datavorous/entrance-exam-dataset huggingface dataset:
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The x,ây coordinates of the centre of mass will be:

Options:
A 712 m, 34 m
B 712 m, 38 m
C 34 m, 512 m
D 38 m, 712 m

The correct answer is:
712 m, 34 m

Solution:
Coordinates of A :0, 0
B:12 m, 32 m
C :(1 m, 0)
Coordinates of center of mass
xcm,ycm=50×0+150×12+100×150+150+100 m, 50×0+150×32+100×050+150+100 m
=712 m, 34 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
Three particles start from the origin at the same time, one with a velocity v1 along the x-axis, second along the negative y-axis with a velocity v2 and third particle moves along the line x=y. The velocity of third particle, so that three may always lie on the same line is :

Options:
A v 1 + v 2 2
B v 1 + v 2
C v 1 v 2 v 1 + v 2
D v=2 v1v2v2-v1

The correct answer is:
v=2 v1v2v2-v1

Solution:
Slope of AC line = Slope of BA line
vt2-0vt2-v1t=0--v2tv1t-0
v=2 v1v2v2-v1

Topic: Physics, Motion In Two Dimensions, NEET
Three pea plants P, Q and R each having inflated green pods are allowed to grow under natural conditions. 1000 seeds from each plant were collected and there were sown to see the properties of their progenies. The following were the observations:

P	Q	R

1000 have inflated green pods	242 constricted green pods 758 inflated green pods	741 inflated green pod 259 inflated yellow pod

The correct genotypes of these three plants are

Options:
A P – IIGG, Q – IiGg, R – IIGg
B P – IIgg, Q – IiGG, R – IiGg
C P – IIGG, Q – IiGG, R – IIGg
D P – IIgg, Q – Iigg, R – IIGg

The correct answer is:
P – IIGG, Q – IiGG, R – IIGg

Solution:
Inflated pods are dominant in pea while constricted pods are recessive. Green pods are dominant in pea while yellow pods are recessive. All three plants have inflated green pods. So they should have one of the following genotypes IIGG, IiGG, IIGg or IiGg. These plants were grown naturally it means that there was self pollination.
IIGG x IIGG will produce all plants which are inflated and green. This is seen in plant P.
IiGG x IiGG will produce IIGG, IiGG and iiGG in the ratio 1 : 2 : 1 . Out of these IIGG and IiGG have inflated green pods while iiGG has constricted green pods. Thus Inflated green pods and Constricted green pods are produced in the ratio 3 : 1 . This is seen in plant Q
IIGg x IIGg will produce IIGG, IIGg and IIgg in the ratio 1 : 2 : 1 . Out of these IIGG and IIGg have inflated green pods while IIgg has inflated yellow pods. Thus Inflated green pods and inflated yellow pods are produced in the ratio 3 : 1 . This is seen in plant R

Topic: Biology, Principles of Inheritance and Variation, NEET
Three plants A, B and C are modified for storage in the given diagram. Identify the correct statement:

Options:
A A, B and C - all are modifications of stem
B A and B are modifications of tap root
C A and C are modifications of adventitious roots
D C is modification of tap root.

The correct answer is:
A and B are modifications of tap root

Solution:
Underground stems may modify to store food.
Examples - Potato, Turmeric, Ginger, Colocasia, Zaminkand, etc.

Topic: Biology, Morphology of Flowering Plants, NEET
Three point charges of +2q, +2q and -4q are placed at the corners A, B and C of an equilateral triangle ABC of side x. The magnitude of the electric dipole moment of this system is

Options:
A 2 qx
B 32 qx
C 3 qx
D 23 qx

The correct answer is:
23 qx

Solution:
According to the equation of magnitude of the electric dipole moment
Pnet=p2+p2+2p pcosâ¡60o
=2p2+2p2Ã12
=3âp
We know that, p=2 qâx âµp=qâ
=3â2 qx=23 qx

Topic: Physics, Electrostatics, NEET
Three point charges of +2q,+2q and -4q are placed at the corners A, B and C of an equilateral triangle ABC of side x. The magnitude of the electric dipole moment of this system is

Options:
A 2 qx
B 32 qx
C 3 qx
D 23 qx

The correct answer is:
23 qx

Solution:
According to the equation of magnitude of the electric dipole moment
Pnet=p2+p2+2p pcosâ¡60o
=2p2+2p2Ã12
=3âp
We know that, p=2 qâx âµp=qâ
=3â2 qx=23 qx

