Physics MCQs for NEET — Practice Questions with Answers

When three identical charged spheres each with charge q and radius R are kept such that they touch each other, the distance between the centers of any two spheres is 2R. The force between any two charges is given by Coulomb's law: $ F = extbackslash frac{1}{4 extbackslash pi extbackslash epsilon_0} extbackslash frac{q^2}{(2R)^2} $. Since there are two such forces acting at an angle of 120 degrees with respect to each other, the resultant force can be calculated using vector addition. The correct expression for this combined force is $ extbackslash frac{1}{4 extbackslash pi extbackslash epsilon_0} extbackslash frac{ extbackslash sqrt 3}{4} extbackslash left ( extbackslash frac{q}{R} extbackslash right ) ^2 $.

To determine the force between a point charge and a uniformly charged rod, we use the principle of superposition and Coulomb's law. Integrating the contributions of small charge elements along the length of the rod gives the expression for the force. The correct formula is $\frac{kqQ}{r(r+L)}$, where k is the Coulomb constant, q is the point charge, Q is the total charge on the rod, and r and L are the given distances. Hence, the correct option is $\frac{kqQ}{r(r+L)}$.

$ { k . q_1 \over (x + 0.08 ) ^2 } = { kq_2 \over x^2 } $ then find x

EP = [k(4 × 10–6)/(PR)2] = [(9 × 109 × 4 × 10–6)/(4 × 10–4)] = 9 ×107 N/C EQ = [k(2 × 10–6)/(QR)2] = [(9 × 109 × 2 × 10–6)/(10–4)] = 18 × 107 N/C in Δ PQR, cos θ = [(10–2)/(2 × 10–2)] = (1/2) i.e. θ = 60° E = √(EP2 + EQ2 + 2EPEQ cos 60) = √[(9 × 107)2 + (18 × 107)2 + {2 × 9 × 107 × 18 × 107 × (1/2)}] = √(81 × 1014 + 324 × 1014 + 162 × 1014) = √(547 × 1014) = 2.38 × 108 N/C tan α = [(EQ sin θ)/(EP + EQ cos θ)] = [(18 × 107 sin 60)/{(9 × 107 + 18 × 107 × cos 60)] = [{(18 × 107 × (√3/2)}/(9 × 107 + 9 × 107)] tan α = (√3/2) α = 40.89°

When the rod makes a small angle with the field, it undergoes simple harmonic motion. The formula for the time period of such motion is $T = 2\pi \sqrt{\frac{I}{qEl}}$, where I is the moment of inertia. For small angles, the time to align is $t = \frac{T}{4} = \frac{\pi}{2} \sqrt{\frac{ml}{2qE}}$.

After connection $ V_1 = V_2 , { KQ_1 \over r_1 } = { KQ_2 \over r_2 } \Rightarrow {Q_1 \over Q_2 } = { r_2 \over r_1 } $

The ratio of electric fields $ {E_1 \over E_2 } = { { KQ_1 \over r_1^2} \over {KQ_2 \over r_2^2 }} = { Q_1 \over r_1^2 } \times {r_2^2 \over Q_2 } $ Now calculate