datavorous/entrance-exam-dataset huggingface dataset:
Diabetes is one of the most common lifestyle disorders in India, which is successfully treated with insulin therapy which reduces the blood glucose level. Which of the following mechanisms correctly shows the action of insulin in lowering blood glucose level?

Options:
A It stimulates the breakdown of glycogen.
B It stimulates the process of gluconeogenesis.
C It acts on adipocytes and hepatocytes and increase cellular glucose uptake and utilization.
D It reduces glucose uptake and utilization.

The correct answer is:
It acts on adipocytes and hepatocytes and increase cellular glucose uptake and utilization.

Solution:
Insulin and Glucagon are two main hormones that bring about glucose metabolism in humans. Glucagon is secreted by alpha cells of islets of langerhans that are small clusters of endocrine cells in pancreas and Insulin is released by beta cells of pancreas. Glucagon brings about breakdown of glycogen into glucose and thus increases the level of glucose in blood. Insulin further carries this glucose to different cells and thus reduces blood glucose level. After cellular uptake too if the glucose levels are high in the blood insulin causes mobilization of glucose from blood into hepatocytes and adipocytes where it gets converted into glycogen which is the reserve form.

Topic: Biology, Chemical Coordination and Integration, NEET
Diagonal relationship is shown by

Options:
A Elements of first period
B Elements of second period
C Elements of third period
D (b) and (c) both

The correct answer is:
(b) and (c) both

Solution:
Elements of second and third period
Diagonal relationship

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Diagram shows the variation in the internal energy U with the volume V of 2.0 moles of an ideal gas in cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K, respectively. Calculate the heat absorbed by the gas during the process.

Options:
A 400 R ln 2
B 500 R ln 2
C 700 R ln 2
D 800 R ln 2

The correct answer is:
400 R ln 2

Solution:
In the process a to b and c to d
as Δ U = 0 , therefore Δ T = 0 or T = constant
W = ∫ V i V f⁡ PdV
We have,
PV = nRT ⇒ P = nRT V
∴ W = ∫ V i V f⁡ nRT RV V
= nRT ln V V i V f⁡ = nRT ln V f⁡ V i
W ab = nRT b ln 2 V 0 V 0 = 2 R × 5 0 0 ln 2 = 1 0 0 0 R ln 2
and W cd = nRT c ln V 0 2 V 0 = 2 R × 3 0 0 ln 1 2 = - 6 0 0 R ln 2
There is no volume change from b to c and from d to a, so
W bc = W da = 0
The work done in the complete cycle
W = W ab + W bc + W cd + W da
=1000R ln 2+0-600R ln 2+0
= 4 0 0 R ln 2

Topic: Physics, Thermodynamics, JEE Main
Diagram shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of gas at c and d are 300 K and 500 K, respectively. Calculate the heat absorbed by the gas during the process.

Options:
A 440 R ln2
B 400 R ln2
C 430 R ln2
D 414 R ln2

The correct answer is:
400 R ln2

Solution:
Change in internal energy for cyclic process Δ U = 0
For process a → b , P - constant
W a → b = P Δ V = n R Δ T = - 4 0 0 R
For process b → c , T - constant
Wb → c=nRTlnPiPf=-2R300ln 2
For process c → d , P - constant
Wc → d=PΔV=nRΔT=+400R
For process d → a , T - constant
Wd → a=nRTlnPiPf=+2R500ln 2
Net work,
Δ W = W a → b + W b + c + W c → d + W d → a
Δ W = 4 0 0 R ln 2
dQ = dU + dW , first law of thermodynamics
dQ = 4 0 0 R ln 2

Topic: Physics, Thermodynamics, JEE Main
Diamagnetic Lanthanoid ions are:

Options:
A Nd3+ and Eu3+
B La3+ and Ce4+
C Nd3+ and Ce4+
D Lu3+ and Eu3+

The correct answer is:
La3+ and Ce4+

Solution:
Lanthanide ion with no unpaired electron is diamagnetic in nature.

Lanthanoid is also called lanthanide, which can be defined as any of the series of 15 consecutive chemical elements in the periodic table from lanthanum to lutetium (atomic numbers 57-71). With the elements scandium and yttrium, they make up the rare-earth metals. Lanthanides are also radioactive in nature.

