As shown in fig. when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e on the surface of the cavity. R can be determined by the equation :

Question

As shown in fig. when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e on the surface of the cavity. R can be determined by the equation : (A) (R2+R + 1) ( 2 - R ) = 1 (B) (R2+ R -1 )(2 -R )= 1 (C) (R2- R - 1) (2 - R) =1 (D) (R2- R + 1) (2 - R )=1

Tardigrade Admin · Accepted Answer

By concept of COM m1R1=m2R2 Remaining mass Ã (2-R) = cavity mass Ã (R-1) ((4/3)π R3 ρ-(4/3)π l3 ρ)(2-R)=(4/3)π l3 ρ×(R-1) (R3-1)(2-R)=R-1 (R2+R+1)(2-R)=1

Q. As shown in fig. when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e on the surface of the cavity. R can be determined by the equation :

$(R^{2} + R + 1) (2 - R) = 1$

$(R^{2} + R - 1) (2 - R) = 1$

$(R^{2} - R - 1) (2 - R) = 1$

$(R^{2} - R + 1) (2 - R) = 1$

By concept of $COM$
$m_{1} R_{1} = m_{2} R_{2}$
Remaining mass $\times (2 - R) =$ cavity mass $\times (R - 1)$
$(\frac{4}{3} π R^{3} ρ - \frac{4}{3} π l^{3} ρ) (2 - R) = \frac{4}{3} π l^{3} ρ \times (R - 1)$
$(R^{3} - 1) (2 - R) = R - 1$
$(R^{2} + R + 1) (2 - R) = 1$

Q. As shown in fig. when a spherical cavity (centred at O) of radius 1 is cut out of a uniform sphere of radius R (centred at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e on the surface of the cavity. R can be determined by the equation :

(R2+R+1)(2−R)=1

(R2+R−1)(2−R)=1

(R2−R−1)(2−R)=1

(R2−R+1)(2−R)=1

By concept of COM m1​R1​=m2​R2​ Remaining mass ×(2−R)= cavity mass ×(R−1) (34​πR3ρ−34​πl3ρ)(2−R)=34​πl3ρ×(R−1) (R3−1)(2−R)=R−1 (R2+R+1)(2−R)=1

$(R^{2} + R + 1) (2 - R) = 1$

$(R^{2} + R - 1) (2 - R) = 1$

$(R^{2} - R - 1) (2 - R) = 1$

$(R^{2} - R + 1) (2 - R) = 1$

By concept of $COM$
$m_{1} R_{1} = m_{2} R_{2}$
Remaining mass $\times (2 - R) =$ cavity mass $\times (R - 1)$
$(\frac{4}{3} π R^{3} ρ - \frac{4}{3} π l^{3} ρ) (2 - R) = \frac{4}{3} π l^{3} ρ \times (R - 1)$
$(R^{3} - 1) (2 - R) = R - 1$
$(R^{2} + R + 1) (2 - R) = 1$