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NEET Practice Test - 11 - NEET MCQ


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180 Questions MCQ Test NEET Mock Test Series 2025 - NEET Practice Test - 11

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NEET Practice Test - 11 - Question 1

What is formed when a primary alcohol undergoes catalytic dehydrogenation?

Detailed Solution for NEET Practice Test - 11 - Question 1

NEET Practice Test - 11 - Question 2

The following amine is:
image

Detailed Solution for NEET Practice Test - 11 - Question 2

According to lewis's theory, the element which donates lone pair of electrons is basic.

When electrons of lone pair are get involved in resonance, the basicity decreases.

because the electrons of the lone pair are not free to donate.

In the above image, the lone pairs of N of NMe2 are not involved in resonance due to steric hindrance. Hence, it is more basic.

Hence, the correct answer is option C.

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NEET Practice Test - 11 - Question 3

The radius of the hydrogen atom in the ground state is 0.53 Å. The radius of Li2+ ion (atomic number = 3) in a similar state is

Detailed Solution for NEET Practice Test - 11 - Question 3
For hydrogen atom (n) = 1 (due to ground state)

Radius of hydrogen atom (r) = 0.53 Å.

Atomic number of Li (Z) = 3.

Radius of Li2+ ion

NEET Practice Test - 11 - Question 4

Match the vitamin of column I with deficiency disease given in column II

Detailed Solution for NEET Practice Test - 11 - Question 4
Vitamin A - Xerophthalmia

Vitamin B12 - Pernicious anemia

Vitamin C - Scurvy

Vitamin E - Sterility

NEET Practice Test - 11 - Question 5

The correct relation between the bond lengths a and b is:

Detailed Solution for NEET Practice Test - 11 - Question 5

The correct relation is b > a because bond ‘a’ has partial double bond characters due to resonance.

NEET Practice Test - 11 - Question 6

An ether is more volatile than an alcohol having the same molecular formula. This is due to

Detailed Solution for NEET Practice Test - 11 - Question 6
In ether, there is no H-bonding while alcohols have intermolecular H-bonding.
NEET Practice Test - 11 - Question 7

An exothermic reaction A → B has an activation energy of 17 kJ per mole of A. The heat of the reaction is 40 kJ. Calculate the activation energy for the reverse reaction B → A.

Detailed Solution for NEET Practice Test - 11 - Question 7
Exothermic reaction

According to graph,

Ea(b) = Ea(f) + DH

= 17 + 40

= 57 kJ mol–1

NEET Practice Test - 11 - Question 8

 

To which of the following is Bohr's theory applicable:

(I) He+ 
(II) Li2+
(III) Tritium  
(IV) Be2+

The correct combination is:

Detailed Solution for NEET Practice Test - 11 - Question 8

Bohr's atomic model explains the spectra of single-electron atoms like H-atom. It does not explain the spectra of multi-electron atoms.

So the atoms which are isoelectronic and have the same electronic configuration are well explained by this atomic model.

He+, Li2+ and tritium 1H3 have same number of electrons as that of hydrogen atom i.e. 1 while Be2+ is formed by losing 2e. So, it is a two-electron system.

NEET Practice Test - 11 - Question 9

Select correct match:

Detailed Solution for NEET Practice Test - 11 - Question 9
⇒ [Co(ox) (H2O)3 (NH3)]Br: Does not exhibit structural and optical isomerism but it can exhibit geometrical isomerism.

⇒ [Cr(SCN) (H2O)3 (en)] (C2O4) : It can exhibit linkage and geometrical isomerism.

⇒ [ZnBr(CN) (SCN) (NH3)] : It exhibits optical isomerism.

⇒ [Co(BrCl) (H2O)4] [Ag(CN)2] : and [CoCl(CN)(H2O)4] [AgBr(CN)] are coordination isomers.

NEET Practice Test - 11 - Question 10

The transition metal ions responsible for color in ruby and emerald are, respectively:

Detailed Solution for NEET Practice Test - 11 - Question 10
More equivalents of AgNO3 aqueous solution will be consumed if the complex will furnish more Cl ions in solution. Hence complex [Cr(H2O)6]Cl3 will consume more equivalents of aqueous solution of AgNO3.
NEET Practice Test - 11 - Question 11

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes.

The metal/metals electrodeposited will be :

Detailed Solution for NEET Practice Test - 11 - Question 11
Millimoles of Au+ = 0.1 × 250 = 25

Mole of Au+ 25/100 = 1/40 = 0.025

Similarly, moles of Ag+ = 0.025

Charge passed = I × t = 1 × 15 × 60 = 900 C

Moles of e passed 900/96500 = 0.0093 mol.

Species with higher value of SRP will get deposited first at cathode.

So, only Au will get deposited.

NEET Practice Test - 11 - Question 12

Molecule which contains 4 bonded pairs and 2 lone pairs of electrons on the central atom is:

Detailed Solution for NEET Practice Test - 11 - Question 12

Xe is a noble gas. It has 8 electrons in its valence shell. Now, it is bonded to 4F atoms. 

So, 4 of its electrons are involved in bonding. 

Hence, bond pairs = 4. Remaining 4 are unbonded electrons. So, lone pairs = 4/2 = 2.
image

NEET Practice Test - 11 - Question 13

When chlorine is passed through propene at 400°C, which of the following is formed?

Detailed Solution for NEET Practice Test - 11 - Question 13

At high temp. I.e 400°C substitution occurs in preference to addition.

NEET Practice Test - 11 - Question 14

Which of the following compounds can best be prepared by Wurtz reaction?

Detailed Solution for NEET Practice Test - 11 - Question 14

Due to stability of 2o− radical, it is easy to synthesize n-butane,

The Wurtz reaction is a coupling reaction in organic chemistry, whereby two alkyl halides are reacted with sodium metal in dry ether solution to form a higher alkane

NEET Practice Test - 11 - Question 15

At a certain temperature, only 50% HI is dissociated at equilibrium in the following reaction: 2HI(g) ⇌ H2(g) + I2(g). The equilibrium constant for this reaction is:

Detailed Solution for NEET Practice Test - 11 - Question 15
2HI(g) ⇌ H2(g) + I2(g)

NEET Practice Test - 11 - Question 16

The vapour pressure of water at 300 K in a closed container is 0.4 atm. If the volume of the container is doubled, its vapour pressure at 300 K will be:

Detailed Solution for NEET Practice Test - 11 - Question 16

Vapour pressure will be the same because it doesn't depend on the volume it depends only on Temperature.

NEET Practice Test - 11 - Question 17

The number of hydrogen atoms present in 25.6 g of sucrose (C12H22O11) which has a molar mass of 342.3 g is:

Detailed Solution for NEET Practice Test - 11 - Question 17
Moles of C12H22O11 = 25.6/342.3 = 0.0747

Number of H-atoms = 0.0747 × 22 × 6.023 × 1023

= 9.91 × 1023

NEET Practice Test - 11 - Question 18

A gas that cannot be collected over water is

Detailed Solution for NEET Practice Test - 11 - Question 18
SO2 is highly soluble in water and therefore cannot be collected over water.
NEET Practice Test - 11 - Question 19

An aqueous solution of the following concentration of Acetic acid is the best conductor.

Detailed Solution for NEET Practice Test - 11 - Question 19

for weak electrolytes


∴  The one with lowest concentration has highest molar conductance.

Hence, option B is correct.

NEET Practice Test - 11 - Question 20

Neo-heptyl alcohol is correctly represented as:

Detailed Solution for NEET Practice Test - 11 - Question 20

In the trivial system of naming organic compounds, the names of compounds are dependent on the source of the organic compounds or their properties.

If the prefix is iso, then there will be one methyl group on the secondary carbon of the parent chain. And if the prefix is neo, then there will be two methyl groups on the secondary carbon of the parent chain.

Also, heptyl is there that means there are a total of 7C present and the alcohol is primary.

Hence, Neo-heptyl alcohol is correctly represented as:

NEET Practice Test - 11 - Question 21

Number of ions present in 1 ml of 0.1 M barium nitrate solution is:

Detailed Solution for NEET Practice Test - 11 - Question 21

1 ml of 0.1 M barium nitrate solution corresponds to 0.001 ×0.1 = 0.0001 moles.


