Drift velocity is typically very small due to the high number of free electrons.
Step 1: Use the formula for drift velocity - Drift velocity is given by: \[ V_d = \frac{I}{n e A}, \] where \(I\) is the current, \(n\) is the number of free electrons per unit volume, \(e\) is the electron charge, and \(A\) is the cross-sectional area.
Step 2: Substitute the values - \[ V_d = \frac{2}{(2 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (25 \times 10^{-6})}. \] Simplifying: \[ V_d = \frac{2}{800 \times 10^{-6}} = 25 \times 10^{-6} \, \text{m/s}. \]
Final Answer: The drift velocity of the electrons is 25 × 10⁻⁶ m/s.
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is:
The velocity (v) - time (t) plot of the motion of a body is shown below :

The acceleration (a) - time(t) graph that best suits this motion is :
A wheel of a bullock cart is rolling on a level road, as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively) ?



For the circuit shown above, the equivalent gate is:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) is: