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Motion question

2017 · Shift 2 · Q47
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Motion question

2017 · Shift 2 · Q47

JEE AdvancedPhysicsMotionMCQ+3 / −0.75
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ\rhoρ remains uniform throughout the volume. The rate of fractional change in density (1ρdρdt)\left( {{1 \over \rho } {{d\rho } \over {dt}}} \right)(ρ1​dtdρ​) is constant. The velocity vvv of any point on the surface of the expanding sphere is proportional to
  1. A
    RRR
  2. B
    R3{R^3}R3
  3. C
    1R{1 \over R}R1​
  4. D
    R2/3{R^{2/3}}R2/3
View written solutionFree

Correct answer: A

Step-by-step Derivation

  1. Relate Mass, Density, and Volume: The mass M of the sphere is related to its density ρ and volume V by the formula: M=ρVM = \rho VM=ρV For a sphere of radius R, the volume is V=43πR3V = \frac{4}{3}\pi R^3V=34​πR3. Substituting this into the mass equation, we get: M=ρ(43πR3)M = \rho \left( \frac{4}{3}\pi R^3 \right)M=ρ(34​πR3)

  2. Use the Condition of Constant Mass: The problem states that the total mass M of the sphere remains constant. Since M, 43\frac{4}{3}34​, and π\piπ are all constants, the product ρR3\rho R^3ρR3 must also be a constant. ρR3=constant\rho R^3 = \text{constant}ρR3=constant

  3. Differentiate with Respect to Time: To find the relationship between the rates of change, we differentiate the equation ρR3=constant\rho R^3 = \text{constant}ρR3=constant with respect to time t. Using the product rule for differentiation (uv)' = u'v + uv': \frac{d}{dt} (\rho R^3) = \frac{d}{dt} (\text{constant}) $$$$ \left( \frac{d\rho}{dt} \right) R^3 + \rho \left( \frac{d(R^3)}{dt} \right) = 0 Applying the chain rule to the R3R^3R3 term: d(R3)dt=3R2dRdt\frac{d(R^3)}{dt} = 3R^2 \frac{dR}{dt}dtd(R3)​=3R2dtdR​. (dρdt)R3+ρ(3R2dRdt)=0\left( \frac{d\rho}{dt} \right) R^3 + \rho \left( 3R^2 \frac{dR}{dt} \right) = 0(dtdρ​)R3+ρ(3R2dtdR​)=0

  4. Incorporate the Given Rate of Fractional Change in Density: The problem states that the rate of fractional change in density, 1ρdρdt\frac{1}{\rho} \frac{d\rho}{dt}ρ1​dtdρ​, is a constant. Let's denote this constant by C: 1ρdρdt=C\frac{1}{\rho} \frac{d\rho}{dt} = Cρ1​dtdρ​=C From this, we can write dρdt=Cρ\frac{d\rho}{dt} = C\rhodtdρ​=Cρ.

  5. Substitute and Solve for Velocity: Substitute dρdt=Cρ\frac{d\rho}{dt} = C\rhodtdρ​=Cρ into the equation from Step 3: (Cρ)R3+ρ(3R2dRdt)=0(C\rho) R^3 + \rho \left( 3R^2 \frac{dR}{dt} \right) = 0(Cρ)R3+ρ(3R2dtdR​)=0 The velocity v of a point on the surface is the rate of change of the radius, v=dRdtv = \frac{dR}{dt}v=dtdR​. So we have: CρR3+3ρR2v=0C\rho R^3 + 3\rho R^2 v = 0CρR3+3ρR2v=0 We can divide the entire equation by ρR2\rho R^2ρR2 (assuming ρ\rhoρ and R are non-zero): CR+3v=0CR + 3v = 0CR+3v=0 Solving for v: 3v = -CR $$$$ v = -\frac{C}{3} R

  6. Conclusion: Since C is a constant, the term −C3-\frac{C}{3}−3C​ is also a constant. Therefore, the velocity v is directly proportional to the instantaneous radius R. v∝Rv \propto Rv∝R This corresponds to option A.

Evaluation of Options

  • A: RRR: Our derivation shows v ∝ R. This is the correct option.
  • B: R3R^3R3: Incorrect.
  • C: 1/R1/R1/R: Incorrect.
  • D: R2/3R^{2/3}R2/3: Incorrect.
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