1 molecule of barium nitrate Ba(NO3)2 ​ will give 3 ions.

Hence, the number of ions present in 1 ml of 0.1 M Ba(NO3)2 solution = 0.0001 × 3 × 6.02 ×1023 

=1.8 × 1020.

NEET Practice Test - 11 - Question 22

In the lowest energy level of hydrogen atom, electron has an angular momentum equal to:

Detailed Solution for NEET Practice Test - 11 - Question 22

The angular momentumL = mevr is on integer multiple of h/2π

mvr = nh/2π

for, n = 1

mvr = h/2π

NEET Practice Test - 11 - Question 23

The size of particles in suspension, true solution and colloidal solution varies in the order

Detailed Solution for NEET Practice Test - 11 - Question 23
Particle size

True solution - Size < 1 nm

Colloidal solution - 1 nm < size < 1000 nm

Suspension - Size > 1000 nm

NEET Practice Test - 11 - Question 24

A system absorbs 10 kJ of heat and does 4 kJ of work. The internal energy of the system

Detailed Solution for NEET Practice Test - 11 - Question 24
q = + 10 kJ, w = – 4 kJ

∵ ΔE = q + w

= 10 – 4 = 6 kJ

So, energy increases by 6 kJ

NEET Practice Test - 11 - Question 25

Acetanilide on nitration followed by alkaline hydrolysis mainly gives –

Detailed Solution for NEET Practice Test - 11 - Question 25

NEET Practice Test - 11 - Question 26

What is the IUPAC name of the compound?

Detailed Solution for NEET Practice Test - 11 - Question 26
The cyclic portion contains more C-atoms than the acyclic portion. Hence it is a derivative of cyclopentane.
NEET Practice Test - 11 - Question 27

Which one of the following octahedral complexes will not show geometric isomerism? (A and B are monodentate ligands)

Detailed Solution for NEET Practice Test - 11 - Question 27
MA3B3 – 2 geometrical isomers

MA2B4 – 2 geometrical isomers

MA4B2 – 2 geometrical isomers

The complexes of general formula MA6 and MA5B having octahedral geometry do not show geometrical isomerism.

NEET Practice Test - 11 - Question 28

Which of the following ions has the maximum magnetic moment?

Detailed Solution for NEET Practice Test - 11 - Question 28
Mn2+ –5 unpaired electrons

Fe2+ – 4 unpaired electrons

Ti2+ – 2 unpaired electrons

Cr2+ – 4 unpaired electrons

Hence the maximum number of unpaired electrons is present in Mn2+.

Magnetic moment ∝ number of unpaired electrons.

NEET Practice Test - 11 - Question 29

Aluminum oxide may be electrolysed at 1000 °C to furnish aluminum metal (At. Mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is Al3+ + 3e– → Al

To prepare 5.12kg of aluminum metal by this method would require

Detailed Solution for NEET Practice Test - 11 - Question 29
27 g of Al is obtained by charge of 3 × 96500 C.

∴ 1 g of Al is obtained by charge of

3 × 96500/27 C.

∴ 5.12 × 103 g of Al is obtained by charge of

= 3 × 96500/27 x 5.12 x 1000 = 5.49 × 107C.

NEET Practice Test - 11 - Question 30

If Co = initial concentration of the reactant, Ct = concentration of the reactant at time t and k = rate constant of the reaction, then the equation applicable for a first order reaction is:

Detailed Solution for NEET Practice Test - 11 - Question 30


Thus, The integrated form for the first order reaction is Ct = Coe−Kt.

Here, Co = represents initial concentration of the reactant, Ct represents the concentration of the reactant at time t and k is the rate constant of the reaction.

