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

### Question 1

Three particles, each of charge 10 μC are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is

Given (14πϵ=9×109 N−m2/C2)

#### SOLUTION

Solution :C

For pair of charge U=14πϵ0[10×10−6×10×10−610100+10×10−6×10×10−610100+10×10−6×10×10−610100]

=3×9×109×100×10−12×10010=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∘ with the X-axis is 4.0, what is the value of E

#### SOLUTION

Solution :D

W=qV=qE.d

⇒4=0.2×E×(2 cos 60∘)

=0.2E×(2×0.5)

∴E=40.2=20 NC−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

#### SOLUTION

Solution :C

Potential at centre O of the square

V0=4(Q4πϵ0(a/√2))

Work done in shifting (– Q) charge from centre to infinity

W=−Q(V∞−V0)=QV0=4√2Q24πϵ0a=√2Q2πϵ0a

### Question 4

Four identical charges +50 μC 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 μC from infinity to the centre of the square (Given 14πϵ0=9×109 Nm2C2)

#### SOLUTION

Solution :A

Potential at the centre of square

V=4×(9×109×50×10−6√2)=90√2×104 V

Work done in bringing a charge (q = 50 μC) from ∞ to centre (O) of the square is

W=q(V0−V∞)=qV0

⇒W=50×10−6×90√2×104=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

#### SOLUTION

Solution :C

After redistribution, charges on them will be different, but they will acquire common potential

i.e. kQ1r1=kQ2r2⇒Q1Q2=r1r2

As σ=Q4πr2⇒σ1σ2=Q1Q2×r22r21⇒σ∝1r

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×m)

#### SOLUTION

Solution :B

p=q×(d)=1.6×10−19×10−10=1.6×10−29 C−m

### Question 7

Two charges +3.2×10−19 and −3.2×10−19 C placed 2.4˙A apart form an electric dipole. It is placed in a uniform electric field of intensity 4×105 volt/m. The electric dipole moment is

#### SOLUTION

Solution :C

Dipole moment p = q (d)

=3.2×10−19×(2.4×10−10)=7.68×10−29C−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

#### SOLUTION

Solution :C

Pnet=√p2+p2+2pp cos60∘=√3p=√3ql (∵p=ql)

### 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

#### SOLUTION

Solution :B

K.E.=q0(VA−VB)=1.6×10−19(70−50)=3.2×10−18J

### 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 WA, WB and WC , then

#### 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. WA=WB=WC=0

### Question 11

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

#### SOLUTION

Solution :A

W=U2−U1=14πϵ0q1q2[1r2−1r1]=9×109×10−12[11−13]=9×23×10−3W=0.006J

### Question 12

Two Charges Q1 and Q2 are as shown in the figure. A third charge Q3 is moved from points A to B along a circular path. Change in potential energy of the system is

#### SOLUTION

Solution :C

ΔU=Uf−Ui Uf=k[Q1Q20.6+Q2Q30.2+Q3Q10.8]ΔU=Uf−Ui=kQ2Q3[10.2−1]=4kQ2Q3

### Question 13

A body of mass 4 gm carrying a charge 1μ 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?

#### SOLUTION

Solution :A

12mv2=qV

12×4×10−3×9×10−2=10−6V

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 14πϵ0 times:

#### SOLUTION

Solution :C

Charge of -q will be induced on the the inner surface and +q on the outer surface

V=V1+V2+V3=14πϵ0[qR−q2R+q3R]

V=14πϵ0[6q−3q+2q6R]

V=14πϵ0[5q6R]

### 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?

#### SOLUTION

Solution :C

V1=V1+V2+V3;V1=V2=V3

=3V1

=314πϵ0Qa√3

=14πϵ03√3Qa