
The specific resistance of manganin is \[50 \times {10^{ - 8}}\Omega m\] . The resistance of a manganin cube having length \[50\,cm\] is
A. ${10^{ - 6}}\Omega $
B. $2.5 \times {10^{ - 5}}\Omega $
C. ${10^{ - 8}}\Omega $
D. $5 \times {10^{ - 4}}\Omega $
Answer
162.6k+ views
Hint: To solve this question, we first need to find out the area of the given manganin cube by using the formula of area of cube. Then we find the value of resistance using the relationship between resistance, resistivity (or specific resistance), length and area of the cube.
Formula Used:
Area of cross section of the cube,
$A = {l^2}$
where $l$ is the length of each side of the cube.
Resistance,
$R = \dfrac{{\rho l}}{A}$
where $\rho $ is the specific resistance (or resistivity), $l$ is the length of each side of the given manganin cube and $A$ is the area of the cross section of the cube.
Complete step by step solution:
Given: specific resistance (or resistivity) of the given manganin cube,
$\rho = 50 \times {10^{ - 8}}\Omega m$
And length of each side of the given manganin cube, $l = 50\,cm$
Converting the length from centimetre to metre, we get $l = 50 \times {10^{ - 2}}m$
Now, we find the area of cross section of the cube using the formula,
$A = {l^2}$
We get,
$A = {\left( {50 \times {{10}^{ - 2}}} \right)^2}{m^2}$
Now, we find the resistance of the given manganin cube using the relationship between resistance, resistivity (or specific resistance), length and area of the cube.That is,
$R = \dfrac{{\rho l}}{A}$
Substituting the values of resistivity, length and area of cross section of the cube we get,
$R = \dfrac{{50 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 2}}}}{{{{\left( {50 \times {{10}^{ - 2}}} \right)}^2}}} \\ $
Solving this further, we get
$R = {10^{ - 6}}\Omega $
Hence, option A is the correct answer.
Note: While solving this question do not take the area of the whole cube which is $6{l^2}$ . Take the area of the cross section of the cube only which is the area of any one side of the cube i.e., ${l^2}$ . Also do not confuse between resistivity and specific resistance because they both mean the same.
Formula Used:
Area of cross section of the cube,
$A = {l^2}$
where $l$ is the length of each side of the cube.
Resistance,
$R = \dfrac{{\rho l}}{A}$
where $\rho $ is the specific resistance (or resistivity), $l$ is the length of each side of the given manganin cube and $A$ is the area of the cross section of the cube.
Complete step by step solution:
Given: specific resistance (or resistivity) of the given manganin cube,
$\rho = 50 \times {10^{ - 8}}\Omega m$
And length of each side of the given manganin cube, $l = 50\,cm$
Converting the length from centimetre to metre, we get $l = 50 \times {10^{ - 2}}m$
Now, we find the area of cross section of the cube using the formula,
$A = {l^2}$
We get,
$A = {\left( {50 \times {{10}^{ - 2}}} \right)^2}{m^2}$
Now, we find the resistance of the given manganin cube using the relationship between resistance, resistivity (or specific resistance), length and area of the cube.That is,
$R = \dfrac{{\rho l}}{A}$
Substituting the values of resistivity, length and area of cross section of the cube we get,
$R = \dfrac{{50 \times {{10}^{ - 8}} \times 50 \times {{10}^{ - 2}}}}{{{{\left( {50 \times {{10}^{ - 2}}} \right)}^2}}} \\ $
Solving this further, we get
$R = {10^{ - 6}}\Omega $
Hence, option A is the correct answer.
Note: While solving this question do not take the area of the whole cube which is $6{l^2}$ . Take the area of the cross section of the cube only which is the area of any one side of the cube i.e., ${l^2}$ . Also do not confuse between resistivity and specific resistance because they both mean the same.
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