NEET PHYSICS question bank, MCQ CBSE Notes, Lectures

1 Electrostatics

Charge q1=+2.0 nC is on Y− axis at y=+2cm and charge q2=−2.0 nC is on Y− axis at y=−2cm . The force on a test charge q0=1μC placed on X−axis at x = 2cm is :

1) $$ -0.01 \hat{j} N $$
2) $$ +0.01 \hat{j} N $$
3) $$ -0.03 \hat{j} N $$
4) $$ +0.03 \hat{j} N $$

Option 1

Option 2

Option 3 Solution:

Solution

Here three charges $${q_1} = + 2nC$$, $${q_2} = - 2nC$$ and $${q_0} = 1\mu C$$ are at $$\left( {0,2} \right)$$, $$\left( {0, - 2} \right)$$ and $$\left( {2,0} \right)$$ respectively. The charges are at vertices of a right angle triangle, and charge $${q_0}$$ is at the right angle vertex.

Now force between $${q_1}$$ and $${q_0}$$which is repulsive in nature is given by

$$F = \dfrac{{9 \times {{10}^9} \times 2 \times {{10}^{ - 9}} \times {{10}^{ - 6}}}}{{{{\left( {2\sqrt 2 \times {{10}^{ - 2}}} \right)}^2}}}$$

$$F = \dfrac{{18 \times {{10}^{ - 2}}}}{8}$$

$$F = 2.25 \times {10^{ - 2}}{\rm{N}}$$

Similarly force between $${q_2}$$ and $${q_0}$$ is also same but attractive in nature and perpendicular to the direction of first force.

Resultant force

$${F_R} = \sqrt {{{2.25}^2} + {{2.25}^2}} \times {10^{ - 2}}$$

$${F_R} = 3.18 \times {10^{ - 2}}\;{\rm{N}}$$

$${F_R} = 0.0318\;{\rm{N}}$$ in downward vertical direction i.e. negative $$j$$ direction.

Thus $$F = - 0.0318\;j\;{\rm{N}}$$

Option 4

2 Ray Optics

A concave mirror is placed over a beaker containing water of (� = 1.33) and image of the object placed at the bottom of beaker is formed at a distance of 25 cm from the top then focal length of the mirror is

(1)10 cm (2) 15 cm

(3) 20 cm (4) 25 cm

Option 1

Option 2

Option 3 Solution:

Option 4

3 Gravitation

The orbital velocity of an artificial satellite in a circular orbit of a height R above earth's surface is v (R is radius of earth). The orbit velocity of a satellite orbiting at an altitude 2R above the earth's surface will be:

1) v/2
2) $$\dfrac {3}{2}v$$
3) $$\dfrac {2}{3}v$$
4) $$\sqrt {\dfrac {2}{3}}v$$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Orbital velocity $$=\sqrt { \dfrac { GM }{ r } } $$

Where $$r$$ is the distance of satellite from the centre of earth.

$$V=\sqrt { \dfrac { GM }{ 2R } } $$

$${ V }^{ 1 }=\sqrt { \dfrac { GM }{ 3R } \times \dfrac { 2 }{ 2 } } $$

$${ V }^{ 1 }=\sqrt { \dfrac { 2 }{ 3 } } \sqrt { \dfrac { GM }{ 2R } } $$

$${ V }^{ 1 }=\sqrt { \dfrac { 2 }{ 3 } } V$$

4 Electrostatics

A point charge $$q$$ is placed at a point on the axis of a non-conducting circular plate of radius r at a distance $$R ( >> r)$$ from its center. The electric flux associated with the plate is

1) $$\dfrac { q{ r }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }{ R }^{ 2 } } $$
2) $$\dfrac { q{ r }^{ 2 } }{ 4{ \varepsilon }_{ 0 }{ R }^{ 2 } } $$
3) $$\dfrac { q{ r }^{ 2 } }{ 4{ \varepsilon }_{ 0 }{ r }^{ 2 } } $$
4) $$\dfrac { q }{ 4{ \varepsilon }_{ 0 }} $$

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Charge $$=q$$

radius $$=r$$

$$R>>r$$

its centre. The electric flux associated with the plate $${ E }_{ x }$$ on the ring of radius $$X=\dfrac { KQ\cos\theta }{ \left( { b }^{ 2 }+{ x }^{ 2 } \right) } $$

$${ E }_{ x }=\dfrac { KQb }{ { \left( { b }^{ 2 }+{ x }^{ 2 } \right) }^{ 3/2 } } $$

$$\varphi =\phi \int { { E }_{ x } } da$$

$$\varphi =\int _{ x=0 }^{ x=R }{ \dfrac { KQ{ b }_{ 2 }\pi xdx }{ { \left( { b }^{ 2 }+{ x }^{ 2 } \right) }^{ 3/2 } } } $$

putting $$x=b\tan\theta $$

$$\varphi =\dfrac { 2\pi KQb }{ { b }^{ 3 } } \int _{ 0 }^{ { \tan }^{ -1 }\left( \dfrac { R }{ b } \right) }{ \dfrac { b\left( \tan\theta \right) b{ \sec }^{ 2 }\theta }{ { \sec }^{ 3 }\theta } } d\theta $$

$$\varphi =2\pi KQ\int _{ 0 }^{ a }{ \sin\theta } d\theta $$ $$\left[ \alpha ={ \tan }^{ -1 }\sqrt { 3 } ={ 60 }^{ 0 } \right] $$

$$\varphi =2\pi KQ\left( 1-\cos\alpha \right) $$

$$\varphi =\dfrac { Q }{ 2{ \epsilon }_{ 0 } } \times \dfrac { 1 }{ 2 } =\dfrac { Q }{ 4{ \epsilon }_{ 0 } } $$

5 Physical world and measurement

The circular divisions of shown screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball is

1) 2.25 mm
2) 2.20 mm
3) 1.20 mm
4) 1.25 mm

Option 1

Option 2

Option 3 Solution:

Solution

Zero error=5 x 0.5/50 = 0.05 mm

Actual measurement

= 2x0.5mm + 25 x 0.5 /50 -0.05 mm

= 1mm + 0.25mm — 0.05mm = 1.20mm

Option 4

6 Oscillations

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum 20 m/s² at a distance of 5 m from the moving position. The time period of oscillation is?

1) 2s
2) 2πs
3) 1s
4) πs

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Since in SHM, magnitude of acceleration is given as

a=ω²x

20=ω²×5

ω=2ω=2

Now, Time Period (T),

T=2π / ω

T=π s

7 Laws of Motion

The maximum speed at which a car can turn around a curve of 20m radius on a level road if the coefficient of friction between the tyres and the road is 0.4, is

1) 10.26 m/s
2) 8.85 m/s
3) 9.16 m/s
4) 11.22 m/s

Option 1

Option 2 Solution:

Option 3

Option 4

8 Thermodynamics

If the temperature of a body is make amount of radiated energy will become :

(1) 16 times
(2) half
(3) two times
(4) four times

Option 1 Solution:

Option 2

Option 3

Option 4

In a triode amplifier phase between input and output voltages is :

(1) zero (2) p

(3) 2 p (4) p/2

Option 1

Option 2 Solution:

Option 3

Option 4

10 Kinematics

An aeroplane is travelling horizontally at a height of 2000 m from the ground. The aeroplane, when at a point P, drops a bomb
to hit a stationary target Q on the ground. In order that the bomb hits the target, what angle θ must the line PQ make with the vertical ? [g = 10ms^–2]

1) 15°
2) 30°
3) 90°
4) 45°

Option 1

Option 2

Option 3

Option 4 Solution:

Solution
Let t be the time taken by bomb to hit the target.

h = 2000 = 1/2gt² => t = 20 sec

R = ut = (100) (20) = 2000 m

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