datavorous/entrance-exam-dataset huggingface dataset:
___ comprises a foreign gene which is purposefully inserted into its genome.

Options:
A Transgenic bacteria
B Transgenic fungus
C Transgenic plant
D transgenic human

The correct answer is:
Transgenic bacteria

Solution:
Transgenic bacteria comprises a foreign gene which is purposefully inserted into its genome.

Topic: Biology, Biotechnology and Its Applications, NEET
____ comprises a group of related species which has more characters in common

Options:
A Phylum
B Class
C Family
D Genus

The correct answer is:
Genus

Solution:
Genus comprises a group of related species which has more characters in common.

Topic: Biology, The Living World, NEET
Concanavalin A is:

Options:
A An alkaloid
B An essential oil
C A lectin
D A pigment

The correct answer is:
A lectin

Solution:
Concanavalin A (ConA) is a lectin (carbohydrate-binding protein) originally extracted from the jack-bean, Canavalia ensiformis.

Topic: Biology, Biomolecules (B), NEET, NEET 2019 (All India)
Concanavalin A is

Options:
A Alkaloids
B Chlorophyll
C Terpenoids
D Lectins

The correct answer is:
Lectins

Solution:
Concanavalin A is a lectin (carbohydrate-binding protein) originally extracted from the jack-bean, Canavalia ensiformis. It is a member of the legume lectin family.

Topic: Chemistry, Biomolecules, NEET
Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL-1. Volume of acid required to make one litre of 0.1 M H2SO4 solution

Options:
A 16.65 mL
B 22.20 mL
C 5.55 mL
D 11.10 mL

The correct answer is:
5.55 mL

Solution:
M= w/w%×d×10 mol. weight
= 98×1.8×10 98
∴ M=18 M
Now,   M 1 V 1 =M 2 V 2
⇒  18×  V 1 =0.1×1000
⇒  V 1 = 100 18
=5.55 mL

Topic: Chemistry, Solutions, NEET
Concentrated nitric acid used for laboratory works is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of solution is 1.504 g mL-1 ?

Options:
A 26.23M
B 16.23M
C 6.23M
D 46.23M

The correct answer is:
16.23M

Solution:
68% mass of HN O 3 means 100 g solution contains 68 g HN O 3
â´ Volume of solution = wt. of solution / density =1001.504 = 66.49 mL
Molarity M=mole of HNO3volume of solution in litre
=68 Ã100063Ã66.49=16.23M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?

Options:
A 21.24 M
B 18.67 M
C 16.23 M
D 23.34 M

The correct answer is:
16.23 M

Solution:
Correct Option is : (C)
16.23 M

Topic: Chemistry, Solutions, NEET
Concentrated of aqueous sulphuric acid is $98 \% \mathrm{H}_2 \mathrm{SO}_4$ by mass and has a density of $1.80 \mathrm{~g} \mathrm{~mL}^{-1}$ Volume of acid required to make one litre of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution is:

Options:
A $16.65 \mathrm{~mL}$
B $22.20 \mathrm{~mL}$
C $5.55 \mathrm{~mL}$
D $11.10 \mathrm{~mL}$

The correct answer is:
$5.55 \mathrm{~mL}$

Solution:
Molarity of $\mathrm{H}_2 \mathrm{SO}_4$ solution
$=\frac{98 \times 1000}{98 \times 100} \times 1.84=18.4$
Suppose $\mathrm{V} \mathrm{ml}$ of this $\mathrm{H}_2 \mathrm{SO}_4$ is used to prepare 1 . of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
$\begin{aligned}
& \therefore \mathrm{V} \times 18.02=1000 \times 0.1 \\
& \mathrm{~V}=\frac{1000 \times 0.1}{18.02}=5.55 \mathrm{ml}
\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2007
Concentration of sulphide ores are generally done by

Options:
A Froth floatation process
B Gravity separation
C Magnetic separation
D By hand picking

The correct answer is:
Froth floatation process

Solution:
Froth floatation method is based on the fact that the surface of sulphide ores is preferentially wetted by oil while that of gangue is wetted by water.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.2×10-4 mol L-1. Solubility product of Ag2C2O4 is

Options:
A 2.42× 10-8
B 2.66×10-12
C 4.5×10-11
D 5.3×10-12

The correct answer is:
5.3×10-12

Solution:
Ag2C2O4s⇌2Ag+aq+C2O42-aq2ss
Ksp=Ag+2 C2O42-
Ag+=2.2×10-4mol/L
∴ C2O42-=2.2×10-42 M=1.1×10-4M
∴ Ksp=2.2×10-42 1.1×10-4 (mol/L)3
=5.324×10-12 (mol/L)3.

