NEET PHYSICS question bank, MCQ CBSE Notes, Lectures

1 Dual Nature of Matter and Radiation

Radius of H-atom in its ground state is 5.03×10⁻¹¹m. After collision with an electron it is found to have a radius of 20.12×10⁻¹¹m.. The principal quantum number ′n′ of the final state of the atom is:

1) 1
2) 2
3) 3
4) 4

Option 1

Option 2 Solution:

Solution

From Bohr's model of the atom:

$${ r }_{ n }=\dfrac { { n }^{ 2 }{ h }^{ 2 } }{ Z{ k }_{ e }{ e }^{ 2 }{ m }_{ e } } $$

where $${ r }_{ n }$$ is radius of the nth orbit

n is principal quantum number

h is reduced Plank's constant

Z is atomic number

$${ k }_{ e }$$ is coulomb's constant

e is electronic charge

$${ m }_{ e }$$ is the mass of electron

$${r}_{1} = 1 \times \dfrac{h^2}{1 \times k_e \times e^2\times m_e }= 5.03 \times 10^{-11} \Rightarrow \dfrac{h^2}{k_e \times e^2 \times m_e}= 5.03 \times 10^{-11} $$

$${ r }_{ n }=\dfrac { { { n }^{ 2 }h }^{ 2 } }{ { k }_{ e }{ e }^{ 2 }{ m }_{ e } } =20.12\times { 10 }^{ -11 }\\ \rightarrow { n }^{ 2 }=20.12\times { 10 }^{ -11 }\times \dfrac { { k }_{ e }{ e }^{ 2 }{ m }_{ e } }{ { h }^{ 2 } } \\ =20.12\times { 10 }^{ -11 }\times \dfrac { 1 }{ 5.03\times { 10 }^{ -11 } } \\ { n }^{ 2 }=4\\ \rightarrow n=2$$

Option 3

Option 4

2 Thermodynamics

The coefficient of volume expansion of glycerin is 34 ´ 10^–5 per °C. The fractional change in its density for a 30°C rise in temperature is:

1) 2.02 ´ 10^–2

2)1.86 ´ 10^–2

3)1.02 ´ 10^–2

4)0.86 ´ 10^–2

Option 1

Option 2

Option 3 Solution:

Solution

$$\displaystyle\gamma=\frac{1}{{\Delta{T}}}{\left(\frac{{\Delta{V}}}{{{V}}}\right)}\Rightarrow\frac{{\Delta{V}}}{{{V}}}=\gamma\times\Delta{T}$$

Option 4

3 Capacitance

A circuit has a self-inductance of 1H and carries a current of 2A. To prevent sparking, when the circuit is switched off, a capacitor which can withstand 400V is used. The least capacitance of capacitor connected across the switch must be equal to

1) 50μF

2) 25μF

3) 100μF

4) 12.5μF

Option 1

Option 2 Solution:

Solution

Energy stored in capacitor = energy stored in inductance

i.e., $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }=\dfrac { 1 }{ 2 } { LI }^{ 2 }$$
$$\Rightarrow C=\dfrac { { LI }^{ 2 } }{ { V }^{ 2 } } =\dfrac { 1\times { \left( 2 \right) }^{ 2 } }{ { \left( 400 \right) }^{ 2 } } =25\mu F$$

Option 3

Option 4

4 Properties of Bulk Matter

The work done in increasing the radius of a soap bubble from 4 cm to 5 cm is ................Joule.(given surface tension of soap water to be 25×10−3N/m)

1) $$0.5657 \times 10^{-3}$$
2) $$5.657 \times 10^{-3}$$
3) $$56.5 \times 10^{-3}$$
4) $$565 \times 10^{-3}$$

Option 1 Solution:

Solution
$$\Delta A=2\times \dfrac{4\pi (5^{2}-4^{2})}{10000}$$
$$\Delta \bigcup =\Delta A\times T$$
$$=\dfrac{8\pi }{10000}\times 9\times 25\times 10^{-3}$$
$$=5.65\times 10^{-4} J$$

Option 2

Option 3

Option 4

5 Atoms and Nuclei

The half-life of $$^{215}At$$ is $$100\ mu s$$. The time taken for the radioactivity of a sample of this nucleus to decay to $$\dfrac {1}{16}th$$ of its initial value is

1) 6.3μs
2) 40μs
3) 300μs
4) 400μs

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

Activity, $$\left (\dfrac {R}{R_{0}}\right ) = \left (\dfrac {1}{2}\right )^{n}$$
$$\Rightarrow \dfrac {1}{16} = \left (\dfrac {1}{2}\right )^{n}$$
$$\Rightarrow \left (\dfrac {1}{2}\right )^{4} = \left (\dfrac {1}{2}\right )^{n}$$
$$\Rightarrow n = 4$$
$$\Rightarrow \dfrac {t}{T} = 4 \ \ $$ $$\left [\because n = \dfrac {t}{T}\right ]$$
$$\Rightarrow t = 4T$$
$$t= 4\times 100\mu s$$
$$\therefore t = 400\mu s$$.

