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

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 Nm2/C2)

A. Zero
B. Infinite
C. 27 J
D. 100 J

#### SOLUTION

Solution : C

For pair of charge U=14πϵ0[10×106×10×10610100+10×106×10×10610100+10×106×10×10610100]
=3×9×109×100×1012×10010=27 J

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

A. 3 N/C
B. 4 N/C
C. 5 N/C
D. None of these

#### SOLUTION

Solution : D

W=qV=qE.d 4=0.2×E×(2 cos 60)
=0.2E×(2×0.5)
E=40.2=20 NC1

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
B. 2Q24πϵ0a
C. 2Q2πϵ0a
D. Q22πϵ0a

#### 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(VV0)=QV0=42Q24πϵ0a=2Q2πϵ0a

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)

A. 64 J
B. 41 J
C. 16 J
D. 10 J

#### SOLUTION

Solution : A

Potential at the centre of square
V=4×(9×109×50×1062)=902×104 V
Work done in bringing a charge (q = 50 μC) from to centre (O) of the square is
W=q(V0V)=qV0
W=50×106×902×104=64 J

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. kQ1r1=kQ2r2Q1Q2=r1r2
As σ=Q4πr2σ1σ2=Q1Q2×r22r21σ1r
i.e. surface charge density on smaller sphere will be more.

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

A. 1.6×1019
B. 1.6×1029
C. 3.2×1019
D. 3.2×1029

#### SOLUTION

Solution : B

p=q×(d)=1.6×1019×1010=1.6×1029 Cm

Two charges +3.2×1019 and 3.2×1019 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

A. 15.36×1029 Coulomb×m
B. 15.36×1019 Coulomb×m
C. 7.68×1029 Coulomb×m
D. 7.68×1019 Coulomb×m

#### SOLUTION

Solution : C

Dipole moment p = q (d)
=3.2×1019×(2.4×1010)=7.68×1029Cm

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

A. ql
B. 2ql
C. 3ql
D. 4ql

#### SOLUTION

Solution : C Pnet=p2+p2+2pp cos60=3p=3ql (p=ql)

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

A. 3.2×1010J

B. 3.2×1018J
C. 1J
D. 1 dyne

#### SOLUTION

Solution : B

K.E.=q0(VAVB)=1.6×1019(7050)=3.2×1018J

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 A. WA=WB=WC

B. WA=WB=WC=0

C. WA>WB>WC

D. WA<WB<WC

#### 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

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

A. 0.006 J
B. 0.008 J
C. 0.0006 J
D. 0.06 J

#### SOLUTION

Solution : A

W=U2U1=14πϵ0q1q2[1r21r1]=9×109×1012=9×23×103W=0.006J

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 A. kQ1Q2Q3
B. 4kQ1Q2
C. 4kQ2Q3
D. 23kQ2Q3

#### SOLUTION

Solution : C

ΔU=UfUi    Uf=k[Q1Q20.6+Q2Q30.2+Q3Q10.8]ΔU=UfUi=kQ2Q3[10.21]=4kQ2Q3

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?

A. 180 V
B. 360 V
C. 90 V
D. 18 V

#### SOLUTION

Solution : A

12mv2=qV
12×4×103×9×102=106V
V=180 V

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:

A. q2R
B. 4q3R
C. 5q6R
D. 2q3R

#### 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[qRq2R+q3R]
V=14πϵ0[6q3q+2q6R]
V=14πϵ0[5q6R]

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?

A. Zero
B. 14πϵ03Qa
C. 14πϵ033Qa
D. 14πϵ0Q3a

#### SOLUTION

Solution : C

V1=V1+V2+V3;V1=V2=V3
=3V1
=314πϵ0Qa3
=14πϵ033Qa