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Monthly Mock Test (October 31) - NEET MCQ


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30 Questions MCQ Test Daily Test for NEET Preparation - Monthly Mock Test (October 31)

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Monthly Mock Test (October 31) - Question 1

A hollow insulated conduction sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere if its radius is 2 metres? [1998]

Detailed Solution for Monthly Mock Test (October 31) - Question 1

Charge resides on the outer surface of a conducting hollow sphere of radius R. We consider a spherical surface of radius r < R.
By Gauss's theorem,

i.e. electric field inside a hollow sphere is zero.

Monthly Mock Test (October 31) - Question 2

A drupe develops in

[2011]

Detailed Solution for Monthly Mock Test (October 31) - Question 2

Some fleshy fruits such as mango, plum etc. usually have a single hard stone that encloses a seed, called drupe.

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Monthly Mock Test (October 31) - Question 3

Megasporangium is equivalent to :

[NEET 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 3

Ovule is also called integumented megasporangium. It develops into seed after fertilisation in spermatophytes. It occurs singly or in a cluster inside ovary with parenchymatous cushions called placenta.

Monthly Mock Test (October 31) - Question 4

Wind pollination is common in  

[2011]

Detailed Solution for Monthly Mock Test (October 31) - Question 4

Wind pollination is common in grasses. Grasses produce large amount of pollen which by the help of wind reach to opposite sex for reproduction.

Monthly Mock Test (October 31) - Question 5

A 0. 00 20 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl freezes at – 0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = – 1.86°C/m)

Detailed Solution for Monthly Mock Test (October 31) - Question 5

ΔTf = 0 – (0.00732°) = 0.00732
ΔTf = i × Kf × m

Monthly Mock Test (October 31) - Question 6

The electric intensity due to a dipole of length 10 cm and having a charge of 500 μC, at a point on the axis at a distance 20 cm from one of the charges in air, is [2001]

Detailed Solution for Monthly Mock Test (October 31) - Question 6

Given : Length of the dipole (2l) =10cm
= 0.1m or l = 0.05 m
Charge on the dipole (q) = 500 μC = 500 × 10–6 C and distance of the point on the axis from the mid-point of the dipole (r) = 20 + 5 = 25 cm = 0.25 m. We know that the electric field intensity due to dipole on the given 

Monthly Mock Test (October 31) - Question 7

Which one of the following statements is correct? [NEET 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 7

Sporogenous tissue is always diploid, endothecium is second layer of anther wall and perform the function of protection and help in dehiscence of anther to release the pollen. Hard outer layer of pollen is called exine but tapetum always nourishes the developing pollen.
Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus (polypoid).

Monthly Mock Test (October 31) - Question 8

The freezing point depression constant for water is – 1.86ºC m–1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by – 3.82ºC. Calculate the van’t Hoff factor for Na2SO4. [2011]

Detailed Solution for Monthly Mock Test (October 31) - Question 8

Given Kf = – 1.86º cm–1, mass of solute = 5.00 g, mass of solvent = 45.0 g

∴ i = 2.63 (Molecular mass of Na2SO4 = 142)

Monthly Mock Test (October 31) - Question 9

Perisperm differs from endosperm in;

[NEET 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 9

Perisperm is remnants of nucellus which is diploid (2n) but endosperm is triploid (3n). Perisperm occurs in the seeds of Black pepper, coffee, castor, cardamum, Nymphaea. Endosperm is the food  laden tissue which is meant for nourishing the embryo in seed plants. In angiosperms the endosperm is formed as a result of vegetative fertilization, triple fusion or fusion of a male gamete with diploid secondary nucleus of the central cell.

Monthly Mock Test (October 31) - Question 10

Animal vectors are required for pollination in

[NEET Kar. 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 10

There are different types of vectors involved in pollination. For example, Maize, mulberry → pollination by wind. Vallisneria → pollination through water (hydrophily). Cucumber → Bees are brought for the commercial plantings of cucumber.

