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Properties of Matter question

2008 · Shift 0 · Q81
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Properties of Matter question

2008 · Shift 0 · Q81

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A spherical solid ball of volume VVV is made of a material of density ρ1{\rho _1}ρ1​. It is falling through a liquid of density ρ2(ρ2<ρ1){\rho _2}\left( {{\rho _2} \lt {\rho _1}} \right)ρ2​(ρ2​<ρ1​). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v,v,v, i.e., Fviscous=−kv2(k>0).{F_{viscous}} = - k{v^2}\left( {k \gt 0} \right).Fviscous​=−kv2(k>0). The terminal speed of the ball is
  1. A
    Vg(ρ1−ρ2)k\sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}}kVg(ρ1​−ρ2​)​​
  2. B
    Vgρ1k{{{Vg{\rho _1}} \over k}}kVgρ1​​
  3. C
    Vgρ1k\sqrt {{{Vg{\rho _1}} \over k}}kVgρ1​​​
  4. D
    Vg(ρ1−ρ2)k{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}kVg(ρ1​−ρ2​)​
View written solutionFree

Correct answer: A

  1. Identify the forces on the falling ball

    Since the ball is moving downward through a liquid, the forces are:

    • Weight downward: W=mg=ρ1VgW = mg = \rho_1 V gW=mg=ρ1​Vg
    • Buoyant force upward: Fb=ρ2VgF_b = \rho_2 V gFb​=ρ2​Vg
    • Viscous force upward (opposes downward motion): Fv=kv2F_v = kv^2Fv​=kv2

  2. Condition for terminal speed

    At terminal speed, acceleration is zero, so net force is zero.

    Taking downward direction as positive: ρ1Vg−ρ2Vg−kvt2=0\rho_1 V g - \rho_2 V g - kv_t^2 = 0ρ1​Vg−ρ2​Vg−kvt2​=0

    Therefore, Vg(ρ1−ρ2)=kvt2Vg(\rho_1 - \rho_2) = kv_t^2Vg(ρ1​−ρ2​)=kvt2​

  3. Solve for terminal speed

    vt2=Vg(ρ1−ρ2)kv_t^2 = \frac{Vg(\rho_1-\rho_2)}{k}vt2​=kVg(ρ1​−ρ2​)​

    Hence, vt=Vg(ρ1−ρ2)kv_t = \sqrt{\frac{Vg(\rho_1-\rho_2)}{k}}vt​=kVg(ρ1​−ρ2​)​​

  4. Match with the options

    This corresponds to: A\boxed{\text{A}}A​

  5. Compare with stored correct answer

    Stored correct answer is A, which matches our derived result.

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