datavorous/entrance-exam-dataset huggingface dataset:
Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron.

Options:

None

Solution:
(a) When a neutral atom gains one electron to form an anion, its radius increases. The reason being that the number of electrons in the anion increases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts greater number of electrons, therefore, force of shells decreases (i.e., effective nuclear charge decreases) and hence the electron cloud expands. In other words, the distance between the centre of the nucleus and the last shell that contains electrons increases thereby increasing the ionic radius. Thus,
(b) When a neutral atom loses one electron to form a cation, its atomic radius decreases. The reason being that the number of electrons in the cation decreases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts lesser number of electrons, therefore, the force of attraction of the nucleus on the electrons of all the shells increases (i.e., effectivenuclear charges increases) and hence the size of cation decreases. Thus

Topic: Chemistry, Classification of Elements and Periodicity in Properties
Describe the two main types of semiconductors and contrast their conduction mechanism.

Options:
A Type I - Majority carriers are electrons, Type II - Majority carriers are holes.
B Type I - Majority carriers are holes, Type II - Majority carriers are electrons.
C Type I - Majority carriers are electrons, Type II - Majority carriers are ions.
D Type I - Majority carriers are ions, Type II - Majority carriers are holes.

The correct answer is:
Type I - Majority carriers are holes, Type II - Majority carriers are electrons.

Solution:
Correct Option is : (B)
Type I - Majority carriers are holes, Type II - Majority carriers are electrons.

Topic: Chemistry, Solid State, NEET
Describe the usefulness of water in the biosphere and biological systems.

Options:
A Water is primarily used as a solvent in biological systems to dissolve various molecules required by plants and animals for various metabolic reactions.
B Water has no significant role in the biosphere or biological systems.
C Water is mainly useful as a structural component in cells and tissues.
D Water is primarily involved in the transport of gases in the atmosphere.

The correct answer is:
Water is primarily used as a solvent in biological systems to dissolve various molecules required by plants and animals for various metabolic reactions.

Solution:
Correct Option is : (A)
Water is primarily used as a solvent in biological systems to dissolve various molecules required by plants and animals for various metabolic reactions.

Topic: Chemistry, Hydrogen, NEET
Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?

Options:

None

Solution:
Transpiration pull theory : The theory was put forward by Dixon and Jolly.
The main features of the theory are :
(i) There is a continuous column of water (present in tracheary element) from root through the stem and into the leaves.
(ii) Water molecule attached to one another, by cohesion force, adhesion force (between their walls and water molecules) and surface tension.
(iii) The water in tracheary element would, come under tension, due to transpiration also called as transpiration pull.
- Factor affecting transpiration : External factor $e . g$. atmospheric pressure, temperature, humidity, $\mathrm{CO}_2$, sunlight, wind velocity, etc. Thickness of cuticle, number, position and closing and opening of stomata, $\mathrm{pH}$ and hormones are internal factors which affects transpiration.
Significance of transpiration :
- Ascent of sap
- Removal of excess of water
- Cooling effect
- Distribution of minerals.
- Transpiration supplies water for photosynthesis.

Topic: Biology, Transport in Plants
$\begin{array}{|l|l|}
\hline \text{Description} & \text{Units} \\
\hline \text{a) Tesla} & \text{1)Electric current} \\
\hline \text{b) Gauss} & \text{2) magnetic induction} \\
\hline \text{c) Weber} & \text{3) magnetic field strength} \\
\hline \text{d) Ampere} & \text{4) magnetic flux} \\
\hline
\end{array}$

Options:
A a) - 3, b) - 2, c) - 4, d) - 1
B a) - 2, b) - 3, c) - 1, d) - 4
C a) - 2, b) - 3, c) - 4, d) - 1
D More than one of the above

The correct answer is:
a) - 3, b) - 2, c) - 4, d) - 1

Solution:
- Tesla (T): Tesla is the unit of measurement for magnetic field strength in the International System of Units (SI).
- It represents the strength of a magnetic field produced by a magnetic source, such as a magnet or an electric current
- Gauss (G): Gauss is another unit of measurement for magnetic induction or magnetic flux density.
- It's named after the German mathematician and physicist Carl Friedrich Gauss.
- Weber (Wb): Weber is the unit of measurement for magnetic flux. It represents the total magnetic field passing through a given area
- . One Weber is equal to one volt-second $(\mathrm{V} \cdot \mathrm{s}$ ) and is equivalent to onetesla square meter $(T \cdot m)$
- Ampere (A): Ampere is the unit of measurement for electric current.
- It represents the flow of electric charge through a conductor in an electrical circuit. One ampere is defined as one coulomb per second $(\mathrm{C} / \mathrm{s})$.

