Sindh Mcat Nts 2013 Duhs And Jsmu — Solved Past Paper with Answers

Q1. Complete the sentence by choosing the most appropriate option, from the given choices (A to E) below each.I _ have to get up early tomorrow morning.

• A. Shall✓
• B. Would
• C. Had
• D. Could

Explanation: Possible words here can “will” or “shall”. Shall is mentioned in options unlike will hence that is our correct answer.

Q2. Complete the sentences by choosing the most appropriate option, from the given lettered choices (A to E) below each.He says he is a _, but he can’t play the piano or any other instrument and he can’t sing.

• A. Musician✓
• B. Magician
• C. Physician
• D. Dietician
• E. Technician

Explanation: The use of the conjunction “but” implies that the following sentence is against the word that should be in place of the blank. Since musicians normally know how to play the piano or any other instrument so it would fit best with this conjunction.

Q3. Identify the word or phrase that needs to be changed for the sentence to be correct. The secretary must type these letters before lunch.

• A. The
• B. Must
• C. These letters
• D. Lunch
• E. No error✓

Explanation: There is no error in this sentence.

Q4. Identify the word or phrase that needs to be changed for the sentence to be correct:I go by the post office every morning on my way too work.

• A. I
• B. Post office
• C. On
• D. Too✓
• E. No error

Explanation: The verb “too” is wrongly used here.” Too” implies intensity or inclusiveness.The correct verb to use here will be “to”, which is a preposition.

Q5. Choose the lettered word or phrase that is most nearly opposite in meaning to the word in capital letters.GRIEVE

• A. Disturb
• B. Hurt
• C. Feel sad
• D. Rejoice✓

Explanation: GRIEVE means to be "saddened" or "depressed".The opposite of this word is Rejoice which means to be happy or elated.

Q6. Choose the lettered word or phrase that is most nearly opposite in meaning to the word in capital letters.BLAZE

• A. Quench✓
• B. Burn
• C. Rage
• D. Shine
• E. Flame

Explanation: A very large or fiercely burning fire is called BLAZE.Quench means to “extinguish” which is the opposite.

Q7. Choose the word most similar in meaning to the capitalized ones.GENTLE

• A. Rough
• B. Expert
• C. Heartless
• D. Calm✓

Explanation: The word most similar’ to the word ‘GENTLE’ is ‘CALM’.Both refer to the relaxed or smooth state.

Q8. Choose the word most similar in meaning to the capitalized ones.SQUASH:

• A. Squeeze✓
• B. Beat
• C. Evolution
• D. Pace
• E. Rapidity

Explanation: ‘SQUASH’ means to “crush or squeeze (something) with force so that it becomes flat, soft, or out of shape.”

Q9. Read the following passage to answer the given question: However, it must be recognized that science has its limitations. Its methods apply only to those things which can be observed, measured, and treated mathematically. It has nothing to do with values - save those of truth and accuracy. It has nothing to do with happiness, goodness, beauty, courage, adventure, justice, altruism, friendship, love of family, and love of country. Yet all these values enter into a man's conception of what is a good personal life within a good society. Honest and intelligent men can differ profoundly on the nature of these values and their respective degrees of importance. Hence the contrast between the modern world's command of material things and its tragic failure to organize a harmonious world society.It can be inferred from the passage that the author thinks

• A. Science to be the necessary element of a good personal life
• B. Science brings happiness, goodness, beauty, courage adventure,justice, altruism, friendship, love of family and love of country to a man's life
• C. Science has successfully brought a balance between the modern world's command of material things and a harmonious world society
• D. A good personal life can be achieved by recognizing the nature of values and their degree of importance✓
• E. The limitations of science are negligible

Explanation: It can be inferred from the passage that a good personal life can be achieved by recognizing the nature of values and their degrees of importance.(Lines 5-7).

Q10. Read the following passage to answer the given question:However, it must be recognized that science has its limitations. Its methods apply only to those things which can be observed, measured, and treated mathematically. It has nothing to do with values- save those of truth and accuracy. It has nothing to do with happiness, goodness, beauty, courage, adventure, justice, altruism, friendship, love of family, and love of country. Yet all these values enter into a man's conception of what is a good personal life within a good society. Honest and intelligent men can differ profoundly on the nature of these values and their respective degrees of importance. Hence the contrast between the modern world's command of material things and its tragic failure to organize a harmonious world society.According to the paragraph, science applies certain values, which of the following describes these values?

• A. Truth and justice
• B. Love of country and accuracy
• C. Truth and accuracy✓
• D. Justice and accuracy

Explanation: Science applies to certain values such as truth and accuracy as mentioned is line 3.

Q11. Two blocks, X and Y, of masses m and 2m respectively, are accelerated along a smooth horizontal surface by a force F applied to block X, as shown in the diagram.What is the magnitude of the force exerted by block Y on block X during this acceleration?

• A. 0
• B. F/3
• C. F/2
• D. 2F/3✓
• E. F

Explanation: Both blocks will be accelerated when the block X is applied by a force of magnitude F. Since, the total mass will be m+2m which is 3m.So, the acceleration of the system will be given as F=ma or a=F/3m. So, from this we can find the force acting on the block Y. So, to calculate the force on the block Y we have that F(y) =2m*a or 2m*F/3m which on solving we will get the force on the block y as F(y)= 2F/3Magnitude of the force exerted by block Y on block X during this acceleration is = 2F/3

Q12. A box of mass m = 6 kg slides with speed v = 4 m/s across a frictionless floor. It explodes into two pieces: one piece, with mass m1 = 2 kg move in the same direction with speed v1 = 8 m/s. Find the velocity of the second piece.

• A. 2 m/s✓
• B. 4 m/s
• C. 8 m/s
• D. 9 m/s
• E. 11 m/s

Explanation: Before explosion m1 = 6 kg , u1 = 4m/sAfter the explosion, it explodes into two pieces, such that m1 = 2kg, v1 = 8m/s, m2 = 4 kg, v2 =?In case of an explosion, momentum will remain conserved.According to the Law of Conservation of Momentum:Momentum before explosion =Momentum after the explosionm1u1 = m1v1 + m2v2(6)(4) = (2)(8)+(4)(v2)24 = 16 + 4v28 = 4v2v2 = 2 m/s

Q13. A generator of e.m.f. 80 V has an internal resistance of 0.04 Ω. If its terminal voltage is 75 V, determine the current.

• A. 125 A✓
• B. 135 A
• C. 145 A
• D. 155 A
• E. 165 A

Explanation: To determine the current, use the formula I = (E - V) / r, where E is the electromotive force (80V), V is the terminal voltage (75V), and r is the internal resistance (0.04Ω). Substitute the values into the formula to get I = (80V - 75V) / 0.04Ω = 125A. Therefore, option A is correct. All other options result from incorrect calculations based on this formula.