Ce:[Xe]4f15 d16 s2;Ce4+ diamagnetic

La:[Xe]4f05 d16 s2;La3+ diamagnetic

Hence, the answer is option B.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Diameter of a plano - convex lens is 6 cm and thickness at the centre is 3 mm. If the speed of light in the material of lens is 2×108 m s-1, the focal length of the lens is:

Options:
A 30 cm
B 10 cm
C 15 cm
D 20 cm

The correct answer is:
30 cm

Solution:
AB=6 cm
r=3 cm
O A 2 = O P 2 + P A 2
R 2 = R - t 2 + r 2
R 2 = R 2 + t 2 - 2 R t + r 2
2 R t = t 2 + r 2
R = t 2 + r 2 2 t = 0 · 3 2 + 3 2 2 × 0 · 3
R≈r22t=322×0·3=90·6=906=15 cm
Use 1 f = μ - 1 1 R 1 - 1 R 2
1 f = 1 . 5 - 1 1 1 5 = 1 3 0
f=30 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (07 Apr)
Diameter of a steel ball is measured using a Vernier calipers which has divisions of 0.1 cm on its main scale (MS) and 10 divisions of its Vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as:

S.No.	MS (cm)	VS divisions

1.	0.5	8

2.	0.5	4

3.	0.5	6

If the zero error is - 0.03 cm, then mean corrected diameter is:

Options:
A 0.53 cm
B 0.56 cm
C 0.59 cm
D 0.52 cm

The correct answer is:
0.59 cm

Solution:
L.C. of Vernier calipers
= 1 main scale divisionTotal divisons vernier scale
=0.110=0.01 cm
Reading of Vernier calipers
=M.S.R. +L.C×vs divisions.
d1=0.5+0.01×8
d2=0.5+0.01×4
d3=0.5+0.01×6
∴ Measured diameter are respectively
0.58 cm 0.54 cm, 0.56 cm
∴ average diameter = 0.58 + 0.54 + 0.563
=1.683=0.56
Negative zero error is added in the average diameter to get the corrected diameter
∴ corrected diameter=0.56-(-0.03)
=0.56+0.03=0.59 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2015 (10 Apr Online)
Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm coming from a distant object, the limit of resolution of the telescope is close to:

Options:
A 1.5×10-7 rad
B 3.0×10-7 rad
C 2.0×10-7 rad
D 4.5×10-7 rad

The correct answer is:
3.0×10-7 rad

Solution:
Resolving limit of telescope =1.22λD
Where, D = Diameter of lens
λ = Wavelength of light
Resolving limit of telescope =1.22×600×10-9250×10-2=3×10-7 rad

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Diameter or aperture of a plano-convex lens is 6 cm and its thickness at the centre is 3 mm. The image of an object formed is real and twice the size of the object. If the speed of light in the material of the lens is 2Ã108 m s-1. The distance where the object is placed from the plano-convex lens is â¦â¦â¦Ã15 cm.

Options:

The correct answer is:
3

Solution:
R-y2+r2=R2
R=r22y=6222×0.3=15 cm
1f=μ-1R
μ=CV=3.0×1082.0×108=1.5
∴f=30 cm converging
Now

f=30 cm
1f=1v-1u
130=1v-1u
m=vu=-2
v=-2u
130=-12u-1u
130=-32u
u=-45 cm

Topic: Physics, Ray Optics, JEE Main
Diamond is hard because

Options:
A all the four valence electrons are bonded to each carbon atoms by covalent bonds
B it is a giant molecule
C it is made up of carbon atoms
D it cannot be burnt

The correct answer is:
all the four valence electrons are bonded to each carbon atoms by covalent bonds

Solution:
In the structure of diamond each carbon atom in $\mathrm{sp}^{3}$ hybridised $\&$ is covalently bonded with four other carbon atom held at the corners of a regular tetrahedron by covalent bonds. This results in a very big three dimensional polymeric structure in which $\mathrm{C}$ - C distance is $154 \mathrm{pm}$ and bond angle is $109.5^{\circ}$. Owing to very strong covalent bonds by which atoms are held to gether diamond is the hardest substance known.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Diamonds are formed from graphite under high pressure in coal mines. Calculate the equilibrium pressure (in atm) at which graphite is converted to diamonds at 25oC (assumed constant) given densities of Ï graphite = 2 g / cc & Ï diamond = 3 g / cc Î G f o for diamonds is 3 kJ m- 1 from graphite

Options:

The correct answer is:
15001

Solution:
Differential equation of free energy dG = VdP - SdT at constant temp. with change in allotropic modification there is a change in molar volume ∴ d Δ G = Δ V dP .
Δ V = M Diamond ρ Diamond - M graphite ρ graphite = 1 2 3 - 1 2 2
Δ V = - 2 cm 3 / mole = - 2 × 1 0 - 6 m 3 mole -1
∫ d Δ G Δ G f o at 2 9 8 K, 1 atm Δ G f o at 2 9 8 K, p atm = 0 = - 2 × 1 0 - 6 m 3 mole - 1 ∫ dP P = 1 atm P = p atm
Because at equilibrium when diamonds are formed both the phases are in equilibrium Δ G f Diamonds o at 2 9 8 K , P atm = 0
∫ 0 d Δ G 3 0 0 0 J mole -1 = - 2 × 1 0 - 6 m 3 mole - 1 × 1 0 5 Nm - 2 ∫ dP P = 1 atm P = p atm
( 1 atm = 105 Nm- 2 )
0 - 3000 Joules mole- 1 = - 2 x 10- 1 J mole- 1 (P - 1)
3 0 0 0 0 . 2 = 1 5 0 0 0 = P - 1
∴ P = 15001 atm.

Topic: Chemistry, Thermodynamics (C), JEE Main
Diaphragms, cervical caps and vaults prevent conception by

Options:
A increasing phagocytosis of sperms within uterus
B suppressing sperm motility
C inhibiting ovulation and implantation
D blocking the entry of sperms through the cervix

The correct answer is:
blocking the entry of sperms through the cervix

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, diaphragms, cervical caps and vaults etc. Barrier methods are popular as they are easy to use without the help of a practioner. These are effective methods, and their use does not interfere with coitus and also prevents sexually transmitted diseases. These are barrier methods because these block the entry of sperms through the cervix. The remaining three options are mechanisms of IUD's.

Topic: Biology, Reproductive Health, NEET
Diatoms are also called as 'pearls of ocean', why? What is diatomaceous earth?

Options:

None

Solution:
Diatoms and desmids are included under chrysophytes, kingdom-Protista. These are the main producers in the ocean. They prepare food for themselves and for the other life forms in the ocean as were a siliceous shell known as frustule cores the body of diatoms, this is the reason they are also called as 'pearls of ocean.
'Diatomaceous earth' is the accumulation of large deposits of diatoms that forms a siliceous covering extending for several 100 metres formed in billions of years. The material obtained from these deposits is used in polishing and filtration of oils and syrups.

Topic: Biology, Biological Classification
Diazonium salts
the reaction is known as

Options:
A Chlorination
B Sandmeyer’s reaction
C Perkin reaction
D Substitution reaction

The correct answer is:
Sandmeyer’s reaction

Solution:
The correct option is B Sandmeyer's reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Diborane B2H6 reacts independently with O2 and H2O to produce, respectively:

Options:
A HBO2 and H3BO3
B H3BO3 and B2O3
C B2O3 and [BH4]-
D B2O3 and H3BO3

The correct answer is:
B2O3 and H3BO3

Solution:
Diborane reacts with oxygen exothermically to form boric oxide (B2O3)
B2H6+3O2→B2O3+3H2O
Diborane gets readily hydrolysed to give boric acid
B2H6+3H2O→2BOH3+3H2(g)

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2019 (08 Apr Shift 1)
Diborane reacts with ammonia to initially forms X which on further heating gives borazine X is

Options:
A BH3.NH3
B B2H6.NH3
C B2H6.2NH3
D NH3BH3NH3

The correct answer is:
B2H6.2NH3

Solution:
B2H6+2NH3âB2H6.2NH3
BH 2 NH 3 2 + BH 4 − → Δ B 3 N 3 H 6 (Borazine)  +  H 2

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Diborane reacts with ammonia under different conditions to give a variety of products. Which one among the following is not formed in these reactions?

