(D) The centripetal force required for circular motion is provided by the tension in the string.For a horizontal circle,the tension $T$ is given by th

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(D) The centripetal force required for circular motion is provided by the tension in the string.For a horizontal circle,the tension $T$ is given by the formula $T = \frac{mv^2}{r}$.Given: Mass $m = 16 \, kg$,Radius $r = 144 \, m$,and Maximum Tension $T_{max} = 16 \, N$.Substituting the values into the formula: $16 = \frac{16 \times v^2}{144}$.Simplifying the equation: $1 = \frac{v^2}{144}$.Therefore,$v^2 = 144$.Taking the square root,we get $v = 12 \, m/s$.

Class 11 Physics 3-2.Motion in Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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$A$ stone of mass $16 \, kg$ is attached to a string $144 \, m$ long and is whirled in a horizontal circle. The maximum tension the string can withstand is $16 \, N$. The maximum velocity of revolution that can be given to the stone without breaking it will be ....... $m/s$.

A
$20$
B
$16$
C
$14$
D
$12$

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3-2.Motion in Plane

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Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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$A$ stone of mass $16 \, kg$ is attached to a string $144 \, m$ long and is whirled in a horizontal circle. The maximum tension the string can withstand is $16 \, N$. The maximum velocity of revolution that can be given to the stone without breaking it will be ....... $m/s$.

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$A$ roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20\, m$. The speed of the car at the top of the hill is between

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$Assertion$ : Centripetal and centrifugal forces cancel each other.
$Reason$ : Centrifugal force is a reaction of centripetal force.