Practice Questions

Q14.Three charges +𝑄, 𝑞, + 𝑄 are placed respectively, at distance, 0, 𝑑/ 2 and 𝑑 from the origin, on the 𝑥 -axis. If the net force experienced by +𝑄, placed at 𝑥= 0, is zero, then value of 𝑞 is: (1) +𝑄/ 2 (2) +𝑄/ 4 (3) -𝑄/ 4 (4) -𝑄/ 2

Q14.One mole of an ideal gas passes through a process where pressure and volume obey the relation 1 Vo 2 P = Po1 - . Here Po and Vo are constants. Calculate the change in the temperature of the gas if its volume 2 V changes from Vo to 2Vo . 1 P0V0 5 P0V0 (1) (2) 4 R 4 R 1 P0V0 3 P0V0 (3) (4) 2 R 4 R

Q14.A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is: (Take ∈0= 8.85 × 10−12 N-m2C 2 ) (1) 8.85 x 10−10C (2) 6.85 x 10−10C (3) 9.85 x 10−10C (4) 7.85 x 10−10C

Q14.A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency ω. If the radius of the bottle is 2.5 cm then ω is close to: ( density of water = 103kg/m3) (1) 5.00 rad sec−1 (2) 2.50 rad sec−1 (3) 7.9 rad sec−1 (4) 3.75 rad sec−1

Q14.Charge is distributed within a sphere of radius R with a volume charge density ρ(r) = A e−2ra , where A and a r2 are constants. If Q is the total charge of this charge distribution, the radius R is: (1) a 1 (2) 1 a Q Q 2 log( 1− 2πaA ) log( 1− 2πaA ) − 2πaA − 2πaA (3) a log(1 Q ) (4) a2 log(1 Q ) q2(−25 μC) are placed on the x -axis at x = 1 m and x = 4 m

Q14.A simple pendulum of length 1 m is oscillating with an angular frequency 10rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1rad/s and an amplitude of 10−2 m . The relative change in the angular frequency of the pendulum is best given by : (1) 10−3rad/s (2) 1rad/s (3) 10−1rad/s (4) 10−5rad/s

Q15.A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V . If the shell is now given a charge of – 4Q, the new potential difference between the same two surfaces is: (1) – 2V (2) 2 V (3) V (4) 4 V

Q15.An electric field of 1000 V/m is applied to an electric dipole at angle of 45∘ . The value of electric dipole moment is 10−29C ⋅m. What is the potential energy of the electric dipole? (1) −20 × 10−18 J (2) −7 × 10−27 J (3) −10 × 10−29 J (4) −9 × 10−20 J

Q15.A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10−8 Ω m ) of radius of cross-section 5 mm . Find the mobility of the charges if their drift velocity is 1.1 × 10−3 ms−1 . (1) 1.5 m2 V−1 s−1 (2) 1.8 m2 V−1 s−1 (3) 1.0 m2 V−1 s−1 (4) 1.3 m2 V−1 s−1

Q15.A charge Q is distributed over three concentric spherical shells of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be: (1) Q (2) Q (a2+b2+c2) 4πϵ0(a+b+c) 4πϵ0(a3+b3+c3) (3) Q ab+bc+ca (4) Q(a+b+c) 12πϵ0 abc 4πϵ0(a2+b2+c2)

Q15.A Point dipole →p = - p0^𝑥 at a distance d are, respectively: (Taken V = 0 at infinity) | (1) |→p | -→p (2) |→p 4πε0d2, 4πε0d3 0, 4πε0d3 -→p (3) |→p | →p (4) 4πε0d3 4πε0d3, 4πε0d3 0,

Q15.Three charges Q, +y and +q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is (1) +q (2) −√2q √2+1 (3) −q (4) −2q 1+√2

Q15.In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 μF. All values in the circuit are in μF. JEE Main 2019 (12 Jan Shift 2) JEE Main Previous Year Paper (1) 7 μF (2) 4μF 10 (3) 7 μF (4) 6 μF 11 5

Q15.When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume its temperature increases by ∆T . The heat required to produce the same change in temperature, at a constant pressure is: (1) 3 (2) 7 2Q 5Q 5 2 (3) 3Q (4) 3Q

