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NEET NEET Questions — Free MCQ Practice

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A point charge Q is placed at a distance d from the centre of an uncharged conducting sphere of radius R. The potential of the sphere is (d > R) –

A.

$\frac{1}{4 π \in_{0}} . \frac{Q}{(d - R)}$

B.

$\frac{1}{4 π \in_{0}} . \frac{Q}{d}$

C.

$\frac{1}{4 π \in_{0}} . \frac{Q}{R}$

D.

zero

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The plates of a parallel plate capacitor are separated by d cm. A plate of thickness t cm with dielectric constant $k_{1}$ is inserted and the remaining space is field with a plate of dielectric constant $k_{2}$ . If Q is the charge on the capacitor and area of plates is $A {cm}^{2}$ each, then potential difference between the plates is –

A.

$\frac{Q}{ε_{0} A} (\frac{t}{k_{1}} + \frac{d - t}{k_{2}})$

B.

$\frac{4 πQ}{A} (\frac{t}{k_{1}} + \frac{d - t}{k_{2}})$

C.

$\frac{4 πQ}{A} (\frac{k_{1}}{t} + \frac{k_{2}}{d - t})$

D.

$\frac{Q}{ε_{0} A} (\frac{k_{1}}{t} + \frac{d - t}{k_{2}})$

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Explanation

Potential difference across the plates,

$v = v_{1} + v_{2} = E_{1} t_{1} + E_{2} t_{2} = \frac{σ t}{K_{1} \in_{0}} + \frac{σ}{K_{2} \in_{0}} (d - t) = \frac{Q}{A \in_{0}} [\frac{t}{K_{1}} + \frac{d - t}{K_{2}}]$

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When a hydrogen atom is excited from ground state to first excited state then the incorrect option is–

A.

its kinetic energy increases by $10.2 eV$

B.

its kinetic energy decreases by $10.2 eV$

C.

its potential energy increases by $20.4 eV$

D.

its angular momentum increases by $1.05 \times 10^{- 34} J$ -s

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Explanation

$In ground state n = 1 and for first exited state n = 2 KE = \frac{1}{4 π \in_{0}} \frac{e^{2}}{2 r} (z = 1) = \frac{14.4 \times 10^{- 10}}{2 r} eV (∵ r = 0.53 n^{2} \overset{\circ}{A}, (z = 1)) {(KE)}_{1} = \frac{14.4 \times 10^{- 10}}{2 \times 0.53 \times 10^{- 10}} eV = 13.58 eV and {(KE)}_{2} = \frac{14.4 \times 10^{- 10}}{2 \times 0.53 \times 10^{- 10} \times 4} eV = 3.39 eV ∴ KE decreases by = 10.2 eV ∴ PE increases by = Excitation energy + Loss in kinetic energy = 10.2 + 10.2 = 20.4 eV Now angular momentum, L = mvr = \frac{nh}{2 π} \Rightarrow L_{2} - L_{1} = \frac{h}{2 π} = \frac{6.6 \times 10^{- 34}}{6.28} = 1.05 \times 10^{- 34} J ‐ \sec$

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A conducting disc of radius R is rotating about its axis with an angular velocity $ω$ . Then the potential difference between the centre of the disc and its edge is (no magnetic field is present)

A.

zero

B.

$\frac{m_{e} ω^{2} R^{2}}{2 e}$

C.

$\frac{m_{e} {ωR}^{3}}{3 e}$

D.

$\frac{{em}_{e} {ωR}^{2}}{2}$

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Explanation

When a conducting disc of radius R rotates about its axis with an angular velocity Ï‰, the electrons in the disc experience a centrifugal force. This leads to a potential difference between the center and the edge given by (meÏ‰^2R^2)/(2e), where me is the mass of an electron and e is the charge of an electron. This is due to the work done in moving electrons against the centrifugal force.

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Electrical potential ‘v’ in space as a function of coordinates is given by, $v = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ . Then the electric field intensity at (1, 1, 1) is given by –

A.

$- (\hat{i} + \hat{j} + \hat{k})$

B.

$\hat{i} + \hat{j} + \hat{k}$

C.

zero

D.

$\frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$

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Explanation

$2 . \vec{E} = - \nabla V = + \frac{1}{x^{2}} \hat{i} + \frac{1}{y^{2}} \hat{j} + \frac{1}{z^{2}} \hat{k} = \hat{i} + \hat{j} + \hat{k}$

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Two concentric, thin metallic spheres of radii $R_{1}$ and $R_{2}$ $(R_{1} > R_{2})$ bear changes $Q_{1}$ and $Q_{2}$ respectively. Then the potential at distance r between $R_{1}$ and $R_{2}$ will be $(k = \frac{1}{4 {πε}_{0}})$

A.

$k (\frac{Q_{1} + Q_{2}}{r})$

B.

$k (\frac{Q_{1}}{r} + \frac{Q_{2}}{R_{2}})$

C.

$k (\frac{Q_{2}}{r} + \frac{Q_{1}}{R_{1}})$

D.

$k (\frac{Q_{1}}{R_{1}} + \frac{Q_{1}}{R_{2}})$

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A parallel plate capacitor with air between the plates is charged to a potential difference of 500V and then insulated. A plastic plate is inserted between the plates filling the whole gap. The potential difference between the plates now becomes 75V. The dielectric constant of plastic is –

A.

10/3

B.

5

C.

20/3

D.

10

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Explanation

$3 . V = \frac{q}{C} \Rightarrow \frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}} \Rightarrow \frac{C_{2}}{C_{1}} = \frac{500}{75} = \frac{20}{3} By definition, C_{2} = {kC}_{1} \Rightarrow k = \frac{20}{3}$

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A capacitor of $1 µ F$ withstands a maximum voltage of 6 kilovolt while another capacitor of $2 µ F$ withstands a maximum voltage of 4 kilovolt. If the two capacitors are connected in series, the system will withstand a maximum voltage of –

A.

2 kV

B.

4 kV

C.

6 kV

D.

9 kV

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Explanation

4. For series combination $C_{s} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} \Rightarrow C_{s} = \frac{2}{3} μF$

When connected in series the maximum charge that can flow through the combination equals

the lower value of charge accommodated by the first capacitor i.e. $6000 µ C$

$∴ Q_{1} = 6000 μC and 8000 μC V_{s} = \frac{Q_{1}}{C_{s}} = \frac{6000 μC}{(2 / 3) μC} \Rightarrow V_{s} = 9 kV$

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The potential at a certain point in an electric field is 200 V. The work done in carrying an electron upto that point will be.

A.

$3.2 \times 10^{- 17} J$

B.

$- 3.2 \times 10^{- 17} J$

C.

$200 J$

D.

$- 200 J$

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Explanation

$1 . W = qV = - 1.6 \times 10^{- 19} \times 200 = - 3.2 \times 10^{- 17} J$

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The electric field lines due to a single negative charge are represented by

A.

B.

C.

D.

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Explanation

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