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NEET Practice Test - 18 - NEET MCQ


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180 Questions MCQ Test NEET Mock Test Series 2025 - NEET Practice Test - 18

NEET Practice Test - 18 for NEET 2024 is part of NEET Mock Test Series 2025 preparation. The NEET Practice Test - 18 questions and answers have been prepared according to the NEET exam syllabus.The NEET Practice Test - 18 MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for NEET Practice Test - 18 below.
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NEET Practice Test - 18 - Question 1

In the Freidel Craft's acylation reaction, the effective electrophile is

Detailed Solution for NEET Practice Test - 18 - Question 1

NEET Practice Test - 18 - Question 2

Consider the following statements

I. The radius of an anion is larger than that of the parent atom.

II. The ionization energy generally increases with increasing atomic number in a period.

III. The electronegativity of an element is the tendency of an isolated atom to attract an electron.

Which of the above statements is/are correct?

Detailed Solution for NEET Practice Test - 18 - Question 2
The tendency of an atom in a compound to attract a pair of bonded electrons towards itself is known as electronegativity of the atom.
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NEET Practice Test - 18 - Question 3

Pick out the wrong statement.

Detailed Solution for NEET Practice Test - 18 - Question 3
Catenation tendency is higher in phosphorus Mien compared with other elements of same group.
NEET Practice Test - 18 - Question 4

Aniline is an activated system for electrophilic substitution. The compound formed on heating aniline with acetic anhydride is

Detailed Solution for NEET Practice Test - 18 - Question 4
Aniline when treated with acetic anhydride forms acetanilide (nucleophilic substitution)

NEET Practice Test - 18 - Question 5

Sewage mostly constitutes :

Detailed Solution for NEET Practice Test - 18 - Question 5
Domestic sewage constitute biodegradable pollutants.
NEET Practice Test - 18 - Question 6

Which of the following can be repeatedly soften on heating?

(i) Polystyrene

(ii) Melamine

(iii) Polyesters

(iv) Polyethylene

(v) Neoprene

Detailed Solution for NEET Practice Test - 18 - Question 6
Polystyrene and polyethylene belong to the category of thermoplastic polymers which are capable of repeatedly softening on heating and harden on cooling.
NEET Practice Test - 18 - Question 7

The correct order of acid strength of oxyacids is:

Detailed Solution for NEET Practice Test - 18 - Question 7
The acid strength of oxyacids of the same halogen increases with increase in oxidation state of the halogen

Thus Cl in HClO4 is in highest oxidation state (+7) and hence strongest acid, while HCIO (Cl is in +1 oxidation state) is weakest.

NEET Practice Test - 18 - Question 8

In which alkyl halide, SN2 mech an ism is favoured maximum ?

Detailed Solution for NEET Practice Test - 18 - Question 8
SN2 mechanism is favoured maximum in methyl chloride among all the given options. Difference in rates are related to the bulk of the substituents. As the number of substituents attached to carbon bearing the halogen is increased, the reactivity towards SN2 substitution decreases. The order of reactivity towards SN2 mechanism is

CH3CI > CH3CH2Cl > (CH3)2CHCl >(CH3)3CCl

NEET Practice Test - 18 - Question 9

Which of the two ions from the list given below that have the geometry that is explained by the same hybridization of orbitals, NO2, NO3, NH2, NH4+, SCN ?

Detailed Solution for NEET Practice Test - 18 - Question 9

Answer: A: NO2 and NO3 Explanation: To determine the geometry of these ions, we need to understand their hybridization. Let's analyze the hybridization of each ion: - NO2: Nitrite ion has a central nitrogen atom surrounded by two oxygen atoms and one lone pair. The nitrogen atom forms two sigma bonds with the oxygen atoms and has one lone pair. This results in sp2 hybridization, and the geometry is bent (approximately 120° bond angles). - NO3: Nitrate ion has a central nitrogen atom surrounded by three oxygen atoms. The nitrogen atom forms three sigma bonds with the oxygen atoms. This results in sp2 hybridization, and the geometry is trigonal planar (120° bond angles). - NH2: This ion has a central nitrogen atom surrounded by two hydrogen atoms and one lone pair. The nitrogen atom forms two sigma bonds with the hydrogen atoms and has one lone pair. This results in sp2 hybridization, and the geometry is bent (approximately 120° bond angles). - NH4 : Ammonium ion has a central nitrogen atom surrounded by four hydrogen atoms. The nitrogen atom forms four sigma bonds with the hydrogen atoms. This results in sp3 hybridization, and the geometry is tetrahedral (approximately 109.5° bond angles). - SCN: Thiocyanate ion has a linear geometry with a central carbon atom surrounded by a sulfur atom and a nitrogen atom. The carbon atom forms two sigma bonds with the sulfur and nitrogen atoms. This results in sp hybridization, and the geometry is linear (180° bond angles). From the analysis above, we can see that NO2 and NO3 have the same sp2 hybridization, which explains their similar geometries.

NEET Practice Test - 18 - Question 10

White lead is

Detailed Solution for NEET Practice Test - 18 - Question 10
White lead is 2PbC3.Pb (OH)2 which is prepared by Pb (NO3)2 & Na2CO3. It is called as basic lead carbonate.

NEET Practice Test - 18 - Question 11

NH4Cl crystallises in a bcc lattice with edge length of unit cell equal to 387 pm. If the radius of the Cl ion is 181 pm, the radius of ions is

Detailed Solution for NEET Practice Test - 18 - Question 11
For bcc lattice

or, rCl- = 335.1 -101.0 = 154.1 pm

NEET Practice Test - 18 - Question 12

When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice in initial temperature, the change in entropy (ΔS) is

Detailed Solution for NEET Practice Test - 18 - Question 12
When there is simultaneously change in temperature and volume (or pressure)

NEET Practice Test - 18 - Question 13

For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, ΔU and w corresponds to

Detailed Solution for NEET Practice Test - 18 - Question 13
Zn + H2SO4 = ZnSO4H2

In bomb calorimeter, there is no expansion in volume, so, work done will be zero. This reaction is exothermic. So, some heat will be evolved which will result in lowering of internal energy. Hence, ΔU < 0 and w = 0

NEET Practice Test - 18 - Question 14

In most cases, for a rise of 10K temperature the rate constant is doubled to tripled. This is due to the reason that

Detailed Solution for NEET Practice Test - 18 - Question 14
For a 10 K rise in temperature, collision frequency increases merely by 1 to 2% but the number of effective collisions increases by 100 to 200%.
NEET Practice Test - 18 - Question 15

The major product obtained in the photo catalysed bromination of 2-methylbutane is:

Detailed Solution for NEET Practice Test - 18 - Question 15
The order of substitution in different alkanes is 3° >2° >1°

Thus the bromination of 2-methyl butane mainly gives 2- Bromo - 2 - methyl butane

NEET Practice Test - 18 - Question 16

Among the following, correct statement is :

Detailed Solution for NEET Practice Test - 18 - Question 16
Brownian movement is the random motion of particles suspended in a fluid (a liquid or gas) resulting from their collision with the fast moving atom or molecules in the liquid or gaseous state of the matter. That means smaller particles are responsible for the Brownian movement than for bigger particles.
NEET Practice Test - 18 - Question 17

In the presence of platinum catalyst, hydrocarbon A adds hydrogen to form n-hexane.

When hydrogen bromide is added to A instead of hydrogen, only a single bromo compound is formed. Which of the following is A?

Detailed Solution for NEET Practice Test - 18 - Question 17

NEET Practice Test - 18 - Question 18

An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules :

Detailed Solution for NEET Practice Test - 18 - Question 18
The vapour pressure of solution will be less than the vapour pressure of pure solvent, so some vapour molecules will get condensed to maintain new equilibrium.
NEET Practice Test - 18 - Question 19

Match the columns

NEET Practice Test - 18 - Question 20

Cl2 changes to Cl and ClO in cold NaOH. Equivalent weight of Cl2 will be

Detailed Solution for NEET Practice Test - 18 - Question 20

equivalent mass of Cl2 in reduction half-reaction.

equivalent mass of Cl2 in oxidation half-reaction.

Thus, equivalent mass in overall reaction

NEET Practice Test - 18 - Question 21

Chlorine cannot displace

Detailed Solution for NEET Practice Test - 18 - Question 21
Chlorine cannot displace F from NaF. The reactivity follows the order F > Cl > Br > I
NEET Practice Test - 18 - Question 22

The formula mass of Mohr's salt is 392. The iron present in it is oxidised by KMnO4 in acid medium. The equivalent mass of Mohr's salt is

Detailed Solution for NEET Practice Test - 18 - Question 22
Mohr's salt is FeSO4.(NH4)2SO4.6H2O.

