Question

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

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

Answer 1

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

Buoyant force will act upwards on the sphere as shown in figure.


From FBD,
Balancing forces along horizontally, we get
$T_1\sin {45}^{\circ }=T_2\sin {45}^{\circ }$
$\Rightarrow T_1=T_2$
Balancing vertical forces,
$Mg+2T_1\cos {45}^{\circ }=\frac{4}{3}\pi r^3{\rho }_{w}g$
$\Rightarrow 50\times 10+\sqrt{2}{T}_1=\frac{4}{3}\pi \times \frac{3}{40\pi }\times {10}^3\times 10$
$\Rightarrow T_1=250\sqrt{2}\text{N}$