Topic: Physics, Electrostatics, NEET
Three point charges $+q,-2 q$ are placed at points $(x=0, y=a, z=0),(x=0, y=0, z$ $=0)$ and $(x=a, y=0, z=0)$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are:

Options:
A $\sqrt{2} q a$ along the line joining points $(x=0, y=0, z=0)$ and $(x=a, y=a$, $z=0)$
B $q a$ along the line joining points $(x=0$, $y=0, z=0)$ and $(x=a, y=a, z=0)$
C $\sqrt{2} q a$ along $+x$ direction
D $\sqrt{2} q a$ along $+y$ direction

The correct answer is:
$\sqrt{2} q a$ along the line joining points $(x=0, y=0, z=0)$ and $(x=a, y=a$, $z=0)$

Solution:
This consists of two dipoles, $-q$ and $+q$ with dipole moment along with the $+y$ direction and $-q$ and $+q$ along the X-direction
$\therefore$ The resultant moment
$\begin{aligned}
& =\sqrt{q^2 a^2+q^2 a^2} \\
& =\sqrt{2} q a
\end{aligned}$
Along the direction $45^{\circ}$ that is along OP where $\mathrm{P}$ is $(+a, a, 0)$

Topic: Physics, Electrostatics, NEET, NEET 2007
Three point masses, each of mass m , are situated at the three corners of an equilateral triangle of each side l . The moment of inertia of this system about one of the sides of the triangle will be

Options:
A ml2
B 3ml2
C 34ml2
D 23ml2

The correct answer is:
34ml2

Solution:
I=0+0+m32l2
=34ml2

Topic: Physics, Rotational Motion, NEET
Three resistances of 4 Î© each are connected as shown in the figure. If point D divides the resistance into two equal halves, the resistance between points A and D will be

Options:
A 12 Ω
B 6 Ω
C 3 Ω
D 13 Î©

The correct answer is:
3 Ω

Solution:
The equivalent circuit is given by
4 Ω and 2 Ω resistances are in series on both sides.
∴4 Ω+2 Ω=6 Ω
Then 6 Ω and 6 Ω resistances are in parallel.
1R=16+16=26=13
R=3 Ω

Topic: Physics, Current Electricity, NEET
Three resistances $P, Q, R$ each of $2 \Omega$ and an unknown resistances $\mathrm{S}$ form the four arms of a Wheatstone bridge circuit. When a resistance of $6 \Omega$ is connected in parallel ot $\mathrm{S}$ the bridge gets balanced. What is the value of $\mathrm{S}$ ?

Options:
A $3 \Omega$
B $6 \Omega$
C $1 \Omega$
D $2 \Omega \quad ।$

The correct answer is:
$3 \Omega$

Solution:
The bridge formed with given resistances is a balanced Wheatstone's bridge
The situation can be depicted as shown in the figure.

As resistances \(S\) and \(6 \Omega\) are in paralle, their effective resistance is \(\frac{6 S}{6+S} \Omega\).
For balancing condition, \(\frac{P}{Q}=\frac{R}{\left(\frac{6 S}{6+S}\right)}\)
or \(\frac{2}{2}=\frac{2(6+\mathrm{S})}{6 \mathrm{~S}}\)
or \(3 \mathrm{~S}=6+\mathrm{S} \Rightarrow \mathrm{S}=3 \Omega\)

Topic: Physics, Current Electricity, NEET, NEET 2007
Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The ratio i3i1 of currents in terms of resistances used in the circuit is :

Options:
A r1r2+r3
B r2r2+r3
C r1r1+r2
D r1r1+r2.

