datavorous/entrance-exam-dataset huggingface dataset:
Statement-1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
Statement-2 : Principle of conservation of momentum holds true for all kinds of collisions.
Of the four choices given after the statements, choose the one that best describes the two statements.
Of the four choices given after the statements, choose the one that best describes the two statements.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

B
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement1

C
Statement-1 is false, Statement-2 is true.

D
Statement-1 is true, Statement-2 is false.

The correct answer is:
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Solution:
If it is a completely inelastic collision then
$$
\begin{gathered}
m_1 v_1+m_2 v_2=m_1 v+m_2 v \\
v=\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\
\mathrm{~K} \cdot \mathrm{E}=\frac{\mathrm{p}_1^2}{2 \mathrm{~m}_1}+\frac{\mathrm{p}_2^2}{2 \mathrm{~m}_2}
\end{gathered}
$$
as $\overrightarrow{\mathrm{p}}_1$ and $\overrightarrow{\mathrm{p}_2}$ both simultaneously cannot be zero therefore total KE cannot be lost.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2010
Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $\mathrm{K}_{\max }$. When the ultraviolet light is replaced by $\mathrm{X}-$ rays, both $\mathrm{V}_0$ and $\mathrm{K}_{\max }$ increase.
Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.
Of the four choices given after the statements, choose the one that best describes the two statements.

Options:
A
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

B
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-
1.

C
Statement-1 is false, Statement-2 is true.

D
Statement-1 is true, Statement- 2 is false.

The correct answer is:
Statement-1 is true, Statement- 2 is false.

Solution:
Since the frequency of ultraviolet light is less than the frequency of $X$-rays, the energy of each incident photon will be more for X-rays
$K.E_{photoelectron} =h v-\varphi$
Stopping potential is to stop the fastest photoelectron
$$
\mathrm{V}_0=
$$
so, $\mathrm{K}_{\mathrm{E}} \mathrm{E}_{\max }$ and $\mathrm{V}_0$ both increases.
But K.E ranges from zero to K.E $E_{\max }$ because of loss of energy due to subsequent collisions before getting ejected and not due to range of frequencies in the incident light.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2010
Statement-I : A point charge is brought in an electric field. The value of electric field at a point near to the charge may increase if the charge is positive.

Statement-II : An electric dipole is placed in a uniform electric field. The net electric force on the dipole will not be zero.

Options:
A Both statement-I and statement-II are true.
B Both statement-I and statement-II are false.
C Statement-I is true but statement-II is false.
D Statement-I is false but statement-II is true.

The correct answer is:
Statement-I is true but statement-II is false.

Solution:
If a positive point charge is brought in an electric field. The value of electric field at a point near to the charge may increase or decreases. Hence, statement I is true.

An electric dipole experiences zero net force in a uniform electric field. Hence, statement II is false.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Statements about heavy water are given below.

A. Heavy water is used in exchange reactions for the study of reaction mechanisms.

B. Heavy water is prepared by exhaustive electrolysis of water.

C. Heavy water has higher boiling point than ordinary water.

D. Viscosity of H2O is greater than D2O.

Choose the most appropriate answer from the options given below:

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

The correct answer is:
A,B and C only

Solution:
Heavy water is used in exchange reactions for study of reaction mechanisms Heavy water is prepared by exhaustive electrolysis of water.

B.P. of D2O=374.4 K

B.P. of H2O=373 K

Viscosity of H2O=0.89 centipoise

Viscosity of D2O=1.107 centipoise

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Statements about heavy water are given below.
A. Heavy water is used in exchange reactions for the study of reaction mechanisms.
B. Heavy water is prepared by exhaustive electrolysis of water.
C. Heavy water has higher boiling point than ordinary water.
D. Viscosity of \( \mathrm{H}_{2} \mathrm{O} \) is greater than \( \mathrm{D}_{2} \mathrm{O} \).
Choose the most appropriate answer from the options given below:

Options:
A \( \mathrm{A} \) and \( \mathrm{C} \) only
B \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) only
C \( \mathrm{A} \) and \( \mathrm{B} \) only
D A and D only

The correct answer is:
\( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) only

Solution:
Heavy water is used in exchange reactions for study of reaction mechanisms Heavy water is prepared by exhaustive electrolysis of water.