Other forms of the integrated rate law are ln Ct/Co = -kt and k

NEET Practice Test - 11 - Question 31

The major product of the following reaction is:

Detailed Solution for NEET Practice Test - 11 - Question 31

NEET Practice Test - 11 - Question 32

The major product obtained in the ph oto catalyzed bromination of 2-methylbutane is

Detailed Solution for NEET Practice Test - 11 - Question 32
The order of substitution in alkanes is 3° > 2° > 1°

Thus the bromination of 2-methylbutane mainly gives 2-bromo - 2 - methylbutane

NEET Practice Test - 11 - Question 33

The following figure denotes the energy diagram for a reaction. Then the activation energy for the reverse reaction is:

Detailed Solution for NEET Practice Test - 11 - Question 33

Activation energy for reverse reaction will be: X + Y

NEET Practice Test - 11 - Question 34

The magnetic moment of [Cu(NH3)4]2+ was found to be 1.73 B.M. The number of unpaired electrons in the complex is:

Detailed Solution for NEET Practice Test - 11 - Question 34

Magnetic moment of a complex having n-unpaired electrons is given by: 

 where n is the number of unpaired electrons.


∴ n = 1.

NEET Practice Test - 11 - Question 35

Amphoteric behaviour is shown by the oxides of:

Detailed Solution for NEET Practice Test - 11 - Question 35

Amphoteric oxides can behave as an acid as well as the base.

They react with both acids and bases to form salts.

Tin oxide although doesn't dissolve in water but it is amphoteric in nature.

Zinc oxide is also amphoteric in nature.

ZnO + HCl → ZnCl2​ + H2​O

ZnO + NaOH → Na2​​ZnO2​​ + H2​O

SnO2​ ​+ HCl → SnCl4​ + H2​O

SnO2​ ​+ NaOH + H2​O → Na2​​[Sn(OH)6​]

Hence, option B is correct.

NEET Practice Test - 11 - Question 36

Maleic acid and fumaric acids are

Detailed Solution for NEET Practice Test - 11 - Question 36
Mal eic acid and fumar ic acids are geometrical isomers.

NEET Practice Test - 11 - Question 37

In BrF3 molecule, the lone pairs occupy equatorial positions to minimize

Detailed Solution for NEET Practice Test - 11 - Question 37
In BrF3, both bond pairs as well as lone pairs of electrons are present. Due to the presence of lone pairs of electrons (lp) in the valence shell, the bond angle is contracted and the molecule takes the T-shape. This is due to greater repulsion between two lone pairs or between a lone pair and a bond pair than between the two bond pairs.
NEET Practice Test - 11 - Question 38

In the reversible reaction

the rate of disappearance of NO2 is equal to

Detailed Solution for NEET Practice Test - 11 - Question 38

For forward reaction rate =

k1[ NO2]2

For backward reaction

NEET Practice Test - 11 - Question 39

Among the following pairs of ions, lower oxidation state in aqueous solution is more stable than the other in

Detailed Solution for NEET Practice Test - 11 - Question 39

Explanation of Oxidation States Stability
The stability of oxidation states depends on a number of factors, including the electron configuration of the ion, the nature of the ligands or other species present in solution, and the balance between ionization energy and lattice or hydration energy.
Titanium Ions (Ti+, Ti3+)
The Ti3+ ion is more stable than the Ti+ ion in aqueous solution. This is because the half-filled d-orbital of Ti3+ (which has an electron configuration of [Ar]3d1) is particularly stable.
Copper Ions (Cu+, Cu2+)
In the case of copper, the Cu2+ ion is more stable than the Cu+ ion in aqueous solution. This is because the Cu2+ ion has a fully-filled d-orbital ([Ar]3d9), which is more stable than the partially-filled d-orbital of Cu+.
Chromium Ions (Cr2+, Cr3+)
For chromium, the Cr3+ ion is more stable than the Cr2+ ion in aqueous solution. The greater stability of Cr3+ is due to the half-filled d-orbital of Cr3+ ([Ar]3d3), which is particularly stable.
Vanadium Ions (V2+, VO2+)
In the case of vanadium, the V2+ ion is more stable than the VO2+ ion in aqueous solution. This is because the V2+ ion has a half-filled d-orbital ([Ar]3d3), which is more stable than the d-orbital of VO2+.

In summary, the lower oxidation state is more stable in aqueous solution for the V2+ ion compared to the VO2+ ion, making the correct answer to the question D.

