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NEET NEET Questions — Free MCQ Practice

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Point charges 4 c and 2 c are placed at the vertices P and Q of a right angle triangle PQR respectively. Q is the right angle,$PR=2 \times10^{–2}m $ and $QR =10^{–2}m$ . The magnitude and direction of the resultant electric field at c is .........

A.

$ 4.28 \times 10^9 NC^{–1} , 45 $

B.

$ 2.38 \times 10^8 NC^{–1} , 40.9 $

C.

$ 1.73 \times 10^4 NC^{-1}, 34.7 $

D.

$ 4.9 \times 10^{10} NC^{–1} , 34.7 $

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Explanation

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°

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A small sphere whose mass is 0.1 gm carries a charge of $ 3 \times 10^{–10}C $ and is tieup to one end of a silk fibre 5 cm long. The other end of the fibre is attached to a large vertical conducting plate which has a surface charge of $ 25 \times 10^{–6}Cm^{–2}$ , on each side. When system is freely hanging the angle fibre makes with vertical is

A.

$ 41.8 ^\circ $

B.

$ 45 ^\circ $

C.

$ 40.2 ^\circ $

D.

$ 45.8 ^\circ $

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A Semicircular rod is charged uniformly with a total charge Q coulomb. The electric field intensity at the centre of curvature is

A. $ 2KQ / \pi r^2 $ B. $ 3KQ / \pi r^2 $ C. $ KQ / \pi r^2 $ D. $ 4KQ / \pi r^2 $

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Two point masses m each carrying charge â€“q and +q are attached to the ends of a massless rigid non-conducting rod of length l. The arrangement is placed in a uniform electric field E such that the rod makes a small angle 50 with the field direction. The minimum time needed by the rod to align itself along the field is ........

A. $ t = \pi \sqrt { 2ml \over 3qE } $ B. $ t = {\pi \over 2 } \sqrt { ml \over 2qE } $ C. $ t = \sqrt { ml \over qE } $ D. $ t = 2 \pi \sqrt { ml \over E } $

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Explanation

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 $t = \frac{T}{4} = \frac{\pi}{2} \sqrt{\frac{ml}{2qE}}$.

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Two uniformaly charged spherical conductors A and B having radius 1mm and 2mm are separated by a distance of 5 cm. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is .........

A. 4:1 B. 1:2 C. 2:1 D. 1:4

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Explanation

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

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Let $ P(r) = {Q \over \pi R ^4} r $ be the charge density distribution for a solid sphere of radius R and total charge Q. For a point â€˜Pâ€™ inside the sphere at distance r1 from the centre of the sphere the magnitude of electric field is

A. $ {Q \over 4 \pi \varepsilon_0 r_1 ^2} $ B. $ {Q r_1^2 \over 4 \pi \varepsilon_0 R^4} $ C. $ {Q r_1^2 \over 3 \pi \varepsilon_0 R^4} $ D. 0

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A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is

A. $ T = 2 \pi \left[ {l \over g - {qE \over m} }\right ] ^ {1 /2 } $ B. $ T = 2 \pi \left( {l \over g }\right )^ {1 /2 } $ C. $ T = 2 \pi \left[ {l \over g + {qE \over m} }\right ] ^ {1 /2 } $ D. $ T = 2 \pi \left[ { ml \over qE } \right ] ^ {1 /2 } $

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In Millikanâ€™s oil drop experiment an oil drop carrying a charge Q is held stationary by a p.d. 2400 v between the plates. To keep a drop of half the radius stationary the potential differ- ence had to be made 600 v. What is the charge on the second drop ?

A. 3Q /2 B. Q / 4 C. Q D. Q/2

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An electric dipole is placed along the x-axis at the origin o. A point P is at a distance of 20 cm from this origin such that OP makes an angle $ \pi / 3$ with the x-axis. If the electric field at P makes an angle $ \theta $ with the x-axis, the value of $ \theta $ would be ...........

A. $ { \pi \over 3 } + tan^{-1} ({ \sqrt 3 \over 2}) $ B. $ { \pi \over 3 } $ C. $ { 2 \pi \over 3 } $ D. $ tan ^{-1} ({ \sqrt 3 \over 2}) $

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A particle having a charge of $ 1.6 \times 10^{-19} C $ enters between the plates of a parallel plate capaciter. The initial velocity of the particle is parallel to the plates. A potential difference of 300v is applied to the capacitor plates. If the length of the capacitor plates is 10cm and they are separated by 2cm, Calculate the greatest initial velocity for which the particle will not be able to come out of the plates. The mass of the particle is $ 12 \times 10^{â€“24} kg $ .

A. $ 10^ 4 m/s $ B. $ 10^ 2 m/s $ C. $ 10^ {-1} m/s $ D. $ 10^ 3 m/s $

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