91Ó°ÊÓ

Buoyancy is an upward force exerted by the fluid in the upward direction that opposes the weight of a fully or partially submerged body. It is equal to the weight of the water displaced. It is expressed in Newtons.

The external diameter of the research chamber is \({D_{{\rm{chamber}}}} = 5.20\;{\rm{m}}\).

The mass of the chamber is \({m_{{\rm{chamber}}}} = 74400\;{\rm{kg}}\).

Let us understand the question better with the help of the diagram shown below:

There are three forces on the chamber:

The buoyant force \({F_{\rm{B}}}\), the tension in the cable \({F_{\rm{T}}}\), and the weight of the chamber \(mg\).

Now, we know that the buoyant force is the weight of water displaced by the chamber.

\(\begin{aligned}{F_{{\rm{buoyant}}}} &= {\rho _{{\rm{water}}}}{V_{{\rm{chamber}}}}g\\{F_{{\rm{buoyant}}}} &= {\rho _{{\rm{water}}}}\frac{4}{3}\pi R_{{\rm{chamber}}}^3g\end{aligned}\)

Here \({\rho _{{\rm{water}}}}\) is the density of seawater, which is \(1.025 \times {10^3}\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}\)(standard), \({V_{{\rm{chamber}}}}\) is the volume of the chamber (equal to the volume of water displaced), and \({R_{{\rm{chamber}}}}\) is the radius of the chamber \(\left( {{R_{{\rm{chamber}}}} = \frac{{{D_{{\rm{chamber}}}}}}{2} = \frac{{5.20}}{2} = 2.60\;{\rm{m}}} \right)\).

Substitute the values in the above formula for buoyant force.

\(\begin{aligned}{F_{{\rm{buoyant}}}} &= \left( {1.025 \times {{10}^3}\;{{{\rm{kg}}} \mathord{\left/{\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.} {{{\rm{m}}^3}}}} \right) \times \frac{4}{3}\pi \left( {2.60\;{{\rm{m}}^3}} \right)\left( {9.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}} \right)\\{F_{{\rm{buoyant}}}} &= 7.3953 \times {10^5}\;{\rm{N}}\\{F_{{\rm{buoyant}}}} &\approx 7.40 \times {10^5}\;{\rm{N}}\end{aligned}\)

Hence, the buoyant force on the spherical undersea research chamber is \(7.40 \times {10^5}\;{\rm{N}}\).

To find the tension, we will be using the equilibrium condition for the chamber, i.e.,

\(\begin{aligned}{F_{{\rm{buoyant}}}} &= mg + {F_T}\\{F_{\rm{T}}} &= {F_{{\rm{buoyant}}}} - mg\end{aligned}\)

Substitute the values in the above formula for buoyant force.

\(\begin{aligned}{F_{\rm{T}}} &= 7.3953 \times {10^5}\;{\rm{N}}\; - \;\left( {7.44 \times {{10}^4}\;{\rm{kg}}} \right)\left( {9.80\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}} \right)\left( {\frac{{1\;{\rm{N}}}}{{1\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right.} {{{\rm{s}}^2}}}}}} \right)\\{F_{\rm{T}}} &= 1.0 \times {10^4}\;{\rm{N}}\end{aligned}\)

Hence, the tension in the cable is \(1.0 \times {10^4}\;{\rm{N}}\).