NEET Practice Test - 11 - Question 40

The van't Hoff factor for a very dilute aqueous solution of K[Ag(CN)2] is:

Detailed Solution for NEET Practice Test - 11 - Question 40

K[Ag(CN)2] dissociate into K+ and [Ag(CN)2]− 
So  i = 1 + 1 = 2.
Hence, the correct option is D

NEET Practice Test - 11 - Question 41

Butene-1 may be converted to butane by reaction with

Detailed Solution for NEET Practice Test - 11 - Question 41

NEET Practice Test - 11 - Question 42

Conjugate acid of NH2 is

Detailed Solution for NEET Practice Test - 11 - Question 42
Because NH3 after losing a proton (H+) gives NH2.

NH3 + H2O ⇌ NH2+ H3O+

(Conjugate acid-base pair differ only by a proton)

NEET Practice Test - 11 - Question 43

Iron carbonyl, Fe(CO)5​ is __________.

Detailed Solution for NEET Practice Test - 11 - Question 43

Iron carbonyl, Fe(CO)5 is mononuclear. In Fe(CO)5, one Fe atom is surrounded by 5CO ligands.

Mononuclear complexes are those complexes in which one metal atom/ion is surrounded by ligands.
image

NEET Practice Test - 11 - Question 44

The shape of XeO2F2 molecule is

Detailed Solution for NEET Practice Test - 11 - Question 44
XeO2F2 has trigonal bipyramidal geometry, but due to presence of lone pairs of electrons on equatorial position, its actual shape is see-saw.

NEET Practice Test - 11 - Question 45

The shape of methyl carbanion is similar to that of –

Detailed Solution for NEET Practice Test - 11 - Question 45

Methyl carbanion is sp3 hybridized, with three bond pairs and one lone pair

Explanation:
Methyl carbanion is a methyl molecule (CH3) with an extra electron, giving it a negative charge. It is a type of carbanion, which are carbon atoms with a negative charge due to the presence of eight valence electrons in their outer shell.
Understanding Molecular Geometry:
Molecular geometry helps to determine the shape of a molecule. The shape of a molecule is determined by the number of bonding sites (number of atoms it is bonded to) and lone pairs of electrons on the central atom. Methyl carbanion, as well as ammonia (NH3), both have a central atom with three bonding sites and one lone pair of electrons.
Comparison with Given Molecules:
A. BF3 (Boron Trifluoride): In BF3, the boron atom is bonded to three fluorine atoms and has no lone pairs, which creates a trigonal planar shape. This does not match the shape of the methyl carbanion.
B. NH3 (Ammonia): In ammonia, the nitrogen atom is bonded to three hydrogen atoms and also has one lone pair of electrons, which gives it a trigonal pyramidal shape. This shape matches the shape of the methyl carbanion.
C. Methyl Free Radical: A methyl free radical has a central carbon atom bonded to three hydrogen atoms, but it also has one unpaired electron. This gives it a trigonal planar shape, which does not match the shape of the methyl carbanion.
D. Methyl Carbocation: A methyl carbocation has a central carbon atom bonded to three hydrogen atoms and has an empty p-orbital due to the positive charge. This gives it a trigonal planar shape, which does not match the shape of the methyl carbanion.

 

NEET Practice Test - 11 - Question 46

An ac source of angular frequency w is fed across a resistor r and a capacitor C in series.

The current registered user is I. If now the frequency of source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency ω

Detailed Solution for NEET Practice Test - 11 - Question 46

NEET Practice Test - 11 - Question 47

One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of the Balmer series in the hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.

Detailed Solution for NEET Practice Test - 11 - Question 47
For 2nd line of Balmer series in hydrogen spectrum

which is satisfied by n = 12 → n = 6.

NEET Practice Test - 11 - Question 48

The potential difference across the terminals of a battery is 50 V when 11A current is drawn and 60 V when 1A current is drawn. The e.m.f. and the internal resistance of the battery are

Detailed Solution for NEET Practice Test - 11 - Question 48

⇒ E − 11r = 50

E − r = 60

⇒ 10r = 10

r = 1Ω

E = 61V

NEET Practice Test - 11 - Question 49

In a photoelectric effect measurement, the stopping potential for a given metal is found to be V0 when radiation of wavelength λ0 is used. If radiation of wavelength 2 λ0 is used with the same metal, the stopping potential (in volt) is

Detailed Solution for NEET Practice Test - 11 - Question 49

Subtracting them we have

NEET Practice Test - 11 - Question 50

The charges on two spheres are +7µC and –5µC respectively. They experience a force F. If each of them is given and additional charge of –2µC, the new force of attraction will be

Detailed Solution for NEET Practice Test - 11 - Question 50

NEET Practice Test - 11 - Question 51

The current in an induction coil varies with time t according to the graph shown in figure.

