datavorous/entrance-exam-dataset huggingface dataset:
With the addition of impurities, surface tension of a liquid

Options:
A increases
B decreases
C remains constant
D may increase or decrease depending on impurities

The correct answer is:
may increase or decrease depending on impurities

Solution:
The addition of impurities in a liquid may increase or decrease the surface tension of liquid depending on the nature of impurities.
If the impurities are highly soluble in liquid, then it will increase the surface tension. But if it is less soluble in liquid, then it will decrease the surface tension of liquid.

Topic: Physics, Mechanical Properties of Fluids, COMEDK, COMEDK 2016
With the disintegration of corpus luteum, a decrease in the secretion of

hormone occurs.

Options:
A LH
B Progesterone
C LTH
D FSH

The correct answer is:
Progesterone

Solution:
Progesterone secreted from corpus luteum prepares uterine endometrium for receiving blastocysts for implementation. Progesterone is also called pregnancy hormone and anti- FSH and anti- LH. It maintains pregnancy and prevents the formation of new follicles and ovulation during the gestation period.

Topic: Biology, Chemical Coordination and Integration, NEET
With the gradual increase in frequency of an a. c. source, the impedance of an LCR series circuit

Options:
A first decreases, becomes minimum and then increases.
B increases.
C decreases.
D remains constant.

The correct answer is:
first decreases, becomes minimum and then increases.

Solution:
With increase in frequency current increases initially, becomes maximum and then decreases with further increase in frequency.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2023 (11 May Shift 1)
With the help of a telescope that has an objective of diameter $200 \mathrm{~cm}$, it is proved that light of wavelengths of the order of $6400 Å$ coming from a star can be easily resolved. Then, the limit of resolution is

Options:
A $39 \times 10^8 \mathrm{deg}$
B $39 \times 10^8 \mathrm{rad}$
C $19.5 \times 10^8 \mathrm{rad}$
D $19.5 \times 10^8 \mathrm{deg}$

The correct answer is:
$39 \times 10^8 \mathrm{rad}$

Solution:
The limit of resolution of a telescope is
$$
\theta=\frac{1.22 \lambda}{d}
$$
where, $\lambda=$ wavelength of light rays
$d$ = aperture of objective lens of telescope.
Here, $\lambda=6400 Å=6400 \times 10^{-10} \mathrm{~m}$
$$
d=200 \mathrm{~cm}=2 \mathrm{~m}
$$
Substitution of values gives
$$
\begin{aligned}
\theta & =\frac{1.22 \times 6400 \times 10^{-10}}{2} \\
& =\frac{1.22 \times 64 \times 10^{-8}}{2} \\
& =1.22 \times 32 \times 10^{-8} \\
& =39 \times 10^{-8} \mathrm{rad}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is

(A) directly proportional to the length of the potentiometer wire

(B) directly proportional to the potential gradient of the wire

(C) inversely proportional to the potential gradient of the wire

(D) inversely proportional to the length of the potentiometer wire

Choose the correct option for the above statements:

Options:
A B and D only
B A and C only
C A only
D C only

The correct answer is:
A and C only

Solution:
The basic principle of the potentiometer is that the potential drop across any section of the wire will be directly proportional to the length of the wire, provided the wire is of the uniform cross-sectional area and a uniform current flows through the wire.

Therefore, sensitivity of potentiometer wire is inversely proportional to potential gradient.

Sensitivity∝1Vl∝lV

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (29 Jan Shift 2)
With the increase in atomic weights, melting points of the alkali metals

Options:
A Increase
B Decrease
C Remain constant
D Do not show definite trend

The correct answer is:
Decrease

Solution:
With increase in atomic mass, melting points of alkali metals decreases due to increase in kernel size and decrease in strength of metallic bond.

Topic: Chemistry, s Block Elements, JEE Main
With the passage of time, in which direction is the equilibrium favored in an acid (or base) dissociation equilibrium?

Options:
A The forward direction is favored.
B The reverse direction is favored.
C Both the forward and reverse directions are equally favored.
D The equilibrium remains static and does not favor either direction.

The correct answer is:
The reverse direction is favored.

Solution:
Over time, the equilibrium in an acid (or base) dissociation equilibrium is favored in the reverse direction.

