A ball follows a motion with constant speed $V_{0}$ & goes from A to E. If normal force at A, B, C, D & E are $N_{A}, N_{B}, N_{C}, N_{D}, & N_{E}$ respectively, then

N_{A} > N_{B} > N_{C} > N_{D} > N_{E}

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N_{D} > N_{B} > N_{A} > N_{C} > N_{E}

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N_{A} > N_{B} > N_{D} > N_{E} > N_{C}

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N_{B} > N_{C} > N_{A} > N_{D} > N_{E}

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Solution

The correct option is B $N_{D} > N_{B} > N_{A} > N_{C} > N_{E}$
$\begin{matrix} a t B & D & N = m g + \frac{m v^{2}}{r} a t C & E & N = m g - \frac{m v^{2}}{r} a t A & N = m g \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ a t {(\frac{m v^{2}}{r})}_{B} < {(\frac{m v^{2}}{r})}_{D} < {(\frac{m v^{2}}{r})}_{C} < {(\frac{m v^{2}}{r})}_{E}$

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A ball follows a motion with constant speed V0 & goes from A to E. If normal force at A, B, C, D & E are NA,NB,NC,ND,&NE respectively, then

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A ball follows a motion with constant speed $V_{0}$ & goes from A to E. If normal force at A, B, C, D & E are $N_{A}, N_{B}, N_{C}, N_{D}, & N_{E}$ respectively, then