Question

A hollow sphere of mass $M=50kg$ and radius $r={\left(\frac{3}{40\pi }\right)}^{1/3}m$ is immersed in a tank of water (density ${\rho }_w={10}^{3}kg/m^3$). The sphere is tied to the bottom of a tank by two wires $A$ and $B$ shown. Tension in wire is $A$ is ($g=10m/s^2$)

• A. $125\sqrt{2}N$
• B. $125N$
• C. $250\sqrt{2}N$
• D. $250N$

Answer 1

The correct option is C $250\sqrt{2}N$

For horizontal equilibrium, $F_{net}$ in horizontal direction $=0$

$\Rightarrow T_1\cos 45=T_2\cos 45\Rightarrow T_1=T_2=T\left(say\right)$

For vertical equilibrium-

$2T\sin 45+mg=F_{b}2T\left(\frac{1}{\sqrt{2}}\right)+\left(50\right)\left(10\right)=8\omega \left(\frac{4}{3}\pi r^3\right)g\sqrt{2}T+500=1000\times \frac{4}{3}\pi \times \frac{3}{40\pi }\times 10\sqrt{2}T=500T=\frac{500}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=250\sqrt{2}N$