Question

A hollow sphere of mass $M=50kg$ and radius $r{\left(\frac{3}{40\pi }\right)}^{\frac{1}{3}}$ 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 as shown. Tension in wire A is $(g=10m/s^2)$

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

Answer 1

Answer: (C)

$250\sqrt{2}N$

When the sphere is tied to the strings, it is in equilibrium. net force on sphere is zero. i.e. net upward force= net downward force
$\Rightarrow F_b=2T_1\cos {45}^{\circ }+Mg$; here $F_b$ is the buoyant force
$\Rightarrow \frac{4}{3}\pi R^3{\rho }_{w}g=2T_1\cos {45}^{\circ }+Mg$
$\Rightarrow T_1=\frac{\frac{4}{3}\pi R^3{\rho }_{w}g-Mg}{\sqrt{2}}\Rightarrow T_1=\frac{\frac{4}{3}\times \pi \times {\left({\left(\frac{3}{40\pi }\right)}^{\frac{1}{3}}\right)}^3\times 1000\times 10-50\times 10}{\sqrt{2}}=\frac{1000-500}{\sqrt{2}}=\frac{500}{\sqrt{2}}N=250\sqrt{2}N$