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 at A and C is zero and resultant charge due to B and D is towards D along the diagonal BD.

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

$F^{'} = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{2}} = \frac{F}{K}$

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

A. Zero having no direction

B. 8F towards +q charge

C. 8F towards -q charge

D. 4F towards +q charge

SOLUTION

Solution : C

Initially, force betwen A and C $F = K \frac{Q^{2}}{r^{2}}$

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

$F_{n e t} = F_{A} + F_{C} = K \frac{Q^{2}}{(\frac{r}{2})} + \frac{K Q^{2}}{(\frac{r}{2})^{2}} = 8 \frac{K Q^{2}}{r^{2}} = 8 F .$

(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 $[\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} \frac{N - m^{2}}{c o u l o m b^{2}}]$

2.5 micro-coulomb

2.0 micro-coulomb

1.0 micro-coulomb

0.5 micro-coulomb

SOLUTION

Solution : D

$E = 9 \times 10^{9} \times \frac{Q}{r^{2}} \Rightarrow 500 = 9 \times 10^{9} \times \frac{Q}{(3)^{2}} \Rightarrow Q = 0.5 μ C$

Question 5

When a glass rod is rubbed with silk, it

A. Gains electrons from silk

B. Gives electrons to silk

C. Gains protons from silk

D. Gives protons to silk

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 $10^{3}$ V/m along y-axis. A body of mass 1g and charge $10^{- 6}$ C is projected into the field from origin along the positive x-axis with a velocity 10m/s. Its speed in m/s after 10s is (Neglect gravitation)

A. 10

5 \sqrt{2}

10 \sqrt{2}

D. 20

SOLUTION

Solution : C

Body moves along the parabolic path.

For vertical motion : By using $v = u + a t$

$\Rightarrow v_{y} = 0 + \frac{Q E}{m} . t = \frac{10^{- 6} \times 10^{3}}{10^{- 3}} \times 10 = 10 m / s e c$

For horizontal motion - it;s horizontal velocity remains the same i.e after the same i.e after 10 sec, horizontal
velocity of body $v_{x} = 10 m / s e c .$
Velocity aftey 10 sec $v = \sqrt{V_{x}^{2} + V_{y}^{2}} = 10 \sqrt{2} m / s e c$

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

A. +,-,+,-,-,+

B. +,-,+,-,+,-

C. +,+,-,+,-,-

D. -,+,+,-,+,-

SOLUTION

Solution : D

If the charges are arranged according to the option (d), the electric fields due to P and S and due to Q and T add to zero, while due to U and R will 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

A. The path of the electron will be a circle

B. The path of the electron will be a parabola

C. The magnitude of velocity of the electron will first decrease and then increase

D. The magnitude of velocity of the electron will first increase and then decrease

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.

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

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

C. If assertion is true but reason is false.

D. If the assertion and reason both are false.

E. 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 $5 \times 10^{- 11}$ C and $- 2.7 \times 10^{- 11}$ C is $0.2 m$ . 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

A. 0.44 m

B. 0.65 m

C. 0.556 m

D. 0.350 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 $- 2.7 \times 10^{- 11} C$ then for it's
equalibrium $| F_{1} | = | F_{2} |$

$\Rightarrow \frac{K Q_{1} q}{(x + 0.2)^{2}} = \frac{K Q_{2} q}{x^{2}} \Rightarrow x = 0.556 m$

$(H e r e K = \frac{1}{4 π ϵ_{0}} a n d Q_{1} = 5 \times 10^{- 11} C, Q_{2} = - 2.7 \times 10^{- 11} C)$

Question 11

Three charges 4q, Q and q are in a straight line in the position of 0, $\frac{l}{2}$ and l respectively. The resultant force on q will be zero, if Q =

A. -q

B. -2q

- \frac{q}{2}

D. 4q

SOLUTION

Solution : A

The force between 4q and q; $F_{1} = \frac{1}{4 π ϵ_{0}} . \frac{4 q \times q}{l^{2}}$

The force between Q and q; $F_{2} = \frac{1}{4 π ϵ_{0}} . \frac{Q \times q}{(\frac{l}{2})^{2}}$

We want $F_{1} + F_{2} = 0$ or $\frac{4 q^{2}}{l^{2}} = - \frac{4 Q q}{l^{2}} \Rightarrow Q = - q$

Question 12

A conductor has $14.4 \times 10^{- 19}$ coulombs positive charge. The conductor has (Charge on electron $1.6 \times 10 - 19$ coulombs )

A. 9 electrons in excess

B. 27 electrons in short

C. 27 electrons in excess

D. 9 electrons in short

SOLUTION

Solution : D

Positive charge shows the deficiency of electrons. number of electrons $= \frac{14.4 \times 10^{- 19}}{1.6 \times 10^{- 19}} = 9$

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 $\frac{π}{3}$ with the x-axis. If the electric field at P makes an angle θ with the x-axis, the value of $θ$ would be

\frac{π}{3}

\frac{π}{3} + t a n^{- 1} (\frac{\sqrt{3}}{2})

\frac{2 π}{3}

t a n^{- 1} (\frac{\sqrt{3}}{2})

SOLUTION

Solution : B

$θ = \frac{π}{3} + α$ where $t a n α = \frac{1}{2} t a n \frac{π}{3}$ $\Rightarrow α = t a n^{- 1} \sqrt{\frac{3}{2}}$ So, $θ = \frac{π}{3} + t a n^{- 1} \sqrt{\frac{3}{2}}$

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

$\frac{25}{9} c m$

10 cm

$\frac{4}{13} c m$

SOLUTION

Solution : A

Suppose neutral point N lies at a distance x from dipole of moment $ρ$ or at a distance $X_{2}$ from
dipole of 64 $ρ .$

$\Rightarrow \frac{1}{4 π ϵ_{0}} \frac{2 p}{x^{3}} = \frac{1}{4 π ϵ_{0}} . \frac{2 (64 P)}{(25 - x)^{3}}$
$\Rightarrow \frac{1}{x^{3}} = \frac{64}{(25 - x)^{3}} \Rightarrow x = 5 c m .$

Question 15

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

A. A force and a torque

B. A force but not a torque

C. A torque but not a force

D. Neither a force nor a torque

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.

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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