The correct answer is:
r2r2+r3

Solution:
Potential difference between B and C,

VBC=i3r3=i1×r2r3r2+r3

⇒i3=i1×r2r2+r3

Topic: Physics, Current Electricity, NEET, NEET 2021
Three rings, each having equal radius R, are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If a current I is flowing through each ring, then the magnitude of the magnetic field at the common centre is

Options:
A 3μ0I2R
B zero
C 2-1μ0I2R
D 3-2μ0I2R

The correct answer is:
3μ0I2R

Solution:
B→=Bxi^+Byj^+Bzk^⇒B=Bx2+By2+Bz2
Since Bx=By=Bz so B=3B0
B=3μ0I2R

Topic: Physics, Magnetic Effects of Current, NEET
Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY′ axis will be

Options:
A 72PQ2
B 27PQ2
C 25PQ2
D 52PQ2

The correct answer is:
72PQ2

Solution:
For ring 1 and 2 using parallel axis theorem,
I1=I2=Icm+Md2
I1=I2=12PQ2+PQ2
I1=I2=32PQ2
For ring 3,
I3=12PQ2
Moment of inertia of the arrangement,
I=I1+I2+I3
I=32PQ2+32PQ2+12PQ2
I=72PQ2

Topic: Physics, Rotational Motion, NEET
Three rods are made from the same material and having the same cross-sectional area and length have been joined as shown in the figure.

The left end and right end are kept 0 Â°C and 100 Â°C, respectively. The temperature of the junction of three rods will be

Options:
A 50 °C
B 60 °C
C 80 °C
D 66.66 °C

The correct answer is:
66.66 °C

Solution:
Let the temperature of the junction is θ. Since, rods B and C parallel to each other. Thus, the thermal resistance by rod B and C,
R1=R2 ∵RC=RB
Therefore,
=θ-0°R=100-θR/2
⇒θR=100-θR/2
⇒θR=2100-θR
⇒3θ=200
⇒θ=66.66 °C

Topic: Physics, Thermal Properties of Matter, NEET
Three rods made of the same material and having the same cross-section have been joined as shown in the figure. Each rod is of the same length. The left and right ends are kept at 0 oC and 90 oC respectively. The temperature of the junction of the three rods will be

Options:
A 45 oC
B 60 oC
C 30 oC
D 20 oC

The correct answer is:
60 oC

Solution:
From Junction law,
KAlT-T1+KAlT-T2+KAlT-T3=0
T=T1+T2+T33
T=0+90+903=60 oC

Topic: Physics, Thermal Properties of Matter, NEET
Three rods of same dimensions have thermal conductivities 3K, 2K,and K , with their ends at 100 °C,50 °C and 0 °C respectively. They are arranged as shown in the diagram. The temperature of the junction J in steady-state is

Options:
A 2003°C
B 1003°C
C 75 °C
D 60 °C

The correct answer is:
2003°C

Solution:
Let T = temperature of the junction J
L = length and A = area of each rod
Heat conducted in first rod = heat conducted by second rod + heat conducted by third rod
3KA100-TL=2KAT-50L+AKT-0L
3100-T=2T-50+T
400 = 6T
T =2003°C

Topic: Physics, Current Electricity, NEET
Three rods of the same dimension have thermal conductivity 3K, 2K and K as shown in the figure. The temperature of the junction in steady-state is

Options:
A 2003âoC
B 1003âoC
C 75°C
D 503°C

The correct answer is:
2003 oC

Solution:
According to the figure

H=H1+H2

3 KA 1 0 0 - T lâ¡ = 2 KA T - 5 0 lâ¡ + KA T - 0 lâ¡

300â3T=2Tâ100+T

6T=400

Or T=2003âoC

Topic: Physics, Thermal Properties of Matter, NEET
Three rods of the same dimension have thermal conductivity 3 K, 2 K and K. They are arranged as shown in the figure below

Then, the temperature of the junction in steady-state is

Options:
A 2003 Â°C
B 1003 °C
C 75 °C
D 503 °C

The correct answer is:
2003 Â°C

Solution:
According to the figure

H = H1 + H2

3KA100-Tlâ¡=2KAT-50lâ¡+KAT-0lâ¡

300 - 3 T = 2 T - 100 + T

6 T = 400

Or T=2003âÂ°C

Topic: Physics, Thermal Properties of Matter, NEET
Three small balls of masses 1 kg, 2 kg and 3 kg are moving in a plane and their velocities are 1 m s-1, 2 m s-1 and 3 m s-1, respectively as shown. The total angular momentum of the system of three balls about point P at the given instant of time is

