SATHEE: Kinematics Question 802

Kinematics Question 802

Question: A stone ties to the end of a string $1 m$ long it’s whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, What is the magnitude and direction of acceleration of the stone [CBSE PMT 2005]

Options:

A) $\frac{π^{2}}{4} m s^{- 2}$ and direction along the radius towards the centre

B) $π^{2} m s^{- 2}$ and direction along the radius away from the centre

C) $π^{2} m s^{- 2}$ and direction along the radius towards the centre

D) $π^{2} m s^{- 2}$ and direction along the tangent to the circle

Show Answer

Answer:

Correct Answer: C

Solution:

a $= \frac{v^{2}}{r} = ω^{2} r$

$= 4 π^{2} n^{2} r = 4 π^{2} {(\frac{22}{44})}^{2} \times 1$

$= π^{2} m / s^{2}$ and it’s direction is always along the radius and towards the centre.

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Kinematics Question 802

Question: A stone ties to the end of a string 1m long it’s whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, What is the magnitude and direction of acceleration of the stone [CBSE PMT 2005]

Options:

Answer:

Solution:

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Question: A stone ties to the end of a string $1 m$ long it’s whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, What is the magnitude and direction of acceleration of the stone [CBSE PMT 2005]