NEET PHYSICS question bank, MCQ CBSE Notes, Lectures

1 Dual Nature of Matter and Radiation

The half life of a radioactive substance is 25 days. The 25 gm. sample of this substance will reduce is 150 days to :

(1) 0.375 gm.
(2) 0.75 gm.
(3) 1.5 gm.
(4) 4 gm.

Option 1 Solution:

Option 2

Option 3

Option 4

2 Gravitation

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

$$\begin{array}{1 1} (1)\;\large\frac{5GmM}{6R}\\ (2)\;\large\frac{2GmM}{3R} \\ (3)\;\large\frac{GmM}{2R}\\ (4)\;\large\frac{GmM}{3R}\end {array}$$

Option 1 Solution:

Solution

The kinetic energy at altitude $$2R= \large\frac{GMm}{6R}$$

The gravitational potential energy at altitude $$2R= \large\frac{-GMm}{3R}$$

Total energy =$$=k \in +PE=-\large\frac{GMm}{6R}$$

Potential energy at the surface is $$\large\frac{-GMm}{R}$$

∴Req. kinetic energy = $$\large\frac{GMm}{R}-\frac{GMm}{6R};=$$

$$\qquad=\large\frac{5GMm}{6R}$$

Option 2

Option 3

Option 4

3 Waves

Sound waves of frequency 16 kHz are emitted by two coherent point sources placed 2 m apart at the centre of a circular train track of large radius. A person riding the train observes 2 maxima per second when the train is running at a speed of 36 km/h. Calculate the radius of the track. [Velocity of sound in air 320 m/s]

1) $$\cfrac{1000}{\pi }\ m$$
2) $$\cfrac{500}{\pi }\ m$$
3) $$\cfrac{250}{\pi }\ m$$
4) $$\cfrac{700}{\pi }\ m$$

Option 1 Solution:

Solution

The wavelength of the sound wave in air is

$$\cfrac{320}{16\times {{10}^{3}}}=2\times {{10}^{-2}}\ m.$$

The positions of maxima on the circumference of the circular track will be given by
= nλ
When d is the separation between the sources and θ is the angular position of n^th maximum as shown in the figure

2 sin θ = n (2 × 10^-2) ⇒ sin θ = n / 100

Since sin θ lies between 0 and 1 there are 400 maxima on the entire circle.

These 400 maxima will be heard by the person in the time t = 400 / 2 = 200 s

Speed of the train = 36 km/h = 36 × 5/ 18 = 10 m/s

From the obtained values so far we get length of the track

= (10 m/s) (200s) = 2000 m

So radius of the track = $$\cfrac{2000}{2\pi }=\cfrac{1000}{\pi }m $$

Option 2

Option 3

Option 4

4 Kinetic Theory of gases

In a closed container the mass of molecule is 3 x 10^-27 kg. and velocity of molecule is 10 m/sec. If the no. of molecules in the container is 10²⁴, the pressure will be :

(1) 100 N/m²
(2) 10 N/m²
(3) 1 N/m²
(4) 0.5 N/m²

Option 1

Option 2

Option 3 Solution:

Option 4

5 Magnetic Effects of Current and Magnetism

Which of the following demonstrates that earth has a magnetic field?

(1) Earth is a planet rotating about the North-South axis.
(2) The intensity of cosmic rays (stream of charged particles coming particle coming from outer space) is more at the poles than at the Equator.
(3) A large quantity of iron-ore is found in the earth.
(4) The earth is surrounded by an ionosphere (a shell of charged particles)

Option 1

Option 2 Solution:

Option 3

Option 4

6 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 $$x=b\tan\theta $$

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

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

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

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

7 Gravitation

If the mass of the earth increases by 80%80% and radius of the earth increases by 40%40% then the percentage charge in acceleration due to gravity on the surface of radius of earth is (where $$g_{s}=\dfrac{GM} {R^{2}} ,M=$$ mass of earth and $$R=$$ radius of earth})

1) zero
2) +8.16%
3) - 8.16%
4) 160%

Option 1

Option 2

Option 3 Solution:

Solution

As per question final radius = 1.4 times initial radius

And final mass =1.8 times initial mass

Using proportionality we can write final g$$_s$$= $$\dfrac{1.8} {1.96} $$ times initial g$$_s$$

Thus acceleration due to gravity decrease by approximately 8.16%

Option 4

8 Laws of Motion

A uniform rope of length l lies on a table. If the coefficient of friction is µ, then the maximum length l₁ of the hanging part of the rope which can overhang from the edge of the table without sliding down is

1) l/µ
2) l/(µ+1)
3) µl/(µ+1)
4) µl/ (l-1)

Option 1

Option 2

Option 3 Solution:

Solution

To prevent sliding of rope
friction due to rope on table = weight of rope hanging
let m be the mass per unit length of rope. l1 be the length hanging
m(l-l₁)gµ = l1mg
lµ - l₁µ = l1
l₁ ( 1+ µ) = lµ
l₁ = lµ / ( 1+ µ)

Option 4

9 Waves

The equation of a stationary wave along a stretched string is given by $$y= 5 \sin\lgroup \dfrac{\pi x}{3} \rgroup \cos(40 \pi t)$$ Where x and y are in cm and t is in sec. The separation between two adjacent nodes is:

1) 3
2) 1.5
3) 6
4) 4

Option 1

Option 2 Solution:

Solution

On comparing the given equation with standard equation

$$y=2a\sin { \cfrac { 2\pi x }{ \lambda } } \cos { \cfrac { 2\pi vt }{ \lambda } } $$

we get

$$\cfrac { 2\pi x }{ \lambda } =\cfrac { 2\pi x }{ 3 } \\ \therefore \lambda =3$$

Separation between two adjacent nodes is

$$\cfrac { \lambda }{ 2 } =\cfrac { 3 }{ 2 } =1.5\quad cm$$

Option 3

Option 4

10 Motion of System of Particles and Rigid Body

A thin rod of length l and mass m is suspended horizontally by two vertical strings, A & B, one attached at each end of the rod. The density of the rod is given by ρ(x)=ρ₀?(x/I)³ corresponds to the position of string A. Give your answer in terms of m, l, and g. What is the tension in string B?

1) 3mg? / 5
2) mg? / 5
3) 4mg? / 5
4) 6mg? / 5

Option 1

Option 2

Option 3 Solution:

Solution

Option 4

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