Fe2+ present in it is oxidized to Fe3+.

So, we have, Fe2+ → Fe3+ +e

= 392

NEET Practice Test - 18 - Question 23

The Langmuir adsorption isotherm is deduced using the assumption

Detailed Solution for NEET Practice Test - 18 - Question 23
Langmuir adsorption isotherm is based on the assumption that every adsorption site is equivalent and the ability of a particle to bind there is in dependent of whether or not nearby sites are occupied.
NEET Practice Test - 18 - Question 24

Which of the following is used for making optical instruments?

Detailed Solution for NEET Practice Test - 18 - Question 24
SiO2 is used for this purpose.
NEET Practice Test - 18 - Question 25

25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ions, Na+ and carbonate ions, CO32– are respectively (Molar mass of Na2CO3 = 106 g mol–1)

Detailed Solution for NEET Practice Test - 18 - Question 25
Concentration of Na2 CO2

NEET Practice Test - 18 - Question 26

Equilibrium constant K changes with temperature.

At 300 K, equilibrium constant is 25 and at 400 K it is 10. Hence, backward reaction will have energy of activation

Detailed Solution for NEET Practice Test - 18 - Question 26
K decreases with increase in temperature. Thus, forward reaction is exothermic.

Ea = activation energy of forward reaction

Ea’ = ...backward...

= (Ea+ΔH)

Thus, E’a ; > Ea

NEET Practice Test - 18 - Question 27

Detailed Solution for NEET Practice Test - 18 - Question 27

NEET Practice Test - 18 - Question 28

The commonest method for the extraction of metals from oxide ores involves :

Detailed Solution for NEET Practice Test - 18 - Question 28
Most metals occur in their combined state. They have to be reduced to become free metals. The choice of reducing agent is decided by the factors of energetics and economics. In this regard C is the most common reducing agent used. Carbon can be used in the form of coal, coke, charcoal and carbon monoxide, e.g:-

NEET Practice Test - 18 - Question 29

IUPAC name of [Pt(NH3)3 (Br) (NO2) Cl] Cl is

Detailed Solution for NEET Practice Test - 18 - Question 29
We know that IUPAC name of [Pt(NH3)3 (Br) (NO2)Cl]Cl is triamminebromochloronitroplatinum (TV) chloride.
NEET Practice Test - 18 - Question 30

If x is amount of adsorbate and m is amount of adsorbent, which of the following relations is not related to adsorption process ?

Detailed Solution for NEET Practice Test - 18 - Question 30

The correct answer is d. The given relation x/m = pt is not related to the adsorption process. Let's discuss the other options to understand why they are related to the adsorption process: A: x / m = K (p) at very low pressure. - This relation is derived from the Freundlich adsorption isotherm, which describes the relation between the amount of adsorbate adsorbed per unit mass of the adsorbent and the pressure of the adsorbate. - At very low pressure, the Freundlich isotherm becomes linear, and the relation x/m = K(p) holds true. B: x / m = Kp1/n at intermediate pressure - This relation is the Freundlich adsorption isotherm, a widely used empirical model to describe the adsorption process. - The Freundlich isotherm is given by x/m = Kp1/n, where x is the amount of adsorbate, m is the amount of adsorbent, K is a constant, p is the pressure of the adsorbate, and n is the heterogeneity factor. C: x/m = K at very high pressure - This relation is derived from the Langmuir adsorption isotherm, which is based on the assumption of monolayer adsorption on a homogeneous adsorbent surface. - At very high pressure, the Langmuir isotherm approaches saturation, and the relation x/m = K holds true, where K is a constant. In summary, options A, B, and C are related to the adsorption process through different adsorption isotherms, while option D is not related to the process.