Options:
A $\mathrm{B}_2 \mathrm{H}_6 \cdot 2 \mathrm{NH}_3$
B $\mathrm{B}_{12} \mathrm{H}_{12}$
C $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$
D $(\mathrm{BN})_n$

The correct answer is:
$\mathrm{B}_{12} \mathrm{H}_{12}$

Solution:
Diborane reacts with ammonia and gives different products under different reaction conditions as
$\mathrm{B}_2 \mathrm{H}_6+\underset{\text { (excess) }}{\mathrm{NH}_3} \underset{\text { temperature }}{\stackrel{\text { at low }}{\longrightarrow}}$ $\underset{\text {addition product} }{\mathrm{B}_2 \mathrm{H}_6 \cdot 2 \mathrm{NH}_3}$
$\mathrm{B}_2 \mathrm{H}_6+\underset{\text { (excess) }}{\mathrm{NH}_3} \stackrel{\text { high temperature }}{\longrightarrow}$ $\underset{\text {boron nitride (resembles with graphite) }}{(\mathrm{BN})_n}$
$\underset{1}{\mathrm{~B}_2 \mathrm{H}_6}\underset{:}{+}\underset{2}{\mathrm{NH}_3 }\stackrel{\text { high temperature }}{\longrightarrow}$ $\underset{\text {borazine (inorganic benzene)}}{\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6}$
Thus, $\mathrm{B}_{12} \mathrm{H}_{12}$ is not the product of reaction between diborane and ammonia.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Dice A has 4 red and 2 white faces whereas dice B has 3 red and 3 white faces. A coin is tossed once, if it falls head then the game continues by throwing the dice A and if it falls tail then the dice B is to be used. If red turns up at first 3 throws, then the probability that dice A is being used is

Options:
A 737
B 6491
C 941
D 2735

The correct answer is:
6491

Solution:
R→ Red turns up at first three throws
A→ Dice A is selected. PA=12
B→ Dice B is selected. PB=12
Using Bayes theorem, we get,
PAR=PA∩RPR=PA×PRAPA×PRA+PBPRB
=12×23312×233+12123=6491

Topic: Mathematics, Probability, JEE Main
Dichromate ion is treated with base, the oxidation number of Cr in the product formed is

Options:

The correct answer is:
6

Solution:
Cr2O72-+OH-⟶CrO42-

Oxidation state of Cr in CrO42- is +6

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Dichromate ion reduces to __________ ion in an acidified solution of potassium dichromate (VI).

Options:
A Chromate (V) ions
B Chromium (III) ions
C Chromium (II) ions
D Chromium (IV) ions

The correct answer is:
Chromium (III) ions

Solution:
Dichromate(VI) ions (for example, in potassium dichromate(VI) solution) can be reduced to chromium(III) ions and then to chromium(II) ions using zinc and either dilute sulfuric acid or hydrochloric acid. Hydrogen is produced from a side reaction between the zinc and acid.

Topic: Chemistry, Redox Reactions, JEE Main
Diels-Alder reaction will not take place with which of the following reactants?

Options:
A

B

C

D

The correct answer is:

Solution:
In Diels-Alder reaction, conjugated dienes react with enes or olefinic compounds and give cycloalkenes.
conjugated alkene, it will not take part in Diels-Alder's reaction.

Topic: Chemistry, Hydrocarbons, JEE Main
Diethyl ether and methyl n-propyl ether are

Options:
A Position isomers
B Functional isomers
C Metamers
D Chain isomers

The correct answer is:
Metamers

Solution:
$\mathrm{C}_2 \mathrm{H}_5-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5$ and $\mathrm{CH}_3-\mathrm{O}-\mathrm{C}_3 \mathrm{H}_7$ are metamers.

Topic: Chemistry, General Organic Chemistry, JEE Main
Diethyl ether can be decomposed by heating with

Options:
A $\mathrm{HI}$
B $\mathrm{NaOH}$
C Water
D $\mathrm{KMnO}_4$

The correct answer is:
$\mathrm{HI}$

Solution:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{HI} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Difference between colloids and crystalloids is of

Options:
A Particle composition
B Particle size
C Concentration
D Ionic character

The correct answer is:
Particle size

Solution:
The main difference between crystalloids and colloids is the size of particles, therefore the answer is Option (2) Particle size.

