NEET PHYSICS question bank, MCQ CBSE Notes, Lectures

1 Electrostatics

Four charges q1=1μC, q2=2μC, q3=3μC and q4=4μC are placed at (0,0,0), (1m,0,0), (0,1m,0), (0,0,1m) respectively. Let $$\vec { { F }_{ i } } $$ be the net force acting on $$i$$th charge of the given charges then $$\sum { \vec { { F }_{ i } } } =$$

1) 0.018
2) 0.02N
3) 0.036N
4) zero

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Let $$F_{ij}$$ be the force on jth charge by ith charge.

we have to find $$F_u=\Sigma_{j=1}^{u}{\Sigma_{i=1}^{u}{F_{ij}}}$$

And By $$F_{12}=F_{21}$$ or say generally

$$F_{ij}=F_{ji}, \; Thus, \; F_{Total}=0$$

2 Kinematics

Which of the following is a scalar quantity?

(1) Velocity
(2) Retardation
(3) Speed
(4) Displacement

Option 1

Option 2

Option 3 Solution:

Option 4

3 Electrostatics

Three small identical neutral metal balls are at the vertices of an equilateral triangle. The balls are in turn connected to an isolated large conducting sphere whose centre is on the perpendicular from the plane of the triangle and passing through its centre. As a result, the first and second balls have acquired charges q₁ and q₂ respectively. Determine the charge q₃ of the third ball.

1) $$\cfrac{2\text{q}_{2}^{2}}{{{\text{q}}_{1}}}$$
2) $$\cfrac{3\text{q}_{2}^{2}}{2{{\text{q}}_{1}}}$$
3) $$\cfrac{\text{q}_{2}^{2}}{2{{\text{q}}_{1}}}$$
4) $$\cfrac{\text{q}_{2}^{2}}{{{\text{q}}_{1}}}$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Let 'r' be the radius of each ball, 'd' be the separation between the balls and 'L' be the distance from the large conducting sphere to each of the small balls.
When we connect the large sphere to the first ball, their potentials become equal:

$$\cfrac{1}{4\,\pi\,{{\varepsilon }_{\text{o}}}}\left( \cfrac{\text{Q}}{L}\,\,+\,\,\cfrac{{{\text{q}}_{1}}}{\text{r}} \right) = V$$

$$\cfrac{\text{Q}}{L}\,\,+\,\,\cfrac{{{\text{q}}_{1}}}{\text{r}}= B$$ ... (1)

Here Q is the charge of the large sphere, and V is its potential. We assume that the charge and the potential of the large sphere changes insignificantly in each charging of the balls.
If the large sphere is connected to the second ball, then again potentials of the large sphere and the second ball should be equal

$$\cfrac{1}{4\,\pi\,{{\varepsilon }_{\text{o}}}} \left( \cfrac{\text{Q}}{L}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{1}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{2}}}{\text{r}} \right)= V$$

$$\cfrac{\text{Q}}{L}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{1}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{2}}}{\text{r}}$$= B ... (2)

Similarly, when the large sphere is connected to the third ball

$$\cfrac{1}{4\,\pi\,{{\varepsilon }_{\text{o}}}} \left( \cfrac{\text{Q}}{L}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{1}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{2}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{3}}}{\text{r}} \right)$$= V

$$\cfrac{\text{Q}}{L}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{1}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{2}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{3}}}{\text{r}}$$ =B

Eq (3) - Eq (2)

$$\left( \,\cfrac{{{\text{q}}_{2}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{3}}}{\text{r}} \right) - \left( \cfrac{{{\text{q}}_{2}}}{\text{r}} \right) = 0$$

$$\cfrac{{{\text{q}}_{\text{3}}}}{\text{r}}\,\,\,\,=\,\,\,\,{{\text{q}}_{2}}\,\,\left( \,\cfrac{1}{\text{r}}\,\,\,-\,\,\,\cfrac{1}{\text{d}} \right)$$ ... (4)

Eq (2) - Eq (1)

$$\left( \,\cfrac{{{\text{q}}_{1}}}{\text{d}}\,\,\,+\,\,\,\cfrac{{{\text{q}}_{2}}}{\text{r}} \right) - \left( \cfrac{{{\text{q}}_{1}}}{\text{r}} \right)= 0$$

$$\cfrac{{{\text{q}}_{2}}}{\text{r}}\,\,\,\,=\,\,\,\,{{\text{q}}_{1}}\,\,\left( \,\cfrac{1}{\text{r}}\,\,\,-\,\,\,\cfrac{1}{\text{d}} \right)$$ ... (5)