NEET Practice Test - 18 - Question 31

Reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the order

Detailed Solution for NEET Practice Test - 18 - Question 31
Reactivity of Hydrogen Atoms in Alkanes

The reactivity of hydrogen atoms in alkanes depends on the type of carbon atom they are attached to. Carbon atoms in alkanes can be classified as primary (1°), secondary (2°), or tertiary (3°) based on the number of carbon atoms they are bonded to:

  • Primary (1°) Carbon: A carbon atom bonded to only one other carbon atom and three hydrogen atoms.
  • Secondary (2°) Carbon: A carbon atom bonded to two other carbon atoms and two hydrogen atoms.
  • Tertiary (3°) Carbon: A carbon atom bonded to three other carbon atoms and one hydrogen atom.
Reactivity Order

The reactivity order for hydrogen atoms attached to different carbon atoms in alkanes is as follows:

  • Tertiary (3°) Hydrogen > Secondary (2°) Hydrogen > Primary (1°) Hydrogen
Explanation

This reactivity order is primarily due to the difference in the stability of carbocations formed during the reaction process. Carbocations are positively charged carbon atoms that are intermediates in various organic reactions.

  • Tertiary Carbocations: These are more stable than secondary and primary carbocations due to the electron-donating (hyperconjugation) effect of the three neighboring carbon atoms. This stabilization increases the likelihood of reactions involving the hydrogen atom attached to the tertiary carbon.
  • Secondary Carbocations: These are less stable than tertiary carbocations but more stable than primary carbocations due to the electron-donating effect of the two neighboring carbon atoms.
  • Primary Carbocations: These are the least stable among the three types of carbocations because they have only one neighboring carbon atom to donate electrons and stabilize the positive charge.

As a result, the reactivity of hydrogen atoms in alkanes follows the order: Tertiary > Secondary > Primary (Option D).

NEET Practice Test - 18 - Question 32

Which of the following colligative property can provide molar mass of proteins (or polymers or colloids) with greatest precision ?

Detailed Solution for NEET Practice Test - 18 - Question 32
Answer: A. Osmotic pressure Explanation: Colligative properties are properties of solutions that depend on the ratio of solute particles to solvent particles, regardless of their chemical nature. The four main colligative properties are: 1. Osmotic pressure 2. Elevation of boiling point 3. Depression of freezing point 4. Relative lowering of vapor pressure The molar mass of proteins (or polymers or colloids) can be determined using any of these colligative properties. However, the osmotic pressure method provides the greatest precision for the following reasons: - Sensitivity: Osmotic pressure is sensitive to very low concentrations of solute particles, which makes it possible to accurately determine the molar mass of proteins even in dilute solutions. - Non-volatility: Unlike boiling point elevation and vapor pressure lowering, osmotic pressure is not affected by the volatility of the solute or the solvent, which makes it more reliable for determining the molar mass of proteins. - Temperature independent: Osmotic pressure is relatively less dependent on temperature compared to other colligative properties, which simplifies the experimental procedure and reduces potential sources of error. Because of these advantages, osmotic pressure is the preferred method for determining the molar mass of proteins (or polymers or colloids) with the greatest precision.
NEET Practice Test - 18 - Question 33

The reason for double helical structure of DNA is the operation of:

Detailed Solution for NEET Practice Test - 18 - Question 33
The two polynucleotide chains of DNA molecules are twisted around a common axis but run in opposite directions to form a right handed helix. The two chains are joined together by specific hydrogen bonds.
NEET Practice Test - 18 - Question 34

What is the correct order of spin only magnetic moment (in BM) of Mn2+, Cr2+ and V2+ ?

Detailed Solution for NEET Practice Test - 18 - Question 34
Number of unpaired electrons in various species.

Mn2+ 3d5 5

Cr2+ 3d4 4

V2+ 3d3 3

i.e., it increases with increase in number of unpaired electrons present in a species. Thus, the correct order is Mn2+ > Cr2+ > V2+.

NEET Practice Test - 18 - Question 35

Dihydrogen of high purity (> 99.95%) is obtained through:

Detailed Solution for NEET Practice Test - 18 - Question 35
Dihydrogen of high purity (>99.95%) is obtained by the electrolysis of Ba(OH)2 using Ni electrodes.
NEET Practice Test - 18 - Question 36

An organic compound X (molecular formula C6H7O2N) has six carbon atoms in a ring system, two double bonds and a nitro group as substituent, X is :

Detailed Solution for NEET Practice Test - 18 - Question 36

NEET Practice Test - 18 - Question 37

The r.m.s velocity of hydrogen is √7 times the r.m.s velocity of nitrogen. If T is the temperature of the gas , then

Detailed Solution for NEET Practice Test - 18 - Question 37

NEET Practice Test - 18 - Question 38

Equivalent conductance of an electrolyte containing NaF at infinite dilution is 90.1 Ohm–1cm2. If NaF is replaced by KF what is the value of equivalent conductance?

