A hollow sphere of mass M and radius R is imm | vedclass

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(A) The forces acting on the sphere are:$1$. The buoyant force $B$ acting vertically upwards,given by $B = V_{disp} \rho_w g = \frac{4}{3} \pi R^3 \rh

Vedclass · Accepted Answer

$\frac{\frac{4}{3}\pi R^3 \rho_w g - Mg}{\sqrt{2}}$

Vedclass · Answer

(A) The forces acting on the sphere are:$1$. The buoyant force $B$ acting vertically upwards,given by $B = V_{disp} ho_w g = \frac{4}{3} \pi R^3 ho_w g$.$2$. The weight of the sphere $W = Mg$ acting vertically downwards.$3$. The tension $T_1$ in each of the two wires,acting at an angle of $45^{\circ}$ with the horizontal.For the sphere to be in equilibrium,the sum of vertical forces must be zero:$2 T_1 \sin 45^{\circ} + Mg = B$$2 T_1 \sin 45^{\circ} = B - Mg$Substituting the values:$2 T_1 	imes \frac{1}{\sqrt{2}} = \frac{4}{3} \pi R^3 ho_w g - Mg$$T_1 \sqrt{2} = \frac{4}{3} \pi R^3 ho_w g - Mg$$T_1 = \frac{\frac{4}{3} \pi R^3 ho_w g - Mg}{\sqrt{2}}$

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