Topic: Chemistry, Surface Chemistry, JEE Main
Difference between the corresponding roots of $x^2+a x+b=0$ and $x^2+b x+a=0$ is same and $a \neq b$, then

Options:
A
a + b + 4 = 0

B
a + b - 4 = 0

C
a - b - 4 = 0

D
a - b + 4 = 0

The correct answer is:
a + b + 4 = 0

Solution:

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2002
Different combination of 3 resistors of equal resistance R are shown in the figures. The increasing order for power dissipation is:

(A)

(B)

(C)

(D)

Options:
A PB<PC<PD<PA
B PC<PD<PA<PB
C PC<PB<PA<PD
D PA<PB<PC<PD

The correct answer is:
PC<PB<PA<PD

Solution:
The resistances can be calculated as

RA=R+R2=3R2

For second circuit

1RB=12R+1R=32R⇒RB=2R3

For third circuit

RC=R3

For fourth circuit

RD=3R

The formula for power is given by P=I2R.

Clearly,

PD>PA>PB>PC

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Different species are represented by A, B, C, D, G, P, Q, R and S . Which of the following represents a community?

Options:
A (i)
B (ii)
C (iii)
D both (ii) and (iii)

The correct answer is:
(ii)

Solution:
A community of organisms interacting with abiotic component in a particular region/area will be called as ecosystem. Various groups of interacting organisms in a particular region are included in community. There are various types of ecosystem like terrestrial, aquatic.
Various groups of interacting organisms in a particular region are included in community. A community is said to be stable when its populations size of all species remains same over a period of time.
Group of organisms belonging to same species in a particular region or habitat is called as population. Carrying capacity is the maximum population that can be sustained is a particular area.

Topic: Biology, Organisms and Populations, NEET
Differential equation of $y=\sec \left(\tan ^{-1} x\right)$ is

Options:
A $\left(1+x^2\right) \frac{d y}{d x}=y+x$
B $\left(1+x^2\right) \frac{d y}{d x}=y-x$
C $\left(1+x^2\right) \frac{d y}{d x}=x y$
D $\left(1+x^2\right) \frac{d y}{d x}=\frac{x}{y}$

The correct answer is:
$\left(1+x^2\right) \frac{d y}{d x}=x y$

Solution:
$\begin{aligned} & y=\sec \left(\tan ^{-1} x\right) \\ & \frac{d y}{d x}=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \cdot \frac{1}{1+x^2}=\frac{x y}{1+x^2} \\ & \Rightarrow\left(1+x^2\right) \frac{d y}{d x}=x y\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
Digit at the unit place of the sum of 1!2+2!2+3!2………+2008!2 is

Options:
A 5
B 7
C 9
D 6

The correct answer is:
7

Solution:
Given the sum S= ( 1! ) 2 + ( 2! ) 2 + ( 3! ) 2 +.......+ ( 2008! ) 2
Since the digit at units place is zero in n! for nâ¥5
Hence, S= ( 1 ) 2 + ( 2 ) 2 + ( 6 ) 2 + ( 24 ) 2 + numbers having zero at units place
=617+ all other numbers having zero at units place

Topic: Mathematics, Permutation Combination, JEE Main
Digital signals
(i) avoid supplying a continuous range of numbers;
(ii) convey values in the form of discrete steps,
(iii) can use the binary system;
(iv) can work with binary and decimal systems.
Which of the claims given above are true?

Options:
A Only (i) and (ii)
B Only (ii) and (iii)
C Only (i), (ii), and (iii)
D (i) (ii), (iii), and (iv)

The correct answer is:
Only (i), (ii), and (iii)

Solution:
A digital signal sends messages using the binary ( 0 and 1 ) system. The decimal system cannot be trusted with this type of system (which correlates to analogue signals). This is because digital signals are made up of discrete values.

Topic: Physics, Communication System, JEE Main
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
\(\mathrm{CH}_4(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_2(\mathrm{~g})\)
(a) Write an expression for \(K_p\) for the above reaction.
(b) How will the values of \(\mathbf{K}_p\) and composition of equilibrium mixture be affected by
(i) increasing the pressure ?
(ii) increasing the temperature ?
(iii) using a catalyst ?

Options:

None

Solution:
(a) \(\mathrm{K}_{\mathrm{p}}=\frac{p_{\mathrm{CO}} \times\left(p_{\mathrm{H}_2}\right)^3}{p_{\mathrm{CH}_4} \times p_{\mathrm{H}_2 \mathrm{O}}}\)
(b) (i) By Le Chatelier's principle, equilibrium will shift in the backward direction, and value of \(K_p\) decreases.
(ii) By Le Chatelier's principle, equilibrium will shift in the forward direction, and value of \(K_p\) increases.
(iii) Equilibrium composition will not be disturbed but equilibrium will be attained quickly and value of \(K_p\) remains unaffected.