Monthly Mock Test (October 31) - Question 11

Point charges + 4q, –q and +4q are kept on theX-axis at points x = 0, x =a and x = 2a respectively.

Detailed Solution for Monthly Mock Test (October 31) - Question 11

Net force on each of the charge due to the other charges is zero. However, disturbance in any direction other than along the line on which the charges lie, will not make the charges return.

Monthly Mock Test (October 31) - Question 12

Intensity of an electric field (E) depends on distance r, due to a dipole, is related as [1996]

Detailed Solution for Monthly Mock Test (October 31) - Question 12

Intensity of electric field due to a Dipole

Monthly Mock Test (October 31) - Question 13

Which one of the following statements is wrong?

[2012M]

Detailed Solution for Monthly Mock Test (October 31) - Question 13

In over 60 per cent of angiosperms, pollen grains are shed at cell 2-celled stage. In the remaining species the generative cell divides mitotically to give rise to the two male gametes before pollen grains are shed (3-celled stage.)

Monthly Mock Test (October 31) - Question 14

Which one of the following statements is correct?

[NEET Kar. 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 14

Cleistogamous flowers do not expose their reproductive parts. Anthers and stigma lie close to each other. Pure autogamy occurs since there is no chance of crosspollination. Cleistogamy is the most efficient floral adaptation for promoting self-pollination. E.g., Viola mirabilis and Oxalis autosella.

Monthly Mock Test (October 31) - Question 15

Wind pollinated flowers are

[2010]

Detailed Solution for Monthly Mock Test (October 31) - Question 15

(b) Pollination by wind is called anemophily. Wind pollinated flowers are small in size, producing large number of dry pollen grains. Pollens are small, dry and light in weight. Grasses are anemophilous plants.

Monthly Mock Test (October 31) - Question 16

During osmosis, flow of water through a semi-permeable membrane  [2006]

Detailed Solution for Monthly Mock Test (October 31) - Question 16

Osmosis is the phenomenon of flow of pure solvent from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane. Common semipermeable membranes are permeable to certain solute particles also. In fact, there is no perfect semipermeable membrane. Therefore we can say that flow of water through a semipermeable membrane takes place on both sides with unequal rates.

Monthly Mock Test (October 31) - Question 17

Nucellar polyembryony is reported in species of 

[2011]

Detailed Solution for Monthly Mock Test (October 31) - Question 17

Nucellar polyembryony is reported in a Citrus species.

Monthly Mock Test (October 31) - Question 18

Which condition is not satisfied by an ideal solution? [NEET Kar. 2013]  

Detailed Solution for Monthly Mock Test (October 31) - Question 18

An ideal solution is that solution in which each component obeys Raoult’s law under all conditions of temperatures and concentrations. For an ideal solution.

Monthly Mock Test (October 31) - Question 19

What is the function of germ pore?

[2012M]

Detailed Solution for Monthly Mock Test (October 31) - Question 19

The germ pores are apertures in the exine layer of the pollen grain where the sporopollenin is absent. The germ pore helps in the formation of the germ tube or pollen tube and which makes its exit on germination. 

Monthly Mock Test (October 31) - Question 20

Albuminous seeds store their reserve food mainly in

[NEET Kar. 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 20

Endosperm is the nutritive tissue which provides nourishment to the embryo in seed plant. Albuminous seeds retain a part of endosperm as it is not completely used up during embryo development (e.g., wheat, maize, barley, castor, sunflower).

Monthly Mock Test (October 31) - Question 21

Even in absence of pollinating agents seed setting is assured in [2012]

Detailed Solution for Monthly Mock Test (October 31) - Question 21

Commelina bears aerial, chasmogamous (stigma and anthers exposed to pollinating agents), insect pollinated flowers and underground cleistogamous flowers. Cleistogamous flowers are bisexual flowers which never open. In such flowers, the anther and stigma lie very close to each other. When anther dehisce in the flower buds, pollen grains come in contact with the stigma of the same flower i.e, autogamy occurs. So, these flowers produce an assured seed set, even in the absence of pollinators.