Topic: Physics, Magnetic Properties of Matter, JEE Main
Detergents are prepared by the action of $\mathrm{H}_2 \mathrm{SO}_4$ on which of the following?

Options:
A cholesterol
B Lauryl alcohol
C Cyclohexanol
D p-Nitrophenol

The correct answer is:
Lauryl alcohol

Solution:
Detergents are prepared by the action of $\mathrm{H}_2 \mathrm{SO}_4$, followed by neutralization by starting with lauryl alcohol.
An example is NaLS which is sodium lauryl sulphonate. Lauryl alcohol $\left(\mathrm{C}_{12} \mathrm{H}_{26} \mathrm{O}\right)$ reacts with sulphuric acid followed by aq. $\mathrm{NaOH}$ to form $\mathrm{NaLS}$, $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{10} \mathrm{OSO}_3^{-} \mathrm{Na}^{+}$.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Determinant of second order is made with the element \( 0 \) and \( 1 \). The number of determinants with non negative values is

Options:
A \( 3 \)
B \( 10 \)
C \( 11 \)
D \( 13 \)

The correct answer is:
\( 13 \)

Solution:
Three determinant of second order negative value are possible 0110 0111 1110
No. of possible determinant with element 0,1 are 24= 16
so determinant with non negative value is 13

Topic: Mathematics, Matrices, JEE Main
Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by:

Options:
A Complex formation
B Association
C Partial ionization
D Dissociation

The correct answer is:
Association

Solution:
Acetic acid in benzene undergoes dimerization
2CH3COOH⇌CH3COOH2

due to which Von't hoff factor i<1 and i=Theoretical molar massExperimental molar mass

Topic: Chemistry, Solutions, JEE Main, JEE Main 2015 (11 Apr Online)
Determine how many of the following statements are true at very high pressure for a real gas:

a Compressibility factor is greater than 1.

b Compressibility factor varies linearly with pressure.

c Molar volume occupied by gas is more as compared to ideal gas at similar pressure and temperature.

d Gas is less compressible as compare to ideal gas.

e Compressibility factor is given by Z = 1 +  Pb RT .

Options:

The correct answer is:
5

Solution:
At high pressure Z > 1i.e. repulsive forces are relatively dominant, so gas is less compressible.
at high pressure, a≅0.

Topic: Chemistry, States of Matter, JEE Main
Determine the activity where α-interferon, a biological response modifier, is involved

Options:
A Activation of immune system
B Suppression of immune system
C Making cancerous cells detectable by immune system
D Both A and C

The correct answer is:
Both A and C

Solution:
Interferon alpha is a chemical given to patients with cancer which helps in stimulating the immune system. Cancer cells generally remain undetected, these interferon help detect such cells and assist the immune system in destroying them

Topic: Biology, Human Health and Diseases, NEET
Determine the charge on the capacitor in the following circuit:

Options:
A 2 μC
B 60 μC
C 10 μC
D 200 μC

The correct answer is:
200 μC

Solution:
After a long time capacitor branch will behave like open circuit so circuit will be like,
Equivalent resistance is
R=6+4||2+10 Ω = 9 Ω
and, the current through battery becomes 729=8 A
From the circuit shown, current through 10 Ω resistor becomes, 4(10+2+4)×8=2 A

Potential available for capacitor is = potential drop across 10 Ω
∴ Potential available for capacitor is: 10×2=20 V
∴ Charge on capacitor Q=10 μF20 V=200 μC

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Determine the correct match for the given columns.

Column I	Column II

(a) Hyperparathyroidism	(i) Saltwater balance

(b) Oxytocin	(ii) Milk let down

(c) Aldosterone	(iii) Demineralization

(d) Luteinizing hormone	(iv) Maintains corpus luteum

Options:
A a-ii, b-i, c-iii, d-iv
B a-iii, b-ii, c-i, d-iv
C a-ii, b-iii, c-iv, d-i
D a-i, b-iv, c-iii, d-ii

The correct answer is:
a-iii, b-ii, c-i, d-iv

Solution:
Parathormone secreted by parathyroid helps in removing mineral from bone. Its hyperactivity will make bones weaker due to demineralization. Oxytocin is necessary for the release of the milk from the nipples during lactation process. Aldosterone is a steroidal hormone produced by the adrenal cortex which helps in maintaining salt balance in kidney, sweat glands and even salivary gland. Lutenising hormone secreted by pituitary gland is promotes and maintains corpus luteum which in turn secretes.

