Question

A household $20$ liter geyser consists of a coil of resistance R, across which an ac source of rms voltage V is connected. The wall thickness of geyser is t ( $\ll \sqrt A$ ) and the inner surface area of geyser is A. the average thermal conductivity of material of walls of geysers is k and S is specific heat capacity of water. Assuming the initial temperature of water is equal to the temperature of water is equal to the temperature of the atmosphere. The time taken to rise the temperature of water by ${50^ \circ }$ is (neglect the specific heat of wall of geyser and radiation losses)
A.\dfrac{{2St}}{{kA}}\ln (\dfrac{{50RkA}}{{r{V^2}}} - 1) \\\ B.\dfrac{{20St}}{{kA}}\ln (\dfrac{{{V^2}t}}{{{V^2}t - 20RkA}} - 1) \\\ C.\dfrac{{2St}}{{kA}}\ln (\dfrac{{50t{V^2}}}{{RkA}}) \\\ D.\dfrac{{2St}}{{kA}}\ln (\dfrac{{50RkA}}{{t{V^2}}}) \\\

Answer 1

Here we use the dimensional formula which is an expression for the unit of a physical quantity in terms of fundamental quantities. It is expressed in terms of M, L and T. Also first we use to write options according to their dimensional formula then we try to look which can match to the dimension of the temperature.

Complete step by step solution:
In the given question, the geyser is of $20$ litre, also the resistance of the coil is R, and the thickness of the wall of the geyser is t.
Inner surface area of the geyser is A, and the thermal conductivity of the geyser is k.
Specific heat capacity is S.
Now to find the time taken of the geyser to rise the temperature of the water by ${50^ \circ }$ , we have the following dimensions;
Dimension of the heat capacity is $M{L^2}{T^{ - 2}}{t^{ - 1}}$
Dimension of resistance is $M{L^2}{T^{ - 3}}{I^{ - 2}}$
Dimension of thickness is $L$
Dimension of area is ${L^2}$
Dimension of conductivity is ${T^3}{I^2}{M^{ - 1}}{L^3}$
Dimension of voltage is ${T^{ - 3}}IM{L^2}$ .
Now we can try the options, trying the option A;
$\dfrac{{2St}}{{kA}}\ln (\dfrac{{50RkA}}{{r{V^2}}} - 1)] \\\ = \dfrac{{{L^2} \times {T^{ - 2}}{t^{ - 1}} \times M{L^2}{T^{ - 3}}{I^{ - 2}} \times {T^3}{I^2}{M^{ - 1}}{L^3} \times {L^2}}}{{{M^{ - 1}}{L^{ - 3}}{T^3}{I^2} \times {L^2} \times {T^{ - 6}}{I^{ - 2}}{M^2}{L^4}}}] \\\ = \dfrac{{M{L^6}{T^3}{I^2} \times {T^6}{I^2}M{L^3}}}{{{M^2}{T^3}{I^2}{L^8} \times {T^5}{I^2}{M^1}{L^3}}}] \\\ = [T]$
Therefore option A is the correct option.

Note:
Thermal conductivity of a material is the ability of a material to conduct heat. This is transferred through conduction. The lower the thermal conductivity, the better the heat energy.
Here we use the dimensional formula which is an expression for the unit of a physical quantity in terms of fundamental quantities. It is expressed in terms of M, L and T.