A) \[\frac{R}{4}\]from the centre
B) \[\frac{R}{3}\]from the centre
C) \[\frac{R}{5}\]from the centre
D) \[\frac{R}{6}\]from the centre
Correct Answer: D
Solution :
Mass of the disc removed \[{{N}_{S}}=\left( \frac{\rho S}{\rho P} \right)\times {{N}_{P}}\] Remaining mass \[=\frac{(4.4\times {{10}^{3}})\,\times 100}{220}=2000\] Let, the origin of the coordinate system coincide with centre of mass of whole disc. Now we know that, \[{{\rho }_{S}}\] \[{{\rho }_{P}}\] will be zero, when\[-h=-u{{t}_{1}}+\frac{1}{2}gt_{1}^{2}\] \[-h=-u{{t}_{2}}+\frac{1}{2}gt_{2}^{2}\] \[0=u({{t}_{2}}-{{t}_{1}})+\frac{1}{2}g(t_{1}^{2}-t_{2}^{2})\] Here, \[u=\frac{1}{2}g({{t}_{1}}+{{t}_{2}})\]and \[h=\frac{g{{t}_{1}}{{t}_{2}}}{2}\] (for remaining mass) Hence, \[\sqrt{2}\,mv\] i.e., \[\sqrt{2}\,mv'=(2m)\,v\] from the centre (on LHS).
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