NEET PHYSICS question bank, MCQ CBSE Notes, Lectures

1 Laws of Motion

A man of mass m = 60 kg is standing on a weighing machine fixed on a triangular wedge of angle θ = 60o with horizontal as shown in the figure.

The wedge is moving up with an upward acceleration a = 2 m/s2. The weight registered by machine is

1) 600 N

2) 1440 N

3) 360 N

4) 240 N

Option 1

Option 2

Option 3 Solution:

Solution

Option 4

2 Alternating Currents

A mains transformer working on 240V has 400 turns on the primary and 100 turns on the secondary. Assuming no power loss, the secondary output will be

(1) 6 V
(2) 12 V
(3) 24 V
(4) 960 V

Option 1

Option 2 Solution:

Option 3

Option 4

3 Physical world and measurement

If light ray is reflected from the denser medium, the path difference produced in the reflected ray will be :

(1) λ/4

(2) λ/2

(3) λ

(4) zero

Option 1

Option 2 Solution:

Option 3

Option 4

The temperature of 100 g of water is to be raised from 24� C to 90� C by adding steam to it. Calculate the mass of the steam required for this purpose (latent heat of steam = 540 k Cal/kg)
1.10 g
2. 11 g
3. 12 g
4. 15 g

Option 1

Option 2

Option 3 Solution:

Option 4

5 Capacitance

In the circuit shown, the charge on the 3μF capacitor at steady state will be:-

1) 6μF

2) 4μc

3) 2/ 3μF

4) 3μF

Option 1

Option 2 Solution:

Solution

In steady state,

i = (1 / 3)A

∴VA+1+1 / 3×1=VB

⇒VB−VA=4 / 3

∴Q=(VB−VA)×3

∴Q=4/ 3 ×3 = 4μc

Option 3

Option 4

6 Current Electricity

Two conductors A and B are joined by a copper wire. If A is positively charged and B is uncharged, the direction of flow of electrons is

1) B to A
2) A to B
3) no flow
4) none of these

Option 1 Solution:

Solution

The current in the wire is due to the drifting of electrons inside a wire in a direction opposite to the flow of electrons. During their drifting they collide with their atoms vibrating about their mean position and lose some of kinetic energy to the vibrating atoms. The electrons are negatively charged particles and so, the electrons move towards the positively charged conductor A.
Hence, the direction of the flow of electrons in the copper wire is from B to A.

Option 2

Option 3

Option 4

7 Electrostatics

A thin semi-circular ring of radius R has a positive charge q distributed uniformly over it. The net field $$\vec E$$ at the centre O is:

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

Option 1

Option 2

Option 3 Solution:

Solution

Given,

$$k=\dfrac{1}{4\pi {{\varepsilon }_{o}}}\,\,and\,\,\lambda =\dfrac{q}{\pi R}\,\,\,\,\,(\,where,\,\lambda =\text{linear}\,\text{charge}\,\text{density})$$

$$\text{Small}\,\text{charge}\,\text{on}\,\text{small}\,\text{length(Rd}\theta )\,\,is\,,\,\,dq=\lambda Rd\theta $$

$$ \vec{E}=\int\limits_{-\pi /2}^{\pi /2}{dE}\cos \theta =2\int\limits_{0}^{\pi /2}{\dfrac{k\left( \lambda Rd\theta \right)}{{{R}^{2}}}}\cos \theta \,\left( -\hat{j} \right) $$

$$ \Rightarrow \vec{E}=2\times \left( \dfrac{1}{4\pi {{\varepsilon }_{0}}} \right)\,\,\,\left( \dfrac{q}{\pi R} \right)\dfrac{R}{{{R}^{2}}}[\sin \theta ]_{0}^{\pi /2}\,\left( -\hat{j} \right) $$

$$ \Rightarrow \dfrac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}[sin90-\sin 0]\left( -\hat{j} \right) $$

$$ \Rightarrow \vec{E}=\dfrac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}(-\hat{j}) $$

Hence, Electric field at point O is $$\dfrac{q}{2{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}(-\hat{j})$$

Option 4

8 Electromagnetic Induction

The equivalent inductance of two inductances is 2.4 henry when connected in parallel and 10 henry when connected in series. The difference between the two inductances is:

1) 4 henry
2) 3 henry
3) 2 henry
4) 8 henry

Option 1

Option 2

Option 3 Solution:

SOLUTION

We have,

Equivalent inductance is 2.4H when connected in the parallel and 10H when connected in the series

Let the two inductor is L1? and L2?

So, when they are connected in series :

Ls?=L1?+L2? =10H

When they are connected in parallel

$$\displaystyle{L}_{{{p}}}=\frac{{{L}_{{{1}}}{L}_{{{2}}}}}{{{L}_{{{1}}}+{L}_{{{2}}}}}={2.4}{H}$$

On solving (i) and(ii)L1L2=24
Also (L1−L2)²=(L1+L2)²−4L1L2
⇒(L1−L2)²=(10)²− 4 × 24 = 4

⇒L1− L2 = 2H

Option 4

9 Thermodynamics

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (ΔS_surr) in JK^-1 is (1 L atm = 101.3 J)

1) 5.763
2) 1.013
3) -1.013
4) -5.763

Option 1

Option 2

Option 3 Solution:

Solution

ΔE = q + w
0 = q - Pext ΔV
q = P_ext ×V = 3 atm (2 - 1) L = 3 atm L
q= (3 × 101.3) Joule

Option 4

10 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

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