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2017
________ concluded that the presence of cell wall is a unique character of plant cells.

Options:
A T. Schwann
B M. Schleiden
C R. Virchow
D R. Brown

The correct answer is:
T. Schwann

Solution:
It was observed by Schleiden that all tissues of the plants are composed of different kinds of cells. The different kinds of animal and plant cells were studied by Schwann. He observed that animal cells had a thin outer layer which later was named as plasma membrane. Through his observations on the plant tissues he also concluded that an unique character found in the plant cells is the presence of cell wall. The cell theory was together formulated by Schleiden and Schwann. Rudolf Virchow was the first person to explain that new cells arise from cell division of pre-existing cells.

Topic: Biology, Cell - The Unit of Life, NEET
Conclusions similar to those drawn by Charles Darwin about evolution were also found by:

Options:
A Lamarck
B A. Wallace
C Oparin and Haldane
D L. Pasteur

The correct answer is:
A. Wallace

Solution:
Darwin drew the conclusions regarding the fitness of a population or an individual and the various adaptations acquired by organisms to survive better. He called it natural selection and implied it is a mechanism of evolution. Similar findings were also proposed by naturalist Alfred Wallace who was working on the Malay Archipelago around the same time.

Topic: Biology, Evolution, NEET
Condoms are one of the most popular contraceptives because of the following reasons,

Options:
A These are effective barriers for insemination
B They do not interfere with coital act
C These help in reducing the risk of STDs
D All the above

The correct answer is:
All the above

Solution:
Contraceptive methods avoid pregnancy. These can be Natural methods, Barrier methods, IUD's, Chemical methods, Surgical methods and Implants. Barrier methods are condoms, femidoms etc. Barrier methods are popular as they are easy to use with the help of a practioner. These are effective methods, and their use does not interfere with coitus and also prevents sexually transmitted diseases.

Topic: Biology, Reproductive Health, NEET
Conductivity of Conductors are of order-

Options:
A 102
B 10
C 105
D 103

The correct answer is:
103

Solution:
Correct Option is : 4
103

Topic: Chemistry, Electrochemistry, NEET
$\left.\left[\mathrm{CoNH}_3\right)_4\left(\mathrm{NO}_2\right)_2\right] \mathrm{Cl}$ exhibits

Options:
A linkage isomerism, geometrical isomerism and optical isomerism
B linkage isomerism, ionization, isomerism and optical isomerism
C linkage isomerism, ionization isomerism and geometrical isomerism
D ionization isomerism, geometrical isomerism and optical isomerism

The correct answer is:
linkage isomerism, ionization isomerism and geometrical isomerism

Solution:
The compound $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right)_2\right]$ $\mathrm{Cl}$ exhibits linkage, ionisation and geometrical isomerism.
Linkage isomers: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right)_2\right]$ $\mathrm{Cl}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4(\mathrm{ONO})_2\right] \mathrm{Cl}$
Ionisation isomers: $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right)_2\right]$ $\mathrm{Cl}$ and $\left.\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_2\right) \mathrm{Cl}\right)\right] \mathrm{NO}_2$

Topic: Chemistry, Coordination Compounds, NEET, NEET 2006
Conifers are adapted to tolerate extreme environmental conditions because of:

Options:
A Broad hardy leaves
B Superficial stomata
C Thick cuticle
D Presence of vessels

The correct answer is:
Thick cuticle

Solution:
In conifers, the needle-like leaves, thick cuticle and sunken stomata help to reduce water loss. One of the most important adaptations of conifer trees is the thick waxy cuticle that waterproof the leaves. It also has the presence of needle-like leaves. These leaves are adapted to survive in harsher and colder conditions compared to broad leaves.