Topic: Chemistry, Ionic Equilibrium, NEET
With the propagation of a longitudinal wave through a material medium,the quantities transmitted in the propagation direction are

Options:
A Energy, momentum and mass
B Energy
C Energy and mass
D Energy and linear momentum

The correct answer is:
Energy and linear momentum

Solution:
The correct option is D Energy and momentum
Whenever any wave travels through a material medium, the particles of the medium start vibrating about their mean positions.
Every vibrating particle transfers its vibration to its immediate next particle. In any vibration there exists Kinetic Energy and Potential Energy. As the vibration gets transfered, the energy also gets transfered. Momentum is defined as the product of mass and velocity. Since the particles have mass and they are in motion, they have momentum. So obviously when vibration is transfered, momentum is also transfered.
In any wave motion, the vibration travels in the forward direction, but no particle actually travels forward. Hence mass doesn't get transfered. We get an illusion that something is moving forward, although nothing moves forward.

Topic: Physics, Waves and Sound, JEE Main
With the usual notation in ∆ABC, if ∠A+∠B=120°, a=3+1 units and b=3-1 units, then the ratio ∠A:∠B is

Options:
A 7:1
B 9:7
C 3:1
D 5:3

The correct answer is:
7:1

Solution:
Let angle B is x.

Using sine rule, we get

3+1sin120°-x=3-1sinx.

⇒3+13-1=sin120°-xsinx

⇒3+13-1=sin120°cosx-cos120°sinxsinx

⇒3+13-1=sin120°cotx-cos120°

⇒3+13-1-12=32cotx

⇒3+232=32cotx

⇒cotx=3+2

⇒tanx=2-3

⇒x=∠B=15°

⇒120-x=∠A=105°

⇒∠A:∠B=7:1

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2019 (10 Jan Shift 2)
With two forces acting at a point, the maximum effect is obtained when their resultant is $4 \mathrm{~N}$. If they act at right angles, then their resultant is $3 \mathrm{~N}$. Then the forces are

Options:
A
$(2+\sqrt{2}) N$ and $(2-\sqrt{2}) N$

B
$(2+\sqrt{3}) \mathrm{N}$ and $(2-\sqrt{3}) \mathrm{N}$

C
$\left(2+\frac{1}{2} \sqrt{2}\right) \mathrm{N}$ and $\left(2-\frac{1}{2} \sqrt{2}\right) \mathrm{N}$

D
$\left(2+\frac{1}{2} \sqrt{3}\right) \mathrm{N}$ and $\left(2-\frac{1}{2} \sqrt{3}\right) \mathrm{N}$

The correct answer is:
$\left(2+\frac{1}{2} \sqrt{2}\right) \mathrm{N}$ and $\left(2-\frac{1}{2} \sqrt{2}\right) \mathrm{N}$

Solution:
$P+Q=4, P^2+Q^2=9 \Rightarrow P=\left(2+\frac{1}{2} \sqrt{2}\right) N$ and $Q=\left(2-\frac{1}{2} \sqrt{2}\right) N$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
With what minimum speed $v$ must a small ball should be pushed inside a smooth vertical tube from a height $\mathrm{h}$ so that it may reach the top of the tube? Radius of the tube is $R$.

Options:
A $\sqrt{g(R-h)}$
B $\sqrt{g(2R-h)}$
C $\sqrt{2g}$
D $\sqrt{2g(2 R-h)}$

The correct answer is:
$\sqrt{2g(2 R-h)}$

Solution:
By using conservation of energy, from the initial point to the highest point
$\begin{aligned}& \frac{1}{2} m v^2=m g(2 R-h)+\frac{1}{2} \mathrm{mV}^2 \\& \text { for minimum } V, V=0 \\& \Rightarrow \frac{1}{2} m v^2=m g(2 R-h) \Rightarrow v=\sqrt{2 g(2 R-h)}
\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890 A∘ is observed at 5896 A∘?

Options:
A 306 km sec-1
B 322 km sec-1
C 296 km sec-1
D 336 km sec-1

The correct answer is:
306 km sec-1

Solution:
f=f01+β1-β; β=vc

ff01+β1-β

1+Δff02=1+β(1-β)-1

β is small compared to 1

1+2Δff0=1+2β

β=Δff0=vc

v=6×c5890=305.6 km sec-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (20 Jul Shift 2)
With which one of the following elements silicon should be doped so as to give ptype of semiconductor?

