A) 16 cm
B) 15 cm
C) 14 cm
D) 13 cm
Correct Answer: B
Solution :
Let the length of the spring is \[l.\] When the system is whirled round in a horizontal circle the centripetal force is given by \[F=\frac{m{{v}^{2}}}{r}=\frac{m{{(r\omega )}^{2}}}{r}=mr{{\omega }^{2}}\] Then, \[r=l+\]elongation Given: elongation =1 cm (in the first case) For angular velocity \[\omega \] the force required is \[{{F}_{1}}=m(l+1){{\omega }^{2}}=kx=k\times 1=k\] or \[k=m(l+1){{\omega }^{2}}\] ?(i) For second case, \[\omega =2\omega ,\]elongation \[=\,5\,cm=x\] radius, \[r=l+5\] So, \[{{F}_{2}}=m(l+5){{(2\omega )}^{2}}=kx=k\times 5=5k\] or \[5k=4m(l+5){{\omega }^{2}}\] ?(ii) Now, dividing Eq. (i) by Eq.(ii), we get \[\frac{k}{5k}=\frac{m(l+1){{\omega }^{2}}}{4m(l+5){{\omega }^{2}}}\] \[\Rightarrow \] \[5(l+1)=4(l+5)\] \[\Rightarrow \] \[5l+5=4l+20\] \[l=20-5=15\,cm\]
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