A spherical cavity of radius r is carved out | vedclass

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(D) Let the centre of the solid sphere of radius $R$ be at the origin $(0,0,0)$.Let $\rho$ be the uniform density of the sphere.The mass of the solid

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$\frac{r^3}{R^2+R r+r^2}$

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(D) Let the centre of the solid sphere of radius $R$ be at the origin $(0,0,0)$.Let $\rho$ be the uniform density of the sphere.The mass of the solid sphere is $M = \frac{4}{3} \pi R^3 \rho$.The mass of the removed spherical cavity of radius $r$ is $m = \frac{4}{3} \pi r^3 \rho$.The centre of mass of the cavity is at a distance $x = R-r$ from the origin along the $x$-axis.The centre of mass of the remaining body is given by the formula:$x_{CM} = \frac{M(0) - m(R-r)}{M - m}$Substituting the values:$x_{CM} = \frac{0 - (\frac{4}{3} \pi r^3 \rho)(R-r)}{\frac{4}{3} \pi R^3 \rho - \frac{4}{3} \pi r^3 \rho}$$x_{CM} = -\frac{r^3(R-r)}{R^3 - r^3}$Using the algebraic identity $R^3 - r^3 = (R-r)(R^2 + Rr + r^2)$:$x_{CM} = -\frac{r^3(R-r)}{(R-r)(R^2 + Rr + r^2)} = -\frac{r^3}{R^2 + Rr + r^2}$The negative sign indicates that the centre of mass shifts in the direction opposite to the cavity.The distance $d$ of the centre of mass from the centre of the original solid sphere is the magnitude of $x_{CM}$:$d = \frac{r^3}{R^2 + Rr + r^2}$

$A$ spherical cavity of radius $r$ is carved out of a uniform solid sphere of radius $R$ as shown in the figure below. The distance of the centre of mass of the resulting body from that of the solid sphere is given by

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