Monthly Mock Test (July 31) - NEET MCQ

# Monthly Mock Test (July 31) - NEET MCQ

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

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Monthly Mock Test (July 31) - Question 1

### Which of the following molecule does not have a linear arrangement of atoms ? [1989]

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

For linear arrangement of atoms the hybridisation should be sp(linear shape, 180° angle). Only H2S has sp3-hybridization and hence has angular shape while C2H2, BeH2 and CO2 all involve sp - hybridization and hence have linear arrangement of atoms.

Monthly Mock Test (July 31) - Question 2

### Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio force of contact between the two block will be :

Detailed Solution for Monthly Mock Test (July 31) - Question 2
• If F is applied force then acceleration of the system is F/2m + m = F/3m
• Now when we apply the force from the left, the force applied on the block m is F/3m = F/3. This will be the force in the contact.
• When we apply the force from the right from on the block will be 2Fm/3 = 2F/3, this will be the force on the contact then.

So the ratio is F/3 : 2F/3 = 1 : 3

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Monthly Mock Test (July 31) - Question 3

### A body of mass 5 kg is suspended by the strings making angles 60º and 30º with the horizontal - (a) T1 = 25 N (b) T2 = 25 N (c) T1 = 25N (d) T2 = 25N

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

As the mass is at rest the resultant of forces acting on it are equal to zero so forces in vertical direction are

T1​sin30o + T2​sin60o − mg = 0

similarly in horizontal direction

T1​sin30− T2​sin60= 0

solving above equations will give us

T1​+ √3​×√3​T1​ = 100

T​= 25N

T2​ = 25√3​N

Monthly Mock Test (July 31) - Question 4

Which one shows the strongest hydrogen bonding?

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

Among the F, O, and N, Fluorine has a higher value of electronegativity, this is the reason that fluorine can form the strongest hydrogen bond with the H atom of another molecule.

Just take an idea from the value of the energy of hydrogen bond that is formed with HF, H2O, and NH3

means to break the hydrogen of one mole HF molecules, 10Kcal energy is required.

Monthly Mock Test (July 31) - Question 5

Which of the following absorb light energy for photosynthesis? [2002]

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

Each pigment has its own absorption spectrum, chlorophyll absorbs light energy in blue and red region.

Monthly Mock Test (July 31) - Question 6

Which of the following species is paramagnetic?

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

Paramagnetism is caused by the presence of atoms, ions or molecules with unpaired electrons. In NO the presence of unpaired electron is clear. Therefore it is paramagnetic.

Monthly Mock Test (July 31) - Question 7

The weakest among the following types of bonds is[1994]

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

H-bond is the weakest.

Monthly Mock Test (July 31) - Question 8

ATP is injected in cyanide poisoning because it is[1994]

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

ATP is injected in cyanide poisoning because it is necessary for cellular functions. ATP makes energy available at a spot away from the area of release of energy so it helps in maintaining cellular functions.

Monthly Mock Test (July 31) - Question 9

Among the following orbital bonds, the angle is minimum between [1994]

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

The angle between the bonds formed by px and py orbitals is the minimum i.e. 90°

Monthly Mock Test (July 31) - Question 10

How many turns of Calvin cycle yield one molecule of glucose ? [1996, 2000]

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

Each turn of Calvin cycle generates one carbon atom hence six turns of the cycle is required to generate one molecule of hexose sugar glucose.

Monthly Mock Test (July 31) - Question 11

The principle of limiting factors was proposed by[1996]

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

The principle of limiting factors was formulated by Blackmann (1905). It states that when a process is conditioned as to its rapidity by a number of separate factors, the rate of process is limited by the pace of slowest factor.

Monthly Mock Test (July 31) - Question 12

End product of citric acid/Krebs cycle is [1993]

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

The two molecules of pyruvate are completely degraded in Krebs cycle to form two molecules of ATP, 8 NADH2, 2 FADH2, CO2 and water.

Monthly Mock Test (July 31) - Question 13

Two masses M1 and M2 are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination a and b. The tension in the string is:

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

If we straighten the string horizontally hypothetically and make the free body diagram for the whole two block system we get that
M1g.sinɑ - M2g.sinβ = (M1 + M2)a
For some net acceleration of the system a
Such that a = M1g.sinɑ - M2g.sinβ / (M1 + M2)
Hence if we make a free body diagram of mass 1 only we get
M1g.sinɑ - T = M1a
Thus we get T = M1 ( g.sinɑ - a )
= M1( g.sinɑ - [M1g.sinɑ - M2g.sinβ / (M1 + M2) ] )
= M1( M2 (sinɑ + sinβ )f / (M1 + M2) )

Monthly Mock Test (July 31) - Question 14

How does pruning help in making the hedge dense?[2006]

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

When an apical bud is present on a plant, it suppresses  the growth of axillary buds, this is called apical dominance. When in pruning apical bud is cut off the axillary buds start growing & hedge become dense

