Free Objective Test 01 Practice Test - 11th and 12th

Question 1

Three particles, each of charge $10 μ C$ are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is
Given $(\frac{1}{4 π ϵ} = 9 \times 10^{9} N - m^{2} / C^{2})$

A. Zero

B. Infinite

C. 27 J

D. 100 J

SOLUTION

Solution : C

For pair of charge $U = \frac{1}{4 π ϵ_{0}} [\frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}} + \frac{10 \times 10^{- 6} \times 10 \times 10^{- 6}}{\frac{10}{100}}]$
$= 3 \times 9 \times 10^{9} \times \frac{100 \times 10^{- 12} \times 100}{10} = 27 J$

Question 2

There is an electric field E along X-direction. If the work done on moving a charge 0.2 C through a distance of 2m along a line making an angle $60^{\circ}$ with the X-axis is 4.0, what is the value of E

\sqrt{3} N / C

B. 4 N/C

C. 5 N/C

D. None of these

SOLUTION

Solution : D

$W = q V = q E . d$

$\Rightarrow 4 = 0.2 \times E \times (2 c o s 60^{\circ})$
$= 0.2 E \times (2 \times 0.5)$
$∴ E = \frac{4}{0.2} = 20 N C^{- 1}$

Question 3

Four equal charges Q are placed at the four corners of a square of each side 'd'. Work done in removing a charge – Q from its centre to infinity is

A. 0

\frac{\sqrt{2} Q^{2}}{4 π ϵ_{0} a}

\frac{\sqrt{2} Q^{2}}{π ϵ_{0} a}

\frac{Q^{2}}{2 π ϵ_{0} a}

SOLUTION

Solution : C

Potential at centre O of the square

$V_{0} = 4 (\frac{Q}{4 π ϵ_{0} (a / \sqrt{2})})$
Work done in shifting (– Q) charge from centre to infinity
$W = - Q (V_{\infty} - V_{0}) = Q V_{0} = \frac{4 \sqrt{2} Q^{2}}{4 π ϵ_{0} a} = \frac{\sqrt{2} Q^{2}}{π ϵ_{0} a}$

Question 4

Four identical charges $+ 50 μ C$ each are placed, one at each corner of a square of side 2m. How much external energy is required to bring another charge of $+ 50 μ C$ from infinity to the centre of the square $(G i v e n \frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} \frac{N m^{2}}{C^{2}})$

A. 64 J

B. 41 J

C. 16 J

D. 10 J

SOLUTION

Solution : A

Potential at the centre of square
$V = 4 \times (\frac{9 \times 10^{9} \times 50 \times 10^{- 6}}{\sqrt{2}}) = 90 \sqrt{2} \times 10^{4} V$
Work done in bringing a charge (q = $50 μ C$ ) from $\infty$ to centre (O) of the square is
$W = q (V_{0} - V_{\infty}) = q V_{0}$
$\Rightarrow W = 50 \times 10^{- 6} \times 90 \sqrt{2} \times 10^{4} = 64 J$

Question 5

Two insulated charged conducting spheres of radii 20 cm and 15 cm respectively and having an equal charge of 10 C are connected by a copper wire and then they are separated. Then

A. Both the spheres will have the same charge of 10 C

B. Surface charge density on the 20 cm sphere will be greater than that on the 15 cm sphere

C. Surface charge density on the 15 cm sphere will be greater than that on the 20 cm sphere

D. Surface charge density on the two spheres will be equal

SOLUTION

Solution : C

After redistribution, charges on them will be different, but they will acquire common potential
i.e. $k \frac{Q_{1}}{r_{1}} = k \frac{Q_{2}}{r_{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{r_{1}}{r_{2}}$
As $σ = \frac{Q}{4 π r^{2}} \Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{Q_{1}}{Q_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}} \Rightarrow σ \propto \frac{1}{r}$
i.e. surface charge density on smaller sphere will be more.

Question 6

An electron and a proton are at a distance of $1 ˙ A$ . The dipole moment will be $(C \times m)$

1.6 \times 10^{19}

1.6 \times 10^{- 29}

3.2 \times 10^{19}

3.2 \times 10^{29}

SOLUTION

Solution : B

$p = q \times (d) = 1.6 \times 10^{- 19} \times 10^{- 10} = 1.6 \times 10^{- 29} C - m$

Question 7

Two charges $+ 3.2 \times 10^{- 19}$ and $- 3.2 \times 10^{- 19}$ C placed $2.4 ˙ A$ apart form an electric dipole. It is placed in a uniform electric field of intensity $4 \times 10^{5}$ volt/m. The electric dipole moment is

15.36 \times 10^{- 29} C o u l o m b \times m

15.36 \times 10^{- 19} C o u l o m b \times m

7.68 \times 10^{- 29} C o u l o m b \times m

7.68 \times 10^{- 19} C o u l o m b \times m

SOLUTION

Solution : C

Dipole moment p = q (d)
$= 3.2 \times 10^{- 19} \times (2.4 \times 10^{- 10}) = 7.68 \times 10^{- 29} C - m$

