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NEET Practice Test - 22 - NEET MCQ


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30 Questions MCQ Test - NEET Practice Test - 22

NEET Practice Test - 22 for NEET 2024 is part of NEET preparation. The NEET Practice Test - 22 questions and answers have been prepared according to the NEET exam syllabus.The NEET Practice Test - 22 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for NEET Practice Test - 22 below.
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NEET Practice Test - 22 - Question 1
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Van der waals, equation of state is (P + a/V2)(V - b) = nRT. The dimensions of a and b are

Detailed Solution for NEET Practice Test - 22 - Question 1

Given,
Van der waals equation.

Where, a/V2 have dimension of pressure and b have dimension of volume i.e., [M0L3T0]
∴  a-pressure x V2
= [ML-1T-2] x [L3]2
= [ML5T-2]

NEET Practice Test - 22 - Question 2
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A particle starting with certain initial velocity and uniform acceleration covers a distance of 12 m in first 3 seconds and a distance of 30 m in next 3 seconds. The initial velocity of the particle is 

Detailed Solution for NEET Practice Test - 22 - Question 2

Let u be the initial velocity that have to find and a be the uniform acceleration of the particle.
For t = 3s, distance travelled S = 12 m and for t = 3 + 3 = 6 s distance travelled 
S' = 12 + 30 = 42 m
From, S = ut + 1/2 at2
12 = u x 3 + 1/2 x a x 32
or 24 = 6u + 9a ...(i)
Similarly, 42 = u x 6 +1/2 x a x 62
or 42 = 6u + 18a ...(ii)
On solving, we get u = 1 ms-1

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NEET Practice Test - 22 - Question 3
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Given A= 4i + 6j and B = 2i + 3j. Which of the following is correct?

Detailed Solution for NEET Practice Test - 22 - Question 3

Given, 
A = 4i + 6j
B = 2i + 3j
Now, A x B = (4i + 6j) x (2i + 3j)
= 12 (i x j) + 12 (j x i)
= 0 (i x j) - 12(i x j)
Again, A • B = (4i + 6j).(2i + 3j)
= 8 + 18 = 26
Again, 
∵ B = 1/2 A
A and B are parallel.

NEET Practice Test - 22 - Question 4
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When current in a coil changes from 5 A to 2 A in 0.1 s, average voltage of 50 V is produced. The self-inductance of the coil is:
Detailed Solution for NEET Practice Test - 22 - Question 4

According to Faraday’s law of electromagnetic induction,
Induced emf, e = Ldi/dt.

⇒ 

NEET Practice Test - 22 - Question 5
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A coin placed on a rotating turn table just slips, if it is placed at a distance of 8 cm from the centre. If angular velocity of the turn table is doubled, it will just slip at a distance of
Detailed Solution for NEET Practice Test - 22 - Question 5

We know that,
F = mrω2
rω2 is constant, ω2 ∝ 1/r

⇒ r2 = 2cm

NEET Practice Test - 22 - Question 6
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The work function of aluminium is 4.2 eV. If two photons each of energy 3.5 eV strike an electron of aluminium, then emission of electron will
Detailed Solution for NEET Practice Test - 22 - Question 6

Work function of aluminium is 4.2 eV. The energy of two photons can not be added at the moment photons collide with electron all its energy will be dissipated or wasted as this energy is not sufficient to knock it out.
Hence emission of electron is hot possible.

NEET Practice Test - 22 - Question 7
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Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get exicited to higher level, after absorbing energy E. If final velocities of particle be v1 and v2, the we must have
Detailed Solution for NEET Practice Test - 22 - Question 7

We have, conservation energy = initial total energy

Total final energy = 

⇒ 

NEET Practice Test - 22 - Question 8
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Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drift speed of electrons in A and B is
Detailed Solution for NEET Practice Test - 22 - Question 8

Current flowing through the conductor, I = nevd A. Hence

NEET Practice Test - 22 - Question 9
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A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is
Detailed Solution for NEET Practice Test - 22 - Question 9

We know that,
L = Iω ...(i)
Kinetic energy, KE = 1/2 Iω2 = 1/2Lω [From Eq (i)]
L=2KE/ω
⇒ 
⇒ L' = L/4

NEET Practice Test - 22 - Question 10
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An object undergoing SHM takes 0.5 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. The period, frequency and amplitude of the motion is
Detailed Solution for NEET Practice Test - 22 - Question 10

Given : T/2 = 0.5 s
∴ T = 1s
Frequency, f = 1/T = 1/1 = 1Hz
If A is the amplitude, then
2A = 50 cm ⇒ A = 25cm.