Q14. An object 4 cm high is located 10 cm from the converging lens, whose focal length is 20 cm. The image that will be formed is:

• A. Virtual
• B. Erect
• C. Real
• D. Both Virtual and Erect✓
• E. Both Erect and Real

Explanation: The object is placed 10 cm from a converging lens with a focal length of 20 cm, meaning it is inside the focal point. For converging lenses, when the object is placed between the focal point and the lens, the image formed is virtual and erect. Thus, the correct answer is 'Both Virtual and Erect'.Options stating the image is real or simply erect without being virtual are incorrect because a real image would be inverted, and a solely erect image implies it must also be virtual.

Q15. A rotating wheel of radius 0.5 m has an angular velocity of 5 rad/s at some instant and 10 rad / s after 5 s. Find the angular acceleration of a point on its rim.

• A. 1 rad/s2✓
• B. 3 rad/s2
• C. 5 rad/s2
• D. 7 rad/s2
• E. 9 rad/s2

Explanation: The angular acceleration of a rotating object can be found using the formula α = (ωf - ωi) / t, where ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time interval. In this case, the initial angular velocity (ωi) is 5 rad/s, the final angular velocity (ωf) is 10 rad/s, and the time interval (t) is 5 seconds. Substituting these values into the formula gives: α = (10 rad/s - 5 rad/s) / 5 s = 1 rad/s2. Therefore, Option A is correct. All other options are incorrect as they do not result from applying this formula to the given values.

Q16. A block of mass 50 kg is pulled on a frictionless floor by a force of 210 N directed at 30 degrees to the horizontal. If the block moves 3.0 m, what is the work done on it by the applied force?

• A. 115√2 J
• B. 215√2 J
• C. 315√3 J✓
• D. 415√2 J
• E. 515√3 J

Explanation: w=f×d×cos anglew=210×3×cos 30w=630×√3/2=315√3

Q17. A ball of mass 2 kg travelling at 8 ms-1 strikes a ball of mass 4kg travelling at 2ms-1. Both balls are moving along the same straight line as shown. After collision, both balls move at the same velocity v. What is the magnitude of the velocity V?

• A. 4 ms-1✓
• B. 6 ms-1
• C. 5 ms-1
• D. 8 ms-1

Explanation: For inelastic collision, momentum is still conversed. Momentum before collision = momentum after collision. m1u1+m2u2=(m1+m2)V (2)(8)+(4)(2)=(2+4)V V=4ms-1

Q18. A shot leaves a gun at the rate of 160 m/s. Calculate the greatest distance to which it could be projected. (Take g = 10 m/s2).

• A. 2460 m
• B. 2560 m✓
• C. 2680 m
• D. 2760 m
• E. 2860 m

Explanation: For maximum distance(range) we assume that it is projected at 45 degrees. So, Rmax = u^2/g. (Following is the correct solution):

Q19. On the ground, the gravitational force on a satellite is W. What is the gravitational force on the satellite when at a height R/50, where R is the radius of the Earth?

• A. 1.04 W
• B. 1.02 W
• C. 0.98 W
• D. 0.96 W✓
• E. 2.13 W

Explanation: Because of equality F = - m. (G.M/r ²) = - m.g one draws G = G.M/r ² = [G.M/R ²] * [1 (1+z/R)]² =g○/(1+z/R) ² ► g (z) = g○/(1+z/R) ² ▬▬▬▬▬ > W/W○ = g/g○ = (1+z/R)⁻ ² = (1 + 1/50)^-2 = 0.961

Q20. When the aircraft Concorde is moving in a horizontal plane at a constant speed of 650 ms-1, its turning circle has a radius of 80 km. What is the ratio of the centripetal force to the weight of the aircraft? (g = 9.8 m/s2)

• A. 8.3 x 10^4
• B. 0.54✓
• C. 1.9
• D. 52
• E. 540

Explanation: The ratio of Centripetal force to the weight of the aircraft=mv^2/r divided by mgm*(650)^2/80*1000 / m*9.81=0.54.

Q21. The amount of heat at constant volume is called as:

• A. Internal energy✓
• B. Enthalpy
• C. Entropy
• D. Temperature
• E. Pressure

Explanation: At constant volume, the heat of the reaction is equal to the change in the internal energy of the system. Meanwhile, the heat of reaction at constant PRESSURE is known as enthalpy

Q22. A parallel beam of white light is incident normally on a diffraction grating. It is noted that the second-order and third-order spectra partially overlap. Which wavelength in the third-order spectrum appears at the same angle as the wavelength of 600 nm in the second-order spectrum?

• A. 300 nm
• B. 400 nm✓
• C. 600 nm
• D. 900 nm
• E. 950 nm

Explanation: In a diffraction grating, the relationship between the order of the spectrum (m), wavelength (λ), and the angle of diffraction (θ) is given by the equation: mλ = dsinθ, where d is the grating spacing.For the second-order spectrum with λ = 600 nm, we have:2(600) = dsinθFor the third-order spectrum, where the wavelength is λ', we have:3λ' = dsinθEquating the two equations (since they appear at the same angle θ):3λ' = 2(600)Solving for λ': λ' = (2 * 600) / 3 = 400 nmTherefore, the wavelength in the third-order spectrum that coincides with the 600 nm wavelength in the second-order spectrum is 400 nm. The other options do not satisfy this condition based on the diffraction grating formula.

Q23. If the frequency of a pendulum is four times greater on an unknown planet than it is on Earth, then the gravitational constant on that planet is

• A. 16 times greater✓
• B. 4 times greater
• C. 4 times lower
• D. 16 times lower
• E. 24 times lower

Explanation: f=1 /2π√g/LIf the frequency is increased by a factor of 4 the impact on the value of g will be by a factor of 16.

Q24. A submarine sends out a sonar signal (sound wave) in a direction directly downward. It takes 2.3 s for the sound wave to travel from the submarine to the ocean bottom and back to the submarine. How high (approx) up from the ocean floor is the submarine? (The speed of sound in water is 1,490 m/s.)

• A. 1,700 m✓
• B. 3,000 m
• C. 5,000 m
• D. 9,000 m
• E. It cannot be determined from the information given.

Explanation: Time of receiving of an echo t=2.3 sDistance between the object and the submarine d=? mThus the sound wave travels a distance of 2d in 2.3 seconds.The velocity of the sound in water = 1490 m/s∴ 2d=1490×2.32×d=1490×2.3 ⟹d=1700 m

Q25. A 40 kg block is resting at a height of 5m off the ground. If the block is released and falls to the ground. What is its total energy at a height of 2m? (g= 10 m/s2).

• A. 0 J
• B. 400 J
• C. 2 kJ✓
• D. 6 kJ
• E. It cannot be determined from the information given.