Detailed Solution for NEET Practice Test - 18 - Question 38
Because at infinite dilution the equivalent conductance of strong electrolytes furnishing same number of ions is same.
NEET Practice Test - 18 - Question 39

Which of the following statements(s) is/are incorrect?

(i) Only 1/8th portion of an atom located at corner of a cubic unit cell is its neighbouring unit cell.

(ii) Total number of atoms per unit cell for a face centered cubic unit cell is 3.

(iii) Atom located at the body center is shared between two adjacent unit cells.

Detailed Solution for NEET Practice Test - 18 - Question 39
Total number of atoms per unit cell for a face centered cubic unit is 4.

The atom at the body center completely belongs to the unit cell in which it is present.

NEET Practice Test - 18 - Question 40

The following equilibrium constants are given:

The equilibrium constant for the oxidation of NH3 by oxygen to give NO is

Detailed Solution for NEET Practice Test - 18 - Question 40

We have to calculate

NEET Practice Test - 18 - Question 41

Among the following compounds (I - III), the ease of their reaction with electrophiles is,

Detailed Solution for NEET Practice Test - 18 - Question 41
-OCH3 activates the benzene ring. -NO2 deactivates the ring. Hence the reaction of the given compounds with electrophiles is in the order, I > II > III.
NEET Practice Test - 18 - Question 42

A compound A with molecular formula C10H13Cl gives a white precipitate on adding silver nitrate solution. A on reacting with alcoholic KOH gives compound B as the main product. B on ozonolysis gives C and D. C gives Cannizzaro reaction but not aldol condensation. D gives aldol condensation but not Cannizzaro reaction. A is :

Detailed Solution for NEET Practice Test - 18 - Question 42
Compound A reacts with alc.KOH to give compound B which on further ozonolysis gives C (does not contains α-H atom) and D (contains α-H atom).This reaction sequence can be achieved by compounds in option (a) and (c). Since compound A gives white ppt. with AgNO3 preferable option will be (c) as tert alkyl reacts with AgNO3 more quickly.

NEET Practice Test - 18 - Question 43

The enthalpy of combustion of C6H6 is – 3250 kJ, when 0.39 gm of C6H6 is burnt in excess of oxygen in an open vessel, the amount of heat evolved is

Detailed Solution for NEET Practice Test - 18 - Question 43

NEET Practice Test - 18 - Question 44

The enthalpies of the following reactions are shown below.

Calculate the O –– H bond energy for the hydroxyl radical.

Detailed Solution for NEET Practice Test - 18 - Question 44
We have to calculate the enthalpy of the reaction OH (g) → O(g)+ H(g)

From the given reactions, this can be obtained as follows.

NEET Practice Test - 18 - Question 45

Given

Fe3+ (aq) + e → Fe2+ (aq); Eo = + 0.77 V

Al3+ (aq) + 3e → Al(s); Eo = – 1.66 V

Br2(aq) + 2e → 2Br; Eo = + 1.09 V

Considering the electrode potentials, which of the following represents the correct order of reducing power?

Detailed Solution for NEET Practice Test - 18 - Question 45
Reducing character decreases down the series. Hence the correct order is Al < Fe2+ < Br
NEET Practice Test - 18 - Question 46

A damped har monic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1/1000 of the original amplitude is close to :

Detailed Solution for NEET Practice Test - 18 - Question 46
Time of half the amplitude is = 2s

Using , A = A0e-kt

A0/2 = Ae e-kx2 …..(i)

and A0/1000 = Ae e-kt …..(ii)

Dividing (i) by (ii) and solving, we get t = 20 s

NEET Practice Test - 18 - Question 47

The velocity of water in a river is 18 km/h near the surface. If the river is 5 m deep, find the shearing stress between the horizontal layers of water. The co-efficient of viscosity of water = 10–2 poise