Monthly Mock Test (July 31) - Question 15

The forces acting on an object are shown in the fig. If the body moves horizontally at a constant speed of 5 m/s, then the values of the forces P and S are, respectively-

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

As there is no net acceleration in horizontal or vertical direction, we can say that
P = 300N
And 2S = 2000N
Thus S = 1000N

Monthly Mock Test (July 31) - Question 16

Which one of the following acids is a derivative of carotenoids ? [2009]

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

Abscisic acid (ABA), also known as abscisin II and dormin, is a plant hormone. It functions in many plant developmental processes, includsing bud dormancy. Abscisic acid is a derivative of carotenoids. It was called “abscisin II” originally because it was thought to play a major role in abscission of fruits. At about the same time another group was calling it “dormin” because they thought it had a major role in bud dormancy. The name abscisic acid (ABA) was coined by a compromise between the two groups.

Monthly Mock Test (July 31) - Question 17

Which enzyme is most abudantly found on earth?[1999]

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

RUBISCO  is the enzyme involved in Calvin cycle. Nitrogenase catalyses nitro- genation. Invertase catalyses breaking of sucrose to glucose and fructose.

Monthly Mock Test (July 31) - Question 18

Strongest hydrogen bond is shown by [1992]

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

H – F shows strongest H-bonds.  Linear combination of two hybridized orbitals leads to the formation of sigma bond.

Monthly Mock Test (July 31) - Question 19

In animal cells, the first stage of glucose breakdown is [1994]

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

Glycolysis is the first step of respiration which occurs without requirement of O2 and is common to both aerobic and anaerobic modes of respiration.

Monthly Mock Test (July 31) - Question 20

A weight can be hung in any of the following four ways by string of same type. In which case is the string most likely to break ?

Detailed Solution for Monthly Mock Test (July 31) - Question 20
• In all the given cases the cos component of the tension in the string would balance the weight of the block while the sine component will cancel themselves as they are in pair.
• The larger would be the angle the smaller would be the cos component but as its value is fixed i.e. is mg, we get the larger the angle the larger the tension and hence the more chances of breaking of the rope.
Monthly Mock Test (July 31) - Question 21

The velocity of end `A' of rigid rod placed between two smooth vertical walls moves with velocity `u' along vertical direction. Find out the velocity of end `B' of that rod, rod always remains in constant with the vertical walls.

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

Let say end b has some velocity v in horizontal direction. Thus by constraint motion we get the component of velocities along the rod of both the ends must be equal thus we get,
u.cos (90-q) = v.cos q
Thus we get v = u.tan q

Monthly Mock Test (July 31) - Question 22

In the arrangement shown in fig. the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed.

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

Thus option D is correct

Monthly Mock Test (July 31) - Question 23

The process of growth is maximum during :    [2020]

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

Period of growth characterized by the number of cells increasing at an exponential rate.

Monthly Mock Test (July 31) - Question 24

Which one of the following generally acts as an antagonist to gibberellins? [2012M]

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

Gibberellins & ABA are antagonistic with each other. ABA counteracts many effects of gibberellins like induction of hydrolases and alpha- amylases in barley seedlings.

Monthly Mock Test (July 31) - Question 25

Out of 36 ATP molecules produced per glucose molecule during respiration [1991]

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

In glycolysis 2ATP are produced at substrate level and NADP produce 6ATP by ETS and 30ATP in mitochondria so there are only two ATP produced in cytoplasm.

Monthly Mock Test (July 31) - Question 26

Which one of the following has the shortest carbon carbon bond length ? [1992]

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

The bond length decreases in the order sp3 > sp2 > sp.
Because of the triple bond, the carbon-carbon bond distance in ethyne is shortest.

Monthly Mock Test (July 31) - Question 27

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

Photosystem I is the second photosystem in the photosynthetic light reactions of algae, plants and some bacteria. Photosystem I is so named because it was discovered before photosystem II. It produces NADPH2.

Monthly Mock Test (July 31) - Question 28

The BCl3 is a planar molecule whereas NCl3 is pyramidal because [1995]

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

As there is no lone pair on boron in BCl3 therefore no repulsion takes place. But there is a lone pair on nitrogen in NCl3. Therefore repulsion takes place. Thus BCl3 is planar molecule but NCl3 is a pyramidal molecule.

Monthly Mock Test (July 31) - Question 29

During seed germination its stored food is mobilized [NEET 2013]

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

Gibberellins stimulate the synthesis of α-amylase and proteases enzyme in germinating grains of cereals. They are involved in the conversion of starch into sugar. The proteases convert an inactive β-amylase to the active form. The active β-amylase and α-amylase together digest starch to glucose which is mobilized to meet the metabolic demands of embryo.

Monthly Mock Test (July 31) - Question 30

Oxidative phosphorylation involves simultaneous oxidation and phosphorylation to finally form [1996]

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

Oxidative phosphorylation is the synthesis of energy rich ATP from ADP and  inorganic phosphate, that is connected to oxidation of reduced coenzymes produced in cellular  respiration.

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