Options:
A Boron
B Germanium
C Arsenic
D Selenium

The correct answer is:
Boron

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET, NEET 2008 (Mains)
With which type of reproduction do we associate the reduction division? Analyse the reasons for it.

Options:

None

Solution:
Reduction division (meiosis) is associated with sexual reproduction. The reasons for this are:
(a) Since sexual reproduction involves the fusion of two types of gametes (male and female), they must have haploid number of chromosomes.
(b) The cell (meiocyte) which gives rise to gametes often has diploid number of chromosomes and it is only by reducing the number by half that we can get haploid gametes.
(c) Reduction division also ensures maintenance of constancy of chromosome number from generation to generation.

Topic: Biology, Reproduction in Organisms
With $X$ or phenol with $Y . X$ and $Y$ are respectively

Options:
A Zinc dust and soda lime
B Soda lime and zinc dust
C Zinc dust and sodium hydroxide
D Soda lime and copper

The correct answer is:
Soda lime and zinc dust

Solution:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O}$ $\xrightarrow{\mathrm{NaOH}} \mathrm{C}_6 \mathrm{H}_6+\mathrm{Na}_2 \mathrm{CO}_3$
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}+\underset{(y)}{\mathrm{Zn}} \xrightarrow{\text { distill }} \mathrm{C}_6 \mathrm{H}_6+\mathrm{ZnO}$

Topic: Chemistry, Hydrocarbons, JEE Main
Within a spherical charge distribution of charge density ρr, N equipotential surfaces of potential V0, V0+∆V, V0+2∆V,…V0+N∆V ∆V>0, are drawn and have increasing radii r0, r1, r2,…rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and ∆V then :

Options:
A Ïr = constant
B ρ( r ) ∝ 1 r 2
C Ï( r ) â 1 r
D ρ( r ) ∝ r

The correct answer is:
Ï( r ) â 1 r

Solution:
E=− ΔV Δr
∴ â³v and â³r are constant
E = constant
E=Kqr2=c
⇒ q ∝r2
∴ q=∫0rρ4πr2dr
⇒ ρ∝1r

Topic: Physics, Electrostatics, JEE Main, JEE Main 2016 (10 Apr Online)
Within each pair of elements F and Cl , S and Se , and Li and Na , respectively, the elements that release more energy upon an electron gain are

Options:
A Cl,Se and Na
B Cl,S and Li
C F,S and Li
D F,Se and Na

The correct answer is:
Cl,S and Li

Solution:
(i) Electron affinity of second period p-block element is less than third period p-block element due to small size of second period p-block element, due to small size electron density is more and upcoming electron face extra repulsion, so energy release is less. Hence, EA order: F<Cl.
(ii) Down the group electron affinity decreases due to size increases. EA order: S>Se Li>Na.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Wood is separated from phloem in dicots by

Options:
A Pericycle
B Cortex
C Vascular cambium
D Interfascicular cambium

The correct answer is:
Vascular cambium

Solution:
Vascular cambium is present in between xylem and phloem. Vascular cambium divides to form secondary xylem towards inner side and secondary phloem towards outer side during secondary growth. The vascular cambium is the main growth tissue in the stems and roots of many plants, particularly in dicots.

Topic: Biology, Anatomy of Flowering Plants, NEET
Wooden artifact and freshly cut tree give 7.7 and 15.4 min-1g-1 of carbon ( t12= 5770 years) respectively. The age of the artifact is

Options:
A 5770 years
B 5770×15.47.7 years
C 5770×7.715.4 years
D None of these.

The correct answer is:
5770 years

Solution:
t=2.303Î»logâ¡NoN
t=2.3030.6935770logâ¡15.47.7
t=2.303Ã5770Ã0.30100.693
= 5770 years.

Topic: Chemistry, Chemical Kinetics, JEE Main
Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is:

Options:
A 166
B 111
C 19
D 211

The correct answer is:
111

Solution:
We have,

EXAMINATION

AAEIIMNNOTX

Given word has A's→2, I's→2, N's→2 and rest other are distinct alphabets.

Total number of ways to form words using the alphabets of EXAMINATION is

=11!2!·2!·2!

Now, putting M at 4th position

. - - - M- - - -

Total words with M at fourth Place =10!2! 2! 2!