NEET Practice Test - 22 - Question 11
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A solid sphere of radius R / 2 is cut-out of a solid sphere of radius R such that the spherical cavity, so formed touches, the surface on one side and the centre of the sphere on the other side, as shown in figure. The initial mass of the solid sphere was M. If a particle of mass m is placed at a distance 2.5 R from the centre of the cavity, then what is the gravitational attraction on the mass m?
NEET Practice Test - 22 - Question 12
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The figure shows the path of a positively charged particle 1 through a rectangular region of uniform electric field as shown in the figure. What is the direction of electric field and the direction of particles 2, 3 and 4?
Detailed Solution for NEET Practice Test - 22 - Question 12

Positive charge particle moves in the direction of field and so the right trend is ; Top, down, top, down.

NEET Practice Test - 22 - Question 13
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The surface temperature of the sun is T K and the solar constant for a plate is S. The sun subtends an angle θ at the Planet. Then
Detailed Solution for NEET Practice Test - 22 - Question 13

Let, radius of the sun = R
Distance of the earth from the sun = d
Power radiated from the sun = (4πR2) σ T4 = P
Energy received area ⇒ S = P/4πd2

Angle subtended by the sun at the earth, α = 2R/d
S = constant x T4 x α2
∴ S ∝ T4

NEET Practice Test - 22 - Question 14
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The threshold frequency for a photosensitive metal is 3.3 x 1014 Hz. If light of frequency 8.2 x 1014 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly
Detailed Solution for NEET Practice Test - 22 - Question 14

K .E.= hv - hVth = eV0 (V0 = cut off voltage)
⇒ V0 = h/e(8.2 x 1014 - 3.3 x 1014)

NEET Practice Test - 22 - Question 15
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One 10 V, bulb 60 W of is to be connected to 100 V line. The required self-inductance of induction coil will be (f = 50 Hz)
Detailed Solution for NEET Practice Test - 22 - Question 15

We know, 
Now, 
⇒ (100)2 = (10)2 + VL2
⇒ VL = 99.5V
⇒ 99.5 = 6 x 2 x 3.14 x 50L
⇒ L = 0.052H

NEET Practice Test - 22 - Question 16
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A rough vertical board has an acceleration a along the horizontal so that a block of mass M pressing against it does not fall. The coefficient of friction between block and the board is
Detailed Solution for NEET Practice Test - 22 - Question 16

Force on M = Mg
Reaction force = Ma
Force of friction = μR = μ.Ma

Force of friction will balance the weight. So μMa ≥ Mg ; μ ≥ g/a.

NEET Practice Test - 22 - Question 17
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A hemispherical bowl of radius R is rotating about its axis of symmetry which is kept vertical. A small ball kept in the bowl rotates with the bowl without sliping on its surface. If the surface of the bowl is smooth and the angle made by the radius through the ball with the vertical is α. Find the angular speed at which the bowl is rotating.
Detailed Solution for NEET Practice Test - 22 - Question 17

Balancing the forces, we get

N sin (90°- α) = mg ...(1)
N cos(90°- α) = mω2r
also, r = R sinα
N = mω2R ...(2)
From eqs (i) and (ii), we get

NEET Practice Test - 22 - Question 18
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Plates of area A are arranged as shown. The distance between each plate is d, the net capacitance is
Detailed Solution for NEET Practice Test - 22 - Question 18

The equivalent circuit is shown in figure
Thus, Cab = C = ε0A/d

NEET Practice Test - 22 - Question 19
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Ratio of amplitude of two interfering waves is 2 : 1, then ratio of amplitude of maxima to minima is 
Detailed Solution for NEET Practice Test - 22 - Question 19