Explanation: The total energy the body possesses is the same at all times: We can calculate the energy of the body when it is 5m off the ground and that’ll be energy when is 2m above the ground too (though in KE form too) PE=mgh PE=(40)(10)(5)=2000J=2kJ

Q26. Gamma-ray (𝝲) can produce ionization in which of the following way/s?I. It may lose all its energy in a single encounter with the electron of an atom (Photoelectric effect).II. It may lose only a part of its energy in an encounter (Compton effect).III. Very few very high energy 𝝲 ray photons may impinge directly on heavy nuclei, be stopped and annihilated giving rise to electron-positron pairs (The materialization of energy).

• A. I only
• B. II only
• C. III only
• D. I and III only
• E. I, II and III✓

Explanation: Gamma ray photons can interact with matter in all of the ways described above. All three statements mentioned above are correct.Gamma rays (𝝲) can produce ionization in the following ways:I. Photoelectric Effect: In this process, a gamma-ray photon interacts with an electron bound to an atom. The gamma-ray transfers all its energy to the electron, causing the electron to be ejected from the atom. This results in the ionization of the atom. The ejected electron is called a photoelectron.II. Compton Effect (Compton Scattering): In this process, a gamma-ray photon collides with an electron in an atom. The gamma-ray transfers only a part of its energy to the electron, changing the direction and energy of both the gamma-ray and the electron. The gamma ray continues its path with reduced energy, and the electron is ejected from the atom, causing ionization.III. Materialization of Energy (Pair Production): For very high-energy gamma-ray photons, when they approach the vicinity of a heavy nucleus, they can spontaneously convert their energy into an electron and a positron (an antimatter counterpart of an electron). The gamma-ray photon ceases to exist, and the newly formed electron-positron pair carries the energy of the original gamma-ray. This process is known as pair production and results in the ionization of the material surrounding the heavy nucleus.So, all three processes (Photoelectric effect, Compton effect, and Pair Production) can lead to ionization by gamma rays. The likelihood of each process occurring depends on the energy of the gamma-ray and the properties of the interacting material.

Q27. The internal energy of an object increases in an adiabatic process. Which of the following must be true regarding this process?

• A. The kinetic energy of the system is changing
• B. The potential energy of the system is changing
• C. Work is done on the system✓
• D. Heat flows into the system
• E. No work is done on the system.

Explanation: According to the first law of thermodynamics: U=q+W. Where Q is heat change; U is the change in internal energy and W is work done on the system. Q=0 because it is an adiabatic process, so the increase in internal energy should cause an increase in work done on the system.

Q28. An electric rod of 2000 watts rating boils a certain quantity of water in 10 minutes, the heat which generated for boiling this water is:

• A. 8 x 10^4 Joules
• B. 12 x 10^5 Joules✓
• C. 19 x 10^5 Joules
• D. 23 x 10^5 Joules
• E. 37 x 10^5 Joules

Explanation: Energy=Power x time Energy=2000 x 10 x 60 Energy=12 x 10^5 J

Q29. A nucleus consists of 19 protons and 20 neutrons. The conventional symbol of this nucleus is

• A. 11 Na 12
• B. 19 K 19
• C. 19 K 39✓
• D. 19 K 20
• E. 12 Na 12

Explanation: The conventional symbol of an element contains the mass number and proton numberThe mass number of the element shown above(20n+19p=39)The only option that has a mass number with it is option C.

Q30. Statements: 1. During the past year, Josh saw more movies than Stephen. 2. Stephen saw fewer movies than Darren. 3.Darren saw more movies than Josh. If the first two statements are true, the third statement is:

• A. True
• B. False
• C. Uncertain✓
• D. None of these

Explanation: The statements tell us that Josh > Stephen and Darren > Stephen. However, there is no information to directly compare the number of movies seen by Josh and Darren. Darren could have seen more, fewer, or the same number of movies as Josh.

Q31. The half-life of 14C is approximately 5,730 years, while the half-life of 12C is essentially infinite. If the ratio of 14Cto 12C in a certain sample is 25% less than the normal ratio in nature, how old is the sample?

• A. Less than 5,730 years✓
• B. Approximately 5,730 years
• C. Significantly greater than 5,730 years, but less than 11,460 years
• D. Approximately 11,460 years
• E. Approximately 15,730 years

Explanation: Because the half-life of carbon-12 is essentially infinite, a 25 percent decrease in the ratio of carbon-14 to carbon-12 means the same as a 25 percent decrease in the amount of carbon-14. If less than half of the carbon-14 has deteriorated, then less than one half-life has elapsed. Therefore, the sample is less than 5730 years old. Be careful with the wording here—the question states that the ratio is 25% less than the ratio in nature, not 25% of the ratio in nature, which would correspond to choice (D).

Q32. Which of the following statements is inconsistent with Bohr's set of postulates regarding the hydrogen atom model regarding the emission and absorption of light?

• A. Energy levels of the electrons are stable and discrete
• B. An electron emits or absorbs radiation only when making a transition from one energy level to another
• C. To jump from lower energy to higher energy, an electron must absorb a photon of precisely the right frequency such that the photon's energy equals the energy difference between the two orbits
• D. To jump from a higher energy to a lower energy, an electron absorbs a photon of frequency such that the photon's energy is exactly the energy difference between the two orbits✓
• E. None of the above

Explanation: Bohr’s Model has quantised angular momentum, which is option A. It is consistent with the model. Energy changes only take place if an electron changes shells, as stated in option B. It is consistent with the model. When an electron jumps from a lower orbit to a higher orbit, it absorbs energy such that ΔE= E2-E1. It is consistent with the model. When an electron jumps from a higher orbit to a lower orbit, it emits energy. In the question statement, it's wrong that “It absorbs energy”. Hence, this is not consistent with the model. Thus, Option D is correct, while others are rendered as incorrect.

Q33. The temperature of a body at 100 °C is increased by Δ𝛉 as measured on the Celsius scale. How is this temperature change expressed on the Kelvin scale?

• A. Δ𝛉 + 373
• B. Δ𝛉 + 273
• C. Δ𝛉 + 100
• D. Δ𝛉✓
• E. Δ𝛉 + 212

Explanation: The change in temperature in the kelvin and Celsius scale is the same hence the answer will be D.

Q34. In an astronomical telescope, the distance between the objective and the eyepiece is called :

• A. Magnifying Power of the telescope
• B. Width of the telescope
• C. Length of the telescope✓
• D. Height of the telescope
• E. Diameter of the lens of the telescope

Explanation: The distance between the eyepiece lens and the objective lens is called the length of the telescope as shown below :

Q35. Two resistors R1 and R2 may be connected either in series or in parallel across a battery of zero internal resistance. It is required that the joule heating for parallel combination be five times that for series combination. If R1 = 100Ω, find possible values of R2.

• A. 38.2 Ω✓
• B. 35.3 Ω
• C. 48.2 Ω
• D. None of the above

Explanation: Correct option is 'A'.