Q15.Two point charges q1(√10 μC) and respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, [Take 4πϵ01 = 9 × 109 N m2C−2] + 81 −81 × 102 (1) (−81 ˆi ˆj) × 102 (2) (81 ˆi ˆj) + 27 × 102 (3) (−63 ˆi ˆj) × 102 (4) (63ˆi −27ˆj)

Q15.Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener is moving from source S1 towards S2 with a constant speed u0 m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u0 equals: (1) 10.0 m/s (2) 2.5 m/s (3) 5.5 m/s (4) 15.0 m/s

Q15.For a uniformly charged ring of radius 𝑅, the electric field on its axis has the largest magnitude at a distance ℎ from its centre. Then value of ℎ is: (1) 𝑅 (2) 𝑅 √2 √5 (3) 𝑅√2 (4) 𝑅

Q15.Four point charges −q, + q, + q and −q are placed on y-axis at y = −2d, y = −d, and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D ≫d, will behave JEE Main 2019 (09 Apr Shift 2) JEE Main Previous Year Paper as: (1) E ∝ 1 (2) E ∝ D1 D4 (3) E ∝ 1 (4) E ∝ 1 D3 D2

Q15.The figure shows a capacitor of capacitance C connected to a battery via a switch, having a total charge Q on it, in steady-state. When the switch S is turned from position A to position B , the energy dissipated in the circuit is (1) 1 Q2 (2) 3 Q2 8 C 8 C (3) 3 Q2 (4) 5 Q2 4 C 8 C JEE Main 2019 (12 Jan Shift 1) JEE Main Previous Year Paper

Q15.The pressure wave, P = 0.01 sin[1000t −3x]N m–2 , corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0°C . On some other day when temperature is T , the speed of sound produced by the same blade and at the same frequency is found to be 336 m s–1 . Approximate value of T is: (1) 12° C (2) 15° C (3) 4° C (4) 11° C

Q16.A system of three charges are placed as shown in the figure: If D >> d, the potential energy of the system is best given by: (1) + q D2 4πϵ0 1 [−q2d −q D2Q d ] (2) 4πϵ01 [+ q2d Q d ] 2q (3) + D2 4πϵ0 1 [−q2d −q2D2Q d ] (4) 4πϵ01 [−q2d Q d ]

Q16.Voltage rating of a parallel plate capacitor is 500 V . Its dielectric can withstand a maximum electric field of 106V / m . The plate area is 10-4 m2 . What is the dielectric constant if the capacitance is 15 pF ? given ϵ0 = 8.86 × 10-12C2 / Nm2 (1) 3.8 (2) 8.5 (3) 4.5 (4) 6.2

Q16.A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be: (1) K = ( K1+K2K1K2 + K3+K4K3.K4 ) (2) K = 2(K1+K2+K3+K4)(K1+K2)(K3+K4) (3) K = (K1+K4)(K2+K3) (4) K = (K1+K2)(K3+K4) 2(K1+K2+K3+K4) K1+K2+K3+K4

Q16.The galvanometer deflection, when key K1 is closed but K2 is open, equals θ0 (see figure). On closing K2 also and adjusting R2 to 5Ω, the deflection in galvanometer becomes θ05 . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]: (1) 12 Ω (2) 22 Ω (3) 5 Ω (4) 25 Ω

Q16.Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d . The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K2 and K3 . The first capaciitor is filled as shown in figure I, and the second one is filled as shown in figure II . If these two modified capacitors are charged by the same potential V , the ratio of the energy stored in the two, would be ( E1 refers to capacitor I and E2 to capacitor ( II ) ): E1 9K1K2K3 ( K1K2K3 ) ( K2K3 + K3K1K1 + K1K2 ) (1) (2) E1 = = E2 ( K1K2K3 ) ( K2K3 + K3K1K1 + K1K2 ) E2 K1K2K3 K1K2K3 ( K1K2K3 ) ( K2K3 (3) E1 + K3K1K1 + K1K2 ) (4) E1 = = E2 ( K1K2K3 ) ( K2K3 + K3K1K1 + K1K2 ) E2 9K1K2K3