If the distance between two charges is doubled, the electrostatic force between them will become:
A. One-fourth
B. Half
C. Four times
D. Double
Answer: Option A
Solution (By JKSSB Mock Tests)
According to Coulomb's Law, the electrostatic force (F) between two point charges is inversely proportional to the square of the distance (r) between them (F ∝ 1/r²). If the distance is doubled (r becomes 2r), the new force F' ∝ 1/(2r)² = 1/4r². Thus, the force becomes one-fourth of its original value.
The velocity-time graph of a particle moving in a straight line is shown. The displacement of the particle from t=0 to t=4s is: [Graph: triangle from (0,0) to (2,10) to (4,0)]
Explanation:
Displacement = area under velocity-time graph. The graph forms a triangle with base 4 s and height 10 m/s. Area = ½ × base × height = ½ × 4 × 10 = 20 m. Since velocity is always positive, displacement equals distance travelled. Graphical analysis is powerful: area gives displacement, slope gives acceleration. This triangle represents motion with uniform acceleration followed by uniform retardation. Exam tip: For piecewise linear v-t graphs, calculate area of each geometric segment separately. Such graph-based questions assess conceptual clarity in motion analysis.
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