Q36. Candela is the luminous intensity, in the perpendicular direction of a surface _ square meter of a black body at the temperature of freezing platinum under a pressure of 101325 newtons per square meter.

• A. 1/300000
• B. 1/600000✓
• C. 1/900000
• D. 1/1200000
• E. 1/1500000

Explanation: The candela is the luminous intensity, in the perpendicular direction, of a surface of 1/600000 square meters of a blackbody at the temperature of freezing platinum under a pressure of 101325 newtons per square meter.

Q37. Find the work done in moving an object along a straight line from (3,2,−1) to (2,−1,4) in a force field given by F=4i^−3j^​+2k^.

• A. 10 units
• B. 12 units
• C. 15 units✓
• D. 17 units
• E. 5 units

Explanation: Correct option is 'C'.

Q38. A constant force acting on a body of mass 5 kg changes its speed from 2 m/s to 7 m/s in 10 s the direction of motion of the body remains unchanged. Find the magnitude of the force.

• A. 0.5 N
• B. 1.5 N
• C. 2.5 N✓
• D. 3.5 N
• E. 4.5 N

Explanation: F= m(v-u)/t= 5(7-2)/10= 2.5N

Q39. What force should be applied on a 10 kg body so that it moves down in a vacuum with an acceleration of 3 m/s2 ? (Take g=9.8 m/s2)

• A. 42 N
• B. 46 N
• C. 68 N✓
• D. 53 N
• E. 58 N

Explanation: The question wants the object to only move down at 3 m/s2, we need to apply an upward force that would accelerate the object downwards at 9.8–3=6.8 m/s2. Using F=ma, where m = 10 kg and a = 6.8 m/s2, we find F to be 68 N upwards.

Q40. A special class of waves that does not need a material medium for their propagation are called

• A. Electric waves
• B. Magnetic waves
• C. Electromagnetic waves✓
• D. Sound waves
• E. Earthquake's shock waves

Explanation: Electromagnetic waves do not require a medium for travel since they are not transmitted by the vibration of particles. The waves are generated by the oscillation of electric and magnetic fields which can exist anywhere.Electric waves do not isolatedly exist rather they depend on the magnetic field to form EM waves.A sound wave is a pattern of disturbance caused by the movement of energy traveling through a medium (such as air, water, or any other liquid or solid ).Seismic waves are usually generated by movements of the Earth's tectonic plates but may also be caused by explosions, volcanoes, and landslides. When an earthquake occurs shockwaves of energy, called seismic waves, are released from the earthquake focus.A changing magnetic field will induce a changing electric field and vice-versa—the two are linked. These changing fields form electromagnetic waves. Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate.

Q41. Dissolution of ammonium nitrate in water is an endothermic process. Which of the following graphs shows how the temperature alters as the ammonium nitrate is added to water and then the solution is left at room temperature?

• A. Option A
• B. Option B✓
• C. Option C
• D. Option D

Explanation: Since it is an endothermic process it takes heat from the surrounding making the temperature fall below room temperature. The temperature then gradually then goes back to room temperature.

Q42. From a mixture of CO2 and H2 gases, CO2 can be separated by passing the mixture through:

• A. Water at high temperature
• B. Water under high pressure✓
• C. Cold water
• D. Acidified water

Explanation: Carbon dioxide is separated from hydrogen by passing the mixture through the water under high pressure. When carbon dioxide is compressed into water at pressure, the CO2 gas dissolves into the liquid. When pressure is released, CO2 releases from the liquid as a gas in the form of gas bubbles. This process is also called carbonation and is the basis of many popular beverages. Since hydrogen will not dissolve in the water, the two gases will be effectively separated. Carbon dioxide will not dissolve in the forms of water given in all the remaining options, so separation can not occur, leaving only option B as the correct option.

Q43. Alkanes having five to seventeen carbon atoms per molecule are

• A. liquids✓
• B. solids
• C. gases
• D. semi solid wax

Explanation: Following are the state of alkanes with given carbon numbers: C1-C4=Gases C5-C17=Liquids C18+=Solids.

Q44. Which type of isomerism depends on the distribution of carbon atoms on each side of the functional group?

• A. Structural isomerism
• B. Functional isomerism
• C. Chain isomerism
• D. Metamerism✓

Explanation: This type of isomerism arises due to the presence of different alkyl chains on each side of the functional group. A structural isomer is a type of isomerism that uses the same formula but different structures. Functional isomerism is a type of isomerism in which two structures have the same formula but different functional groups. Chain isomers have the same molecular formula, but the way their carbon atoms are joined together differs from isomer to isomer.

Q45. Non-stoichiometric compounds are formed by:

• A. Only alkali metals
• B. Only transition elements✓
• C. Only noble gases
• D. None of the above

Explanation: Non-stoichiometric compounds are chemical compounds, almost always solid inorganic compounds, having elemental composition whose proportions cannot be represented by a ratio of small natural numbers.

Q46. Which of the following statements are correct about noble gases?

• A. They belong to 18th group of the modern periodic table.✓
• B. Except Rn, all the remaining noble gases are present in the air.
• C. These elements belong to 17th group in the IUPAC system.
• D. These are highly reactive elements.
• E. None of the above

Explanation: Noble gases are elements located in the 18th group of the modern periodic table. This group is characterized by elements having a full valence electron shell, which makes them highly stable and largely unreactive. Therefore, Option A is the correct statement about noble gases. Options B, C, and D are incorrect because they misstate either the presence of noble gases in the air, their group number, or their reactivity. Option E is incorrect because Option A is indeed correct.

Q47. Magnesium oxide is used in the making of the lining of blast furnaces. It is extracted from seawater as follows. Aqueous calcium hydroxide is added to seawater. Ca(OH)2 (aq) + MgCl2 (aq) + Mg(OH)2(s) + CaCl2 (aq)The magnesium hydroxide is then filtered off and roasted. Which of the following comparisons between calcium and magnesium explains why magnesium hydroxide forms?

• A. Magnesium is less electropositive than calcium.
• B. Magnesium is lower than calcium in the reactivity series
• C. The enthalpy change of hydration for Mg2+ is less exothermic than for Ca2+
• D. The solubility product for Mg(OH)2 is lower than that for Ca(OH)2✓
• E. The magnitude of the lattice energy of Mg(OH)2 is less than that of Ca(OH)2

Explanation: The lesser the solubility product the more likely the precipitation will be. Magnesium hydroxide forms as a precipitate as compared to aqueous calcium hydroxide because the solubility product of Mg(OH)2 is lower than Ca(OH)2

Q48. When methylbenzene is treated with bromine in the presence of a catalyst, a mixture of two monobromo isomers is formed. What are the structures of these two isomers?