Topic: Biology, Chemical Coordination and Integration, NEET
Determine the correct order of hybridization of the central atom in the following species NH3 , [PtCl4]2-, PCl5 and BCl3 .

Options:
A dsp2, dsp3, sp2 and sp3
B sp3, dsp2, dsp3, sp2
C dsp2, sp2, sp3, dsp3
D dsp2, sp3, sp2, dsp3

The correct answer is:
sp3, dsp2, dsp3, sp2

Solution:
Hybridization of N in NH3 is sp3 that of Pt in [PtCl4]2- is dsp2 that P in PCl5 is sp3d and that of B in BCl3 is sp2 .

Topic: Chemistry, Coordination Compounds, JEE Main
Determine the correct statement from the given option with respect to an electrostatic precipitator.

Options:
A the velocity of air is very fast so that a large volume of air can be purified at a time.
B the electrode wires maintained at several thousand volts produces a corona which releases positively charged particles.
C air is passed through a spray of water and lime.
D collecting plates are grounded and collect negatively charged dust particles.

The correct answer is:
collecting plates are grounded and collect negatively charged dust particles.

Solution:
Electrostatic precipitator has electrodes maintained at many thousand volts. They produce a corona that releases negatively charged electrons which combine with dust particles to give them a negative charge. Grounded collecting plates collect these negatively charged dust particles if the air velocity is low allowing the dust to fall.

Topic: Biology, Environmental Issues, NEET
Determine the degree of association (polymerization) for the reaction in aqueous solution

6 HCHO ⇌ C 6 H 1 2 O 6

If observed (mean) molar mass of HCHO and C 6 H 12 O 6 is 150:

Options:
A 0.50
B 0.833
C 0.90
D 0.96

The correct answer is:
0.96

Solution:
6 HCHO ⇌ C 6 H 1 2 O 6 Conc. at equilibrium C 1 - α C α 6
α= M Th − M obs ( n−1 ) M obs =   30−150 [ 1 6 −1 ]150 =   120 5 6 ×150 =0.96

Topic: Chemistry, Chemical Equilibrium, JEE Main
Determine the domain and range of the relation $\mathrm{R}$ defined by $R=\{(x, x+5): x \in\{(0,1,2,3,4,5)\}$.

Options:

None

Solution:
Here,
$R=\{(x, x+5): x \in(0,1,2,3,4,5)\}$
$\therefore \quad$ Domain $=\{0,1,2,3,4,5\}$
[On putting $x=0,1,2,3,4,5$ ]
and Range $=\{5,6,7,8,9,10\}$
[On putting $x=0,1,2,3,4,5$ in $x+5]$.

Topic: Mathematics, Relations and Functions (Class 11)
Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure:

Options:
A 3 qlj^-i^2
B -3 ql j^
C 2 ql j^
D qli^+j^2

The correct answer is:
-3 ql j^

Solution:
P→1=ql
P→2=ql
Angle between them 60o
∴ Resultant dipole moment P=ql2+ql2+2ql2cos⁡60o
=3 ql
As P→1=P→2 direction of resultant is along negative y axis thus option is -3 ql j^

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Determine the element from the following description.

(I) Activate enzyme of photosynthesis and respiration

(II) Involved in the synthesis of DNA and RNA

(III) Help to maintain structure of ribosome

Options:
A Ca
B Mg
C S
D Fe

The correct answer is:
Mg

Solution:
Magnesium is one of the macronutrient significant for plant growth and development. Magnesium activates the enzymes RubisCO and PEP carboxylase which are important in photosynthesis and respiration process. It is a constituent of the green pigment chlorophyll and actively involved in making DNA and RNA. It is also required to maintain the integrity of ribosome.

Topic: Biology, Mineral Nutrition, NEET
Determine the lengths of an iron rod and a copper ruler at 0 Â°C if the difference in their lengths at 50 Â°C and 450 Â°C is the same and is equal to 2 cm. The coefficient of linear expansion of iron = 12Ã10-6 K-1 and that of copper = 17Ã10-6 K-1.

