Q117 / 150 · MHT CET 2008 · Physics

A roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20 \mathrm{~m}$. The speed of the car at the top of the hill is between

  • Correct answer
    A
    $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$
  • B
    $15 \mathrm{~m} / \mathrm{s}$ and $16 \mathrm{~m} / \mathrm{s}$
  • C
    $16 \mathrm{~m} / \mathrm{s}$ and $17 \mathrm{~m} / \mathrm{s}$
  • D
    $13 \mathrm{~m} / \mathrm{s}$ and $14 \mathrm{~m} / \mathrm{s}$

Solution

Balancing the forces, we get Figure
$$ M g-N=M \frac{v^{2}}{R} $$ For weightlessness, $N=0$ $\therefore$ $$ \frac{M v^{2}}{R}=M g $$ where $R$ is the radius of curvature and $v$ is the speed of car. Therefore, $$ v=\sqrt{\mathrm{Rg}} $$ Putting the values, $R=20 \mathrm{~m}, g=10.0 \mathrm{~m} / \mathrm{s}^{2}$ So, $\quad v=\sqrt{20 \times 10.0}=14.14 \mathrm{~m} / \mathrm{s}^{2}$ Thus, the speed of the car at the top of the hill is $$ \text { between } 14 \mathrm{~m} / \mathrm{s} \text { and } 15 \mathrm{~m} / \mathrm{s} \text { . } $$ Note : The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.

Hence, the correct answer is option: 'A'.

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