Topic: Biology, Plant Kingdom, NEET, NEET 2016 (Phase 2)
Conifers differ from grasses in the:

Options:
A formation of endosperm before fertilization
B production of seeds from ovules
C lack of xylem tracheids
D absence of pollen tubes

The correct answer is:
formation of endosperm before fertilization

Solution:
In gymnosperms, the endosperm is formed, before fertilization. In both conifers and grasses seeds are produced from ovules. Xylem tracheid and pollen tubes are present in both conifers and grasses.

Topic: Biology, Plant Kingdom, NEET, NEET 2006
Conifers differ from grasses in the

Options:
A Lack of xylem tracheids
B Absence of pollen tubes
C Formation of endosperm before fertilisation
D Production of seeds from ovules

The correct answer is:
Formation of endosperm before fertilisation

Solution:
Conifers (gymnosperms) differ from grasses (angiosperms) because in gymnosperms, the female gametophyte is actually endosperm which is made before fertilisation. However, in grasses, endosperm is a tissue formed by the fertilization of second male gamete to polar nuclei. Moreover, in gymnosperms, the endosperm is a haploid tissue, whereas in angiosperms it is triploid.

Topic: Biology, Plant Kingdom, NEET
Conjugate acid of a weak base is always stronger. What will be the decreasing order of basic strength of the following conjugate bases? OH–, RO–, CH3COO–, Cl–

Options:
A OH–, RO–, CH3COO–, Cl–
B RO–, OH–, CH3COO–, Cl–
C OH–, RO–, Cl–, CH3COO–
D OH–, CH3COO–, Cl–,RO–

The correct answer is:
RO–, OH–, CH3COO–, Cl–

Solution:
Correct Option is : (B)
RO–, OH–, CH3COO–, Cl–

Topic: Chemistry, Ionic Equilibrium, NEET
Conjugate base for Bronsted acids H 2 O and HF are:

Options:
A OH- and H2F+, respectively.
B H3O+ and F-, respectively.
C OH- and F-, respectively.
D H3O+ and H2F+, respectively.

The correct answer is:
OH- and F-, respectively.

Solution:
Conjugate base for Bronsted acid H2O, HF are:
H2OBronsted acid→OH-+H+Conjugate base
HFBronsted acid→F-+H+Conjugate base.

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2019 (All India)
Connection between the membranes of a neuron and muscle is termed as:

Options:
A Neuromuscular junction
B Synaptic cleft
C Cyton
D Axon hillock

The correct answer is:
Neuromuscular junction

Solution:
A typical neuron consists of 3 major parts :
Cell body
It contains cytoplasm, typical cell organelles and Nissl's granules (granular bodies)
Dendrites
These are short fibres arising from the cell body and branching repeatedly.
These also possess Nissl's granules.
These fibres transmit impulses towards the cell body.
Axon
It is a long fibre which is branched at its distal end. Each branch ends into a bulb-like structure, which is termed as synaptic knob. Synaptic knob is having synaptic vesicles filled with neurotransmitters

Topic: Biology, Locomotion and Movement, NEET
Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance of 2 m. Assume that the metal surface has a work function of 2 eV, and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å. Estimate the number of photons emitted by the bulb per second. [Assume no other losses]

Options:
A 1×1018/s
B 2×1019/s
C 5×1019/s
D 1×1020/s

The correct answer is:
5×1019/s

Solution:
Correct Option is : (C)
5×1019/s

Topic: Chemistry, Dual Nature of Matter, NEET
Consider a car moving along a straight horizontal road with a speed of 72 km h-1 . If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g=10 m s-2 )

Options:
A 30 m

B 40 m
C 72 m
D 20 m

The correct answer is:
40 m

Solution:
Here u=72 km h-1=20 m s-1; v=0;
a=-μg=-0.5×10=-5 m s-2
As v2=u2+2 as,
∴ s=v2-u22a=(0-202)2×-5=40 m

