A stone of mass 0.5 kg is attached to a string of length 2 m and is whirled in a horizontal circle. If the string can withstand a tension of 9 N, the maximum velocity with which the stone can be whirled is :A 6 ms–1 B 8 ms–1 C 4 ms–1 D 12 ms–1
Question
A stone of mass 0.5 kg is attached to a string of length 2 m and is whirled in a horizontal circle. If the string can withstand a tension of 9 N, the maximum velocity with which the stone can be whirled is :A 6 ms–1 B 8 ms–1 C 4 ms–1 D 12 ms–1
Solution
The tension in the string is equal to the centripetal force when the stone is whirled in a circle. The formula for centripetal force is:
F = m*v^2/r
where: F is the force (in Newtons), m is the mass (in kg), v is the velocity (in m/s), and r is the radius (in m).
We know that the tension the string can withstand is 9 N, the mass of the stone is 0.5 kg, and the length of the string (which is the radius of the circle) is 2 m. We can substitute these values into the formula and solve for v:
9 N = 0.5 kg * v^2 / 2 m
Rearranging the equation to solve for v gives:
v^2 = 9 N * 2 m / 0.5 kg
v^2 = 36
Taking the square root of both sides gives:
v = 6 m/s
So, the maximum velocity with which the stone can be whirled is 6 m/s. The correct answer is A.
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