Required probability =10!11!=111

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Work done by a Carnot engine operating between temperatures 127°C and 27°C is 2 kJ. The amount of heat transferred to the engine by the reservoir is:

Options:
A 8 kJ
B 2.67 kJ
C 2 kJ
D 4 kJ

The correct answer is:
8 kJ

Solution:
The temperatures in Kelvin scales are

T2=27 °C+273=300 KT1=127 °C+273=400 K

The efficiency of the Carnot engine is

η=1-T2T1=1-300400=14
Using the relation, η=WQ
⇒2 kJQ=14⇒Q=8 kJ

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Work done by an external agent to move slowly a charge Q from rim of a uniformly charged horizontal disc of radius a and charge per unit area R, to center of this disc is -

Options:
A σRQε02π-12
B σRQε012-1π
C σaQε01π-12
D σaQε012-2π

The correct answer is:
σRQε012-1π

Solution:
2Rcosθ=r
dr=−2Rsinθdθ
when r=0  θ= 90 o = π 2
r=2R  θ=0
dV= Kdθ r
dV= K( σr2θ )dr r
dV=2Kσθdr
∫ dV= ∫ 2Kσθ( −2R sinθdθ )
V=−4KσR ∫ π 2 0 θsinθdθ
V=4KσR ∫ 0 π 2 θ sinθdθ
V=4KσR[ −θcosθ− ∫ 1( −cosθ )dθ ]
V=4KσR [ −θcosθ+sinθ ] 0 π 2
V=4KσR= σ π ε 0 R
For Rim point V R =V= σR π ε 0
W=Q( V f − V i )
W=Q( V C = V R )
W=Q( σR 2 ε 0 − σR π ε 0 )
W=QσRε0(12−1π)

Topic: Physics, Electrostatics, JEE Main
Work done during isothermal expansion of one mole of an ideal gas from 10 atm. to $1 \mathrm{~atm}$ at $300 \mathrm{~K}$ is

Options:
A $4938.8 \mathrm{~J}$
B $4138.8 \mathrm{~J}$
C $5744.1 \mathrm{~J}$
D $6257.2 \mathrm{~J}$

The correct answer is:
$5744.1 \mathrm{~J}$

Solution:
Given that
$P_1=10 \mathrm{~atm}, P_2=1 \mathrm{~atm}, T=300 \mathrm{~K}, n=1$
$R=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$
Now, by using
$W=2.303 n R T \log _{10} \frac{P_2}{P_1}$
$=2.303 \times 1 \times 8.314 \times 300 \log _{10} \frac{1}{10}$
$W=5744.1 \mathrm{Joule}$

Topic: Chemistry, Thermodynamics (C), JEE Main
Work done in increasing the size of a soap bubble from a radius of $3 \mathrm{~cm}$ to $5 \mathrm{~cm}$ is nearly (Surface tension of soap solution $=0.03 \mathrm{Nm}^{-1}$ ):

Options:
A
$0.2 \pi \mathrm{mJ}$

B
$2 \pi \mathrm{mJ}$

C
$0.4 \pi \mathrm{mJ}$

D
$4 \pi \mathrm{mJ}$

The correct answer is:
$0.4 \pi \mathrm{mJ}$

Solution:
$\mathrm{W}=\mathrm{T} \times \Delta \mathrm{A}=\mathrm{T} \times 8 \pi\left(\mathrm{r}_2{ }^2-\mathrm{r}_1{ }^2\right)=0.4 \pi \mathrm{mJ}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2011
Work done in stretching a wire through 1mm is 2J. What amount of work will be done for elongating another wire of same material, with half the length and double the radius of cross section, by 1mm?

Options:
A 2J
B 4J
C 8J
D 16J

The correct answer is:
16J

Solution:
k=YAl
âµâÏâ=12kx2
2=12YÏR2L1mm2 (1)
Ï1=12YÏ2R2L21mm2 â¦(2)
Equation (2) Ã· (1)
Ï12=8⇒Ï'=16

Topic: Physics, Mechanical Properties of Solids, JEE Main
Work done on a ideal gas in a cylinder when it is compressed by an external pressure in a single step is shown below:

Which of the following graphs will show the work done on the gas?

Options:
A
B
C
D

The correct answer is:

Solution:
The pressure is constant and equal to Pex
Also the process is a single step
Therefore, it is a straight line parallel to volume axis and has a value of Pex
Also, work done =Pex⋅ΔV= Area under theP−Vcurve.

