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 In meiosis, the daughter cells are not similar to the parent because of
  • a)
    Crossing over
  • b)
    Synapsis
  • c)
    Both 1 and 2 above
  • d)
    None of the above
Correct answer is 'A'. Can you explain this answer?

Rajeev Saxena answered
Meiosis is basically an extended and complex version of mitosis. The term “Reductional division” gives this process the underlying definition, a sequence of events that results in reduction of the total number of chromosomes (from a diploid state of 46 to a haploid state of 23; in general, from ‘2n’ to ‘n’). 

However, in the initial steps of meiosis, there is an eventful period wherein the previously duplicated chromosomes (in the S-Phase or Synthesis phase of cell cycle) exchange a part of their respective genetic material, a process termed as Recombination through crossing over (of the genetic material). The further steps are the same as in mitosis, but because of the exchange, the daughter chromosomes will have a different identity than their parents.

 Crossing over may result in
  • a)
    Addition of genetic material
  • b)
    Deletion of genetic material
  • c)
    Exchange of genetic material
  • d)
    All of the above
Correct answer is option 'C'. Can you explain this answer?

Vijay Bansal answered
Crossing over is the exchange of genes between two chromosomes, resulting in non-identical chromatids that comprise the genetic material of gametes. This process occurs during Prophase I of Meiosis, just prior to chromosome alignment and splitting of the cell.

Meiosis results in
  • a)
    Production of gametes
  • b)
    Reduction in Chromosome number
  • c)
    Introduction of variation
  • d)
    All of the above
Correct answer is option 'D'. Can you explain this answer?

Anjali Iyer answered
As previously mentioned, the first round of nuclear division that occurs during the formation of gametes is called meiosis I. It is also known as the reduction division because it results in cells that have half the number of chromosomes as the parent cell.

Match List I and List II, and select the correct answer.
  • a)
    1, 2 and 3 are correct.
  • b)
    1 and 3 are correct.
  • c)
    2 and 4 are correct.
  • d)
    Only 1 is correct.
Correct answer is option 'D'. Can you explain this answer?

Correct Answer :- d
Explanation : Genetic variation comes from crossing over, which may occur during prophase I of meiosis.
In prophase I of meiosis, the replicated homologous pair of chromosomes comes together in the process called synapsis, and sections of the chromosomes are exchanged.

Standard entropies of X2, Y2 and XY3 are given below the reaction
Q. At what temperature, reaction would be in equilibrium? 
  • a)
    500 K
  • b)
    750 K
  • c)
    1000 K
  • d)
    1250 K
Correct answer is option 'B'. Can you explain this answer?

Suresh Reddy answered
1/2X + 3/2Y2 ⟶XY3,
ΔH= −30 kJ
ΔSreaction = ∑ΔSproduct−∑ΔSreactant 
X+ 3Y→ 2XY3
​ΔH=−60 kJ
ΔSreaction = 2×50−3×40−1×60 =100−120−60=−80 JK−1mol−1
 ΔG=ΔH−TΔS=0
ΔH=TΔS
1000×(−60)=−80×T
T=750 K

The escape velocity for the moon is nearly
  • a)
    11.2km/s
  • b)
    2.4km/s
  • c)
    24km/s
  • d)
    10km/s
Correct answer is option 'B'. Can you explain this answer?

Rohit Shah answered
About 11.2 km/s
In common usage, the initial point is on the surface of a planet or moon. On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s).

Synapsis occurs in which of the following stages of meiosis?
  • a)
    Diakinesis
  • b)
    Pachytene
     
  • c)
    Leptotene
  • d)
    Zygotene
Correct answer is option 'D'. Can you explain this answer?

Vijay Bansal answered
Most of the events that function to differentiate meiosis from mitosis occur in Prophase I

Homologous chromosomes form bivalents (or tetrads) and crossing over occurs between non-sister chromatids

Prophase I is divided into 5 distinctive sub-stages:

1. Leptotene – The chromosomes begin to condense and are attached to the nuclear membrane via their telomeres
2. Zygotene – Synapsis begins with a synaptonemal complex forming between homologous chromosomes
3. Pachytene – Crossing over of genetic material occurs between non-sister chromatids
4. Diplotene – Synapsis ends with disappearance of synaptonemal complex; homologous pairs remain attached at chiasmata
5. Diakinesis – Chromosomes become fully condensed and nuclear membrane disintegrates prior to metaphase I

 Synaptonemal complex is observed during cell division in
  • a)
    Meiotic prophase
  • b)
    Mitotic prophase
  • c)
    Meiotic metaphase
  • d)
    Mitotic telophase
Correct answer is option 'A'. Can you explain this answer?

