Question 1 of 10 Divergence of gold leaf is used to study A charge B potential C energy D none 💡 Explanation In a gold-leaf electroscope, when charge is transferred to the instrument, the gold leaves repel each other (both acquire the same sign of charge) and diverge. The extent of divergence indicates the amount of charge present. Question 2 of 10 When a wooden sweater worn over a nylon shirt is removed, sparking is observed due to A static electricity B current electricity C none D both 💡 Explanation Friction between the sweater and nylon shirt causes charge transfer (triboelectric effect), building up static charge. When separated, this charge discharges suddenly through the air, producing visible sparks — a static electricity phenomenon. Question 3 of 10 A charge q placed at the centre of the line joining two equal charges Q. The system of three charges will be in equilibrium if q is equal to A [latex]-\dfrac{Q}{2}[/latex] B [latex]-\dfrac{Q}{4}[/latex] C [latex]-Q[/latex] D [latex]\dfrac{Q}{2}[/latex] 💡 Explanation Let the two charges Q be at [latex]x=-d[/latex] and [latex]x=+d[/latex], with q at the origin. By symmetry, q always experiences zero net force. For the charge Q at [latex]x=d[/latex] to also be in equilibrium, the repulsive force from the other Q must balance the force from q: [latex]\frac{kQ^2}{(2d)^2} = \frac{kQ|q|}{d^2}[/latex] [latex]\frac{Q}{4} = |q| \implies q = -\frac{Q}{4}[/latex] (negative, so the force is attractive) Question 4 of 10 The distance between two charges is increased by 25% then the percentage change in force is A 36% B 56% C 65% D 50% 💡 Explanation New distance [latex]d' = 1.25d[/latex] [latex]F' = \frac{kQ_1Q_2}{(1.25d)^2} = \frac{F}{1.5625} = 0.64F[/latex] Percentage change [latex]= \frac{F-F'}{F}\times100% = (1-0.64)\times100% = 36%[/latex] decrease Question 5 of 10 Coulomb force between two charges kept at certain distance is F. If half of separation between them is filled with dielectric of constant [latex]\epsilon_r = 4[/latex], then the new coulomb's force is A F B [latex]\dfrac{3F}{8}[/latex] C [latex]\dfrac{4F}{9}[/latex] D [latex]\dfrac{2F}{9}[/latex] 💡 Explanation When a slab of thickness [latex]t[/latex] and dielectric constant [latex]\epsilon_r[/latex] partially fills the gap [latex]d[/latex], the effective separation becomes [latex]d_{eff} = d + t(\sqrt{\epsilon_r}-1)[/latex]. Here [latex]t = d/2[/latex], [latex]\sqrt{\epsilon_r}=2[/latex]: [latex]d_{eff} = d + \frac{d}{2}(2-1) = \frac{3d}{2}[/latex] [latex]F' = \frac{kQ_1Q_2}{d_{eff}^2} = \frac{kQ_1Q_2}{(1.5d)^2} = \frac{F}{2.25} = \frac{4F}{9}[/latex] Question 6 of 10 A polythene piece, rubbed with wool is found to have negative charge of [latex]4\times10^{-7}[/latex] C. The number of electrons transferred from wool to polythene is A [latex]15\times10^{12}[/latex] B [latex]2.5\times10^{12}[/latex] C [latex]2.5\times10^{13}[/latex] D [latex]3.5\times10^{13}[/latex] 💡 Explanation [latex]n = \dfrac{q}{e} = \dfrac{4\times10^{-7}}{1.6\times10^{-19}} = 2.5\times10^{12}[/latex] Question 7 of 10 Three charges each of [latex]+1\mu C[/latex] are placed at the corners of an equilateral triangle. If the force between any two charges be F, then, the net force on either charge will be A [latex]\sqrt{2}F[/latex] B [latex]\sqrt{3}F[/latex] C 2F D 3F 💡 Explanation Each charge experiences two forces of equal magnitude F (from the other two charges), directed along the two sides of the triangle meeting at that corner, with an angle of 60° between them: [latex]F_{net} = \sqrt{F^2+F^2+2F^2\cos60^\circ} = \sqrt{2F^2+F^2} = \sqrt{3},F[/latex] Question 8 of 10 Coulomb's law is analogous to A Charge conservation law B Newton's second law of motion C Law of conservation of energy D Newton's law of gravitation 💡 Explanation Both laws follow an inverse-square relationship: [latex]F \propto \dfrac{1}{r^2}[/latex]. Coulomb's law describes the force between charges ([latex]F=k\dfrac{q_1q_2}{r^2}[/latex]), while Newton's law of gravitation describes the force between masses ([latex]F=G\dfrac{m_1m_2}{r^2}[/latex]) — both are central, inverse-square force laws. Question 9 of 10 Two-point charges [latex]Q_1[/latex] and [latex]Q_2[/latex] exert a force F on each other when kept certain distance apart. If the charge on each particle is halved and the distance between the two particles is doubled, then the new force between the two particles would be A [latex]\dfrac{F}{2}[/latex] B [latex]\dfrac{F}{4}[/latex] C [latex]\dfrac{F}{8}[/latex] D [latex]\dfrac{F}{16}[/latex] 💡 Explanation [latex]F' = \frac{k\left(\dfrac{Q_1}{2}\right)\left(\dfrac{Q_2}{2}\right)}{(2d)^2} = \frac{kQ_1Q_2/4}{4d^2} = \frac{1}{16}\cdot\frac{kQ_1Q_2}{d^2} = \frac{F}{16}[/latex] Question 10 of 10 Two charges are placed at a certain distance apart in air. If the glass slab is introduced between them, the force between them will A Increase B decrease C Remain same D be zero 💡 Explanation Glass has a dielectric constant [latex]\epsilon_r > 1[/latex]. Introducing it between the charges reduces the electric force by a factor of [latex]\epsilon_r[/latex] compared to air/vacuum: [latex]F_{medium} = \dfrac{F_{air}}{\epsilon_r}[/latex], so the force decreases. ← Prev Check Next → Finish & See Score 0 of 10 checked ← Back 1 / 10 1 2 3 4 5 6 7 8 9 10