B.P. of D2O=374.4 K

B.P. of H2O=373 K

Viscosity of H2O=0.89 centipoise

Viscosity of D2O=1.107 centipoise

Topic: Chemistry, Hydrogen, JEE Main
Stationary waves are formed when

Options:
A Two waves of equal amplitude and equal frequency travel along the same path in opposite directions
B Two waves of equal wavelength and equal amplitude travel along the same path with equal speeds in opposite directions
C Two waves of equal wavelength and equal phase travel along the same path with equal speed
D Two waves of equal amplitude and equal speed travel along the same path in opposite direction

The correct answer is:
Two waves of equal wavelength and equal amplitude travel along the same path with equal speeds in opposite directions

Solution:
Two waves of equal wavelength and equal amplitude travel along the same path with equal speeds in opposite directions

Topic: Physics, Waves and Sound, JEE Main
Statues and monuments in India are affected by acid rain. How?

Options:

None

Solution:
The air around the statues and monuments in India contains fairly high levels of the oxides of sulphur and nitrogen. This is mainly due to a large number of industries and power plants in the nearby areas. The problem has been further aggravated due to use of poor quality of coal, kerosene and firewood as fuel for domestic purposes. The resulting acid rain attacks the marble of these statues and monuments
\(\mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{CaSO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2\)
As a result, these monuments are being slowly eaten away and marble is getting discolored and lusterless.

Topic: Chemistry, Environmental Chemistry
Steam at 100 Â°C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15 Â°C till the temperature of the calorimeter and its contents rise to 80 Â°C . The mass of the steam condensed in kg is

Options:
A 0.130
B 0.065
C 0.260
D 0.200

The correct answer is:
0.130

Solution:
mL+m100-80=1.1Ã1Ã80-15+0.02Ã80-15
540 m+20 m=71.5+1.30
âm=0.130

Topic: Physics, Thermodynamics, JEE Main
Steam at 100 oC is passed into 20 g of water at 10 oC. When water acquires a temperature of 80oC, the mass of water present will be:
[Take specific heat of water =1 cal g-1 oC-1 and latent heat of steam =540 cal g-1]

Options:
A 24Â g
B 31.5Â g
C 42.5Â g
D 22.5 g

The correct answer is:
22.5 g

Solution:
Given: Specific heat of water Sw=1 cal g-1°C-1, Latent heat of steam Ls=540 cal g-1

Let the mass of steam =m g

So heat lost in change of state from steam to water equal to sum of latent heat and heat use in raise of the temperature of water.

Q1 =mLs +msw∆T

Q1=m×540 +m×1×(100-80)

Q1 =540m +20m =560m

Heat gained by water to change its temperature

Q2=mwSw∆T

Q2=20×1×(80-10)=1400.

Now we know from the principle of calorimetry is that total heat lost by the hot body is equal to total heat gained by cold body i.e it follows law of conservation energy. Hence

Heat lost Q1= Heat gained Q2

560m =1400m =1400560 =2.5 g
Therefore, total mass of water at 80°C

m=(20+2.5)g=22.5 g

Topic: Physics, Thermal Properties of Matter, JEE Main
Steel ruptures when a shear of $3.5 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$ is applied. The force needed to punch a $1 \mathrm{~cm}$ diameter hole in a steel sheet $0.3 \mathrm{~cm}$ thick is nearly:

Options:
A
$1.4 \times 10^4 \mathrm{~N}$

B
$2.7 \times 10^4 \mathrm{~N}$

C
$3.3 \times 10^4 \mathrm{~N}$

D
$1.1 \times 10^4 \mathrm{~N}$

The correct answer is:
$3.3 \times 10^4 \mathrm{~N}$

Solution:
Shearing strain is created along the side surface of the punched disk. Note that the forces exerted on the disk are exerted along
the circumference of the disk, and the total force exerted on its center only.
Let us assume that the shearing stress along the side surface of the disk is uniform, then
$$
\begin{aligned}
\mathrm{F}>& \int_{\text {surface }} \mathrm{dF}_{\max }=\int_{\text {surface }} \sigma_{\max } \mathrm{dA}=\sigma_{\max } \int_{\text {surface }} \mathrm{dA} \\
&=\int \sigma_{\max } \cdot \mathrm{A}=\sigma_{\max } \cdot 2 \pi\left(\frac{\mathrm{D}}{2}\right) \mathrm{h} \\
&=3.5 \times 10^8 \times\left(\frac{1}{2} \times 10^{-2}\right) \times 0.3 \times 10^{-2} \times 2 \pi \\
&=3.297 \times 10^4 \simeq 3.3 \times 10^4 \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2014 (12 Apr Online)
Step taken by the Government of India to control air pollution include :

