(A) The tension $T$ in the string provides the necessary centripetal force for the horizontal circular motion.$T = \frac{mv^2}{r}$Given: $m = 0.25 \,

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(A) The tension $T$ in the string provides the necessary centripetal force for the horizontal circular motion.$T = \frac{mv^2}{r}$Given: $m = 0.25 \, kg$,$r = 1.96 \, m$,and $T_{max} = 25 \, N$.Substituting the values into the formula:$25 = \frac{0.25 \times v^2}{1.96}$$v^2 = \frac{25 \times 1.96}{0.25}$$v^2 = 100 \times 1.96 = 196$$v = \sqrt{196} = 14 \, m/s$.Thus,the maximum speed is $14 \, m/s$.

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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$A$ ball of mass $0.25 \, kg$ attached to the end of a string of length $1.96 \, m$ is moving in a horizontal circle. The string will break if the tension is more than $25 \, N$. The maximum speed with which the ball can be moved is .......... $m/s$.

AIPMT 1998 All AIPMT

A
$14$
B
$3$
C
$3.92$
D
$5$

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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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$A$ ball of mass $0.25 \, kg$ attached to the end of a string of length $1.96 \, m$ is moving in a horizontal circle. The string will break if the tension is more than $25 \, N$. The maximum speed with which the ball can be moved is .......... $m/s$.

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$A$ person with his hands in his pockets is skating on ice at a velocity of $10 \, m/s$ and describes a circle of radius $50 \, m$. What is his inclination with the vertical?

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