A) 9 cm
B) 3 cm
C) 1.5 cm
D) 1 cm
Correct Answer: D
Solution :
Centripetal force on coin placed on a rotating table is \[=mr{{\omega }^{2}}\] In both case the force required is same. Hence, \[m{{r}_{1}}\omega _{1}^{2}=m{{r}_{2}}\omega _{2}^{2}\] Given: \[{{r}_{1}}=9cm\] \[{{\omega }_{2}}=3{{\omega }_{1}}\] \[9\times \omega _{1}^{2},={{r}_{2}}{{(3{{\omega }_{1}})}^{2}}\] \[9\omega _{1}^{2}={{r}_{2}}9\omega _{1}^{2}\] \[{{r}_{2}}=\frac{9}{9}=1cm\]
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