Question

A beaker of water rests on an electronic balance that reads 975.0 g. A 2.6-cm-diameter solid copper ball attached to a string is submerged in the water, but does not touch the bottom. What are the tension in the string and the new balance reading?

Short answer

The tension of the string and new balance reading are \(0.717\;{\rm{N}}\)and the reading of the new balance is \(984.2\;{\rm{g}}\), respectively.

Step-by-step solution

Understanding the tension

Whenever a solid object attaches to a wire and enters a specific liquid, there would be tension in the wire due to different densities of the object and liquid. The relation between the tension force and the densities of solid and liquid would be a linear one.

Identification of given data

The mass of beaker of water is\(m = 975.0\;{\rm{g}}\).

The diameter of solid copper ball is \(d = 2.6\;{\rm{cm}}\).

Determination of the tension in the string

The expression of the tension in the string is given by,

\(\begin{array}{c}T = \left( {{\rho _c} - {\rho _w}} \right)Vg\\ = \left( {{\rho _c} - {\rho _w}} \right)\left( {\frac{4}{3}\pi {{\left( {\frac{d}{2}} \right)}^3}} \right)g\end{array}\)

Here, \(T\) is the tension in the string, \({\rho _c}\) is the density of copper \(\left( {{\rho _c} = 8940\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\), \({\rho _w}\) is the density of water \(\left( {{\rho _w} = 1000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\)and \(g\) is the gravitational acceleration \(\left( {g = 9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\).

Substitute all the known values in the above expression.

\(\begin{array}{c}T = \left( {8940\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}} - 1000\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {\frac{4}{3}\pi {{\left( {\frac{{2.6 \times {{10}^{ - 2}}\;{\rm{m}}}}{2}} \right)}^3}} \right]\left( {9.81\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\ \approx 0.717\;{\rm{N}}\end{array}\)

Thus, the tension in the string is \(0.717\;{\rm{N}}\).

Determine the new balance reading

The expression of the new balance reading is given by,

\({m_f} = m + {\rho _w}V\)

Here, \({m_f}\) is the new balance reading.

Substitute all the known values in the above expression.

\(\begin{array}{c}{m_f} = \left( {975\;{\rm{g}}} \right) + ({10^6}\;{\rm{g/}}{{\rm{m}}^{\rm{3}}}) \times \frac{4}{3}\pi \times {\left( {\frac{{2.6 \times {{10}^{ - 2}}\;{\rm{m}}}}{2}} \right)^3}\\ \approx 984.2\;{\rm{g}}\end{array}\)

Thus, the new balance reading is \(984.2\;{\rm{g}}\).