An electron falls from rest through a vertical distance \(h\) in a uniform and vertically upward-directed electric field \(E\). The direction of the electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest through the same vertical distance \(h\). The fall time of the electron in comparison to the fall time of the proton is:

A toy car with charge \(q\) moves on a frictionless horizontal plane surface under the influence of a uniform electric field \(\vec {E}.\) Due to the force \(q\vec {E},\) its velocity increases from \(0\) to \(6~\text{m/s}\) in a one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between \(0\) to \(3\) seconds are respectively:
1. \(2~\text{m/s}, ~4~\text{m/s}\)
2. \(1~\text{m/s}, ~3~\text{m/s}\)
3. \(1~\text{m/s}, ~3.5~\text{m/s}\)
4. \(1.5~\text{m/s},~ 3~\text{m/s}\)

Suppose the charge of a proton and an electron differ slightly. One of them is \(-e\), the other is (\(e+\Delta e\)). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance \(d\) (much greater than atomic size) apart is zero, then$$ \(\Delta e\) is of the order of? (Given mass of hydrogen ${\mathrm{}}_{}$\(m_h = 1.67\times 10^{-27}~\text{kg}\)$$)
1. \(10^{-23}~\text{C}\)
2. \(10^{-37}~\text{C}\)
3. \(10^{-47}~\text{C}\)
4. \(10^{-20}~\text{C}\)

Two identical charged spheres suspended from a common point by two massless strings of lengths l are initially at a distance d(d < < l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then, v varies as a function of the distance x between the sphere, as:

(a) \(v \propto x\)

(b) \(v \propto x^{\frac{-1}{2}}\)

(c) \(v \propto x^{-1}\)

(d) \(v \propto x^{\frac{1}{2}}\)${\frac{\mathrm{}}{}}^{}$