Options:
A 8 kg m2 s-1
B 9 kg m2 s-1
C 36 kg m2 s-1
D 7 kg m2 s-1

The correct answer is:
7 kg m2 s-1

Solution:
The particle of mass 2 kg is situated perpendicular to the axis of rotation. So, the angle will be 90°. Similarly, the particle of mass 1 kg is situated opposite side of the circular path. So, the angle will be 90° and angular momentum will act in opposite direction.
According to the diagram, total angular momentum
=2×2×2sin⁡90°+3×3×3sin⁡0°-1×1×1sin⁡90°
=8-1
=7  kg  m2 s-1s [∵sin⁡0°=0 andsin⁡90°=1]

Topic: Physics, Rotational Motion, NEET
Three sound waves of equal amplitudes have frequencies, (n – 1), n, (n + 1). They superimpose to give beats. The number of beats produced per second will be

Options:
A 1
B 4
C 3
D 2

The correct answer is:
2

Solution:
Number of beats =2.

Maximum difference between the given frequencies.

Topic: Physics, Waves and Sound, NEET, NEET 2016 (Phase 2)
Three stars A, B, C have surface temperatures TA, TB,TC respectively. Star A appears bluish, star B appears reddish and star C yellowish. Hence,

Options:
A TA>TB>TC
B TB>TC>TA
C TC>TB>TA
D TA>TC>TB

The correct answer is:
TA>TC>TB

Solution:
Stars approximate black body radiators and their visible colour depend upon the temperature of the radiator. The curves show blue, white, and red stars. The white star is adjusted to 5270 K so that the peak of its blackbody curve is at the peak wavelength of the sun, 550 nm. From the wavelength at the peak, the temperature can be deduced from the Wien displacement law.

TA>TC>TB

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2020 (Phase 2)
Three students, Manish, Ramesh and Rajni were determining the extra elements present in an organic compound given by their teacher. They prepared the Lassaigne’s extract (L.E.) independently by the fusion of the compound with sodium metal. Then they added solid FeSO4and dilute sulphuric acid to a part of Lassaigne’s extract. Manish and Rajni obtained prussian blue colour but Ramesh got red colour. Ramesh repeated the test with the same Lassaigne’s extract, but again got red colour only. They were surprised and went to their teacher and told him about their observation. Teacher asked them to think over the reason for this. Can you help them by giving the reason for this observation?

Options:
A In Lassaigne’s test SCN–ions are formed due to the presence of sulphur and nitrogen both. These give red colour with Fe3+ions. This happens when fusion is not carried out in the excess of sodium.
B In Lassaigne’s test Na+ions are formed due to the presence of sulphur and nitrogen both. These give red colour with Fe3+ions. This happens when fusion is not carried out in the excess of sodium.
C In Lassaigne’s test Fe2+ions are formed due to the presence of sulphur and nitrogen both. These give red colour with Fe3+ions. This happens when fusion is not carried out in the excess of sodium.
D In Lassaigne’s test CN–ions are formed due to the presence of sulphur and nitrogen both. These give red colour with Fe3+ions. This happens when fusion is not carried out in the excess of sodium.

The correct answer is:
In Lassaigne’s test SCN–ions are formed due to the presence of sulphur and nitrogen both. These give red colour with Fe3+ions. This happens when fusion is not carried out in the excess of sodium.

Solution:
Correct Option is : (A)
In Lassaigne’s test SCN–ions are formed due to the presence of sulphur and nitrogen both. These give red colour with Fe3+ions. This happens when fusion is not carried out in the excess of sodium.

Topic: Chemistry, General Organic Chemistry, NEET
Three thermochemical equation are given below:
(i) $\underset{\mathrm{kJJ} \mathrm{mol}^{-1}}{\mathrm{C}_{\text {(graphite) }}}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(g)} \Delta \mathrm{H}^{\circ}=x$
(iii) $\mathrm{CO}_{(\mathrm{gJ})}+1 / 2 \mathrm{O}_{2(\mathrm{gk}} \rightarrow \mathrm{CO}_{2(\mathrm{~g}) i} \Delta \mathrm{H}^{\circ}=\mathrm{z}$
Based on the above equations, find out which of the relationship given below is correct.

