\frac{22}{7}\times 10^{-3}\right)v^20.11×0.7=21​(722​×10−3)v2

    0.077=117×10−3v20.077=\frac{11}{7}\times 10^{-3} v^20.077=711​×10−3v2

    v2=0.077×711×10−3v^2=\frac{0.077\times 7}{11\times 10^{-3}}v2=11×10−30.077×7​

    v2=49v^2=49v2=49

    v=7 m s−1v=7\,\text{m s}^{-1}v=7m s−1

    So B is correct.

  6. Option C: Height risen above water surface

    The statement says the ball emerges from water surface at speed vvv, without getting wet. That means after leaving the water, it moves upward in air only under gravity.

    Hence maximum height above water surface is: H=v22g=4920=2.45 mH=\frac{v^2}{2g}=\frac{49}{20}=2.45\,\text{m}H=2gv2​=2049​=2.45m

    This is not 1.4 m1.4\,\text{m}1.4m.

    So C is incorrect.

  7. Option D: Ratio of net force excluding viscosity to maximum viscous force

    Maximum viscous force in water occurs at maximum speed, just before emerging, i.e. at v=7 m s−1v=7\,\text{m s}^{-1}v=7m s−1.

    Using Stokes' law: Fv=6πηrvF_v=6\pi\eta rvFv​=6πηrv

    Fv=6⋅227⋅10−3⋅1.5×10−2⋅7F_v=6\cdot \frac{22}{7}\cdot 10^{-3}\cdot 1.5\times 10^{-2}\cdot 7Fv​=6⋅722​⋅10−3⋅1.5×10−2⋅7

    Simplify: Fv=6⋅22⋅1.5×10−5F_v=6\cdot 22\cdot 1.5\times 10^{-5}Fv​=6⋅22⋅1.5×10−5

    Fv=198×10−5=1.98×10−3 NF_v=198\times 10^{-5}=1.98\times 10^{-3}\,\text{N}Fv​=198×10−5=1.98×10−3N

    Therefore, FnetFv,max⁡=0.111.98×10−3=1101.98=550099=5009\frac{F_{\text{net}}}{F_{v,\max}}=\frac{0.11}{1.98\times 10^{-3}}=\frac{110}{1.98}=\frac{5500}{99}=\frac{500}{9}Fv,max​Fnet​​=1.98×10−30.11​=1.98110​=995500​=9500​

    So D is also correct.

  8. Final evaluation of options

    • A: Correct
    • B: Correct
    • C: Incorrect
    • D: Correct

  9. Comparison with stored answer

    Stored correct answer: A, B

    My derived answer: A, B, D

    I disagree with the stored answer because option D follows directly from Stokes' law using the given viscosity and the speed found in part B.

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