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A stone tied to the end of a string of 1 m long is 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?
- a)π2/4 m−s⁻2and direction along the radius
- b)π2 m−s⁻2and direction along the radius towards the centre
- c)π2 m−s⁻2and direction along the tangent to the circle
- d)π2 m−s⁻2 and direction along the radius away from the centre
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A stone tied to the end of a string of 1 m long is whirled in a horizo...
As the speed is constant throughout the motion, so it is a case of uniform circular motion. Therefore, its radial acceleration
a = rω^2 = r(2πn/t)^2 = r x 4π^2n^2/t^2
= 1 x 4 x π^2 x (22)^2/(44)^2 = π^2m/s^2
This acceleration is directed along radius of circle towards center of circle.
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a = rω^2 = r(2πn/t)^2 = r x 4π^2n^2/t^2
= 1 x 4 x π^2 x (22)^2/(44)^2 = π^2m/s^2
This acceleration is directed along radius of circle towards center of circle.
Most Upvoted Answer
A stone tied to the end of a string of 1 m long is whirled in a horizo...
The magnitude of the acceleration of the stone can be found using the formula:
a = v^2 / r
where v is the speed of the stone and r is the radius of the circle (which is equal to the length of the string). We can find v by dividing the distance traveled by the time taken:
distance traveled = circumference of circle * number of revolutions
circumference of circle = 2πr
distance traveled = 2πr * 22 = 44πr
time taken = 44 seconds
v = distance traveled / time taken = (44πr) / 44 = πr
Therefore, the speed of the stone is π m/s. The radius of the circle is 1 m, so the magnitude of the acceleration is:
a = (π)^2 / 1 = π^2 m/s^2
The direction of the acceleration is towards the center of the circle, which is perpendicular to the velocity vector at any point on the circle.
a = v^2 / r
where v is the speed of the stone and r is the radius of the circle (which is equal to the length of the string). We can find v by dividing the distance traveled by the time taken:
distance traveled = circumference of circle * number of revolutions
circumference of circle = 2πr
distance traveled = 2πr * 22 = 44πr
time taken = 44 seconds
v = distance traveled / time taken = (44πr) / 44 = πr
Therefore, the speed of the stone is π m/s. The radius of the circle is 1 m, so the magnitude of the acceleration is:
a = (π)^2 / 1 = π^2 m/s^2
The direction of the acceleration is towards the center of the circle, which is perpendicular to the velocity vector at any point on the circle.
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A stone tied to the end of a string of 1 m long is 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?a)π2/4 m−s2and direction along the radiusb)π2 m−s2and direction along the radius towards the centrec)π2 m−s2and direction along the tangent to the circled)π2 m−s2 and direction along the radius away from the centreCorrect answer is option 'B'. Can you explain this answer?
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A stone tied to the end of a string of 1 m long is 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?a)π2/4 m−s2and direction along the radiusb)π2 m−s2and direction along the radius towards the centrec)π2 m−s2and direction along the tangent to the circled)π2 m−s2 and direction along the radius away from the centreCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A stone tied to the end of a string of 1 m long is 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?a)π2/4 m−s2and direction along the radiusb)π2 m−s2and direction along the radius towards the centrec)π2 m−s2and direction along the tangent to the circled)π2 m−s2 and direction along the radius away from the centreCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone tied to the end of a string of 1 m long is 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?a)π2/4 m−s2and direction along the radiusb)π2 m−s2and direction along the radius towards the centrec)π2 m−s2and direction along the tangent to the circled)π2 m−s2 and direction along the radius away from the centreCorrect answer is option 'B'. Can you explain this answer?.
A stone tied to the end of a string of 1 m long is 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?a)π2/4 m−s2and direction along the radiusb)π2 m−s2and direction along the radius towards the centrec)π2 m−s2and direction along the tangent to the circled)π2 m−s2 and direction along the radius away from the centreCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A stone tied to the end of a string of 1 m long is 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?a)π2/4 m−s2and direction along the radiusb)π2 m−s2and direction along the radius towards the centrec)π2 m−s2and direction along the tangent to the circled)π2 m−s2 and direction along the radius away from the centreCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone tied to the end of a string of 1 m long is 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?a)π2/4 m−s2and direction along the radiusb)π2 m−s2and direction along the radius towards the centrec)π2 m−s2and direction along the tangent to the circled)π2 m−s2 and direction along the radius away from the centreCorrect answer is option 'B'. Can you explain this answer?.
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