• 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

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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