JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    A uniform metal disc of radius R is taken and out of it a disc of diameter R/2 is cut-off from the end. The centre of mass of the remaining part will be

    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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