A particle is moving with constant speed in a circular path. When the particle turns by an angle $90^{o}$ , the ratio of instantaneous velocity to its average velocity is π: x $\sqrt2$ . The value of x will be -
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Question

A particle is moving with constant speed in a circular path. When the particle turns by an angle $90^{o}$ , the ratio of instantaneous velocity to its average velocity is π: x $\sqrt2$ . The value of x will be -

Options

G1 Physics · Accepted Answer

For a 90° turn on a circle of radius R, the distance travelled is one quarter of the circumference: s = (πR)/2 Time taken: t = s/v = πR/(2v) The displacement between the initial and final points is the chord for 90°: Δr = √(R² + R²) = R√2 Average velocity magnitude: v(avg) = Δr/t = R√2 ÷ [πR/(2v)] = 2√2 v/π Therefore, instantaneous speed / average velocity = v / (2√2v/π) = π/(2√2) This is written as π : x√2, giving x = 2. Hence option B follows.

A particle is moving with constant speed in a circular path. When the particle turns by an angle 90 o 90^{o} 9 0 o , the ratio of instantaneous velocity to its average velocity is π: x 2 \sqrt2 2 ​ . The value of x will be -

A particle is moving with constant speed in a circular path. When the particle turns by an angle 90 o 90^{o} 9 0 o , the ratio of instantaneous velocity to its average velocity is π: x 2 \sqrt2 2 . The value of x will be -