Imk Pathsala answered
The formation of tetrad is a special characteristic of Prophase 1 of meiosis 1.
The homologous pair aligns with each other and gets ready for crossing over.
So, the correct option is 'Meiotic Prophase'.

When synapsis is complete all along the chromosome, the cell is said to have entered a stage called
  • a)
    Diakinesis
  • b)
    Zygotene
  • c)
    Diplotene
  • d)
    Pachytene
Correct answer is option 'D'. Can you explain this answer?

Pachytene Stage in Meiosis

The process of meiosis is divided into several stages, one of which is the Pachytene stage. This stage occurs during the first meiotic division or meiosis I. During this stage, homologous chromosomes pair up and form structures called bivalents or tetrads. The Pachytene stage is characterized by the following events:

Synapsis

During the Pachytene stage, homologous chromosomes pair up and form structures called bivalents or tetrads. This process is called synapsis. The synaptonemal complex plays a crucial role in this process. The complex is composed of proteins that hold the homologous chromosomes together.

Crossing over

Crossing over occurs during the Pachytene stage. It is the exchange of genetic material between homologous chromosomes. This process results in the creation of new combinations of genes, which contributes to genetic diversity.

Chiasmata formation

During the Pachytene stage, the homologous chromosomes that have paired up undergo crossing over. This process results in the formation of chiasmata, which are visible under a microscope. Chiasmata are the points where the homologous chromosomes crisscross and exchange genetic material.

Conclusion

When synapsis is complete all along the chromosome, the cell is said to have entered a stage called Pachytene. During this stage, homologous chromosomes pair up and form bivalents or tetrads. Synapsis, crossing over, and chiasmata formation occur during this stage. The Pachytene stage is an essential process in meiosis as it contributes to the genetic diversity of offspring.

The gravitational potential due to the gravitational force on the earth is defined as the
  • a)
    potential energy multiplied by the mass of the object
  • b)
    potential energy of the mass placed at that point
  • c)
    numerically equal to the potential energy
  • d)
    potential energy of a unit mass at that point.
Correct answer is option 'D'. Can you explain this answer?

Nandini Patel answered
Gravitational Potential
Gravitational Potential is dened as the potential energy of a particle of unit mass at that point due to the gravitational force exerted byearth. Gravitational potential energy of a unit mass is known as gravitational potential.

Gravitational Potential is:
  • a)
    negative, scalar quantity , unit JKg-1
  • b)
    positive, vector quantity , unit JKg-1
  • c)
    positive, scalar quantity , unit JKg-1
  • d)
    negative, vector quantity , unit JKg-1
Correct answer is option 'A'. Can you explain this answer?

Ayush Joshi answered
Gravitational potential (radial fields) at a point in a radial field is the work done per unit mass against the field, in bringing a small mass from infinite distance to the point. Since gravitational fields are attractive and the potential at infinite distance is zero, all points within the field have negative values of potential. Gravitational potential is a scalar quantity with SI unit J kg^-1. The symbol used is mostly V but sometimes or Vr or V(r). A radial gravitational field is one in which the field strength has the same magnitude at all points at a given distance from the center. 

Which reaction, with the following values of ΔH and ΔS at 400 K is spontaneous and endothermic?
 
  • a)
    a
  • b)
    b
  • c)
    c
  • d)
    d
Correct answer is option 'C'. Can you explain this answer?

Om Desai answered
∆G = ∆H - T∆S
For opt (c), ∆G = 48000 - 400(135)
= 48000 - 54000
= -6000
∆G is -ve
Therefore reaction is spontaneous.

When a satellite moves in a circular orbit, the _______acceleration is provided by the gravitational attraction of the earth
  • a)
    tangential
  • b)
    centrifugal
  • c)
    centripetal
  • d)
    fictitious
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered
When any body or particle moves in circular orbit centripetal force acts on it and to move the object centripetal acceleration is necessary. So, when a satellite moves in circular orbit the centripetal acceleration is provided by the gravitational attraction of the earth.

An earth satellite is moved from one stable circular orbit to a farther stable circular orbit. which one of following quantity increases
  • a)
    gravitational potential energy
  • b)
    centripetal acceleration
  • c)
    gravitationl force
  • d)
    linear orbital speed
Correct answer is option 'A'. Can you explain this answer?