Options:
A $z=x+y$
B $x=y+z$
C $y=2 z-x$
D $x=y-z$

The correct answer is:
$x=y+z$

Solution:
By considering the Hess's law equation (i) is equal to the sum of equation (ii) and (iii).
Related Theory
Hess's law states that heat absorbed or evolved in any chemical reaction is fixed quantity and is independent of the path of the reaction or the number of steps taken 0 obtain the reaction.
$$
\Delta H_{n e t}=\Sigma \Delta H_r
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2013 (Karnataka)
Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are μr,μy and μv respectively and those for the flint glass are μr′,μy′ and μv′ respectively. Find the ratio A′A for which, there is no net angular dispersion (A and A′ are respective deviations)

Options:
A Î¼y-1Î¼y′-1
B 2(Î¼v-Î¼r)(Î¼′v-Î¼′r)
C Î¼v-Î¼rÎ¼′v-Î¼′r
D 2(Î¼y-1)(Î¼′y-1)

The correct answer is:
2(Î¼v-Î¼r)(Î¼′v-Î¼′r)

Solution:
For no net angular dispersion
Î¸net=Î¸1-Î¸2+Î¸3=0
Î¸2=Î¸1+Î¸3
Î¼v′-Î¼r′A′=Î¼v-Î¼rA+Î¼v-Î¼rA
A′A=2(Î¼v-Î¼r)(Î¼′v-Î¼′r)

Topic: Physics, Ray Optics, NEET
Three thin uniform rods each of mass M and length L and placed along the three axes of a Cartesian coordinate system with one end of each rod at the origin. The M.I. of the system about z-axis is-

Options:
A ML23
B 2ML23
C 2ML26
D ML2

The correct answer is:
2ML23

Solution:
I = Ix + Iy +Iz =ML23+ML23+0=2ML23
As moment of inertia of rod about its own axis is zero.

Topic: Physics, Rotational Motion, NEET
Three types of muscles are given as $\mathrm{a}, \mathrm{b}$ and c . Identify the correct matching pair along with their location in human body:
Name of muscle/location

Options:
A (a) Skeletal - Triceps
(b) Smooth - Stomach
(c) Cardiac-Heart
B (a) Skeletal - Biceps
(b) Involuntary - Intestine
(c) Smooth - Heart
C (a) Involuntary - Nose tip
(b) Skeletal - Bone
(c) Cardiac - Heart
D (a) Smooth - Toes
(b) Skeletal-Legs
(c) Cardiac - Heart

The correct answer is:
(a) Skeletal - Triceps
(b) Smooth - Stomach
(c) Cardiac-Heart

Solution:
The correct answer is option (1) as
Figure (a) represents skeletal muscle fibres which are closely attached to skeletal bones. In a typical muscle such as triceps and biceps, striated muscle fibres are bundled together in a parallel fashion.
Figure (b) represents smooth muscle fibres which are present in the wall of internal organs such as the blood vessels, stomach and intestine.
Figure (c) represents cardiac muscle fibres which are exclusively present in the heart.

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2024
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains helium (monoatomic), the second contains fluorine (diatomic) and the third contains sulfur hexafluoride (polyatomic). The correct statement, among the following is:

Options:
A All vessels contain unequal number of respective molecules
B The root mean square speed of molecules is same in all three cases
C The root mean square speed of helium is the largest
D The root mean square speed of sulfur hexafluoride is the largest

The correct answer is:
The root mean square speed of helium is the largest

Solution:
As, $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}$
Hence, $\quad v_{\mathrm{rms}} \propto \frac{1}{\mathrm{~m}}$
So, lower the mass, higher/greater the velocity. (where, the symbols have their usual meanings)

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2022 (Phase 2)
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Which vessel’s Vrms is the largest ?

Options:
A Neon
B Chlorine
C Uranium hexafluoride
D It is same for every vessel

The correct answer is:
Neon

Solution:
Correct Option is : (A)
Neon

Topic: Chemistry, Kinetic Theory of Gases, NEET
Three waves of equal frequency having amplitudes 10 mm, 4 mm, and 7 mm arrive at a given point with a successive phase difference of Ï2, the amplitude of the resulting wave in mm is given by

Options:
A 7
B 6
C 5
D 4

The correct answer is:
5

Solution:
As phase difference between waves of amplitudes 10 mm and 7 mm is Ï, therefore, their resultant amplitude =10-7=3 mm.
Now amplitudes 3 mm and 4 mm have a phase difference =Ï2.
â´ Resultant amplitude =32+42=5 mm

Topic: Physics, Waves and Sound, NEET
Threshold frequency, ν0is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency 1.0×1015s–1was allowed to hit a metal surface, an electron having 1.988 × 10–19J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

Options:
A Threshold frequency = 5 × 1014Hz; No electron will be emitted.
B Threshold frequency = 1.0 × 1015Hz; No electron will be emitted.
C Threshold frequency = 1.0 × 1014Hz; An electron will be emitted.
D Threshold frequency = 2.988 × 1015Hz; An electron will be emitted.

