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(D) The tension $T$ in a string providing centripetal force for a mass $m$ moving in a circle of radius $r$ with frequency $n$ (in revolutions per sec

Vedclass · Accepted Answer

$7$

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(D) The tension $T$ in a string providing centripetal force for a mass $m$ moving in a circle of radius $r$ with frequency $n$ (in revolutions per second) is given by $T = m \omega^2 r = m(2\pi n)^2 r = 4\pi^2 n^2 mr$.Since $m$ and $r$ are constant,we have $T \propto n^2$.Given the initial frequency $n_1 = 5 \, rpm$ and the final tension $T_2 = 2T_1$,we can write the ratio:$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{2T_1}{T_1}} = \sqrt{2}$.Therefore,$n_2 = n_1 	imes \sqrt{2} = 5 	imes 1.414 = 7.07 \, rpm$.Rounding to the nearest integer,the new speed is approximately $7 \, rpm$.

$A$ mass of $2 \, kg$ is whirled in a horizontal circle by means of a string at an initial speed of $5$ revolutions per minute. Keeping the radius constant,the tension in the string is doubled. The new speed is nearly ....... $rpm$.

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