TG EAMCET 4 May 2025 Shift 1 Paper

Same material: This implies that the resistivity ( $\rho$ ) and the number density of free electrons ( $n$ ) are the same for both wires. Also, the charge of an electron $(e)$ is a constant.Lengths ratio: $L_1: L_2=2: 3$, which means $\frac{L_1}{L_2}=\frac{2}{3}$.Radii ratio: $r_1: r_2=1: 2$.Connected in parallel to a battery: This means the potential difference (voltage $V$ ) across both wires is the same, i.e., $V_1=V_2=V$.We need to find the ratio of the drift velocities $(v_{d 1}: v_{d 2})$.The relationship between current ( $I$ ), number density of free electrons ( $n$ ), cross-sectional area ( $A$ ), charge of an electron ( $e$ ), and drift velocity $(v_d)$ is given by:$I=n A e v_d$From this, the drift velocity can be expressed as:$v_d = \frac{I}{n A e} \; \text{(Equation 1)} \;$According to Ohm's Law, the current in a wire is related to the potential difference $(V)$ and resistance $(R)$ :$I = \frac{V}{R} \;(\text{Equation 2})$The resistance of a wire is given by:$R = \frac{\rho L}{A} \; \text{(Equation 3)} \;$where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area. Now, substitute Equation 3 into Equation 2:$I = \frac{V}{\frac{\rho L}{A}} = \frac{V A}{\rho L} \; \text{(Equation 4)} \;$Finally, substitute Equation 4 into Equation 1 for $I$ :$v_d = \frac{\frac{V A}{\rho L}}{n A e}$Notice that the cross-sectional area $(A)$ appears in both the numerator and the denominator, so it cancels out:$v_d = \frac{V}{n e \rho L}$This derived formula shows that the drift velocity $(v_d)$ depends on the applied voltage $(V)$, the number density of free electrons $(n)$, the electron charge $(e)$, the resistivity $(\rho)$, and the length of the wire $(L)$.Since the wires are made of the same material, $n$ and $\rho$ are constant. The charge of an electron $e$ is also a constant. When connected in parallel, the voltage $V$ across both wires is the same.Therefore, for these two wires, $V, n, e$ and $\rho$ are all constant. This means the drift velocity is inversely proportional to the length of the wire:$v_d \propto \frac{1}{L}$So, the ratio of the drift velocities will be:$\frac{v_{d 1}}{v_{d 2}} = \frac{\frac{1}{L_1}}{\frac{1}{L_2}} = \frac{L_2}{L_1}$We are given the ratio of lengths $L_1: L_2=2: 3$.This means $\frac{L_1}{L_2}=\frac{2}{3}$.Therefore, $\frac{L_2}{L_1}=\frac{3}{2}$.Substituting this into the ratio of drift velocities:$\frac{v_{d 1}}{v_{d 2}} = \frac{3}{2}$The ratio of the drift velocities is $3: 2$.Note that the information about the radii ratio was not needed for this calculation, as the cross-sectional area cancels out in the final expression for drift velocity when the voltage across the wire is constant.The final answer is $3: 2$.