Dividing eq (4) by eq (5)

$$\cfrac{{{\text{q}}_{3}}}{{{\text{q}}_{2}}}\,\,\,\,=\,\,\,\,\cfrac{{{\text{q}}_{2}}}{{{\text{q}}_{1}}}$$

q₃ =$$\cfrac{\text{q}_{2}^{2}}{{{\text{q}}_{1}}}$$

A thin circular sing of area A is held perpendicular to a uniform magnetic field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is:

1) ABR

2) AB/R

3 )AB . Dt/R

4) AB/Dt . R

Option 1

Option 2 Solution:

Option 3

Option 4

5 Kinematics

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx-2n, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by
1) -2nβ² x^-4n+1
2) -2nβ² x^-2n-1
3) -2nβ² x^-4n-1
4) -2β² x^-2n+1

Option 1

Option 2

Option 3 Solution:

Solution

Option 4

6 Physical world and measurement

The velocity of a freely falling body changes as g^ph^q where g is acceleration due to gravity and h is the height. The values of p and q are

1) 1, 1/2
2) 1/2, 1/2
3) 1/2, 1
4) 1, 1

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

v α g^ph^q

By substituting the dimension of each quantity and comparing the powers on both sides we get

7 Waves

Equations of a stationary and a travelling waves are as follows, $${ y }_{ 1 }=a\sin { kx } \cos { \omega t } $$ and $${ y }_{ 2 }=a\sin { \left( \omega t-kx \right) }$$. The phase difference between two points $${ x }_{ 1 }=\dfrac { \pi }{ 3k } $$ and $${ x }_{ 2 }=\dfrac { 3\pi }{ 2k } $$ are $${ \phi }_{ 1 }$$ and $${ \phi }_{ 2 }$$ respectively for two waves. The ratio $$\dfrac { { \phi }_{ 1 } }{ { \phi }_{ 2 } } $$ is :

1) 1
2) $${ 5 }/{ 6 }$$
3) $${ 3 }/{ 4 }$$
4) $${ 6 }/{ 7 }$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

At $${ x }_{ 1 }=\dfrac { \pi }{ 3k } $$ and $${ x }_{ 2 }=\dfrac { 3\pi }{ 2k }$$

$$\sin{k{ x }_{ 1 }}$$ or $$\sin{k{ x }_{ 2 }}$$ is not zero.

Therefore, neither of $${ x }_{ 1 }$$ nor $${ x }_{ 2 }$$ is a node.

$$\Delta x={ x }_{ 2 }-{ x }_{ 1 }=\left( \dfrac { 3 }{ 2 } -\dfrac { 1 }{ 3 } \right) \dfrac { \pi }{ k } =\dfrac { 7\pi }{ 6k } $$

Since, $$\dfrac { 2\pi }{ k } > \Delta x > \dfrac { \pi }{ k } $$

$$\lambda > \Delta x> \dfrac { \lambda }{ 2 } $$ $$\left\{ k=\dfrac { 2\pi }{ \lambda } \right\} $$

Therefore, $${ \phi }_{ 1 }=\pi $$

and $${ \phi }_{ 2 }=k$$

$$\Delta x=\dfrac { 7\pi }{ 6 } $$

Therefore, $$\dfrac { { \phi }_{ 1 } }{ { \phi }_{ 2 } } =\dfrac { 6 }{ 7 } $$

8 Atoms and Nuclei

In terms of Rydberg constant R, the shortest wavelength in Balmer series of hydrogen atom spectrum will have wavelength:

(1) 3/2 R
(2) 9/4 R
(3) 1/R
(4) 4/R

Option 1

Option 2

Option 3

Option 4 Solution:

9 Wave Optics

The numerical aperture of objective of a microscope is 0.12. The limit of resolution, when light of wavelength 6000?A is used to view an object is:

1) 0.25×10⁻⁷m
2) 2.5×10⁻⁷m
3) 250×10⁻⁷m
4) 25×10⁻⁷m

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Resolving power of a microscope of objective with numerical aperture $$NA$$ for a light of wavelength $$\lambda$$ is given by:

$$R = \cfrac{\lambda}{2 NA}$$

$$=\cfrac{6000 \times 10^{-10}}{2 \times 0.12}$$

$$= 25 \times 10^{-7}\ m$$

10 Gravitation

Radius of the Earth were increased by a factor of 3 and its mass remained the same, then the acceleration due to gravity on the Earth would

1) reduce by a factor of 3
2) reduce by a factor of 9
3) increase by a factor of 3
4) increase by a factor of 9

Option 1

Option 2 Solution:

The acceleration due to gravity can be expressed as

g = GM/R²
If the mass Mof the Earth remains constant and the radius R were increased by a factor of 3, then the acceleration g would reduce by a factor of 9 .

Option 3

Option 4

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