• A. Ortho-bromotoluene and meta-bromotoluene
• B. Meta-bromotoluene and para-bromotoluene
• C. Ortho-bromotoluene and para-bromotoluene✓
• D. Ortho-bromotoluene and ortho-bromotoluene
• E. Para-bromotoluene and para-bromotoluene

Explanation: The correct answer is Ortho-bromotoluene and para-bromotoluene. The methyl group (-CH3) on the benzene ring acts as an ortho-para directing group, meaning it directs incoming substituents (like bromine in this case) to the ortho (positions 2 and 6) and para (position 4) positions on the ring. Thus, when methylbenzene (toluene) is treated with bromine in the presence of a catalyst, bromine substitutes at these positions, forming ortho-bromotoluene and para-bromotoluene as the two monobromo isomers.Other options are incorrect because they either suggest meta substitution, which is not favored by the -CH3 group, or imply duplication of the same isomer, which does not align with the formation of two distinct isomers.

Q49. Which one of the following formulae represents the organic compound formed when methylbenzene is heated under reflux with alkaline potassium manganate (VII) solution and the mixture then acidified?

• A. A
• B. B
• C. C
• D. D
• E. E✓

Explanation: Explanation is given below.

Q50. What substrates are required to add a NO2 group to benzene?

• A. Concentrated HNO3/Concentrated H2SO4/100˚C
• B. Concentrated HNO3/Concentrated H2SO4/50˚C✓
• C. Concentrated HNO3/50˚C
• D. Concentrated HNO3/100˚C
• E. Concentrated HNO3/Concentrated HCl/50˚C

Explanation: Nitration of benzene involves the substitution of a hydrogen atom on the benzene ring by a nitro group, NO2. The reaction requires the presence of both concentrated nitric acid and concentrated sulfuric acid as reagents, and it is conducted at a temperature not exceeding 50°C. This temperature control is important to avoid side reactions and to ensure the formation of the desired nitrobenzene. The sulfuric acid acts as a catalyst by generating the nitronium ion, NO2+, which is the active electrophile in the reaction. The other options are incorrect because they either exceed the temperature limit, omit the necessary sulfuric acid, or replace it with an ineffective substitute.

Q51. The false statement about lithium is:

• A. It is softer than other alkali metals✓
• B. It is least reactive
• C. It possesses higher melting and boiling points
• D. It forms chloride which is soluble in alcohol

Explanation: Lithium is actually harder than the other alkali metals (it can even scratch glass), while the others are very soft. So the statement “It is softer than other alkali metals” is false.

Q52. Diamond and Graphite

• A. are isotopes
• B. are isomers
• C. are allotropes✓
• D. have the same structure
• E. are equally hard

Explanation: Diamond and Graphite are allotropes of carbon. The term allotrope refers to one or more physical forms of a chemical element that occurs in the same physical state. Allotropes may show differences in chemical and physical properties.

Q53. Borax exists in nature as:

• A. Na2B4O7.10H2O✓
• B. Na2B4O7.7H2O
• C. Na2B4O7.5H2O
• D. Na2B4O7.3H2O
• E. Na2B4O7.H2O

Explanation: Na2B4O7.10H2O . This is just a fact that should be recalled in the exam. Borax, also called sodium tetraborate, is a powdery white mineral that has been used as a cleaning product for several decades. It has many uses: It helps get rid of stains, mold, and mildew around the house. It can kill insects such as ants.

Q54. Outer transition elements belong to:

• A. s-block
• B. p-block
• C. d-block✓
• D. f-block
• E. None of the above

Explanation: The d-block elements are the outer transition elements as opposed to the f-block, which are the inner transition elements.

Q55. Transition elements have coloured compounds because:

• A. Their bond energy is low
• B. They easily absorb energy
• C. Splitting of the five degenerate d-orbitals take place✓
• D. The d-orbitals are very close to p-orbitals
• E. Degenerate p-orbitals are present

Explanation: The reason why transition metals, in particular, are colorful is that they have unfilled or half-filled d orbitals. The Crystal Field Theory explains the splitting of the d orbital, which splits the d orbital into a higher and lower orbital.

Q56. In a double-bonded carbon atom (C=C)

• A. Hybridization occurs between the s-orbital and one p-orbital
• B. Hybridization occurs between the s-orbital and two p-orbitals✓
• C. Hybridization occurs between the s-orbital and three p-orbitals
• D. No hybridization occurs between the s-and p-orbitals
• E. Hybridization occurs between two s-orbitals and one p-orbital

Explanation: Since there is a double bond formed between two carbon atoms, this implies that the hybridization is sp2, and sp2 hybridization involves one s and two p orbitals.One bond=sp3Two bonds=sp2Three bonds=sp

Q57. The radii of the second orbit of the hydrogen atom calculated from Bohr's model is:

• A. 0.529 Å
• B. 4.8 Å
• C. 2.116 Å✓
• D. 3.4 Å
• E. 1 Å

Explanation: In Bohr's model, the radius of the nth orbit of a hydrogen atom is given by the formula rn = 0.529 x n²/Z Å, where n is the principal quantum number and Z is the atomic number. For hydrogen, Z = 1. For the second orbit (n=2), the radius is calculated as:r2 = 0.529 x 2²/1 = 0.529 x 4 = 2.116 Å.Therefore, the correct answer is 2.116 Å. The other options are incorrect because they either represent the radius of a different orbit (e.g., 0.529 Å for n=1) or are arbitrary values that do not match the calculation for n=2.

Q58. The series limit for the Balmer series of hydrogen spectrum occurs at 3664 A°. Calculate Ionization energy of hydrogen atom.

• A. 21.7 x 10-19 J
• B. 6.626 x 10-34 J
• C. 5.425 x 10-19 J✓
• D. 3664 x 10-10 J
• E. 3 x 108 J

Explanation: The Balmer emission lines correspond to transitions down to the level for which n=2. Given series is the Balmer series, so n1=2 and n2=∞ λ=3664A° Change in ionisation energy= Einfinity−E2= −E2 -E2 =-hc/λ -E2=6.626×10-34 ×3×10⁸/(3664x10-10) -E2 =−5.42×10−19 J so option C is the only correct option.

Q59. The amount of energy released by adding an electron in the valence shell is

• A. Ionization Energy
• B. Electron Affinity✓
• C. Electronegativity
• D. Atomic Radius
• E. Atomisation Energy

Explanation: Electron affinity is defined as the amount of energy released when an electron is added to the valence shell of a gaseous, neutral, and isolated atom.

Q60. Rate = K [N2O5 ] has _ of reaction.

• A. First order✓
• B. Pseudo first order
• C. Second order
• D. Third order
• E. Pseudo order

Explanation: The order of the reaction is the sum of all the exponents of reactants in the rate equation. Since the sum of exponents of reactants in this equation is 1. The order of reaction, in this case, is equal to 1.