Which of the following graphs shows the induced emf (E) in the coil with time?

Detailed Solution for NEET Practice Test - 11 - Question 51

Correct option is C)

When the current is constant, the induced emf is zero as emf=−Ldi​/dt and when the current decreases at constant rate the emf is a positive constant and when it increase at constant rate it is a negative constant.

NEET Practice Test - 11 - Question 52

A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly:

Detailed Solution for NEET Practice Test - 11 - Question 52
Wavelength of monochromatic green light

= 5.5 × 10–5 cm

Now, half of this intensity (I) belongs to electric field and half of that to magnetic field, therefore,

∴ E0 = 2.68 V/m

NEET Practice Test - 11 - Question 53

Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are +q and –q. The potential difference between the centers of the two rings is

Detailed Solution for NEET Practice Test - 11 - Question 53

At (1) using, potential (V1) = Vself + Vdue to (2)

At (2) using potential (V2) = Vself + Vdue to (1)

NEET Practice Test - 11 - Question 54

The variation of acceleration due to gravity g with distance d from center of the earth is best represented by (R = Earth's radius):

Detailed Solution for NEET Practice Test - 11 - Question 54
Variation of acceleration due to gravity, g with distance 'd ' from center of the earth

If d < R,g = Gm/R2 d i.e., g ∝ d (straight line)

If d = R, gs = Gm/R2

If d > R, g = Gm/d2 i.e., g ∝ 1/d2

NEET Practice Test - 11 - Question 55

Two spherical bodies of mass M and 5M and radii R & 2R respectively are released in free space with initial separation between their centers equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

Detailed Solution for NEET Practice Test - 11 - Question 55

Distance between their surfaces = 12R − R − 2R = 9R

Since,

P ∝ mass

a ∝ mass

Distance ∝ acceleration

So, we can write;

a1/a2 ​= m/5m ​= s1/s2

s1/s2​ ​ = 1/5

5s1 ​= 5s2 ​ .......................(1)

s1 ​+ s2 ​ = 9R .......................(2)

On solving these equations;

s1 ​= 1.5R

s2 ​= 7.5R

Since smaller balls have more acceleration in the same time interval, smaller balls will cover more distance.

NEET Practice Test - 11 - Question 56

A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (given g = 10 ms–2)

Detailed Solution for NEET Practice Test - 11 - Question 56

NEET Practice Test - 11 - Question 57

When a piece of a ferromagnetic substance is put in a uniform magnetic field, the flux density inside it is four times the flux density away from the piece. The magnetic permeability of the material is

Detailed Solution for NEET Practice Test - 11 - Question 57
The magnetic permeability of the material

NEET Practice Test - 11 - Question 58

Two tubes of radii r1 and r2, and lengths l1 and l2, respectively, are connected in series and a liquid flows through them in streamline conditions. P1 and P2 are pressure differences across the two tubes. If P2 is 4P1 and l2 is l1/4, then the radius r2 will be equal to:

Detailed Solution for NEET Practice Test - 11 - Question 58
The volume of liquid flowing through both the tubes i.e., rate of flow of liquid is same.