The correct answer is:
Threshold frequency = 5 × 1014Hz; No electron will be emitted.

Solution:
Correct Option is : (A)
Threshold frequency = 5 × 1014Hz; No electron will be emitted.

Topic: Chemistry, Structure of Atom, NEET
Through photosynthesis, the lungs of earth, Amazon forest is estimated to produce ______of the total oxygen in the earth's atmosphere.

Options:
A One-third
B One- fourth
C One-fifth
D One-sixth

The correct answer is:
One-fifth

Solution:
Amazon rain forest produces approximately 20% of oxygen in earth's atmosphere which is about one fifth of the total oxygen.

Topic: Biology, Biodiversity and its Conservation, NEET
Through which cell of the embryo sac, does the pollen tube enter the embryo sac ?

Options:
A Egg cell
B Persistant synergid
C Central cell
D Degenerated synergids

The correct answer is:
Degenerated synergids

Solution:
The pollen tube enters into the embryo sac at the microplyar end. This entry may be between egg And one synergid or between wall of embryo sac and synergid or through one synergid. So one synergid is always degenerated to allow the entry of the pollen tube.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Throughout the country, since October 2010, all 2 wheelers and 3 wheelers have to follow ............ and............. respectively.

Options:
A Bharat stage III and IV
B Bharat stage IV and III
C Bharat stage II and III
D Bharat stage III and III

The correct answer is:
Bharat stage III and IV

Solution:
Bharat Stage II or Euro-II norms (both are almost equivalent) are no more applicable in any of the cities
of India.

2 Wheelers	Bharat Stage III	Throughout the country since October 2010
3 Wheelers	Bharat Stage III	Throughout the country since October 2010
4 Wheelers	Bharat Stage III	Throughout the country since October 2010 except 13 Mega cities
Bharat Stage IV	13 Mega cities (Delhi and NCR, Mumbai, Kolkata, Chennai, Bangalore, Surat, Kanpur, Agra, Lucknow and Sholapur) since April, 2010

Topic: Biology, Environmental Issues, NEET
Thurioside is

Options:
A insecticide
B fungicide
C antibiotic
D weedicide

The correct answer is:
insecticide

Solution:
Thurioside is toxic to several insects and produced by bacterium Bacillus thuringiensis.

Topic: Biology, Microbes in Human Welfare, NEET
Thylakoids removed from chloroplasts were kept in an illuminated culture having CO2 and water. However, it did not produce sugar because of

Options:
A absence of enzymes
B non-linking of PS I and PS II
C absence of light trapping molecules
D non formation of assimilatory power

The correct answer is:
absence of enzymes

Solution:
When thylakoids removed from chloroplast were kept in illuminated culture, having carbon dioxide and water would not be able to synthesize hexose sugar in the absence of an enzyme. These are the sites for light dependent reaction of photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Thymosin is responsible for

Options:
A Raising the blood sugar level
B Raising the blood calcium level
C Differentiation of T lymphocytes
D The decrease in blood RBC

The correct answer is:
Differentiation of T lymphocytes

Solution:
Option (c) is the answer.

Topic: Biology, Chemical Coordination and Integration, NEET
Thymosins play role in

Options:
A Cell-mediated immunity only
B Humoral immunity only
C Both cell-mediated and humoral immunity
D Neither cell-mediated and nor humoral immunity

The correct answer is:
Both cell-mediated and humoral immunity

Solution:
The hormone secreted by the thymus gland is thymosins (peptide hormones).
The functions of thymosins are,
A. Thymosins play an important role in the differentiation of T-lymphocytes. These provide cell-mediated immunity.
B. These also provide humoral immunity by promoting the production of antibodies.
In old individuals, the production of thymosins is decreased as the thymus degenerates during old age. Thus the immune responses of old persons also become weak.

Topic: Biology, Chemical Coordination and Integration, NEET