Options:
A 20.06 m, 20.08 m

B 21.0 m, 7.00 m

C 60.30 cm, 10.10 m

D 11.09 m, 15.10 m

The correct answer is:
20.06 m, 20.08 m

Solution:
Let x be the length of the iron rod at 0 °C, y that of the copper rod at 0 °C, and I the difference in lengths at t1 and t2 oC.
Then l=x1+α1t1-y1+α2t1 ...(i)
and ±l=x1+α1t2-y1+α2t2 ...(ii)
Taking + 1
l=x1+α1t2-y1+α2t2 ...(iii)
from Eqs. (i) and (iii), we get
x α 1 = y α 2 ...(iv)
from Eqs. (i) and (iv), we get
y = l⁡ α 1 α 2 - α 1 = 2 × 1 2 × 1 0 - 6 1 7 - 1 2 × 1 0 - 6 = 4.8 cm
and x = l⁡ α 2 α 2 - α 1
⇒ x = 2 × 1 7 × 1 0 - 6 1 7 - 1 2 × 1 0 - 6 cm = 6.8 cm
Taking - 1
y=2l+lα1t1+t2t2-t1α2-α1;x=2l+lα2t1+t2t2-t1α2-α1
∴ y=2×2+2×12×10-6450+50450-5017-12×10-6cm=2006cm=20.06 m
x=2×2+2×12×10-6450+50450-5017-12×10-6cm=2008.5cm=20.08 m

Topic: Physics, Thermal Properties of Matter, JEE Main
Determine the number of planes of symmetry of the given compound.

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
One plane along Cl.
Therefore the number of planes of symmetry of the given compound is 1.

Topic: Chemistry, General Organic Chemistry, JEE Main
Determine the number of stereoisomers in the given compound

Options:
A 2
B 4
C 8
D 16

The correct answer is:
4

Solution:
When the cyclic ring is containing two or more chiral centre, GI is included in combination of chiral unit.
Hence, n =2 ( nâ= no. of stereo centres)
Total stereoisomer 2 n â=â 4

Topic: Chemistry, General Organic Chemistry, JEE Main
Determine the order of basic strength of the given molecules

Options:
A i > iii > ii
B ii > i > iii
C iii > ii > i
D i > ii > iii

The correct answer is:
iii > ii > i

Solution:
Basic strength depends on the electron donating capacity of the atom. Hence as +I increases, basicity increases. iii has most +I, so iii > ii > i

Topic: Chemistry, General Organic Chemistry, JEE Main
Determine the order of stability of the following resonating structure.

Options:
A i > ii > iii
B iii > ii > i
C i > iii > ii
D ii > i > iii

The correct answer is:
i > ii > iii

Solution:
Charge separation reduces the stability of the resonating structures. If charge separation exists, the structure with opposite charges closer is more stable. So, iis most stable followed by iiand iii

Topic: Chemistry, General Organic Chemistry, JEE Main
Determine the point of application of force, when forces are acting on the rod as shown in figure.

Options:
A $4.625 \mathrm{~cm}$
B $\ 7.658 \mathrm{~cm}$
C $5.255 \mathrm{~cm}$
D $4.375 \mathrm{~cm}$

The correct answer is:
$4.375 \mathrm{~cm}$

Solution:
Since the body is in equillibrium so we conclude Vector $F_{\text {net }}=0$ and torque about any point is zero i.e., $\vec{\tau}_{\text {net }}=0$
Let us assume that we apply $F$ force downward at $A$ angle $\theta$ from the horizontal, at $x$ distance from $B$
$\begin{array}{ll}
\text { From } & \vec{F}_{\text {net }}=0 \\
\Rightarrow & F_{\text {net }} x=0 \\
& F_2=8 \mathrm{~N} \\
\text { From } & F_{\text {net } y}=0 \Rightarrow 5+6=F_1+3 \\
\Rightarrow & F_1=8 \mathrm{~N}
\end{array}$
If body is in equillibrium then torque about point $B$ is zero.
$\begin{array}{ll}
\Rightarrow \quad & 3 \times 5+F_1 \cdot x-5 \times 10=0 \\
\Rightarrow \quad & 15+8 x-50=0 \\
& x=\frac{35}{9} \Rightarrow \quad x=4.375 \mathrm{~cm}
\end{array}$

