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

### Question 1

Four charges are arranged at the corners of a square ABCD , as shown in the adjoining figure. The force on the charge kept at the centre *O* is

Zero

Along the diagonal AC

Along the diagonal BD

Perpendicular to side AB

#### SOLUTION

Solution :C

We put a unit positive charge at

O. Resultant force due to the charge placed atAandCis zero and resultant charge due toBandDis towardsDalong the diagonalBD.

### Question 2

When air is replaced by a dielectric medium of constant k , the maximum force of attraction between two charges separated by a distance

Decreases k times

Remains unchanged

Increases k times

Increases k^{-1} times

#### SOLUTION

Solution :A

F′=14πϵ0q1q2r2=FK

If F is the force in air, then F is less than F since K>1.

### Question 3

Two similar spheres having +q and -q charge are kept at a certain distance. F force acts between the two. If in the middle of two spheres, another similar sphere having +q charge is kept, then it experiences a force whose magnitude and direction are

#### SOLUTION

Solution :C

Initially, force betwen A and C F=KQ2r2

When a similar sphere B having charge +Q is kept at the mid point of line joining A and

C, then Net force on B is

Fnet=FA+FC=KQ2(r2)+KQ2(r2)2=8KQ2r2=8F.

(Direction is shown in figure)

### Question 4

The electric field due to a charge at a distance of 3 *m* from it is 500 *N*/coulomb. The magnitude of the charge is [14πϵ0=9×109 N−m2coulomb2]

2.5 *micro-coulomb*

2.0 *micro-coulomb*

1.0 *micro-coulomb*

0.5 *micro-coulomb*

#### SOLUTION

Solution :D

E=9×109×Qr2⇒500=9×109×Q(3)2⇒Q=0.5μC

### Question 5

When a glass rod is rubbed with silk, it

#### SOLUTION

Solution :B

On rubbing glass rod with silk, excess electron transferred from glass to silk. So glass rod becomes positive and silk becomes negative.

### Question 6

There is a uniform electric field of strength 103 V/m along *y*-axis. A body of mass 1*g* and charge 10−6*C* is projected into the field from origin along the positive *x*-axis with a velocity 10*m*/*s*. Its speed in *m*/*s* after 10*s* is (Neglect gravitation)

#### SOLUTION

Solution :C

Body moves along the parabolic path.

For vertical motion : By using v=u+at

⇒ vy=0+QEm.t=10−6×10310−3×10=10m/sec

For horizontal motion - it;s horizontal velocity remains the same i.e after the same i.e after 10 sec, horizontal

velocity of body vx=10 m/sec.

Velocity aftey 10 sec v=√V2x+V2y=10√2m/sec

### Question 7

Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at *O* is double the electric field when only one positive charge of same magnitude is placed at *R*. Which of the following arrangements of charges is possible for *P*, *Q*, *R*, *S*, *T* and *U* respectively

#### SOLUTION

Solution :D

If the charges are arranged according to the option (d), the electric fields due to

PandSand due toQandTadd to zero, while due toUandRwill be added up.

### Question 8

An electron enters with its velocity in the direction of the uniform electric lines of force. Consider lines of force to be straight. Then

#### SOLUTION

Solution :C

As electric field applies the force on electron in the direction opposite to it’s motion, the velocity of the electron will decrease.

### Question 9

**Assertion:** The coulomb force is the dominating force in the universe.

**Reason :**The coulomb force is weaker than the gravitational force.

*If both assertion and reason are true and the reason is the correct explanation of the assertion.*

*If both assertion and reason are true but reason is not the correct explanation of the assertion.*

*If assertion is true but reason is false.*

*If the assertion and reason both are false.*

*If assertion is false but reason is true.*

#### SOLUTION

Solution :D

Gravitational force is the dominating force in nature and not coulomb's force. Gravitational force is the weakest force. Also, Coulomb's force>> gravitational force.

### Question 10

The distance between charges 5×10−11C and −2.7×10−11C is 0.2m. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is

*m*

*m*

*m*

*m*

#### SOLUTION

Solution :C

If two opposite charges are separated by a certain distance, then for it's equalibrium a third

charge should be kept outside and near the charge which is smaller in magnitude.

Here, suppose third charge q is placed at a distance x from −2.7×10−11C then for it's

equalibrium |F1|=|F2|

⇒ KQ1q(x+0.2)2=KQ2qx2⇒ x=0.556 m

(Here K=14πϵ0 and Q1=5×10−11C, Q2=−2.7×10−11C)

### Question 11

Three charges 4q, Q and q are in a straight line in the position of 0,l2 and l respectively. The resultant force on q will be zero, if Q =

#### SOLUTION

Solution :A

The force between 4q and q; F1=14πϵ0.4q×ql2

The force between Q and q; F2=14πϵ0.Q×q(l2)2

We want F1+F2=0 or 4q2l2=−4Qql2⇒Q=−q

### Question 12

A conductor has 14.4×10−19 coulombs positive charge. The conductor has (Charge on electron 1.6×10−19coulombs )

#### SOLUTION

Solution :D

Positive charge shows the deficiency of electrons. number of electrons =14.4×10−191.6×10−19=9

### Question 13

An electric dipole is placed along the x - axis at the origin O . A point P is at a distance of 20cm from this origin such that OP makes an angle π3 with the *x-*axis. If the electric field at P makes an angle θ with the *x*-axis, the value of θ would be

#### SOLUTION

Solution :B

θ=π3+α where tanα=12tanπ3⇒α=tan−1√32 So, θ=π3+tan−1 √32

### Question 14

Two electric dipoles of moment *P* and 64 *P* are placed in opposite direction on a line at a distance of 25 *cm*. The electric field will be zero at point between the dipoles whose distance from the dipole of moment *P* is

5 *cm*

259cm

10 *cm*

413cm

#### SOLUTION

Solution :A

Suppose neutral point N lies at a distance x from dipole of moment ρ or at a distance X2 from

dipole of 64 ρ.

⇒14πϵ02px3=14πϵ0.2(64P)(25−x)3

⇒ 1x3=64(25−x)3⇒ x=5cm.

### Question 15

An electric dipole is kept in non-uniform electric field. It experiences

#### SOLUTION

Solution :A

As the dipole will feel two forces which are although opposite but not equal.

Therefore, A net force will be there and as these forces act at different points of a body. A torque is also there.