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A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stone
- a)493 cm/s2
- b)720 cm/s2
- c)860 cm/s2
- d)990 cm/s2
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A stone is tied to one end of a spring 50 cm long is whirled in a hori...
Radius =r=50cm=0.5m
In 20 sec revolution =10
In 1 sec revolution =10/20=1/2 Hz
N=1/2Hz
Now,
Angular velocity =w=2πn=w=2×3.14×0.5
Radial Acceleration =w2r
Radial Acceleration =(3.14)2×0.5m/s2
=4.93m/s2
=493cm/s2
In 20 sec revolution =10
In 1 sec revolution =10/20=1/2 Hz
N=1/2Hz
Now,
Angular velocity =w=2πn=w=2×3.14×0.5
Radial Acceleration =w2r
Radial Acceleration =(3.14)2×0.5m/s2
=4.93m/s2
=493cm/s2
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Community Answer
A stone is tied to one end of a spring 50 cm long is whirled in a hori...
Given:
- Length of the spring, L = 50 cm
- Number of revolutions made by the stone, n = 10
- Time taken to complete the revolutions, t = 20 s
To find:
The magnitude of acceleration of the stone
Solution:
Step 1: Calculating the total distance traveled by the stone
The total distance traveled by the stone in one revolution is equal to the circumference of the circle formed by the stone. The circumference of a circle is given by the formula:
C = 2πr
Where r is the radius of the circle. In this case, the radius is equal to the length of the spring, L.
C = 2πL
The total distance traveled by the stone in 10 revolutions can be calculated as:
Total distance = 10 * C = 10 * 2πL
Step 2: Calculating the speed of the stone
The speed of the stone can be calculated by dividing the total distance traveled by the time taken.
Speed = Total distance / Time taken = (10 * 2πL) / t
Step 3: Calculating the magnitude of acceleration
The magnitude of acceleration can be calculated using the formula:
Acceleration = (Speed)^2 / Radius
In this case, the radius is equal to the length of the spring, L.
Acceleration = (Speed)^2 / L
Substituting the value of speed from Step 2, we get:
Acceleration = [(10 * 2πL) / t]^2 / L
Simplifying the equation, we get:
Acceleration = (400π^2L^2) / t^2
Now, substituting the given values, we get:
Acceleration = (400π^2 * 50^2) / 20^2
= (400 * 3.14^2 * 50^2) / 20^2
≈ 492.9 cm/s^2
Therefore, the magnitude of acceleration of the stone is approximately 493 cm/s^2, which corresponds to option A.
- Length of the spring, L = 50 cm
- Number of revolutions made by the stone, n = 10
- Time taken to complete the revolutions, t = 20 s
To find:
The magnitude of acceleration of the stone
Solution:
Step 1: Calculating the total distance traveled by the stone
The total distance traveled by the stone in one revolution is equal to the circumference of the circle formed by the stone. The circumference of a circle is given by the formula:
C = 2πr
Where r is the radius of the circle. In this case, the radius is equal to the length of the spring, L.
C = 2πL
The total distance traveled by the stone in 10 revolutions can be calculated as:
Total distance = 10 * C = 10 * 2πL
Step 2: Calculating the speed of the stone
The speed of the stone can be calculated by dividing the total distance traveled by the time taken.
Speed = Total distance / Time taken = (10 * 2πL) / t
Step 3: Calculating the magnitude of acceleration
The magnitude of acceleration can be calculated using the formula:
Acceleration = (Speed)^2 / Radius
In this case, the radius is equal to the length of the spring, L.
Acceleration = (Speed)^2 / L
Substituting the value of speed from Step 2, we get:
Acceleration = [(10 * 2πL) / t]^2 / L
Simplifying the equation, we get:
Acceleration = (400π^2L^2) / t^2
Now, substituting the given values, we get:
Acceleration = (400π^2 * 50^2) / 20^2
= (400 * 3.14^2 * 50^2) / 20^2
≈ 492.9 cm/s^2
Therefore, the magnitude of acceleration of the stone is approximately 493 cm/s^2, which corresponds to option A.
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A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stonea)493 cm/s2b)720 cm/s2c)860 cm/s2d)990 cm/s2Correct answer is option 'A'. Can you explain this answer?
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A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stonea)493 cm/s2b)720 cm/s2c)860 cm/s2d)990 cm/s2Correct answer is option 'A'. 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 is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stonea)493 cm/s2b)720 cm/s2c)860 cm/s2d)990 cm/s2Correct answer is option 'A'. 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 is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stonea)493 cm/s2b)720 cm/s2c)860 cm/s2d)990 cm/s2Correct answer is option 'A'. Can you explain this answer?.
A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stonea)493 cm/s2b)720 cm/s2c)860 cm/s2d)990 cm/s2Correct answer is option 'A'. 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 is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stonea)493 cm/s2b)720 cm/s2c)860 cm/s2d)990 cm/s2Correct answer is option 'A'. 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 is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stonea)493 cm/s2b)720 cm/s2c)860 cm/s2d)990 cm/s2Correct answer is option 'A'. Can you explain this answer?.
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