Topic: Physics, Rotational Motion, JEE Main
Determine the point of application of force, when forces of $20 \mathrm{~N}$ and $30 \mathrm{~N}$ are acting on a rod as shown in figure

Options:
A $60 \mathrm{~cm}$
B $70 \mathrm{~cm}$
C $65 \mathrm{~cm}$
D $80 \mathrm{~cm}$

The correct answer is:
$70 \mathrm{~cm}$

Solution:
Net force acting on the rod, $\mathrm{F}_{\text {rel }}=10 \mathrm{~N}$
Net torque acting on the rod about point $\mathrm{C}$ is
$\tau_c=(20 \times 0)+(30 \times 20)=600 \text { clockwise }$
Let the point of application be at a distance $x$ from point $C$
$600=10 \mathrm{x} \Rightarrow \mathrm{x}=60 C M$
Therefore, $70 \mathrm{CM}$ from $\mathrm{A}$ is the point of application of force.

Topic: Physics, Rotational Motion, JEE Main
Determine the product formed when phosphine gas is mixed with chlorine gas

Options:
A The mixture only cools down
B PH 3 ⋅ Cl 2 adduct is formed with warming up
C PCl5 and HCl are formed and the mixture cools down
D PCl3 and HCl are formed and the mixture warms up

The correct answer is:
PCl5 and HCl are formed and the mixture cools down

Solution:
The reaction of phosphine gas with chlorine gas is given as,
PH3+4Cl2→PCl5+3HCl
Therefore, the products formed when phosphine gas is mixed with chlorine gas are PCl 5 and HCl.
The mixture will cool down later.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Determine the value of the equilibrium constant for the reaction
2Br-(aq)+I2(s)⇌Br2(l)+2I-(aq)
Ecello=-0.54V

Options:
A 5.01×10-19
B 18.30
C 1.7×1054
D 1.9×1018

The correct answer is:
5.01×10-19

Solution:
The relation between equilibrium constant (K) and Ecello is given as
log⁡K=Ecello.n0.059V
here, n = 2
=-0.54V×20.059V
=-1.080.059=-18.3
On taking antilog
K=5.01×10-19

Topic: Chemistry, Electrochemistry, JEE Main
Determine the values of \( \mathrm{A}, \mathrm{B}, \mathrm{C} \) and \( \mathrm{D} \) shown in the figure.

Options:
A \( \mathrm{SnO}_{3} \), \( { }_{2} \mathrm{PbO}_{3} \), \( { }_{2} \mathrm{ZnO}_{3} \), \( \mathrm{AlO}_{2} \)
B \( 2 \mathrm{SnO}_{2} { }_{2} \mathrm{PbO}_{2} 2 \mathrm{ZnO}_{2} \mathrm{AlO}_{2} \)
C \( { }_{13} \mathrm{SnO}_{3} \)
D \( 3 \mathrm{SnO}_{3} { }_{2} \mathrm{PbO}_{3} 2 \mathrm{ZnO} \mathrm{AlO}_{2} \)

The correct answer is:
\( 2 \mathrm{SnO}_{2} { }_{2} \mathrm{PbO}_{2} 2 \mathrm{ZnO}_{2} \mathrm{AlO}_{2} \)

Solution:
All the given metals are amphoteric in nature and thus, can react with aqueous alkali to give hydrogen gas and double salts. The reactions are:

Sn+2NaOH→H2+Na2SnO2

Pb+2NaOH→H2+Na2PbO2

Zn+2NaOH→H2+Na2ZnO2

Al+2NaOH+2H2O→3H2+NaAlO2

Topic: Chemistry, Hydrogen, JEE Main
Determine the work done by an ideal gas undergoing a cyclic process from 1 â 4 â 3 â 2 â 1 . Given P1=105 Pa, P0 = 3Ã105 Pa, P3=4Ã105 Pa and V2-V1=10 L.