Therefore, V = V1 = V2

NEET Practice Test - 11 - Question 59

What will be the volume contraction of a solid copper cube, 10 cm on an edge when subjected to a hydraulic pressure of 7 × 106 Pa? Bulk modulus of copper = 140 GPa

Detailed Solution for NEET Practice Test - 11 - Question 59
Given

L = 10 cm = 0.1 m; P = 7 × 106 Pa;

B = 140 GPa = 140 × 109 Pa

NEET Practice Test - 11 - Question 60

An object is projected with a velocity of 20 m/s making an angle of 45° horizontally. The equation for the trajectory is h = Ax – Bx2 where h is height, x is horizontal distance, A and B are constants. The ratio A : B is (g = 10 ms–2)

Detailed Solution for NEET Practice Test - 11 - Question 60
Standard equation of projectile motion

Comparing with given equation

(As θ = 45°, u = 20 m/s, g = 10 m/s2)

NEET Practice Test - 11 - Question 61

A particle accelerates from rest at a constant rate for some time and attains a velocity of 8 m/sec. Afterwards it decelerates at a constant rate and comes to rest. If the total time taken is 4 sec, the distance traveled is

Detailed Solution for NEET Practice Test - 11 - Question 61
8 = a t1 and 0 = 8 – a (4 – t1)

8 = 4 a – 8 or a = 4 and t1 = 8/4 = 2 sec

Now, s1 = 0 x 2 + ½ x 4(2)2 or s1 = 8 m

s2 = 8 x 2 - 1/2 x 4 x (2)2 or s2 = 8m

∴ s1 + s2 = 16 m

NEET Practice Test - 11 - Question 62

The distance between the wires of electric mains is 12 cm. These wires experience 4 mg wt. per unit length. The value of current flowing in each wire will be [assume equal current flows in both wires]

Detailed Solution for NEET Practice Test - 11 - Question 62

NEET Practice Test - 11 - Question 63

The radius of germanium (Ge) nuclide is measured to be twice the radius of 94 Be . The number of nucleons in Ge are

Detailed Solution for NEET Practice Test - 11 - Question 63
We use the formula,

R = R0 A1/3

This represents the relation between atomic mass

and radius of the nucleus.

For beryllium, R1 = R0 (9)

For germanium, R2 = R0 A1/3

NEET Practice Test - 11 - Question 64

A simple harmonic wave having an amplitude and time period T is represented by the equation y = 5 sin p(t + 4)m. Then the value of amplitude (a) in (m) and time period (T) in second are [y is in (m) and t is in (sec)]

Detailed Solution for NEET Practice Test - 11 - Question 64
y = 5 sin(pt+ 4p), comparing it with standard equation

NEET Practice Test - 11 - Question 65

For a prism kept in air it is found that for an angle of incidence 60°, the angle of prism A, angle of deviation δ and angle of emergence ‘e’ become equal. Then the refractive index of the material of prism is

Detailed Solution for NEET Practice Test - 11 - Question 65

NEET Practice Test - 11 - Question 66

A diode having potential difference 0.5 V across its junction which does not depend on current, is connected in series with resistance of 20Ω across source. If 0.1 A current passes through resistance then what is the voltage of the source?

Detailed Solution for NEET Practice Test - 11 - Question 66
V' = V + IR = 0.5 + 0.1 × 20 = 2.5 V

NEET Practice Test - 11 - Question 67

The angular velocity of a body changes from ω1 to ω2 without applying a torque but by changing the moment of inertia about its axis of rotation.

The ratio of its corresponding radii of gyration is

Detailed Solution for NEET Practice Test - 11 - Question 67
I1ω1 = I2ω2

NEET Practice Test - 11 - Question 68

A waterfall is 84 m high. If half of the potential energy of the falling water gets converted to heat, the rise in temperature of water will be

Detailed Solution for NEET Practice Test - 11 - Question 68

Let m gram be the mass of water.

Then potential energy of water is:

P.E.= mgh

= m × 980 × 8400

Since potential energy is converted to kinetic energy

P.E. = K.E.

K.E.= m × 980 × 8400

Half of kinetic energy is converted into heat.

Then heat energy:

Heat energy = 1/2 ​ × m × 980 × 8400

Convert joule into calories

heat = m x 980 x 4200/4.2 x 107

= m × 0.098

But heat is:

Heat = msΔT

∴ msΔT = m × 0.098

m × 1 × ΔT = m × 0.098

ΔT = 0.098C

Rise in temperature of water is = 0.098C.

NEET Practice Test - 11 - Question 69

4 kg of oxygen gas is heated so as to raise its temperature from 20 to 120°C. If the heating is done at constant pressure, the external work done by the gas is (Cp = 0.219 cal/g°C and Cv = 0.157 cal/g°C)