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A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.?
Verified Answer
A stone tied at the end of the string 80cm. Long , is whirled in a hor...
Length of string = 80cm =0.8m
Stone makes 14 revolutions in 25sec.
Hence, stone makes 14/25 revolution in 1 sec
e.g frequency = 14/25 sec^-1
We know,
Angular speed (ω)= 2πx frequency
= 2π x 14/25
= 28π/25 rad/sec
Now, acceleration (a) = ω^2r
where r is the radius { length of string}
Acceleration = 0.8 x {28π/25}^2
= 9.91 m/s^2
Stone makes 14 revolutions in 25sec.
Hence, stone makes 14/25 revolution in 1 sec
e.g frequency = 14/25 sec^-1
We know,
Angular speed (ω)= 2πx frequency
= 2π x 14/25
= 28π/25 rad/sec
Now, acceleration (a) = ω^2r
where r is the radius { length of string}
Acceleration = 0.8 x {28π/25}^2
= 9.91 m/s^2
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
A stone tied at the end of the string 80cm. Long , is whirled in a hor...
Given:
- Length of the string: 80 cm
- Number of revolutions made by the stone: 14
- Time taken to make the revolutions: 25 sec
To find:
- Magnitude of acceleration of the stone
Formula:
The formula to calculate the magnitude of acceleration for an object moving in a circular path is given by:
a = (4π²r) / T²
Where:
- a is the magnitude of acceleration
- r is the radius of the circular path
- T is the time taken to complete one revolution
Solution:
Step 1: Calculate the radius of the circular path
The length of the string is equal to the circumference of the circle. Therefore, the radius (r) can be calculated as:
Circumference = 2πr
80 cm = 2πr
r = 40 / π cm
Step 2: Calculate the time taken to complete one revolution
The stone makes 14 revolutions in 25 seconds. Therefore, the time taken to complete one revolution (T) can be calculated as:
T = 25 sec / 14
Step 3: Calculate the magnitude of acceleration
Using the formula mentioned above, we can substitute the values of r and T to calculate the magnitude of acceleration (a):
a = (4π²r) / T²
a = (4 * π² * (40 / π)) / (25 / 14)²
a = (4 * 40 * 14 * π) / (25²)
a = 560π / 625
Final Step: Approximate the value of π and calculate the magnitude of acceleration
Using the approximation π ≈ 3.14, we can calculate the magnitude of acceleration:
a ≈ (560 * 3.14) / 625
a ≈ 2.8 cm/s²
Therefore, the magnitude of acceleration of the stone is approximately 2.8 cm/s².
- Length of the string: 80 cm
- Number of revolutions made by the stone: 14
- Time taken to make the revolutions: 25 sec
To find:
- Magnitude of acceleration of the stone
Formula:
The formula to calculate the magnitude of acceleration for an object moving in a circular path is given by:
a = (4π²r) / T²
Where:
- a is the magnitude of acceleration
- r is the radius of the circular path
- T is the time taken to complete one revolution
Solution:
Step 1: Calculate the radius of the circular path
The length of the string is equal to the circumference of the circle. Therefore, the radius (r) can be calculated as:
Circumference = 2πr
80 cm = 2πr
r = 40 / π cm
Step 2: Calculate the time taken to complete one revolution
The stone makes 14 revolutions in 25 seconds. Therefore, the time taken to complete one revolution (T) can be calculated as:
T = 25 sec / 14
Step 3: Calculate the magnitude of acceleration
Using the formula mentioned above, we can substitute the values of r and T to calculate the magnitude of acceleration (a):
a = (4π²r) / T²
a = (4 * π² * (40 / π)) / (25 / 14)²
a = (4 * 40 * 14 * π) / (25²)
a = 560π / 625
Final Step: Approximate the value of π and calculate the magnitude of acceleration
Using the approximation π ≈ 3.14, we can calculate the magnitude of acceleration:
a ≈ (560 * 3.14) / 625
a ≈ 2.8 cm/s²
Therefore, the magnitude of acceleration of the stone is approximately 2.8 cm/s².
Community Answer
A stone tied at the end of the string 80cm. Long , is whirled in a hor...
Length of string = 80cm =0.8m
Stone makes 14 revolutions in 25sec.
Hence, stone makes 14/25 revolution in 1 sec
e.g frequency = 14/25 sec^-1
We know,
Angular speed (ω)= 2π� frequency
= 2π � 14/25
= 28π/25 rad/sec
Now, acceleration (a) = ω^2r
where r is the radius { length of string}
Acceleration = 0.8 � {28π/25}^2
= 9.91 m/s^2
Stone makes 14 revolutions in 25sec.
Hence, stone makes 14/25 revolution in 1 sec
e.g frequency = 14/25 sec^-1
We know,
Angular speed (ω)= 2π� frequency
= 2π � 14/25
= 28π/25 rad/sec
Now, acceleration (a) = ω^2r
where r is the radius { length of string}
Acceleration = 0.8 � {28π/25}^2
= 9.91 m/s^2
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A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.?
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A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.?.
A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.?.
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A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.?, a detailed solution for A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.? has been provided alongside types of A stone tied at the end of the string 80cm. Long , is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 sec. What is the magnitude of acceleration of the stone? Ans.is = 990cm/s^2. But how .??.? theory, EduRev gives you an
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