A body is moving with constant speed, in a circle of radius $10 m$ .The body completes one revolution in 4 s .At the end of 3 rd second, the displacement of body (in mi) from its starting point is :

From the given figure above , we get
Speed, v = constant
Radius, R = 10 m
T = Time period = 4s
At the end of 3rd second, particle will be at D (Starts from A)
\(\therefore\) displacement S = \(\sqrt2R\)
\(=\sqrt2\times10\)
\(=10\sqrt2\)
So, the correect answer is (B) : $10 \sqrt{2}$
Given: The circle of the radius (R) = 10m,
Speed is constant,
Total time given = 4s
Displacement is the shortest distance between the initial and the final position.
From the above diagram it is clear that the initial position of the body is at A and the final position is D.
Total time of 4 seconds is evenly distributed for each segment of the orbit:
At the end of the 3rd second, the particle will be at D (when Starts from A).
As from the figure, it is clear AOD is right angled triangle, applying Pythagoras theorem, Displacement = S
\(S=\sqrt{AQ^2+OD^2}\)
\(S=\sqrt{R^2+R^2}\)
\(S=R\sqrt{2}\)
\(S=10\sqrt{2}\)
Therefore, At the end of 3 rd second, the displacement of body (in mi) from its starting point is \(10\sqrt{2}.\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

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Consider a series of steps as shown. A ball is thrown from 0. Find the minimum speed to directly jump to 5th step
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Linear motion is also known as the Rectilinear Motion which are of two types: