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The electrical resistance of a mercury column in a cylindrical container is R. When the same mercury is poured into another cylindrical container twice the radius of the first, then resistance of mercury column is
- a)R/2
- b)R/4
- c)R/8
- d)R/16
Correct answer is option 'D'. Can you explain this answer?
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The electrical resistance of a mercury column in a cylindrical contain...
Explanation:
The electrical resistance of a conductor is given by the formula:
\[ R = \frac{{\rho \cdot L}}{A} \]
Where,
- \( R \) = Resistance
- \( \rho \) = Resistivity of the material (constant for a given material)
- \( L \) = Length of the conductor
- \( A \) = Cross-sectional area of the conductor
Effect of changing radius on resistance:
When the mercury is poured into another cylindrical container with twice the radius of the first one, the cross-sectional area of the mercury column changes. As the resistance is inversely proportional to the cross-sectional area, the resistance will change accordingly.
Let's assume the radius of the first container is \( r \) and the radius of the second container is \( 2r \).
The cross-sectional area of the first container is:
\[ A_1 = \pi r^2 \]
The cross-sectional area of the second container is:
\[ A_2 = \pi (2r)^2 = 4\pi r^2 \]
So, the resistance of the mercury column in the second container is:
\[ R' = \frac{{\rho \cdot L}}{4\pi r^2} \]
Comparing the resistances:
We need to find the ratio of the new resistance to the original resistance:
\[ \frac{R'}{R} = \frac{\frac{{\rho \cdot L}}{4\pi r^2}}{\frac{{\rho \cdot L}}{\pi r^2}} = \frac{1}{4} \]
Therefore, the resistance of the mercury column in the second container is \( \frac{1}{4} \) times the resistance of the mercury column in the first container. This corresponds to option D: \( R/16 \).
The electrical resistance of a conductor is given by the formula:
\[ R = \frac{{\rho \cdot L}}{A} \]
Where,
- \( R \) = Resistance
- \( \rho \) = Resistivity of the material (constant for a given material)
- \( L \) = Length of the conductor
- \( A \) = Cross-sectional area of the conductor
Effect of changing radius on resistance:
When the mercury is poured into another cylindrical container with twice the radius of the first one, the cross-sectional area of the mercury column changes. As the resistance is inversely proportional to the cross-sectional area, the resistance will change accordingly.
Let's assume the radius of the first container is \( r \) and the radius of the second container is \( 2r \).
The cross-sectional area of the first container is:
\[ A_1 = \pi r^2 \]
The cross-sectional area of the second container is:
\[ A_2 = \pi (2r)^2 = 4\pi r^2 \]
So, the resistance of the mercury column in the second container is:
\[ R' = \frac{{\rho \cdot L}}{4\pi r^2} \]
Comparing the resistances:
We need to find the ratio of the new resistance to the original resistance:
\[ \frac{R'}{R} = \frac{\frac{{\rho \cdot L}}{4\pi r^2}}{\frac{{\rho \cdot L}}{\pi r^2}} = \frac{1}{4} \]
Therefore, the resistance of the mercury column in the second container is \( \frac{1}{4} \) times the resistance of the mercury column in the first container. This corresponds to option D: \( R/16 \).
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The electrical resistance of a mercury column in a cylindrical container is R. When the same mercury is poured into another cylindrical container twice the radius of the first, then resistance of mercury column isa)R/2b)R/4c)R/8d)R/16Correct answer is option 'D'. Can you explain this answer?
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