Question

A particle starts from center O towards OA then moves along AB and stop at B. if
R=100m then displacement of the particle is
A) 100m
B) 100/√2m
C) 100√2 m
D) None​

Answer 1

Answer:

D) None

Explanation:

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Answer 2

Answer: The correct answer is option C)  100√2 m

Explanation:

Let us assume that a particle is describing a circular motion ( In the figure given below)with a anticlockwise with a constant angular displacement $\theta$. Now let the Centre of the Circular track is O and radius is R. Now if the particle starts from center O towards OA then OA=R ( radius of the circular track) .Then moves along AB and stop at B. then according to the geometry OB=R. Connecting the points O,A and B we obtain a right angled triangle. and the

displacement of the particle can be determined via applying Pythagoras'  Theorem to the ΔOAB as follows-

$Pythagoras' \,\,Theorem-\\\\Base^{2} +Perpendicular^{2} =Hypotenus^{2} \\\\OA^{2} +OB^{2} =AB^{2} \\\\R^{2} +R^{2} = AB^{2} \\\\AB^{2} =2R^{2}\\\\AB=\sqrt{2} R\\\\AB=\sqrt{2} \times 100 m\\\\AB=100\sqrt{2} \,m$

The displacement of the particle is AB= 100√2 m