Question 8

Point charges q, q, -2q are placed at the corners of an equilateral triangle ABC of side l. The magnitude of net electric dipole moment of the system is

q l

2 q l

\sqrt{3} q l

4 q l

SOLUTION

Solution : C

$P_{n e t} = \sqrt{p^{2} + p^{2} + 2 p p c o s 60^{\circ}} = \sqrt{3} p = \sqrt{3} q l$ $(∵ p = q l)$

Question 9

If an electron starts from rest from a point at which potential is 50 volt to another point at which potential is 70 volt, then its kinetic energy in the final state will be

3.2 \times 10^{- 10} J

3.2 \times 10^{- 18} J

1 J

1 d y n e

SOLUTION

Solution : B

$K . E . = q_{0} (V_{A} - V_{B}) = 1.6 \times 10^{- 19} (70 - 50) = 3.2 \times 10^{- 18} J$

Question 10

In the following diagram the work done in moving a point charge from point P to point A, P to B and P to C is respectively $W_{A}$ , $W_{B}$ and $W_{C}$ , then

W_{A} = W_{B} = W_{C}

W_{A} = W_{B} = W_{C} = 0

W_{A} > W_{B} > W_{C}

W_{A} < W_{B} < W_{C}

SOLUTION

Solution : B

According to the figure, there is no other charge. A single charge when moved in a space of no field, does not experience any force hence no work is done. $W_{A} = W_{B} = W_{C} = 0$

Question 11

Two point charges each of +1 $μ$ C are at a distance of 3 m apart. The work done in bringing the two charges closer to a distance 1 m apart is

A. 0.006 J

B. 0.008 J

C. 0.0006 J

D. 0.06 J

SOLUTION

Solution : A

$W = U_{2} - U_{1} = \frac{1}{4 π ϵ_{0}} q_{1} q_{2} [\frac{1}{r_{2}} - \frac{1}{r_{1}}] = 9 \times 10^{9} \times 10^{- 12} [\frac{1}{1} - \frac{1}{3}] = 9 \times \frac{2}{3} \times 10^{- 3} W = 0.006 J$

Question 12

Two Charges $Q_{1}$ and $Q_{2}$ are as shown in the figure. A third charge $Q_{3}$ is moved from points A to B along a circular path. Change in potential energy of the system is

k Q_{1} Q_{2} Q_{3}

4 k Q_{1} Q_{2}

4 k Q_{2} Q_{3}

\frac{2}{3} k Q_{2} Q_{3}

SOLUTION

Solution : C

$Δ U = U_{f} - U_{i} U_{f} = k [\frac{Q_{1} Q_{2}}{0.6} + \frac{Q_{2} Q_{3}}{0.2} + \frac{Q_{3} Q_{1}}{0.8}] Δ U = U_{f} - U_{i} = k Q_{2} Q_{3} [\frac{1}{0.2} - 1] = 4 k Q_{2} Q_{3}$

Question 13

A body of mass 4 gm carrying a charge 1 $μ$ C starts from rest at the positive plate and reaches the negative plate of a parallel plate condenser with final velocity of 0.3 m/s. The potential difference between the plates is?

A. 180 V

B. 360 V

C. 90 V

D. 18 V

SOLUTION

Solution : A

$\frac{1}{2} m v^{2} = q V$
$\frac{1}{2} \times 4 \times 10^{- 3} \times 9 \times 10^{- 2} = 10^{- 6} V$
$V = 180 V$

Question 14

A point charge q is placed inside a conducting spherical shell of inner radius 2R and outer radius 3R at a distance of R from the centre of the shell. The electric potential at the centre of shell will be $\frac{1}{4 π ϵ_{0}}$ times:

\frac{q}{2 R}

\frac{4 q}{3 R}

\frac{5 q}{6 R}

\frac{2 q}{3 R}

SOLUTION

Solution : C

Charge of -q will be induced on the the inner surface and +q on the outer surface
$V = V_{1} + V_{2} + V_{3} = \frac{1}{4 π ϵ_{0}} [\frac{q}{R} - \frac{q}{2 R} + \frac{q}{3 R}]$
$V = \frac{1}{4 π ϵ_{0}} [\frac{6 q - 3 q + 2 q}{6 R}]$
$V = \frac{1}{4 π ϵ_{0}} [\frac{5 q}{6 R}]$

Question 15

Potential at the centroid of an equilateral triangle of side ‘a’, when equal charges each of magnitude ‘Q’ are placed at the vertices of the triangle is?

Z e r o

\frac{1}{4 π ϵ_{0}} \frac{3 Q}{a}

\frac{1}{4 π ϵ_{0}} \frac{3 \sqrt{3} Q}{a}

\frac{1}{4 π ϵ_{0}} \frac{Q}{\sqrt{3} a}

SOLUTION

Solution : C

$V^{1} = V_{1} + V_{2} + V_{3}; V_{1} = V_{2} = V_{3}$
$= 3 V_{1}$
$= 3 \frac{1}{4 π ϵ_{0}} \frac{Q}{\frac{a}{\sqrt{3}}}$
$= \frac{1}{4 π ϵ_{0}} \frac{3 \sqrt{3 Q}}{a}$

Free Objective Test 01 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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