A ball of mass 0.1, Kg. is whirled in a horiz | vedclass

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Question

(a) $T = m{\omega ^2}r$ $&rArr;$ $\omega \propto \sqrt T $

$	herefore \frac{{{\omega _2}}}{{{\omega _1}}} = \sqrt {\frac{1}{4}} $ $\Rightarrow {\o

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(a) $T = m{\omega ^2}r$ $&rArr;$ $\omega \propto \sqrt T $

$	herefore \frac{{{\omega _2}}}{{{\omega _1}}} = \sqrt {\frac{1}{4}} $ $\Rightarrow {\omega _2} = \frac{{{\omega _1}}}{2} = 5\,rpm$

A ball of mass $0.1\, Kg$. is whirled in a horizontal circle of radius $1\, m$. by means of a string at an initial speed of $10\, R.P.M.$ Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is ....... $r.p.m.$

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