Neha Joshi answered
We know that gravitational potential is negative in sign and its magnitude decreases when distance from the massive attracting object increases, hence when considered with sign we can say that gravitational potential increases with increases in distance.

What could be the maximum value for gravitational potential energy?
  • a)
    1
  • b)
    zero
  • c)
    infinity
  • d)
    1000
Correct answer is option 'B'. Can you explain this answer?

Riya Banerjee answered
As, U=−Gm1m2/r, 
it can be used to describe the potential energy in a system of point charges (or radially symmetric spheres) as a function of their separation distance r, then the maximum value is zero at infinite separation.
Hence, its maximum value is zero at infinity.
 

The energy required to be spent by a satellite of mass m and speed v and orbital radius r in completing a circular orbit once round the earth of mass M is
  • a)
    GMm/r
  • b)
    2GMm/r
  • c)
    GMm/2r
  • d)
    zero
Correct answer is 'D'. Can you explain this answer?

Suresh Iyer answered
As the satellite is bounded by earth's gravitational field and has some velocity, it would completely revolve around the circular orbit without any external energy required.

Velocity of geostationary satellite with respect to earth is
  • a)
    10 ms-1
  • b)
    15 ms-1
  • c)
    zero
  • d)
    1 ms-1
Correct answer is option 'C'. Can you explain this answer?

Neha Joshi answered
Because geostationary satellite ( completes it's rotation in about 24hrs).and it's rotation cycle is same as that of Earth .thus it's velocity is zero with respect to earth

 Cross-like configurations when non-sister chromatids of a bivalent come in contact during the first meiotic division are
  • a)
    Chiasmata
  • b)
    Chromomeres
  • c)
    Bivalents
  • d)
    Centromeres
Correct answer is option 'A'. Can you explain this answer?

Rohan Singh answered
Chiasmata) is the point of contact, the physical link, between two (non-sister) chromatids belonging to homologous chromosomes. At a given chiasma, an exchange of genetic material can occur between both chromatids, what is called a chromosomal crossover, but this is much more frequent during meiosis than mitosis.

Direction (Q. No. 9) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.
Q. 
Statement I :Every endothermic reaction is spontaneous if TΔS > ΔH.
Statement II : Sign of ΔG is the true criterion for deciding spontaneity of a reaction.
  • a)
    Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  • b)
    Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
  • c)
    Statement I is correct but Statement II is incorrect
  • d)
    Statement II is correct but Statement I is incorrect
Correct answer is option 'B'. Can you explain this answer?

Statement I: Every endothermic reaction is spontaneous if TΔS > ΔH.
Endothermic reactions have a positive ΔH (heat absorbed). For spontaneity, we use the Gibbs free energy equation:
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS For a process to be spontaneous, ΔG<0\Delta G < 0ΔG<0. If TΔS>ΔHT\Delta S > \Delta HTΔS>ΔH, then ΔG\Delta GΔG becomes negative, indicating spontaneity. Hence, Statement I is correct.
Statement II: Sign of ΔG is the true criterion for deciding spontaneity of a reaction.
This is a fundamental thermodynamic principle. The sign of ΔG\Delta GΔG indeed determines whether a reaction is spontaneous (ΔG<0\Delta G < 0ΔG<0) or non-spontaneous (ΔG>0\Delta G > 0ΔG>0). Hence, Statement II is also correct.
Correct Answer: Since both statements are correct and the second statement explains the first, the correct option is:
A: Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.

What is gravitational potential?
  • a)
    It is defined as the work done in taking a unit mass from one point to that point
  • b)
    It is defined as the work done in taking a unit mass from infinity to that point
  • c)
    It is defined as the energy spent in taking a unit mass from infinity to that point.
  • d)
    none
Correct answer is option 'B'. Can you explain this answer?

Pranavi Chavan answered
The concept of Gravitational Potential:
The gravitational potential at a point is dened as the work done in bringing the unit mass from innity to that point without acceleration.

Gravitational potential is a scalar quantity. It is denoted by V. Its S.I. unit is J kg^-1 . Its c.g.s. unit is erg g^-1

Dimensions of gravitational potential are [M^0 L^2 T^-2 ].

