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A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revolved in a horizontal circle at the rate of 10/p revolution per second. Calculate the tension of the string ?
- a)30 N
- b)40 N
- c)50 N
- d)60 N
Correct answer is option 'B'. Can you explain this answer?
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A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revol...
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A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revol...
Given:
Mass of stone, m = 0.1 kg
Length of string, l = 1.0 m
Rate of revolution, ω = 10/π rev/s
To find: Tension in the string
Formula Used:
Centripetal force, F = mv²/r
Tension in the string, T = mv²/r
Where,
m = mass of the stone
v = linear velocity of the stone
r = radius of the circular motion
Calculation:
The linear velocity of the stone can be found using the formula:
v = ωr
where r is the length of the string.
v = ωr = 10/π × 1 = 10/π m/s
The radius of the circular motion is equal to the length of the string.
r = l = 1.0 m
The centripetal force acting on the stone can be found using the formula:
F = mv²/r
F = m(10/π)²/1 = 100m/π N
The tension in the string is equal to the centripetal force acting on the stone.
T = 100m/π N = 100 × 0.1/π N ≈ 40 N
Therefore, the tension in the string is 40 N.
Hence, option (b) is the correct answer.
Mass of stone, m = 0.1 kg
Length of string, l = 1.0 m
Rate of revolution, ω = 10/π rev/s
To find: Tension in the string
Formula Used:
Centripetal force, F = mv²/r
Tension in the string, T = mv²/r
Where,
m = mass of the stone
v = linear velocity of the stone
r = radius of the circular motion
Calculation:
The linear velocity of the stone can be found using the formula:
v = ωr
where r is the length of the string.
v = ωr = 10/π × 1 = 10/π m/s
The radius of the circular motion is equal to the length of the string.
r = l = 1.0 m
The centripetal force acting on the stone can be found using the formula:
F = mv²/r
F = m(10/π)²/1 = 100m/π N
The tension in the string is equal to the centripetal force acting on the stone.
T = 100m/π N = 100 × 0.1/π N ≈ 40 N
Therefore, the tension in the string is 40 N.
Hence, option (b) is the correct answer.
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Community Answer
A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revol...
Atq, necessary centripetal force is supplied by tension in the string.Therefore,T=mrw^2 =0.1kg*1m*(10/π * 2π)^2 =0.1*(20^2) :. T =40N
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A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revolved in a horizontal circle at the rate of 10/p revolution per second. Calculate the tension of the string ?a)30 Nb)40 Nc)50 Nd)60 NCorrect answer is option 'B'. Can you explain this answer?
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