Q61. Which one of the following molecules has the shortest distance of carbon atoms?

• A. CH3 - CH3
• B. CH2 = CH2
• C. CH ≡ CH✓
• D. CH3 - CH2 - CH3
• E. CH2 = CH2 - CH3

Explanation: The bond lengths of ethane, ethene and ethyne are 154pm, 133pm and 120pm respectively. Hence, ethyne has shortest bond length.

Q62. The most dense element is

• A. Li
• B. K
• C. Ca
• D. Ba✓
• E. Rb

Explanation: Down the group, the density of the elements increases, and Group II elements have a higher density than Group I elements because of their higher mass and lesser volumes. Ba is the only group II in all the options above.

Q63. The isomers must have the same:

• A. Physical properties
• B. Molecular Formula✓
• C. Structural Formula
• D. Chemical properties
• E. Both Molecular and Structural formula

Explanation: Isomers have the same molecular formula but different structures.

Q64. For a reaction 2A + B ⇌ C + D the active mass of B is kept constant and that of A is tripled. It is observed that the rate of reaction:

• A. Decreases three times
• B. Decreases nine times
• C. Increases six times
• D. Increases nine times✓

Explanation: By the law of mass action, the rate ∝ (active mass of A)² × (active mass of B). Since B is constant and A is tripled, the rate becomes 3² = 9 times larger. Therefore, the reaction rate increases nine times.

Q65. This is is a structure of

• A. Menadione✓
• B. 𝛂 - Tocopherol
• C. Calciferol
• D. Thiamine
• E. Pyridoxine

Explanation: The above structure is of Menadione.

Q66. The final answer of the expression is:

• A. 1
• B. 2✓
• C. 3
• D. 4

Explanation: In calculations involving multiplication and division, the significant figures in the final result are determined by the input number with the fewest significant figures.In this case, the number 9.0 has only two significant figures, which constrains the final answer to also have two significant figures.Options 1, 3, and 4 provide incorrect numbers of significant figures, which do not adhere to the rule for significant figures based on the least precise measurement.

Q67. If we take 2.2g of CO2, 6.02 x 1021 atoms of nitrogen and 0.03g atoms of sulfur, then the molar ratio of CO2, N and S atoms will be:

• A. 1:2:5
• B. 5:1:2
• C. 2:5:3
• D. 5:1:3✓

Explanation: We have to calculate moles for each of the elements givenFor CO2 mass if given so use the formulamoles = mass/mr (Mr = 12 + 32 = 44)moles = 2.2/44 moles = 0.05Nitrogen atoms are given so using the formulaatoms = moles x avogadros constantmoles = atoms/ avogadros constant moles = 6.02 x1021 / 6.02x1023moles = 0.01For sulfur, gram atoms are given and gram atoms = molesSo, moles = 0.03Now calculating the molar ratioCO2 : N : S0.05 : 0.01 : 0.03 (Multiply by 100 to convert to whole numbers ratios already in lowest form)5 : 1 : 3 This is the following solution:

Q68. A system at equilibrium can be disturbed by:

• A. Concentration change
• B. Pressure change
• C. Temperature
• D. All of the above✓

Explanation: Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction.

Q69. For a multi-electron atom, the highest energy level among the following is

• A. n = 5, l = 0, m = 0 , s = +1/2
• B. n= 4 , l = 0 , m = 0 , s = - 1/2
• C. n= 3 , l = 1, m = 1 , s = - ½
• D. n= 5 , l = 1 , m = 0 , s = +1/2✓

Explanation: Higher the n + l value higher is the energy. If the n+ l value is same then higher energy corresponds to the greater n value orbital.

Q70. Equal weights of methane and hydrogen are mixed in an empty container at 25°C. The fraction of total pressure exerted by hydrogen is:

• A. ½
• B. 8/9✓
• C. 1/9
• D. 16/17

Explanation: According to Dalton's law of partial pressure, the total pressure of a mixture of gases is the sum of the partial pressure of the gases in the mixture. We can calculate the partial pressure by the following formula: Partial pressure H₂/ Partial pressure CH4 =Moles of H2/ Moles of CH4 pH2/ pCH4 = nH2/nCH4 We assume that x is the mass of hydrogen and methane each. The molecular mass of H2=2 The molecular mass of CH4=16 Mole = mass/Mr Moles of H2= x/2 Moles of CH4= x/16 Total moles= x/2 + x/16 = 9x/16 The partial pressure of H2= moles of H2/total moles =(x/2)÷(9x/16) =(x/2) x (16/9x) =8/9 so option B is the right answer.

Q71. The diagram shows a simplified nitrogen cycle. Which row shows the correct labels for P, Q, R, and S?

• A. Option A✓
• B. Option B
• C. Option C
• D. Option D

Explanation: The conversion of nitrates in the soil to nitrogen is called the denitrifying(P) process, which is carried out by denitrifying bacteria. Then this nitrogen is fixed(Q) into the legume plants, which can form ammonia either by ammonification by saprotrophic fungi(S) or decay of leaf tissue(R). Hence, the answer is A. ( p ) Denitrification. Denitrification is the process that converts nitrate to nitrogen gas, thus removing bioavailable nitrogen and returning it to the atmosphere. Dinitrogen gas (N2) is the ultimate end product of denitrification, but other intermediate gaseous forms of nitrogen exist. ( q ) Legumes can form a symbiotic relationship with nitrogen-fixing soil bacteria called rhizobia. The result of this symbiosis is to form nodules on the plant root, within which the bacteria can convert atmospheric nitrogen into ammonia that can be used by the plant. ( r ) The process of ammonification is carried out by certain bacteria, just as Bacillus ramosus, B. vulgaris, Clostridium, Actinomyces, etc. In this process, these organisms release a specific amount of nitrogen compounds like ammonia. However, saprophytic bacteria will not be involved in this process. ( s ) Plant and animal wastes decompose, adding nitrogen to the soil. Bacteria in the soil convert those forms of nitrogen into forms plants can use. Plants use the nitrogen in the soil to grow. People and animals eat the plants; then animal and plant residues return nitrogen to the soil again, completing the cycle.

Q72. In the commercial manufacture of insulin, a human gene is inserted into which one of the following?

• A. A chromosome of a human cell
• B. A protein molecule in a yeast cell
• C. The DNA of a bacterium✓
• D. The nucleic acid in a virus

Explanation: In the commercial manufacture of a human cell, a human gene is inserted into the DNA of a bacterium which then nurtures itself in an environment of its liking producing insulin which is utilized by humans.

Q73. The diagram shows a model to demonstrate the mass flow hypothesis of translocation. In a plant, what are the structures W, X, Y, and Z, and what is the direction of the flow of solution along W?

