
Write the S.I unit of specific conductance.
Answer
570.9k+ views
Hint: Specific conductance is denoted by kappa \[\left( \kappa \right)\] and formula for it is \[\kappa = \dfrac{1}{\rho }\] where \[\rho \] is having formula \[\rho = \dfrac{{R \times A}}{L}\], so to calculate S.I units substitute S.I unit of \[R\](Resistance), \[A\](Area) and \[L\](Length).
Complete step by step answer
As we know the formula of kappa is given as, \[\kappa = \dfrac{1}{\rho }\]
And we know the formula of \[\rho \] that is given as, \[\rho = \dfrac{{R \times A}}{L}\]
And we know here \[R\] is resistance and S.I. unit of \[R\] is \[\Omega \]
And \[A\] is area and S.I. units of \[A\] is \[{m^2}\]
And \[L\] is length and S.I. units of \[L\] is \[m\]
So now substitute the S.I. units in the formula of kappa.
\[\kappa = \dfrac{m}{{{m^2} \times \Omega }}\]
\[\kappa = {m^{1 - 2}}{\Omega ^{ - 1}}\]
\[\kappa = {m^{ - 1}}{\Omega ^{ - 1}}\]
Hence S.I. units of specific conductance are \[{m^{ - 1}}{\Omega ^{ - 1}}\]
Additional information
Explicit Conductance is a proportion of how well water can lead an electrical flow. Conductivity increments with expanding sum and versatility of particles. These particles, which originate from the breakdown of mixes, direct power since they are contrarily or decidedly charged when disintegrated in water. Thus, SC is a roundabout proportion of the presence of broken solids, for example, chloride, nitrate, sulphate, phosphate, sodium, magnesium, calcium, and iron, and can be utilized as a marker of water pollution. Specific Conductance gauges how well water can direct an electrical momentum for a unit length and unit cross-area at a specific temperature.
Note: As we have calculated S.I. units of specific conductance above taking S.I. units of length as \[m\] and S.I. units of area as \[{m^2}\] and resultant S.I. unit was \[{m^{ - 1}}{\Omega ^{ - 1}}\] and we can also take S.I. units of length as \[cm\] and S.I. units of area as \[{\left( {cm} \right)^2}\] and resultant S.I. units of specific conductance will be \[{\left( {cm} \right)^{ - 1}}{\Omega ^{ - 1}}\].
Complete step by step answer
As we know the formula of kappa is given as, \[\kappa = \dfrac{1}{\rho }\]
And we know the formula of \[\rho \] that is given as, \[\rho = \dfrac{{R \times A}}{L}\]
And we know here \[R\] is resistance and S.I. unit of \[R\] is \[\Omega \]
And \[A\] is area and S.I. units of \[A\] is \[{m^2}\]
And \[L\] is length and S.I. units of \[L\] is \[m\]
So now substitute the S.I. units in the formula of kappa.
\[\kappa = \dfrac{m}{{{m^2} \times \Omega }}\]
\[\kappa = {m^{1 - 2}}{\Omega ^{ - 1}}\]
\[\kappa = {m^{ - 1}}{\Omega ^{ - 1}}\]
Hence S.I. units of specific conductance are \[{m^{ - 1}}{\Omega ^{ - 1}}\]
Additional information
Explicit Conductance is a proportion of how well water can lead an electrical flow. Conductivity increments with expanding sum and versatility of particles. These particles, which originate from the breakdown of mixes, direct power since they are contrarily or decidedly charged when disintegrated in water. Thus, SC is a roundabout proportion of the presence of broken solids, for example, chloride, nitrate, sulphate, phosphate, sodium, magnesium, calcium, and iron, and can be utilized as a marker of water pollution. Specific Conductance gauges how well water can direct an electrical momentum for a unit length and unit cross-area at a specific temperature.
Note: As we have calculated S.I. units of specific conductance above taking S.I. units of length as \[m\] and S.I. units of area as \[{m^2}\] and resultant S.I. unit was \[{m^{ - 1}}{\Omega ^{ - 1}}\] and we can also take S.I. units of length as \[cm\] and S.I. units of area as \[{\left( {cm} \right)^2}\] and resultant S.I. units of specific conductance will be \[{\left( {cm} \right)^{ - 1}}{\Omega ^{ - 1}}\].
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