6 Laws of Motion

A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed of 2v at the bottom, how many times as high must a new ramp be?

1) Ö2

2) 2

3) 4

4) 8

Option 1

Option 2

Option 3 Solution:

We know that the kinetic energy at the bottom of the ramp plus any friction losses sliding down the ramp sum to the potential energy at the top of the ramp. We can find the relation between the height h of the ramp and the velocity V at the bottom of the ramp. Neglecting fricion

mgh = 1/2 mV² => V²= 2gh

V = sqrt(2gh)

Vx = 2V = 2*Ö(2gh) = Ö (2gh*2^2*h)

So h has to be 4 times as high to double the speed.

Option 4

7 Laws of Motion

A cart of mass has a block of mass m attached to it as shown in figure. The coefficient of friction between the block and the cart is µ. What is the minimum acceleration of the cart so that the block m does not fall

1)µg
2) g/µ
3)µ/g
4)Mµg/m

Option 1

Option 2 Solution:

Solution

Let a be the acceleration of trolley

From the FBD of block
N=ma
∴ µN=mg
µma=mg
a=g/µ

Option 3

Option 4

8 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

9 Waves

A stopper is attached in the middle of glass tube. Two bubbles of radius 2 cm. and 4 cm. are formed at the end of the glass tube. If one opens the stopper :

(1) small bubble will reduce and large will increase

(2) both will increase

(3) both will reduce

(4) small will increase and large will reduce

Option 1 Solution:

Option 2

Option 3

Option 4

10 Waves

A composite string is made by joining two strings of different masses per unit length μ and 4μ. The composite string is under the same tension. A transverse wave pulse Y = (6 mm) sin (5t + 40x), where t is in seconds and x is in meters, is sent along the lighter string towards the joint. The joint is at x = 0. The equation of the wave pulse reflected from the joint is

1) Y = (2 mm) sin (5t – 10x)
2) Y = (4 mm) sin (40x – 5t)
3) Y = –(2 mm) sin (5t – 40x)
4) Y = (2 mm) sin (5 t – 40x)

Option 1

Option 2

Option 3

Option 4 Solution:

Solution

$$\sqrt{\cfrac{T}{\mu }}\ ;\ \ {{v}_{2}}=\sqrt{\cfrac{T}{4\mu }}$$

v₁ = 2 v₂

v₂ < v₁
2^nd medium is denser
The wave reflected from the denser medium has phase change of π.

$$\cfrac{{{v}_{2}}-{{v}_{1}}}{{{v}_{2}}+{{v}_{1}}}\times 6=\cfrac{\cfrac{{{v}_{1}}}{2}-{{v}_{1}}}{\cfrac{{{v}_{1}}}{2}+{{v}_{1}}}\times 6=-2\ mm$$

equation of reflected wave pulse is
y = -(2 mm) sin (5t - 40x)

NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

Question Bank

Solution

Solution

i.e., $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }=\dfrac { 1 }{ 2 } { LI }^{ 2 }$$
$$\Rightarrow C=\dfrac { { LI }^{ 2 } }{ { V }^{ 2 } } =\dfrac { 1\times { \left( 2 \right) }^{ 2 } }{ { \left( 400 \right) }^{ 2 } } =25\mu F$$

Solution
$$\Delta A=2\times \dfrac{4\pi (5^{2}-4^{2})}{10000}$$
$$\Delta \bigcup =\Delta A\times T$$
$$=\dfrac{8\pi }{10000}\times 9\times 25\times 10^{-3}$$
$$=5.65\times 10^{-4} J$$

Solution

Solution

Solution

MOCK TEST

NSEP PHYSICS

DAILY PHYSICS TEST QUESTIONS

Question Bank

Solution

Solution

i.e., $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }=\dfrac { 1 }{ 2 } { LI }^{ 2 }$$ $$\Rightarrow C=\dfrac { { LI }^{ 2 } }{ { V }^{ 2 } } =\dfrac { 1\times { \left( 2 \right) }^{ 2 } }{ { \left( 400 \right) }^{ 2 } } =25\mu F$$

Solution $$\Delta A=2\times \dfrac{4\pi (5^{2}-4^{2})}{10000}$$$$\Delta \bigcup =\Delta A\times T$$$$=\dfrac{8\pi }{10000}\times 9\times 25\times 10^{-3}$$$$=5.65\times 10^{-4} J$$

Solution

Solution

Solution

MOCK TEST

i.e., $$\dfrac { 1 }{ 2 } C{ V }^{ 2 }=\dfrac { 1 }{ 2 } { LI }^{ 2 }$$
$$\Rightarrow C=\dfrac { { LI }^{ 2 } }{ { V }^{ 2 } } =\dfrac { 1\times { \left( 2 \right) }^{ 2 } }{ { \left( 400 \right) }^{ 2 } } =25\mu F$$

Solution
$$\Delta A=2\times \dfrac{4\pi (5^{2}-4^{2})}{10000}$$
$$\Delta \bigcup =\Delta A\times T$$
$$=\dfrac{8\pi }{10000}\times 9\times 25\times 10^{-3}$$
$$=5.65\times 10^{-4} J$$