Detailed Solution for NEET Practice Test - 18 - Question 47
η = 10-2 poise

v = 18km/h = 18000/3600 = 5m/s

l = 5m

Stram rate = v/l

Coefficient of viscosity,

∴ Shearing stress = η x strain rate

NEET Practice Test - 18 - Question 48

A hospital uses an ultrasonic scanner to locate tumours in a tissue. The operating frequency of the scanner is 4.2 MHz. The speed of sound in a tissue is 1.7 km/s. The wavelength of sound in tissue is close to

Detailed Solution for NEET Practice Test - 18 - Question 48
Frequency (n) = 4.2 MHz = 4.2 x 106 Hz and speed of sound (v) = 1.7 km/s = 1.7 x 103 m/s. Wave length of sound in tissue

NEET Practice Test - 18 - Question 49

A metallic rod of length ‘ℓ’ is tied to a string of length 2ℓ and made to rotate with angular speed w on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is

Detailed Solution for NEET Practice Test - 18 - Question 49
Here, induced e.m.f.

NEET Practice Test - 18 - Question 50

The transformer voltage induced in the secondary coil of a transformer is mainly due to

Detailed Solution for NEET Practice Test - 18 - Question 50
Explanation: The transformer voltage induced in the secondary coil of a transformer is mainly due to: A varying magnetic field: - Transformers work on the principle of electromagnetic induction. - In a transformer, an alternating current (AC) is passed through the primary coil, which creates a varying magnetic field around it. - This varying magnetic field then induces an electromotive force (EMF) in the secondary coil. - The voltage induced in the secondary coil is directly proportional to the number of turns in the coil and the rate of change of the magnetic field. - Hence, the induced voltage in the secondary coil of a transformer is mainly due to the varying magnetic field created by the primary coil.
NEET Practice Test - 18 - Question 51

When sound waves travel from air to water, which one of the following remains constant?

Detailed Solution for NEET Practice Test - 18 - Question 51
When sound travels from one medium to another its speed changes due to change in wavelength. Its frequency remains constant.
NEET Practice Test - 18 - Question 52

The electrostatic potential inside a charged spherical ball is given by ϕ = ar2 + b where r is the distance from the centre and a, b are constants.

Then the charge density inside the ball is:

Detailed Solution for NEET Practice Test - 18 - Question 52
Electric field

…..(i)

By Gauss’s theorem

…….(ii)

From (i) and (ii),

Q = -8πε0ar3

⇒ dq = - 24πε0ar2 dr

Charge density,

NEET Practice Test - 18 - Question 53

The moment of inertia of disc about a tangent axis in its plane is

Detailed Solution for NEET Practice Test - 18 - Question 53

Moment of inertia of a disc about a diameter

Applying theorem of parallel axis.

NEET Practice Test - 18 - Question 54

Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2 × 10–3 W. The number of photons emitted, on he average, by the source per second is

Detailed Solution for NEET Practice Test - 18 - Question 54
Sincep = nhv

NEET Practice Test - 18 - Question 55

The most appropriate magnetization M versus magnetising field H curve for a paramagnetic substance is

Detailed Solution for NEET Practice Test - 18 - Question 55
For paramagnetic substance magnetization M is proportional to magnetising field H, and M is positive.
NEET Practice Test - 18 - Question 56

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ, then θ is close to :

Detailed Solution for NEET Practice Test - 18 - Question 56

The tower PQ in figure (19.5) subtends an angle α on the objective. As uo​ is very large, the first image P Q' is formed in the focal plane of the objective.

tanα = α = P′Q′/f∘​ ..(i)

 

The final image P"Q" subtends an angle β  on the eyepiece (and hence on the eye). We have from the triangle P Q'E:

tanβ = β = ′P′Q′/EP​ ...(ii)

 

The telescope is set for normal adjustment so that the final image is formed at infinity, the first image P'Q' must be in the focal plane of the eyepiece.Then EP=fe​ .

Thus, equation (ii) becomes

tanβ=β=f∘/fe​​​ ...(iii)

dividing equation (iii) by (i)

β/α​=​f∘/fe​​...(iv)

 

from equation (i) The α=P′Q′/f∘​

P′Q′=mh1​ ...(v) where m is magnification of objective lens m=uv​ 

where v is position of first image P'Q' and u is position of tower v=f∘=150cm,u=1000m substituting value of m in equation (v) m

 

P′Q′= 0.15/1000 x ​50m=.0075m

α= 0.0075/0.15 ​= 0.05

 

value of β can be obtained from equation (v)

 

tanβ=.05.15​×α=30×.05=1.5rad

β=56.3 which is closer to 60°

Fig 19.5

NEET Practice Test - 18 - Question 57

The distance travelled by a body moving along a line in time t is proportional to t3.