Direction (Q. Nos. 10 and 11) This section contains 2 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or  MORE THANT ONE  is correct.
Q. Which of the following statements is/are true?
  • a)
    The entropy of a substance increases on going from the liquid to the vapour state at any temperature
  • b)
    An exothermic reaction will always be spontaneous
  • c)
    Reactions with a positive (ΔH° and ΔS°) can never be product favoured
  • d)
    If ΔG° for a reaction is negative, the reaction will have an equilibrium constant i greater than one
Correct answer is option 'A,D'. Can you explain this answer?

Pallabi Rane answered
The entropy of a substance increases on going from the liquid to the vapour state at any temperature. It is true. As randomness increases, entropy also increases.
∆G = ∆H - T∆S. So for spontaneous reaction, all 3(T, ∆H and ∆S) are needed.
An exothermic reaction will always be spontaneous. THis is false
Reactions with a positive (ΔH° and ΔS°) can never be product favoured. False, for a larger value of T, ∆G might be negative. 
If ΔG° for a reaction is negative, the reaction will have an equilibrium constant greater than one. True, for ΔG° less than 0, reaction is spontaneous and so, ith=s eqm constant will have value greater than 1.

Calculate the Gibbs free energy for the reaction of conversion of ATP into ADP at 293 Kelvin the change in enthalpy is 19.07 Kcal and the change in entropy is 90 cal per Kelvin.
  • a)
    7.3 cal
  • b)
    -5.3 Kcal
  • c)
    7.3 Kcal
  • d)
     -7.3 Kcal
Correct answer is option 'D'. Can you explain this answer?

Krithika Ahuja answered

Calculation of Gibbs Free Energy for the Conversion of ATP into ADP:

- Given data:
- Change in enthalpy (ΔH) = 19.07 kcal
- Change in entropy (ΔS) = 90 cal/K

- Temperature (T) = 293 K

Formula to calculate Gibbs Free Energy (ΔG):
ΔG = ΔH - TΔS

Calculation:
ΔG = 19.07 kcal - (293 K * 90 cal/K)
ΔG = 19.07 kcal - 26370 cal
ΔG = -26350 cal
ΔG = -26.35 kcal (Converting cal to kcal)

Therefore, the Gibbs free energy for the conversion of ATP into ADP at 293 K is -26.35 kcal. Since the options are in Kcal, the answer is -26.35 Kcal, which is closest to -7.3 Kcal (rounded to one decimal place). So, the correct answer is option 'D' (-7.3 Kcal).

Earth attracts a body with a force equal to its
  • a)
    weight
  • b)
    area
  • c)
    volume
  • d)
    pollution
Correct answer is option 'A'. Can you explain this answer?

Nandini Iyer answered
Weight is a consequence of the universal law of gravitation: any two objects, because of their masses, attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

When HCI(g)and NH3(g)come in contact, they react producing a white cloud of solid NH4CI
For this,
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A,B'. Can you explain this answer?

Upasana Roy answered
According to me, correct options are b and d.
For a reversible reaction, ∆S°(Universe) will be zero. ∆S°(Universe) is greater than zero for an irreversible reaction. SInce the randomness decreases, ∆S°(system) becomes less than zero and ∆S°(surrounding) becomes more than zero. BOth becomes equal with sign opposite and thus for reversible reaction, ∆S°(universe) = ∆S°(System) + ∆S°(surrounding) = 0

Direction (Q. Nos. 11-14) This section contains 2 paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d).
Passage l
Sulphur undergoes a phase transition between 80 and 110°C
S(rhombic) S (monoclinic); ΔH° = 3.213 kJ mol-1; ΔS° = 8.71 JK-1 mol-1
 
Q. Select the correct alternate(s).
  • a)
    At 80°C SR is as stable as SM 
  • b)
    At 80°C SR is less stable than SM
  • c)
    At 110°C SR is more stable than SM
  • d)
    At 110°C SR is less stable than SM
Correct answer is option 'A'. Can you explain this answer?

We have ∆G = ∆H - T∆S
For spontaneous reaction ∆G<0 and vice versa,
∆G at 80° C:-
3213 - (80+273)8.73 = 141.9 kJ mol-1
As it comes positive, the conversion of rhombic and monoclinic is non- spontaneous at 80° C.In other words, rhombic is more stable than monoclinic at this temperature.
∆G at 110° C:-
3213 - (110+273)8.73 = -119.1 kJ mol-1
As it comes positive, the conversion of rhombic and monoclinic is spontaneous at 110° C.In other words, rhombic is less stable than monoclinic at this temperature.

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