Topic: Chemistry, Thermodynamics (C), JEE Main
World summit on Sustainable Development (2002) was held in

Options:
A South Africa
B Brazil
C Sweden
D Argentina

The correct answer is:
South Africa

Solution:
It was held in South Africa (Johnnesberg)

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2008 (Mains)
Would the Bohr formula for the $\mathrm{H}$-atom remain unchanged if proton had a charge $(+4 / 3)$ e and electron a charge $(-3 /$ 4)e, where $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$. Give reasons for your answer.

Options:

None

Solution:
As there is no interchange in the position of proton and electron only the magnitude of charge change as coulombian force will be same.
If proton had a charge $(+4 / 3)$ e and electron a charge $(-3 /$ 4)e, then the Bohr formula for $\mathrm{H}$-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

Topic: Physics, Atomic Physics
Write a short note on
(a) Adaptations of desert plants and animals
(b) Adaptations of plants to water scarcity
(c) Behavioural adaptations in animals
(d) Importance of light to plants
(e) Effect of temperature or water scarcity and the adaptations of animals.

Options:

None

Solution:
(a) Desert plants have very small leaves or no leaves at all, and carrying out photosynthesis through the stems. Their stem could become succulent, and can store and retain water. Animals living in hot climatic region tend to be smaller than those living in cold climates. This can be explained by the fact that the amount of heat gained from the environment is approximately proportional to the body surface area. The majority of animals living in desert are small, like kangaroo rat. It feed on dry seeds and other dry plant material and does not drink.
(b) The evergreen trees such as Rhododendron, show water scarcity by an inward curling of the leaves. A more significant response is the closure of stomata, which reduces transpiration, but raises the internal temperature of the leaf affecting the rate of synthesis of proteins and photosynthesis.
Deciduous trees of the temperature region drop their leaves in autumn, avoiding winter drought. Some water stressed plant's accumulate excessive amounts of inorganic ions.
(c) Migrating temporarily to a less stressful habitat forms a more stressful habitat is a kind of behavioural adaptation in animals which enables them to survive in better environmental conditions. Desert lizards regulate their body temperature constant by behavioural means. They bask in the sun and absorb
heat when their body temperature decreases below the optimum, but move into shady or underground places when the temperature of the surrounding area starts increasing.
(d) Light affects plants through its quality, intensity and duration. Duration of light affects phenology, photosynthesis, growth, reproduction, flowering. Quality of light influences flowering, seed germination and movements. Light is required for the production of chlorophyll in chloroplasts. Plants germinated under insufficient illumination causes the destruction of chlorophyll.
(e) Animals mainly are of two types as they are adapted to controlling their body temperature. Poikilothermous are the animals whose temperature fluctuates with that of the environment, e.g., invertebrates and vertebrates, other than birds and animals. In contrast, homeotherms are the animals which can maintain their body temperature at a constant level, e.g., birds and mammals.

Topic: Biology, Organisms and Populations
Write down the common features of the connective tissue. On the basis of structure and function, differentiate between bones and cartilages.

Options:

None

Solution:
Features of connective tissue:
Connective tissues are most abundant and widely distributed in the body of complex animals. They are named connective tissues because of their special function of linking and supporting other tissues/organs of the body. They range from soft connective tissues to specialised types, which include cartilage, bone, adipose, and blood. In all connective tissues except blood, the cells secrete fibres of structural proteins called collagen or elastin. The fibres provide strength, elasticity and flexibility to the tissue. These cells also secrete modified polysaccharides, which accumulate between cells and fibres and act as matrix (ground substance).
Difference between Bone and cartilage:
The intercellular material of cartilage is solid and pliable and resists compression. Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them. Most of the cartilages in vertebrate embryos are replaced by bones in adults. Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column, limbs and hands in adults,
- Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres which give bone its strength. It is the main tissue that provides structural frame to the body. Bones support and protect softer tissues and organs. The bone cells (osteocytes) are present in the spaces called lacunae. Limb bones, such as the long bones of the legs, serve weight-bearing functions. They also interact with skeletal muscles attached to them to bring about movements. The bone marrow in some bones is the site of production of blood cells.

Topic: Biology, Structural Organisation in Animals