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

JIPMER 2016 · Physics · Laws of Motion

A particle slides down on a smooth incline of inclination 30^ ,...

A particle slides down on a smooth incline of inclination 30∘30^{\circ}, fixed in an elevator going up with an acceleration 2 m/s22 \mathrm{~m} / \mathrm{s}^2. The box of incline has a length 4 m . The time taken by the particle to reach the bottom will be A particle slides down on a smooth incline of inclination 30^ , fixed in an elevator going up with an acceleration 2 ~m

Options

  1. A893 s\frac{8}{9} \sqrt{3} \mathrm{~s}
  2. B983 s\frac{9}{8} \sqrt{3} \mathrm{~s}
  3. C4332 s\frac{4}{3} \sqrt{\frac{\sqrt{3}}{2}} \mathrm{~s}
  4. D3432 s\frac{3}{4} \sqrt{\frac{\sqrt{3}}{2}} \mathrm{~s}

Correct answer

C. 4332 s\frac{4}{3} \sqrt{\frac{\sqrt{3}}{2}} \mathrm{~s}

Step-by-step solution

In the frame of elevator, a=a= acceleration of the particle with respect to the elevator ∴msin⁡30(g+2)=maa=(g+2)sin⁡30∘=(10+2)⋅12=6m/s2\begin{aligned} \therefore \quad m \sin 30(g+2) & =m a \\ a & =(g+2) \sin 30^{\circ} \\ & =(10+2) \cdot \frac{1}{2} \\ & =6 \mathrm{ m} / \mathrm{s}^2\end{aligned}\thereforeThedistancetravelledbytheparticlefromthetoptothebottom,The distance travelled by the particle from the top to the bottom,d=4cos⁡30∘=43/2=833m Using s=ut+12at2,\begin{aligned} d & =\frac{4}{\cos 30^{\circ}}=\frac{4}{\sqrt{3} / 2}=\frac{8 \sqrt{3}}{3} \mathrm{ m} \\ \text{ Using } s & =u t+\frac{1}{2} a t^2,\end{aligned}$\begin{aligned} & & \frac{8 \sqrt{3}}{3} & =0 \times t+\frac{1}{2} 6 t^2 \\ \Rightarrow & & t^2 & =\frac{16 \sqrt{3}}{3 \times 6} \\ \therefore & & t & =\frac{4}{3} \sqrt{\frac{\sqrt{3}}{2}} \mathrm{ s}\end{aligned}$
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