Options:
A Use of non-polluting Compressed Natural Gas (CNG) only as fuel by all buses and trucks
B Compulsory mixing of $20 \%$ ethyl alcohol with petrol and $20 \%$ biodiesel with diesel
C Compulsary PUC (Pollution Under Control) certification of petrol driven vehicles which tests for carbon monoxide and hydrocarbons
D Permission to use only pure diesel with a maximum of $500 \mathrm{ppm}$ sulphur as fuel for vehicles

The correct answer is:
Compulsary PUC (Pollution Under Control) certification of petrol driven vehicles which tests for carbon monoxide and hydrocarbons

Solution:
No Solution Available

Topic: Biology, Environmental Issues, NEET, NEET 2009 (Mains)
Step-up transformers are used for

Options:
A increasing electrical power.
B decreasing electrical power.
C decreasing voltage.
D increasing voltage.

The correct answer is:
increasing voltage.

Solution:
In the step-up transformer, the output (secondary Vs) voltage is more than the input (primary Vp) voltage. The ratio of the secondary and primary voltage is given by,

VsVp=NsNp; where Ns number of turns in the secondary coil and Np is the number of turns in the primary coil.

In step-up transformer Ns>Np, hence Vs>Vp.

Thus, step up transformers are used for increasing voltage.

Topic: Physics, Alternating Current, JEE Main
Stoke's law states that the viscous drag force $F$ experienced by a sphere of radius a, moving with a speed $V$ through a fluid with coefficient of viscosity $\eta$, is given by $F=6$ inav. If this fluid is flowing through a cylindrical pipe of radiusr, leng th $/$ and a pressure difference of Pacross its two ends, then the volume of water $V$ which
flows through the pipe in time $t$ can be written as $\frac{V}{t}=k\left(\frac{P}{\ell}\right)^{a} \eta^{b} r^{c}$, where $k$ is a dimensional constant. Correct values of $a, b$ and $c$ are

Options:
A $a=1, b=-1, c=4$
B $a=-1, b=1, c=4$
C $a=2, b=-1, c=3$
D $a=1, b=-2, c=-4$

The correct answer is:
$a=1, b=-1, c=4$

Solution:
$\begin{array}{l}
\frac{V}{t}=k\left(\frac{P}{\ell}\right)^{a} \cdot \eta^{b} \cdot \gamma^{c} \\
{[r]=L ;[l]=L} \\
{[P]=M L^{-1} T^{-2} ;[\eta]=M L^{-1} T^{-1}} \\
{[\mathrm{~V}]=L^{3}} \\
{[t]=T}
\end{array}$
By calculation
$a=1 ; b=-1 ; c=4$

Topic: Physics, Units and Dimensions, JEE Main
Stolon is a type of vegetative propagule which is majorly observed in strawberries. It is a modified

Options:
A sub-aerial stem
B underground stem
C aerial stem
D root

The correct answer is:
sub-aerial stem

Solution:
Vegetative propagules are the plant parts or organs that are involved in the process of reproduction even though their primary function is not it. The reproductive process is called as vegetative propagation as the non-reproductive parts/organs (leaf/stem/roots) are involved.
Stolon is a slender lateral branch that arises from the base of main axis. It is a modified sub-aerial stem observed in jasmine, strawberry. Initially, stolon grows upwards and then bends down and touches the soil where its terminal end gives rise to a new shoot and roots.

Topic: Biology, Morphology of Flowering Plants, NEET
Stomata of CAM plants:

Options:
A are always open
B open during the day and close at night
C open during the night and close during the day
D never open

The correct answer is:
open during the night and close during the day

Solution:
Stomata usually open when leaves are transferred from darkness to light, i.e. at night. Plants such as cacti and Opuntia achieve their high water use efficiency by opening their stomata during the cool, desert nights and closing them during the hot, dry days.
Related Theory
In CAM plants or xerophytes the stomata open in night and remain closed during day as an adaptation to conserve water. The opening of stomata in these plants occurs at night to collect carbon dioxide and allow it to diffuse to the mesophyll cells. Carbon dioxide is stored in the form of malic acid in vacuoles at night.

Topic: Biology, Transport in Plants, NEET, NEET 2003
Stomata opens

Options:
A in day time - for gaseous exchange
B in day time - no gaseous exchange
C in night time only - for gaseous exchange
D in night time only - no gaseous exchange.

The correct answer is:
in day time - for gaseous exchange

Solution:
Stomata are meant for gaseous exchange but are also the main source of transpiration. In majority of plants stomata open in the presence of light i.e., in daytime and close in darkness (night).