Topic: Chemistry, Equilibrium
Dihydrogen of high purity >99.95 % is obtained through :

Options:
A The reaction of Zn with dilute HCl.
B The electrolysis of acidified water using Pt electrodes.
C The electrolysis of brine solution.
D The electrolysis of warm BaOH2 solution using Ni electrrodes.

The correct answer is:
The electrolysis of warm BaOH2 solution using Ni electrrodes.

Solution:
Dihydrogen of high degree of purity >99.95% is obtained by the electrolysis of warm aqueous barium hydroxide solution between nickel electrodes.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Dihydrogen reacts with CuO to give

Options:
A CuOH2
B Cus
C Cu2O
D CuH2

The correct answer is:
Cus

Solution:
CuO→H2Cu+H2O

Copper oxide undergoes reduction to form elemental copper and water. This is a redox reaction. Since oxygen is removed from the CuO it is undergoing reduction. And hydrogen is getting oxidised.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (28 Jun Shift 1)
$\xrightarrow{\text { dil } \mathrm{HNO}_3} \mathrm{C}_6 \mathrm{H}_5-\mathrm{CHO}$
In the above reaction product is

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5$
B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OC}_6 \mathrm{H}_5$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{I}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5$

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Dimension of $\frac{1}{\mu_0 \varepsilon_0}$, where symbols have their usual meaning, are

Options:
A
$\left[\mathrm{L}^{-1} \mathrm{~T}\right]$

B
$\left[\mathrm{L}^{-2} \mathrm{~T}^2\right]$

C
$\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$

D
$\left[\mathrm{LT}^{-1}\right]$

The correct answer is:
$\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$

Solution:
$\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=\mathrm{C} \Rightarrow \frac{1}{\mu_0 \varepsilon_0}=\mathrm{C}^2 \Rightarrow\left[\frac{1}{\mu_0 \varepsilon_0}\right]=[\mathrm{C}]^2$
$[\mathrm{C}]=\mathrm{LT}^{-1} \quad$ or $\quad[\mathrm{C}]^2=\mathrm{L}^2 \mathrm{~T}^{-2}$

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2003
Dimension of 1μ0ε0 should be equal to

Options:
A L T-1
B T2 L-2
C L2 T-2
D T L-1

The correct answer is:
L2 T-2

Solution:
The speed of light can be written as

c=1μ0ε0 ...1

Rearrange equation (1) to obtain the required quantity.

1μ0ε0=c2 ...2

Use the method of dimensional analysis to obtain the dimension of the required quantity.

1μ0ε0=LT-12=L2 T-2

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Dimension of which base quantity corresponds to that Ghc3=?

Options:
A Time
B Length
C Mass
D Temperalure

The correct answer is:
Length

Solution:
As, gravitational constant,

G=M-1 L3 T-2

Planck's constant, h=ML2 T-1

Hence, Gh=M-1L3T-2ML2T-1

=M0L5 T-3

Velocity of light, [c]=LT-1

Now, Ghc31/2=L5T-31/2L3T-31/2=L21/2=L

Hence [L]= length

Topic: Physics, Units and Dimensions, JEE Main
Dimensional analysis of the equation
velocityx=pressure difference3/2Ãdensity-3/2,
gives the value of x as _______.

Options:

The correct answer is:
3

Solution:
(velocity)x=P3/2×ρ-3/2=Pρ2
=M L-1T-2M L-332
=L2T-23/2
=L3T-3
=[L1T-1]3
But L1T-1 is the dimension of velocity.
⇒x=3

Topic: Physics, Units and Dimensions, JEE Main
Dimensional formula for angular momentum is

Options:
A $M L^2 T^{-2}$
B $M {L}^2 T^{-1}$
C $M L T^{-1}$
D $M^0 L^2 T^{-2}$

The correct answer is:
$M {L}^2 T^{-1}$

Solution:
Angular momentum for a point mass about an axis with velocity component of $v$ perpendicular to the radial direction is given by:
$\mathrm{L}_{\mathrm{p}}=\mathrm{mvr}$
Hence, dimensions of angular momentum is:
$\begin{aligned}& {\left[L_p\right]=\left[M^1\right]\left[L^1 T^{-1}\right][L]} \\& {\left[L_p\right]=\left[M^1 L^2 T^{-1}\right]}\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main