Topic: Physics, Laws of Motion, NEET
Consider a car moving on a straight road with a speed of 100 m sâ1. The distance at which car can be stopped is [ Î¼ k = 0.5 ]

Options:
A

400 m

B

100 m

C 1000 m
D 800 m

The correct answer is:
1000 m

Solution:
u=100 m s−1, v=0,  a=−μg

Using v 2 − u 2 =2aS,

02−100×100=−2×0.5×10×S

or S=100×10010 m=1000 m

Topic: Physics, Laws of Motion, NEET
Consider a car moving on a straight road with a speed of 100 m s-1. The distance at which the car can be stopped is [ Î¼ k =0.5]

Options:
A 100 m
B 1000 m
C 800 m
D 400 m

The correct answer is:
1000 m

Solution:
s=u22μg=(100)22×0.5×10=1000 m

Topic: Physics, Laws of Motion, NEET
Consider a cell given below : Cu | Cu2+|| Cl-| Cl2, Pt Which species are getting oxidized and reduced at anode and cathode?

Options:
A In cathode,Cu is getting reduced and in anode, Cl2is getting oxidized.
B In anode, Cu is getting oxidized and in cathode, Cl2is getting reduced.
C In anode, Cu is getting reduced and in cathode, Cl2is getting oxidized.
D In cathode,Cu is getting oxidized and in anode, Cl2is getting reduced.

The correct answer is:
In anode, Cu is getting oxidized and in cathode, Cl2is getting reduced.

Solution:
Correct Option is : (B)
In anode, Cu is getting oxidized and in cathode, Cl2is getting reduced.

Topic: Chemistry, Electrochemistry, NEET
Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities $K$ and $2 K$, respectively. The equivalent thermal conductivity of the slab is:

Options:
A $\frac{4}{3} K$
B $\sqrt{2} K$
C $3 \mathrm{~K}$
D $\frac{2}{3} K$

The correct answer is:
$\frac{4}{3} K$

Solution:
$$
\therefore \frac{2 l}{K_{e g}(A)}=\frac{l}{2 K A}+\frac{l}{K A}
$$
for series connection $\mathrm{R}=\mathrm{R}_1+\mathrm{P}_2$
$$
\Rightarrow K_{e g}=\frac{4}{3} K
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2003
Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1), as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image, formed by this combination, has magnification M1 . When the set-up is kept in a medium of refractive index 76 , the magnification becomes M2. The magnitude M2M1 is

Options:
A 7
B 5
C 3
D 6

The correct answer is:
7

Solution:
For reflection from a concave mirror,
1v+1u=1fâ1v-115= -110
1v=115-110=-130
â´v=-30
Magnification m1=-vu=-2
Now for refraction from lens,
1v-1u=1fâ1v=110-120=120
â´ Magnification m2=vu=-1
â´M1=m1m2=2
Now when the set-up is immersed in liquid, no effect for the image formed by mirror.
We have Î¼L-11R1-1R2=110
â1R1-1R2=15
When lens is immersed in liquid,
1flens=Î¼LÎ¼S-1 1R1-1R2=27Ã15=235
â´1v-1u=1fLiquid
â1v=235-120=8-7140=1140
â´ Magnification =-14020=-7
â´M2=2Ã7=14
â´M2M1=7

Topic: Physics, Ray Optics, NEET
Consider a cube of body-centered cubic unit cell of edge length a. Now atom at the body centre can be viewed to be lying on the corner of another cube 2 . Find the volume common to cube 1 and cube 2 .

Options:
A a327
B a364
C a322
D a38

The correct answer is:
a38

Solution:
Volume of cube =a3 , body centred atom can share to 8 such cubes. Hence volume common to cube 1 and 2 is one-eighth of its volume.

Topic: Chemistry, Solid State, NEET
Consider a cylinder of mass M resting on a rough horizontal rug that is pulled out from under it with acceleration 'a' perpendicular to the axis of the cylinder. What is Ffriction at point P ? It is assumed that the cylinder does not slip.