Options:
A 740 J

B 750 J
C 730 J
D 745 J

The correct answer is:
750 J

Solution:
From figure
V 4 - V 3 V 2 - V 1 = P 3 - P 0 P 0 - P 1 ⇒ V 4 - V 3 1 0 = 4 × 1 0 5 - 3 × 1 0 5 3 × 1 0 5 - 1 0 5
V4- V3 =5 L
Now, work done
W = 1 2 × 1 × 2 × 1 0 5 - 1 2 × 5 × 1 × 1 0 5 × 1 0 - 3 = 7 5 0 J

Topic: Physics, Thermodynamics, JEE Main
Determine the work done by an ideal gas undergoing a cyclic process from \( 1 \rightarrow 4 \rightarrow 3 \rightarrow 2 \rightarrow 1 \). Given
\( P_{1}=10^{5} \mathrm{~Pa}, P_{0}=3 \times 10^{5} \mathrm{~Pa}, P_{3}=4 \times 10^{5} \mathrm{~Pa} \) and \( V_{2}-V_{1}=10 \mathrm{~L} \)

Options:
A \( 740 \mathrm{~J} \)
B \( 750 \mathrm{~J} \)
C \( 730 \mathrm{~J} \)
D \( 745 \mathrm{~J} \)

The correct answer is:
\( 750 \mathrm{~J} \)

Solution:
From figure
V4-V3V2-V1=P3-P0P0-P1⇒V4-V310=4×105-3×1053×105-105

⇒V4- V3 =5 L
Now, work done,
W=12×10×2×105-12×5×1×105×10-3=750J

Topic: Physics, Thermodynamics, JEE Main
Detritivores break down detritus into smaller particles. This process is called

Options:
A fragmentation
B leaching
C humification
D calcification

The correct answer is:
fragmentation

Solution:
Dead plant remains such as leaves flower, bark and dead remains of animals including fecal matter constitute detritus
which is the raw material for decomposition. Earthworm breakdown detritus into smaller fragments to increase surface area for further enzymatic activity.

Topic: Biology, Ecosystem, NEET
Detritivores breakdown detritus into smaller particles. This process is called:

Options:
A Fragmentation
B Humification
C Decomposition
D Catabolism

The correct answer is:
Fragmentation

Solution:
Decomposers break down complex organic matter into inorganic substances like carbon dioxide, water and nutrients, and the process is called decomposition. Decomposition is one of the important processes in the ecosystem which helps in nutrient cycling. Dead plant remains such as leaves, bark, flowers and dead remains of animals, including faecal matter of animals constitute detritus. This is the raw material for decomposition.

The important steps in the process of decomposition are fragmentation, leaching, catabolism, humification, and mineralization.

The detritivores, like earthworms, break down detritus into smaller particles. This process is called fragmentation.

Topic: Biology, Ecosystem, NEET, NEET 2022 (Phase 1)
Deuterium resembles hydrogen in properties but:

Options:
A reacts vigorously than hydrogen
B reacts just as hydrogen
C emits β+ particles
D reacts slower than hydrogen

The correct answer is:
reacts slower than hydrogen

Solution:
Reactivity of Deuterium is less than hydrogen due to high bond dissociation energy of D2

Bond dissociation:H2D2435.88443.35

Energy (kJ/mole)

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Deuteron and \(\alpha\)-particle are put \(1 \mathrm{~A}^{\circ}\) apart in air. Magnitude of intensity of electric field due to deuteron of \(\alpha\)-particle is (N/C)

Options:
A Zero
B $2.88 \times 10^{11}$ newton/coulomb
C $1.44 \times 10^{11}$ newton / coulomb
D $5.76 \times 10^{11}$ newton $/$ coulomb

The correct answer is:
$1.44 \times 10^{11}$ newton / coulomb

Solution:
Due to deutron, intensity of electric field at 1 A distance,
$E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{e}{r^2}=$ $=\frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{10^{-20}}=1.44 \times 10^{11} \mathrm{~N} / \mathrm{C}$.

Topic: Physics, Electrostatics, JEE Main
Deviation of $5^{\circ}$ is observed from a prism whose angle is small and whose refractive index is 1.5 . The angle of prism is

Options:
A $7.5^{\circ}$
B $10^{\circ}$
C $5^{\circ}$
D $3.3^{\circ}$

The correct answer is:
$10^{\circ}$

Solution:
$5=(\mu-1) A=(1.5-1) A \Rightarrow A=10^{\circ}$

Topic: Physics, Ray Optics, JEE Main
D-Glucose and D-mannose are:

Options:
A enantiomers
B functional isomers
C epimers
D metamers

The correct answer is:
epimers

Solution:
D -mannose, is a C -2 epimer of D -glucose

Topic: Chemistry, Biomolecules, JEE Main