Question

The dimensions of resistivity in terms of $M, L, T$ and $Q$, where $Q$ stands for the dimensions of charge is

A. $M L^{3} T^{-1} Q^{-2}$ B. $M L^{3} T^{-2} Q^{-1}$

C. $M L^{2} T^{-1} Q^{-1}$ D. $M L T^{-1} Q^{-1}$.

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Answer

To determine the dimensions of resistivity ($\rho$) in terms of $M$, $L$, $T$, and $Q$ (where $Q$ represents the dimensions of charge), let's first understand the relationship among the physical quantities:

Resistivity ($\rho$) can be expressed as: $$ \rho = \frac{R \cdot A}{L} $$ where:

  • $R$ is the resistance,
  • $A$ is the cross-sectional area,
  • $L$ is the length.

We need the dimensions for each of these terms.

  1. Dimensions of Resistance ($R$)

    • Ohm's law states that $R = \frac{V}{I}$, where $V$ is voltage and $I$ is current.
    • Voltage ($V$) has dimensions of $\left[ M L^2 T^{-3} Q^{-1} \right]$.
    • Current ($I$) has dimensions of $\left[ Q T^{-1} \right]$.

    So, the dimensions of resistance $R$ are: $$ \left[ R \right] = \frac{[V]}{[I]} = \frac{\left[ M L^2 T^{-3} Q^{-1} \right]}{\left[ Q T^{-1} \right]} = M L^2 T^{-3} Q^{-2} $$

  2. Dimensions of Area ($A$)

    • Area is length squared, so: $$ \left[ A \right] = L^2 $$
  3. Dimensions of Length ($L$)

    • Length has dimensions $[L]$.

Combining these, the dimensions of resistivity $\rho$ are: $$ \rho = \frac{R \cdot A}{L} = \frac{[R] \cdot [A]}{[L]} $$ Substituting the dimensions obtained: $$ \left[ \rho \right] = \frac{\left[M L^2 T^{-3} Q^{-2}\right] \cdot \left[L^2\right]}{[L]} = \frac{M L^4 T^{-3} Q^{-2}}{L} = M L^3 T^{-3} Q^{-2} $$

Thus, the correct dimensional formula for resistivity is: $$ M L^3 T^{-3} Q^{-2} $$

Hence, the correct option is:

A. $M L^{3} T^{-1} Q^{-2}$


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