Options:
A Ma
B Ma 2
C Ma 3
D Mg

The correct answer is:
Ma 3

Solution:
For no slipping at P
a 1 + Î± R = a â (1)
f = Ma 1 = M a - Î± R â (2)
f. R = I Â· Î± = MR 2 2 Â· Î± â (3)
Solve the equations (2) and (3)
f = Ma 3 â

Topic: Physics, Rotational Motion, NEET
Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m s-1. Take g constant with a value 10 m s-2. The work done by the (i) gravitational force and the (ii) resistive force of air is

Options:
A i-10 Jii-8.25 J
B iÂ 1.25Â Jii-8.25Â J
C iÂ 100Â JiiÂ 8.75Â J
D i 10 Jii-8.75 J

The correct answer is:
i 10 Jii-8.75 J

Solution:
By Work-Energy Theorem
Wg+Wa=Kf-Ki
mgh+Wa=12mv2-0
10-3Ã10Ã103+wa=12Ã10-3Ã502
Wa= -8.75 J
work done due to gravity =âmgh
= 10 â3 Ã10Ã 10 3
=10J

Topic: Physics, Work Power Energy, NEET, NEET 2017
Consider a hydrogen-like sample of gas whose energy in the nth state is given by En=-13.6Z2n2 eV. When the electron makes transitions from an excited state to the ground state, most energetic photons can have energy Emax=52.224 eV and the least energetic photons can have energy Emin=1.224 eV. The atomic number of the atom and the state of excitation are respectively

Options:
A 4, 5

B 2, 3
C 2, 5
D 2, 4

The correct answer is:
2, 4

Solution:
52.224=13.6z2 112-1n2 .....(1)
1.224=13.6z2 1n-12-1n2 .....(2)
On solving we get z = 2 and n = 5 which means 4 th excited state

Topic: Physics, Atomic Physics, NEET
Consider a long steel bar under a tensile stress due to forces F acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle θ with the length.For what angle is the tensile stress a maximum?

Options:
A θ = 0
B θ = π/4
C θ = π/2
D θ = π

The correct answer is:
θ = π/2

Solution:
Correct Option is : (C)
θ = π/2

Topic: Chemistry, Mechanical Properties of Solids, NEET
Consider a metallic pipe with an inner radius of 1 cm. A cylindrical bar magnet of radius 0.8 cm is dropped through the pipe. Why does it take more time to come down than it takes for a similar unmagnetized cylindrical iron bar dropped through the metallic pipe?

Options:
A Eddy currents produced by the magnet oppose the motion, reducing the downward acceleration.
B The magnet's magnetic field interacts with the metallic pipe, causing a delay in its descent.
C The iron bar has a higher density, leading to faster descent compared to the magnet.
D The metallic pipe's frictional force is stronger when the magnet is dropped through it.

The correct answer is:
Eddy currents produced by the magnet oppose the motion, reducing the downward acceleration.

Solution:
Correct Option is : (A)
Eddy currents produced by the magnet oppose the motion, reducing the downward acceleration.

Topic: Chemistry, Electromagnetic Induction, NEET
Consider a particle, moving in a circle with constant speed. P is a point outside the circle, in the same plane. Then, the angular momentum of the particle about the point P is

Options:
A Never zero
B Zero at exactly two points
C Zero at exactly three points
D Zero at exactly four points

The correct answer is:
Zero at exactly two points

Solution:
At two positions, the velocity vector will pass through point P. For these positions, the angular momentum about point P is zero.

Topic: Physics, Rotational Motion, NEET
Consider a P-V diagram in which the path followed by one mole of a perfect gas in a cylindrical container is shown in Fig. 12.9. What is the ratio of temperature T1/T2, if V2 = 2V1?