• A. Option A
• B. Option B✓
• C. Option C
• D. Option D

Explanation: Since Y represents concentrated sucrose, it means it has higher amount of sucrose. Leaves would have higher amount of sucrose, as sugar is being synthesized and converted into sucrose in leaves. Sucrose is transported through phloem so W represents phloem, Z represents root because direction of flow of sucrose is from source to sink, where source is leaves and sink are roots in this case. So direction of flow is from Y to Z.

Q74. Many scientists believe that one of the following is evolutionary origin of animals, plants and fungi?

• A. Protists
• B. Algae
• C. Bacteria✓
• D. Protozoans

Explanation: Bacteria are considered one of the earliest and most primitive forms of life. They are believed to be the ancestors of all complex eukaryotic life forms, including plants, animals, and fungi, through a process of evolution that involved endosymbiosis and other mechanisms of genetic exchange. Protists, algae, and protozoans are all more complex forms of life that evolved later and are not considered the direct ancestors of plants, animals, and fungi.

Q75. Blood circulating from the gut to the heart passes through the:

• A. Aorta
• B. Kidneys
• C. Liver✓
• D. Lungs

Explanation: The hepatic portal system carries nutrient-rich blood from the gut to the liver before it enters systemic circulation.

Q76. The diagram shows the four types of human teeth. Which teeth are used for cutting rather than grinding food?

• A. 1 and 2✓
• B. 2 and 3
• C. 3 and 4
• D. 4 and 1

Explanation: Most adults have 32 permanent teeth, including eight incisors, four canines, eight premolars, and 12 molars. Incisors and canines are used to cut and chew. Premolars and molars are used for grinding the food.

Q77. What are the functions of the inter, motor and sensory neurons in a reflex response?

• A. Option A✓
• B. Option B
• C. Option C
• D. Option D

Explanation: To relay nerve impulses from within the CNS, interneurons are used. To conduct the nerve impulses from the CNS to the effector, motor neurons are used. To conduct the nerve impulses from the receptors to the CNS, sensory neurons are used.

Q78. The diagram shows the human urinary system.Which substance is not found In the liquid at x In a healthy person?

• A. Glucose✓
• B. Salt
• C. Toxins
• D. Water

Explanation: Option A is correct.When the glomerular filtrate comes into the proximal convoluted tubule, reabsorption of glucose and amino acids takes place through active transport. So, there is no glucose found in the urine. Ordinarily, urine contains no glucose because the kidneys can reabsorb all of the filtered glucose from the tubular fluid back into the bloodstream.In a healthy person, liquid X does not contain glucose. In a diabetic person, glucose does occur in urine but the question asks about the healthy person.

Q79. Male and female sea urchins release their sperm and eggs Into the water where fertilization takes place. How can their reproduction be described?

• A. Asexual reproduction which results in genetically dissimilar offspring
• B. Asexual reproductlon whlch results In genetically identical offspring
• C. Sexual reproduction which results in genetically dissimilar offspring✓
• D. Sexual reproduction which results in genetically identical offspring

Explanation: A type of reproduction in which gametes join to form a new offspring is called sexual reproduction. Sexual reproduction involves the mixing of two genetic materials resulting in genetically dissimilar offspring.

Q80. Which vertebrate groups have scaly skin?

• A. Amphibians and fish
• B. Amphibians and mammals
• C. Fish and mammals
• D. Fish and reptiles✓

Explanation: Fish and reptiles have scaly skins, as shown below:

Q81. The following reaction occurs in the human alimentary canal. What is the enzyme and the product?

• A. Option A
• B. Option B
• C. Option C✓
• D. Option D

Explanation: Amylase is an enzyme that plays a crucial role in the digestion of carbohydrates, specifically starch, in the alimentary canal (digestive system). The alimentary canal includes the mouth, where the digestion process begins, as well as the stomach, small intestine, and large intestine. Starch is a complex carbohydrate made up of long chains of glucose molecules. The digestion of starch begins in the mouth with the action of salivary amylase. Salivary amylase is produced by the salivary glands and released into the mouth along with saliva. Its primary function is to break down starch into simpler sugars, such as maltose and glucose.

Q82. Archaeopteryx is a transitional stage between the members of which one of the following pairs?

• A. Amphibian-Bird
• B. Fish-Amphibian
• C. Reptile-Mammal
• D. Reptile-Bird✓
• E. Mammal-Man

Explanation: It has long been accepted that Archaeopteryx was a transitional form between birds and reptiles and that it is the earliest known bird.

Q83. In the Krebs cycle, substrate-level phosphorylation accompanies the formation of:

• A. Citrate
• B. Alpha-ketoglutarate
• C. Succinate✓
• D. Fumarate

Explanation: Coenzyme A is removed from Succinyl CoA to form succinate. The reaction releases sufficient energy which is used to combine GDP and Pi forming GTP. GTP reacts with ADP to form ATP while GTP is again converted into GDP. In this way a molecule of ATP is generated in this reaction

Q84. When a physician elicits the knee-jerk reflex by tapping deep tendons in the knee, the normal response is for the lower leg to swing forward. When this happens:

• A. Muscles in the front of the thigh are contracting, and muscles in the back of the thigh are relaxing✓
• B. Muscles in the front of the lower leg are contracting, and muscles in the back of the lower leg are relaxing
• C. Muscles in the back of the thigh are contracting, and muscles in the front of the thigh are relaxing
• D. Muscles in the back of the lower leg are contracting and muscles in the front of the lower leg are relaxing
• E. None of the above

Explanation: In the knee-jerk reflex, tapping the patellar tendon causes the quadriceps, located at the front of the thigh, to contract. This contraction results in the extension of the lower leg, moving it forward. Simultaneously, the hamstrings at the back of the thigh relax to facilitate this movement. This reflex is an example of a monosynaptic reflex arc, where sensory neurons are directly connected to motor neurons in the spinal cord, allowing for a rapid response. The incorrect options involve muscles of the lower leg or the opposite muscle action, which do not contribute to the observed reflex action.

Q85. Which is the incorrectly paired one?

• A. Robert Hooke ... cell wall
• B. Schlelden and Schwann ... cell theory
• C. Robert Brown... nucleus
• D. Watson and Crick ... DNA model
• E. Virchow... Mosaic model of plasma membrane✓

Explanation: The Fluid Mosaic hypothesis was formulated by Singer and Nicolson in the early 1970s, not by Virchow. All others are correctly paired.

Q86. Identify the phylum in which the larvae is bilaterally symmetrical but the adult Is radially symmetrical

• A. Ctenophora
• B. Coelenterata
• C. Echinodermata✓
• D. Sipunculoidea

Explanation: In Phylum Echinodermata, the adult echinoderms are radially symmetrical but the larvae are bilaterally symmetrical.