Topic: Biology, Transport in Plants, AIIMS, AIIMS 2001
Stomata which opens up in night and remains close during day time shows

Options:
A C3 pathway
B CAM pathway
C C4â pathway
D Both (B) and (C)

The correct answer is:
CAM pathway

Solution:
In plants growing in deserts, commonly known as succulent xerophytes, the stomata opens during the night and closes during day time, such stomata are known as scotoactive stomata. It is an adaptation to reduce transpiration during day time when the temperature is very high. Such plants show CAM metabolism as a method of photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Stomatal movement is not affected by

Options:
A O2 concentration
B Light
C Temperature
D CO2 concentration

The correct answer is:
O2 concentration

Solution:
Stomatal movement is not affected by O 2 concentration. Stomata are tiny pore found in the epidermis of leaves and other soft aerial parts. They are meant for the gaseous exchange and are also the main source of transpiration. Stomatal movements are affected by many factors like light, temperature and C O 2 concentration, In the majority of plants, the stomata are open in light and close in darkness. Normally high temperature above 30°C reduces stomatal opening in many species. Low C O 2 concentration usually induces opening of stomata while high C O 2 concentration closes the same.

Topic: Biology, Transport in Plants, NEET, NEET 2018
Stratified epithelial cells have limited role in secretion. Justify their role in our skin.

Options:

None

Solution:
Stratified epithelium consists of epithelial cells in which the innermost layer is made up of columnar or cuboidal cells. It is a type of compound epithelium and a waterproof protein called keratin is present few outer layers.
These layers of dead cells is called horny layer which is shed at intervals due to frictions and thus has a limited role in secretions and absorption. The main function of stratified epithelium is to provide protection to the body against mechanical and chemical stresses.

Topic: Biology, Structural Organisation in Animals
String AB of unstretched length L is stretched by applying a force F at the mid-point C such that the segments AC and BC make an angle θ with AB as shown in the figure. The string may be considered as an elastic element with a force to elongation ratio K. The force F is given by

Options:
A KL1-tanθsinθ
B 2KL1-cosθtanθ
C KL1-cosθtanθ
D 2KL1-sinθtanθ

The correct answer is:
KL1-cosθtanθ

Solution:
Refer the diagram applying Lami's thorem,

Fsin180-2θ=Tsin90+θ

∴ Fsin2θ=Tcosθ⇒F2sinθcosθ=Tcosθ

∴ F=2sinθ×T ...1

but T=K×∆L=K×L2×1-cosθcosθ

put in equation 1,

F=2sinθ×K×L2×1-cosθcosθ

∴ F=KL1-cosθtanθ

Topic: Physics, Mechanical Properties of Solids, JEE Main
Stroma in the chloroplast of higher plant contains :

Options:
A Chlorophyll
B Light-independent reaction enzymes
C Light-dependent reaction enzymes
D Ribosomes

The correct answer is:
Light-independent reaction enzymes

Solution:
No Solution Available

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2009 (Mains)
Strongest acid is

Options:
A $\mathrm{HC} \equiv \mathrm{CH}$
B $\mathrm{C}_2 \mathrm{H}_6$
C $\mathrm{C}_6 \mathrm{H}_6$
D $\mathrm{CH}_3 \mathrm{OH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{OH}$

Solution:
Because hydrogen is attached with high electronegative element.

Topic: Chemistry, General Organic Chemistry, JEE Main
Strongest nucleophile is

Options:
A $\mathrm{RNH}_2$
B $\mathrm{ROH}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}$
D $\mathrm{CH}_3 \mathrm{O}^{-}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{O}^{-}$

Solution:
$\mathrm{CH}_3-\mathrm{O}^{-}$is the strongest nucleophile which is capable of acting as doner of electron pair.

Topic: Chemistry, General Organic Chemistry, JEE Main
Strontium crystallizes in a fcc unit cell with edge length a. It contains 0.2% Frenkel defect and another crystal of Sr contains 0.1% Schottky defect. Density of solid with Frankel defect = d f and density with Schottky defect = d s , then

Options:
A df=ds
B df>ds
C df<ds
D df=2ds

The correct answer is:
df>ds

Solution:
No effect on density due to Frenkel defect but due to Schottky defect density of solid decrease

Topic: Chemistry, Solid State, JEE Main
Structurally biodegradable detergent should contain

Options:
A Normal alkyl chain
B Branched alkyl chain
C Alkyl side chain
D Cyclohexyl side chain

The correct answer is:
Alkyl side chain

Solution:
Detergents with straight chain alkyl groups are biodegradable.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Structure of 4-Methylpent-2-enal is:

Options:
A
B
C
D

The correct answer is:

Solution:
"4-Methyl" indicates that there is a methyl group (CH3) attached to the fourth carbon atom.