Options:
A 1/2
B 1
C √2
D 2

The correct answer is:
√2

Solution:
Correct Option is : (C)
√2

Topic: Chemistry, Thermodynamics, NEET
Consider a reaction aG+bH→ Products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
3

Solution:
aG+bH→Product
rate∝[G]a[H]b
As when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled it means order with respect to G is 1. When concentration of both the reactants G and H is doubled, the rate increases by eight times it means order with respect to H is 2.
Hence a=1,b=2
The overall order of reaction =1+2=3

Topic: Chemistry, Chemical Kinetics, NEET
Consider a reaction X + Y → Products. If the initial concentration of X is increased to four times of its original value, keeping the concentration of Y constant, the rate of reaction increases four-fold. When the concentration of both X and Y becomes four times their original values, the rate of reaction becomes 16 times its original value. The observed rate law is

Options:
A k[X]2[Y]2
B k[X]1[Y]2
C k[X]1[Y]1
D k[X]2[Y]1

The correct answer is:
k[X]1[Y]1

Solution:
r1=k[X]m[Y]n …(i)
r2=4r1=k[4X]m[Y]n …(ii)
r3=16r1=k[4X]m[4Y]n …(iii)
Divide equation (ii) by equation (i)
4=(4)m
(4)1=(4)m
m=1
Divide equation (iii) by equation (ii),
4=(4)n
(4)1=(4)n
n=1 .
Hence rate =k[X]1[Y]1

Topic: Chemistry, Chemical Kinetics, NEET
Consider a rubber ball freely falling from a height h= 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time and the height as a function of time will be

Options:
A
B
C
D

The correct answer is:

Solution:
When rubber ball strikes the elastic surface, its velocity becomes negative after collision without change is speed. Since h is positive, thus initial velocity will be negative. Hence option (3) is correct.

Topic: Physics, Center of Mass Momentum and Collision, NEET
Consider a system of two particle having masses $m_1$ and $m_2$. If the particle of mas $m_1$ is pushed towards the mass centre of particle through a distance $d$, by what distance would the particle of mass $m_2$ move so as to keep the mass centre of particles at the original position?

Options:
A $\frac{m_1}{m_1+m_2} d$
B $\frac{m_1}{m_2} d$
C $d$
D $\frac{m_2}{m_1} d$

The correct answer is:
$\frac{m_1}{m_2} d$

Solution:
We know that
$$
\mathrm{CM}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}
$$
After changing a position of $m_1$ and to keep the position of C.M. same.
C.M. $=\frac{m_1\left(x_1 d\right)+m_2\left(\mathrm{x}_2-d_2\right)}{m_1+m_2}$
$$
\begin{aligned}
0 & =\frac{m_1 d-m_2 d_2}{m_1+m_2} \\
\Rightarrow d_2 & =\frac{m_1}{m_2} d
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2004
Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenyl amine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
6

Solution:
Cr2O72-+Fe2+→Fe3++Cr3+
n factor of Cr2O72-=6
n factor of Fe2+=1
So to reduce one mole of dichromate 6 moles of Fe2+ is required.

Topic: Chemistry, Redox Reactions, NEET
Consider a toroid, having a circular cross-section of radius b, major radius R R>>b, having N turns and carrying current I. Find the total energy stored in the toroid.

Options:
A μ0N2I2b22R
B μ0N2I2b23R
C μ0N2I2b26R
D μ0N2I2b24R

The correct answer is:
μ0N2I2b24R

Solution:
The magnetic field inside a toroid is,
B=μ0Ni2πR
Flux linked with the toroid,
ϕ=πb2×B×N
Also, flux is related to self-inductance as,
ϕ=Li
L=ϕi=μ0N2b22R with b<<<R
Energy =12Li2=μ0N2I2b24R

Topic: Physics, Magnetic Effects of Current, NEET
Consider a uniform electric field E = 3 × ${{10}^{3}}$ î N/C. What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?

Options:
A 0.09 Nm²/C
B 0.3 Nm²/C
C 9 Nm²/C
D 30 Nm²/C

The correct answer is:
30 Nm²/C

Solution:
Correct Option is : (D)
30 Nm²/C

Topic: Chemistry, Electrostatics, NEET
Consider a wire with density d and stress Ï. For the same density, if the stress increases 2 times, the speed of the transverse waves along the wire changes by

Options:
A 2
B 12
C 2
D 12

The correct answer is:
2

Solution:
∵ Speed of transverse wave
v=Stressdensity of wire or v=σd
According to question, if σ increases 2 times
then v1=2σd=2σd=2v
Hence, speed of transverse waves along with wire changes by 2 times.

Topic: Physics, Waves and Sound, NEET