Two wires A and B made of same material having length 10 cm and 40 cm respectively are connected in parallel to the same source of emf 10 V . What will be the ratio of the drift velocity of the electrons in the wire $A$ to the drift velocity in the wire $B$ ?

The drift velocity $v_{d}$ of electrons in a conductor is given by the formula $v_{d} = \dfrac{I}{neA}$, where $I$ is the current, $n$ is the electron density, $e$ is the elementary charge, and $A$ is the cross-sectional area. Since the wires are made of the same material, $n$ is the same for both.

The current $I$ in a wire is given by $I = \dfrac{V}{R}$, where $V$ is the potential difference and $R$ is the resistance. The resistance $R$ is given by $R = \rho \dfrac{l}{A}$, where $\rho$ is the resistivity and $l$ is the length. Substituting this into the expression for current, $I = \dfrac{VA}{\rho l}$.

Substituting $I$ into the drift velocity formula, $v_{d} = \dfrac{VA}{neA\rho l} = \dfrac{V}{ne\rho l}$.

Since both wires are connected to the same source of emf, $V$ is the same for both. The material is the same, so $n$ and $\rho$ are constant. Therefore, $v_{d} \propto \dfrac{1}{l}$.

The ratio of drift velocities is $\dfrac{v_{dA}}{v_{dB}} = \dfrac{l_{B}}{l_{A}}$.

Given $l_{A} = 10$ cm and $l_{B} = 40$ cm, we have $\dfrac{v_{dA}}{v_{dB}} = \dfrac{40}{10} = \dfrac{4}{1}$.

Answer: $\dfrac{V_{d A}}{V_{d B}}=4: 1$