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Byju's Answer

Standard XII

Physics

Newton's Second Law

A car moves a...

Question

A car moves at a constant speed on a road as shown in figure. The normal force by the road on the car is $N_{A}$ and $N_{B}$ when it is at the points A and B -

N_{A} = N_{B}

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N_{A} > N_{B}

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N_{A} < N_{B}

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Insufficient information

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Solution

The correct option is C $N_{A} < N_{B}$

At A, $m g - N_{A} = \frac{m v^{2}}{R_{A}}$
$N_{A} = m g - \frac{m v^{2}}{R_{A}}$
Similarly at $B, N_{B} = m g - \frac{m v^{2}}{R_{B}}$
$R_{A} < R_{B}$
So, $N_{A} < N_{B}$

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A car moves at a constant speed on a road as shown in figure. The normal force by the road on the car is N_A and N_B when it is at the points A and B respectively. Then?

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A car moves at a constant speed on a road as shown in figure. The normal force by the road on the car is NA and NB when it is at the points A and B -

The correct option is C NA<NB At A, mg−NA=mv2RA NA=mg−mv2RA Similarly at B,NB=mg−mv2RB RA<RB So, NA<NB

A car moves at a constant speed on a road as shown in figure. The normal force by the road on the car is $N_{A}$ and $N_{B}$ when it is at the points A and B -

The correct option is C $N_{A} < N_{B}$

At A, $m g - N_{A} = \frac{m v^{2}}{R_{A}}$
$N_{A} = m g - \frac{m v^{2}}{R_{A}}$
Similarly at $B, N_{B} = m g - \frac{m v^{2}}{R_{B}}$
$R_{A} < R_{B}$
So, $N_{A} < N_{B}$