Monthly Mock Test (October 31) - Question 22

Which of the following statements is correct?    

[NEET Kar. 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 22

Pollen grains are generally spherical and a prominent two-layered wall. The hard outer layer called the exine is made up of sporopollenin which is one of the most resistant organic material known. It can withstand high temperatures and strong acids and alkali.

Monthly Mock Test (October 31) - Question 23

Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25ºC are 200 mm Hg and 415 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be : (Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85 u).             [2012 M]

Detailed Solution for Monthly Mock Test (October 31) - Question 23

The correct answer is option C
Molar mass of CH2​Cl2​ =12×1+1×2+35.5×2=85 g mol–1
Molar mass of CHCl3 ​  =12×1+1×1+35.5×3=119.5 g mol−1
Moles of CH2​Cl2​= 40 g /85 g mol−1 = 0.47 mol
Moles of CHCl3​ = 25.5 g /119.5 g mol−1 = 0.213 mol
Total number of moles = 0.47+0.213=0.683 mol
Mole fraction of component 2
 =0.47mol/0.683mol= 0.688
Mole fraction of component 1
=1.00–0.688=0.312 
We know that:
PT​=p10​+(p20​−p10​)x2​
=200+(415–200)×0.688
=200+147.9
=347.9 mm Hg

Monthly Mock Test (October 31) - Question 24

The formation of a dipole is due to two equaland dissimilar point charges placed at a[1996]

Detailed Solution for Monthly Mock Test (October 31) - Question 24

Dipole is formed when two equal and unlike charges are placed at a short distance.

Monthly Mock Test (October 31) - Question 25

1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]

Detailed Solution for Monthly Mock Test (October 31) - Question 25

Monthly Mock Test (October 31) - Question 26

A point Q lies on the perpendicular bisector of an electrical dipole of dipole moment p. If the distance of Q from the dipole is r (much larger than the size of the dipole), then the electric field at Q is proportional to [1998]

Detailed Solution for Monthly Mock Test (October 31) - Question 26

Monthly Mock Test (October 31) - Question 27

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be : [2011 M]

Detailed Solution for Monthly Mock Test (October 31) - Question 27

Given α = 30% i.e., 0.3

i = 1.3
ΔTf = 1.3 × 1.86 × 0.1 = 0.2418
Tf = 0 – 0.2418 = – 0.2418 °C

Monthly Mock Test (October 31) - Question 28

2 5. 3 g of sodium carbonate, Na2 CO3 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) [2010]

Detailed Solution for Monthly Mock Test (October 31) - Question 28

Concentration of

[Na+] = 2 × 0.955 = 1.91 M

= 0.955 M

Monthly Mock Test (October 31) - Question 29

An electric dipole, consisting of two opposite charges of 2 × 10-6 C each separated by a distance 3 cm is placed in an electric field of 2 × 105 N/C. Torque acting on the dipole is    [1995]

Detailed Solution for Monthly Mock Test (October 31) - Question 29

Charges (q) = 2 × 10-6 C,

Distance (d) = 3 cm = 3 × 10-2 m

and electric field (E) = 2 × 105 N/C. Torque (t) = q.d.

E = (2 × 10-6) × (3 × 10-2) × (2 × 105)

= 12 × 10-3N-m.

Monthly Mock Test (October 31) - Question 30

Advantage of cleistogamy is :

[NEET 2013]

Detailed Solution for Monthly Mock Test (October 31) - Question 30

Cleistogamy favours no dependence on pollinator because flowers never open. In such flowers, the anthers and stigma lie close to each other. When anthers dehisce in flower buds pollen grains come in contact with the stigma to effect pollination.

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