Q87. The botanical name of gum tree is:

• A. Acacia nilotica✓
• B. Mimosa pudica
• C. Acacia catechu
• D. Prosopis glandulosa
• E. Albizia lebbeck

Explanation: The botanical name for the gum tree commonly known as Acacia nilotica. Acacia nilotica belongs to the genus Acacia and is a deciduous tree native to Africa, the Middle East, and the Indian subcontinent. It is also known by various other names, including gum arabic tree, Egyptian thorn, babul, or prickly acacia. The species is characterized by its compound leaves, thorny branches, and small, yellow, puffball-like flowers.

Q88. A pure breeding plant with the dominant phenotype of character P and the recessive phenotype of character Q was crossed with another pure breeding plant with the recessive phenotype of character P and the dominant phenotype of Q. The offspring of this cross were crossed with a double homozygous recessive for P and Q and the following results were obtained:22 were phenotypically dominant for P and recessive for Q5 were phenotypically dominant for both P and Q4 were phenotypically recessive for both P and Q24 were phenotypically recessive for P and dominant for QWhich one of the following types of Inheritance is illustrated by these results?

• A. Gene linkage of P and Q✓
• B. Independent segregation of P and Q
• C. Mendelian dihybrid Inheritance
• D. Multiple alleles
• E. Polygenic Inheritance

Explanation: When a pure breeding phenotypic dominant character P and recessive character Q is crossed with pure breeding recessive character P and dominant character Q, the cross is below:- PPqq X ppQQ will give PpQq. Now when this PpQq is crossed with recessive for both:- PpQq X ppqq will give PpQq ; Ppqq ; ppqq ; ppQq The ratio when there is a dihybrid cross between heterozygous character and homozygous recessive character is 9:3:3:1. However, the above ratio is not consistent with what it should be. This happens when there is a gene linkage between two characters

Q89. How many metacarpals are present in the hand?

• A. 4
• B. 3
• C. 6
• D. 5✓
• E. 8

Explanation: There are five metacarpal bones in the palm, as shown below (marked red):

Q90. Which of the following is NOT a difference that would allow one to distinguish between a prokaryotic and a eukaryotic cell? Presence or absence of the nucleus Presence or absence of the cell wall Membrane bound versus no membrane bound organelles

• A. I only
• B. II only✓
• C. III only
• D. I and II only
• E. I, II and III

Explanation: Eukaryotic cells have nucleus and membrane-bound organelles, which is not the property of prokaryotes.The cell wall is not a distinguishing feature between prokaryotes or eukaryotes.

Q91. Some enzymes require the presence of a non protein molecule to behave catalytically. An enzyme devoid of this molecule is called a:

• A. Holoenzyme
• B. Apoenzyme✓
• C. Coenzyme
• D. Zymoenzyme

Explanation: An enzyme devoid of the necessary non-protein molecule for its catalytic activity is often referred to as an "apoenzyme." The non-protein molecule required by an enzyme for its catalytic activity is called a "cofactor" or "coenzyme." When the cofactor is tightly and permanently bound to the apoenzyme, forming a complete and active enzyme, the entire structure is termed a "holoenzyme."

Q92. The events shown below occur during different phases of mitosis: Spiralization of DNA Hydration of DNA Centromeres split Centromeres attach to spindle fibresDNA replicates Which one of the following correctly identifies each of the phases described?

• A. Option A
• B. Option B
• C. Option C✓
• D. Option D
• E. Option E

Explanation: Explanation to this question is given below:DNA replicates in interphase.Spiralization(condensation) of DNA occurs in prophase. Centromeres attach to spindle fibers in metaphase.Centromeres start splitting in anaphase. DNA hydrates back in telophase.

Q93. When a fetus is in the uterus, what carries oxygen away from the placenta?

• A. The amniotic fluid
• B. The amniotic sac
• C. The lining of the uterus
• D. The umbilical cord✓

Explanation: The umbilical cord (Latin: Funiculus umbilicalis) is a tube-like structure that connects the developing fetus to the placenta. It has one umbilical vein and two umbilical arteries, which are used by the fetal heart to pump blood to and from the placenta, which exchanges nutrients and waste products with the mother's circulatory system.

Q94. The floral formula of family Mimosaceae is:

• A. Option A
• B. Option B
• C. Option C
• D. Option D✓

Explanation: The floral formula of family Mimosaceae is ⊕,⚥,K(5), C5 or (5), A α or 10 ,G.

Q95. Which type of protein structure contains the three dimensional structure?

• A. Primary
• B. Secondary
• C. Tertiary✓
• D. Quaternary

Explanation: The primary structure is a sequence of amino acids in the protein/polypeptide. The secondary structure is a 2D structure. The tertiary structure is a 3D structure. Quaternary structure is when two or more 3D structures join together.

Q96. Which of the following describes the movements involved in breathing out?

• A. Option A
• B. Option B✓
• C. Option C
• D. Option D

Explanation: Breathing out involves a decrease in the volume of the chest this occurs when ribs move downwards and inwards, the diaphragm moves upwards. The opposite of these movements increases in chest volume and involves movements of ribs upwards and outwards.

Q97. The diagram shows a cell of an organism formed by reduction division. What is the diploid number for this organism?

• A. 10
• B. 30
• C. 40
• D. 44✓

Explanation: Haploid cells are formed by reduction division, which has half the number of chromosomes than diploid cells. To calculate the diploid number of cells just multiply the chromosome number in the haploid number by 2. So, 22x2=44.

Q98. The inhibitor which closely resembles the substrate in its molecular structure and inhibits the enzyme activity by binding to the active site of the enzyme is called

• A. Feedback inhibitor
• B. Non-competitive inhibitor
• C. Allosteric modulator
• D. Competitive inhibitor✓

Explanation: In competitive inhibition, the inhibitor (I) closely resembles the real substrate (S) and is regarded as a substrate analog. The inhibitor competes with the substrate and binds at the active site of the enzyme but does not undergo any catalysis. As long as the competitive inhibitor holds the active site, the enzyme is not available for the substrate to bind. A competitive inhibitor diminishes the rate of catalysis by reducing the proportion of enzyme molecules bound to a substrate. At any given inhibitor concentration, competitive inhibition can be relieved by increasing the substrate concentration. So, the correct answer is 'Competitive inhibitor'.

Q99. Which processes are essential in making nitrogen from dead plant material available to growing plants? ammonification deamination nitrification nitrogen fixation

• A. I, II and III only
• B. I, II and IV only
• C. I, III and IV only✓
• D. II, III and IV only

Explanation: Nitrification, nitrogen fixation, and ammonification are essential in making nitrogen for dead plant material available to the growing.

Q100. The diagram represents two liquids, separated by a membrane through which osmosis can occur. What movement of molecules will occur?

• A. Molecules of dissolved substance move from left to right
• B. Molecules of dissolved substance move from right to left
• C. Overall, water molecules move from left to right✓
• D. Overall, water molecules move from right to left

Explanation: Water molecules move from the left to right as water potential is higher on the left side than the right side.