"Pent-2-enal" indicates that there is an aldehyde functional group (─CHO) attached to the five-carbon chain, and there is a double bond (─C=C─) between the second and third carbon atoms.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal $A$ and the octahedral voids are occupied by a monovalent metal $B$. The formula of the oxide is

Options:
A $\mathrm{ABO}_2$
B $\mathrm{A}_2 \mathrm{BO}_2$
C $\mathrm{A}_2 \mathrm{~B}_3 \mathrm{O}_4$
D $\mathrm{AB}_2 \mathrm{O}_2$

The correct answer is:
$\mathrm{AB}_2 \mathrm{O}_2$

Solution:
According to ccp,
Number of $\mathrm{O}^{2-}$ ions $=4$
So, tetrahedral void $=8$
and $\quad$ octahedral void $=4$
Since, $A$ ions occupied $\frac{1}{4}$ th of tetrahedral void.
$\therefore$ Number of $A$ ions $=\frac{1}{4} \times 8=2$
Again, $B$ ions occupied all octahedral void.
$\therefore$ Number of $B$ ions $=4$
$$
\begin{aligned}
A: B: \mathrm{O} & =2: 4: 4 \\
& =1: 2: 2
\end{aligned}
$$
$\therefore$ Structure of oxide $=A B_2 \mathrm{O}_2$

Topic: Chemistry, Solid State, NEET, NEET 2012 (Mains)
Structure of \( \mathrm{NH}_{3} \) is:

Options:
A Trigonal
B Tetrahedral
C Pyramidal
D Trigonal bipyramidal

The correct answer is:
Pyramidal

Solution:
Ammonia molecule is trigonal pyramidal with the nitrogen at the apex. The N atom in NH3 is sp3 hybridised containing lone pair of electrons due to which the bond angle H-N-H is 107.5°. It has three bond pairs and one lone pair of electrons

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Structure of some important polymers are given. Which one represents Buna-S ?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Polymers, JEE Main, JEE Main 2014 (09 Apr Online)
Structures of some common polymers are given which one is not correctly presented?

Options:
A

B

C

D

The correct answer is:

Solution:
Neoprene is

Topic: Chemistry, Amines, NEET, NEET 2009 (Mains)
Students A,B and C measure the length of a room using 25 m long measuring tape of least count (LC) 0.5 cm, meter-scale of LC0.1 cm and a foot-scale of LC0.05 cm, respectively. If the specified length of the room is 9.5 m, then which of the following students will report the lowest relative error in the measured length?

Options:
A Student A
B Student B
C Student C
D Both, student B and C

The correct answer is:
Student A

Solution:
Student A: Length of scale =25 m
Least count=0.5 cm=0.005 m
Student A can measure the length of 9.5 m by using the scale only once so there will be an error of 0.005 m in 9.5 m
∴ Relative error =0.0059.5=0.0005
Student B: Length of scale:1 m=100 cm
Least count =0.05 cm
To measure 9.5 m, student B has to use this meter scale atleast 10 times
∴Relative error =0.05100×10=0.005 cm
Student C: Length of scale:1 foot =30.48 cm
Least count=0.05 cm

To measure 9.5 m, student C has to use this scale
approximately 31 times
∴Relative error =0.0530.48×31=0.05 cm
∴ Relative error is least for Student A

Topic: Physics, Mathematics in Physics, JEE Main
Study the cycle shown below and select the option which gives correct words for all the four blanks A, B, C and D.

Options:
A (a)
B (b)
C (c)
D (d)

The correct answer is:
(b)

Solution:
A-Denitrification
B-Ammonification
C-Plants
D-Animals

Topic: Biology, Mineral Nutrition, NEET, NEET 2010 (Mains)
Study the diagram given below and mark the INCORRECT statement related to this.

Options:
A It helps in removal of up to 90% particulate matter from exhaust of thermal power plant.
B It is negatively charged wire maintained at several thousand volts.
C It releases electrons.
D It is place where dust particles get collected.

The correct answer is:
It helps in removal of up to 90% particulate matter from exhaust of thermal power plant.

Solution:
The diagram is of Electrostatic precipitator.
An electrostatic precipitator is one of the many ways of removing particulate matter from the exhaust of thermal plants. It is efficient in removing up to 99% of particulate matter. It consists of electrode wires, maintained at several thousand volts, which discharges corona which in turn releases electrons. These electrons get attached dust particles resulting in negatively charged particles. These charged particles get collected in collecting plates.

Topic: Biology, Environmental Issues, NEET