The acceleration-time (a, t) graph for the motion of the body will be

Detailed Solution for NEET Practice Test - 18 - Question 57
Distance along a line i.e ,, displacement (s)

= t3 (∵ s ∝ t3 given)

By double differentiation of displacement, we get acceleration.

Hence graph (b) is correct.

NEET Practice Test - 18 - Question 58

A motor car is travelling at 30 m/s on a circular road of radius 500 m. It is increasing in speed at the rate of 2 m/s2. Then the acceleration of the car will be

Detailed Solution for NEET Practice Test - 18 - Question 58
Radial acceleration = centripetal acceleration

Tangential acceleration (at) = 2 m/ sec2

Total acceleration

NEET Practice Test - 18 - Question 59

An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle a with the vertical, the maximum possible value of a so that the insect does not slip is given by

Detailed Solution for NEET Practice Test - 18 - Question 59
The insect crawls up the bowl up to a certain height h only till the component of its weight along the bowl is balanced by limiting frictional force.

For limiting condition at point A

R = mg cosα …..(i)

F1 = mg sinα …...(ii)

Dividing eq. (ii) by (i)

NEET Practice Test - 18 - Question 60

A charged particle q is placed at the centre O of a cube of length L (A B C D E F G H). Another same charge q is placed at a distance L from O.

Then the electric flux through ABCD is

Detailed Solution for NEET Practice Test - 18 - Question 60
The flux lor both the charges exactly cancels the effect of each other.
NEET Practice Test - 18 - Question 61

A fully charged capacitor C with initial charge Q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic field is

Detailed Solution for NEET Practice Test - 18 - Question 61
Solution: Let's analyze the given problem step by step: 1. Initial conditions: - At t = 0, the capacitor has an initial charge Q0 and is fully charged. - The inductor has no current flow through it, so it has no magnetic field. 2. Charging process: - As time progresses, the charge on the capacitor starts decreasing, and the current starts flowing through the inductor. - The energy stored in the capacitor's electric field decreases, and the energy stored in the inductor's magnetic field increases. 3. Equal energy condition: - The energy is considered equal when the energy stored in the capacitor's electric field (UC) is equal to the energy stored in the inductor's magnetic field (UL). - UC = (1/2)C(Q0^2 - Q^2), where Q is the charge on the capacitor at any time t. - UL = (1/2)L(I^2), where I is the current through the inductor at any time t. 4. Equating energies: - For equal energy, UC = UL. - (1/2)C(Q0^2 - Q^2) = (1/2)L(I^2) 5. Relation between charge and current: - The current through the inductor is related to the rate of change of charge on the capacitor: I = -dQ/dt. 6. Integrating the equation: - To find the time at which the energies are equal, we can integrate the equation -dQ/dt = √((Q0^2 - Q^2)/LC) with respect to time. 7. Result: - After integrating and solving the equation, we get the time at which the energies are equal as: - t = Hence, the correct answer is Option A.
NEET Practice Test - 18 - Question 62

The combination of gates shown below yields

Detailed Solution for NEET Practice Test - 18 - Question 62
The final boolean expression is,

NEET Practice Test - 18 - Question 63

Two coherent plane light waves of equal amplitude makes a small angle α (< < 1) with each other. They fall almost normally on a screen.

If λ is the wavelength of light waves, the fringe width Δx of interference patterns of the two sets of waves on the screen is

Detailed Solution for NEET Practice Test - 18 - Question 63
Δx = λ/(2α)
NEET Practice Test - 18 - Question 64

A driver in a car, approaching a vertical wall notices that the frequency of his car horn, has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is 345 m/s, then the speed of the car is :

Detailed Solution for NEET Practice Test - 18 - Question 64
Let f1 be the frequency heard by wall,

Here, v = Velocity of sound,

vc = Velocity of Car,

f0 = actual frequency of car horn

Let f2 be the frequency heard by the driver after reflection from the wall.