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A circular road of radius r is banked for a speed \( \mathbf{v}=40 \mathrm{km} / \mathrm{hr} . \) A car of mass \( \mathrm{m} \) attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.

  1. The car cannot make a turn without skidding
  2. if the car turns at a speed less than \( 40 \mathrm{km} / \mathrm{hr} \), it will slip down.
  3. If the car turns at the correct speed of \( 40 \mathrm{km} / \mathrm{hr} \), the force by the road on the car is equal to \( m v^{2} / r \)
  4. If the car turns at the correct speed of \( 40 \mathrm{km} / \mathrm{hr} \), the force by the road on the car is greater than mg as well as greater than \( m v^{2} / r \)

Correct answer: If the car turns at the correct speed of \( 40 \mathrm{km} / \mathrm{hr} \), the force by the road on the car is greater than mg as well as greater than \( m v^{2} / r \)

Solution

For a frictionless banked curve at the design speed, the road’s force is the normal reaction. Its vertical component balances weight, while its horizontal component provides centripetal force. Because the normal is tilted, its magnitude must exceed each component, so it is greater than both mg and mv^2/r.

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