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3-2.Motion in Plane
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A boy ties a stone of mass $100 \,g$ to the end of a $2$ $m$ long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of $80 \,N$. If the maximum speed with which the stone can revolve is $\frac{ K }{\pi} rev$. / $min$. The value of $K$ is (Assume the string is massless and unstretchable)
A$400$
B$300$
C$600$
D$800$
(JEE MAIN-2022)
Solution
$T = M \omega^{2} R$
$T =80 N \quad M =0.1 \quad \omega=? \quad R =2 m$
$80=0.1 \omega^{2}(2)$
$\omega^{2}=400$
$\omega=20$
$2 \pi f =20$
$f =\frac{10}{\pi} \frac{ rev }{ s }$
$=\frac{600}{\pi} \frac{ rev }{ min }$
$T =80 N \quad M =0.1 \quad \omega=? \quad R =2 m$
$80=0.1 \omega^{2}(2)$
$\omega^{2}=400$
$\omega=20$
$2 \pi f =20$
$f =\frac{10}{\pi} \frac{ rev }{ s }$
$=\frac{600}{\pi} \frac{ rev }{ min }$
Standard 11
Physics
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