We have, a2/a1 = 2/1

∴ 

NEET Practice Test - 22 - Question 20
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The heat radiated per unit area in 1 hour by a furnace whose temperature is 3000 K is (σ = 5.7 x 10-8 W m-2 K-4)
Detailed Solution for NEET Practice Test - 22 - Question 20

According to Stefan's law 
E = σT4
Heat radiated per unit area in 1 hour (3600s) is
= 5.7 x 10-8 x (3000)4 x 3600 = 1.7 x 1010J

NEET Practice Test - 22 - Question 21
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When n identical mercury droplets charged to the same potential V coalescence to form a single bigger drop. The potential of the new drop will be
Detailed Solution for NEET Practice Test - 22 - Question 21

Volume of n mercury droplets = Volume of bigger drop 
⇒ 
⇒ nr3 = R3
⇒ R = n1/3r ...(i)
Now, charge on bigger drop = n (charge on each mercury drop) 
q = nq ...(ii)
Therefore, potential of bigger drop become 

using eq (i) and (ii), we get

NEET Practice Test - 22 - Question 22
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A generator has an e.m.f. of 440 Volt and internal resistance of 400 Ohm. It's terminals are connected to a load of 4000 Ohm the voltage across the load is
Detailed Solution for NEET Practice Test - 22 - Question 22

Total resistance of the circuit
= 4000 + 400 = 4400 Ω 
Current flowing i = V/R = 440/4400 = 0.1 amp.
Voltage across load = R i
= 4000 x 0.1 = 400 volt.

NEET Practice Test - 22 - Question 23
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Two cells of emfs approximately 5V and 10V are to be accurately compared using a potentiometer of length 400 cm.
Detailed Solution for NEET Practice Test - 22 - Question 23

In a potentiometer experiment, the Emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf to be determined. As values of emfs is of two cells are approximately 5 V and 10 V. There fore the potential drop along the potentiometer wire must be more than 10 V. So option (b) is correct.

NEET Practice Test - 22 - Question 24
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A large number of liquid drops each of radius r coalesce to from a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given, surface tension of liquid T, density r)
Detailed Solution for NEET Practice Test - 22 - Question 24

When drops combine to form a single drop of radius R.
Then energy released, E = 
If this energy is converted into kinetic energy then

NEET Practice Test - 22 - Question 25
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The variation of induced emf (E) with time (t) in a coil, if a short bar magnet is moved along its axis with a constant velocitly is best represented as
Detailed Solution for NEET Practice Test - 22 - Question 25

Polarity of emf will be opposite in the two cases while entering and while leaving the coil, only in (b) polarity is charging.

NEET Practice Test - 22 - Question 26
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The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
Detailed Solution for NEET Practice Test - 22 - Question 26

E = Rhc 
E will be maximum for the transition for which  is maximum. Here n2 is the higher energy level.
Clearly,   is maximum for the third transition, i.e. 2 →1. I transition
represents the absorption of energy.

NEET Practice Test - 22 - Question 27
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An electric field E and magnetic field each B exist in a region. If these fields are not perpendicular to each other, then the electromagnetic wave  
Detailed Solution for NEET Practice Test - 22 - Question 27

The Electromagnetic wave being packets of energy moving with speed of light may pass through the jj region.

NEET Practice Test - 22 - Question 28
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A body of mass 10 kg and velocity 10 m/s collides with a stationary body of mass 5 kg. After collision both bodies stick to each other, velocity of the bodies after collision will be
Detailed Solution for NEET Practice Test - 22 - Question 28

Velocity after the collision 

NEET Practice Test - 22 - Question 29
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If the equations of two light waves are y1 = 8 sin ωt and y2 = 6 sin (ωt + φ). Then, ratio of maximum and minimum intensity will be
Detailed Solution for NEET Practice Test - 22 - Question 29

Here,
a1 = 8
a2 = 6 

⇒ 49 : 1

NEET Practice Test - 22 - Question 30
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Which of the given graphs proves Newton’s law of cooling?
Detailed Solution for NEET Practice Test - 22 - Question 30

When hot water temperature (T) and surrounding temparature (T0) readings are noted, and log(T - T0) is plotted versus time, we get a straight line having a negative slope; as a proof of Newton’s law of cooling.

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