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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Physics
Chapter
Laws of Motion

BITSAT 2016 · Physics · Laws of Motion

A frictionless wire AB is fixed on a sphere of radius R . A very...

A frictionless wire AB\mathrm{AB} is fixed on a sphere of radius R\mathrm{R}. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B\mathrm{B} is A frictionless wire AB is fixed on a sphere of radius R . A very small spherical ball slips on this wire. The time

Options

  1. A2gRgcos⁡θ\frac{\sqrt{2 g R}}{g \cos \theta}
  2. B2gR⋅cos⁡θg2 \sqrt{g R} \cdot \frac{\cos \theta}{g}
  3. C2Rg2 \sqrt{\frac{R}{g}}
  4. DgRgcos⁡θ\frac{g R}{\sqrt{g \cos \theta}}

Correct answer

C. 2Rg2 \sqrt{\frac{R}{g}}

Step-by-step solution

Acceleration of body along AB\mathrm{AB} is g cos⁡θ\cos \theta Distance travelled in time t\mathrm{t} sec =AB=12(gcos⁡θ)t2=A B=\frac{1}{2}(g \cos \theta) t^{2} From ΔABC,AB=2Rcos⁡θ\Delta \mathrm{ABC}, \mathrm{AB}=2 \mathrm{R} \cos \theta Thus, 2Rcos⁡θ=12gcos⁡θt2⇒t2=4Rg⇒t=2Rg2 \mathrm{R} \cos \theta=\frac{1}{2} g \cos \theta t^{2}\Rightarrow t^{2}=\frac{4 R}{g} \Rightarrow t=2 \sqrt{\frac{R}{g}}
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