datavorous/entrance-exam-dataset huggingface dataset:
Minimum distance between the curves y2=x-1 and x2=y-1 is equal to

Options:
A 324 units
B 524 units
C 724 units
D 24 units

The correct answer is:
324 units

Solution:
General point on the curve y2=x-1 is ( t 1 2 +1,ââ t 1 ) and the general point on the curve x2=y-1 is ( t 2 ,ââ t 2 2 +1 ). Since both the curves are symmetrical about the line y=x, for the nearest point on the curve y2=x-1 from the line y=x
Let, Qt12+1, t1
P(t2,t22+1)
slopes of the tangents are same
1 2 t 1 =2ât2
t1t2=14 ......(1)
Since PQ is perpendicular to line x=y
t1-t22-1t12+1-t2=-1 ......(2)
Solving (1) & (2), we get t1=t2=12
P12,54, Q54,12
therefore, PQ=324

Topic: Mathematics, Application of Derivatives, JEE Main
Minimum length of antenna required to transmit a radio signal of frequency 10 MHz is equal to

Options:
A 8.5 m
B 6 m
C 4 m
D 7.5 m

The correct answer is:
7.5 m

Solution:
L min=λ4
λ=cv
=3×108107=30 m

L min=7.5 m

Topic: Physics, Communication System, JEE Main
Minimum moles of NH3 required to be added to 1L solution so as to dissolve 0.1mol of AgCl(Ksp=1.0Ã10â10) by the reaction is:

AgCl(s) + 2 NH 3 â Ag NH 3 2 + + Cl -
Given K f âofâ Ag(NH 3 ) 2 + =10 8

Options:
A 0.5mol
B 1.0mol
C 1.1mol
D 1.2mol

The correct answer is:
1.2mol

Solution:
AgCl s + aq â Ag aq + + Cl aq - ; K sp Ag aq + + 2 NH 3 aq â Ag NH 3 2 aq + ; K f
---------------------------------------------------------------------------------------------------
AgCl s + 2 NH 3 â Ag NH 3 2 aq + + Cl aq - K = K sp Ã K f
C-0.2M 0.1 M 0.1 M Equilibrium concentration
K = 1 0 - 1 0 Ã 1 0 8 = 1 0 - 2 = Ag [ NH 3 ) 2 + Cl - NH 3 2
1 0 - 2 = 1 0 - 1 Ã 1 0 - 1 M - 0 . 2 2
â´ââMâ0.2 =1
â´ââM=1.2M

Topic: Chemistry, Ionic Equilibrium, JEE Main
Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is:

Options:
A 8
B 6
C 5
D 7

The correct answer is:
7

Solution:
Let coin is tossed n times.

The probability of getting a head in a single throw of a die is 12.

By using binomial theorem, we know that the probability that an event X appear r times in n trials is PX=r=Crnpn-rqr, q=1-p, r=0, 1, 2, ..., n.

Now, P(at least are head)=1-P(no head)

=1-12n

Using, the given condition, we get

1-12n≥0.99

⇒12n≤1-0.99

⇒12n≤1100

As, 26=64 and 27=128, hence, the minimum value of n is 7.

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Minimum value of $5 \sin ^2 \theta+4 \cos ^2 \theta$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Let $f(\theta)=5 \sin ^2 \theta+4 \cos ^2 \theta=4+\sin ^2 \theta$
$\therefore f(\theta) \geq 4+0 \quad\left(\sin ^2 \theta \geq 0\right)$
$\therefore$ The minimum value of $f(\theta)$ is 4

Topic: Mathematics, Trigonometric Equations, JEE Main
Mischmetal is an alloy consisting mainly of :

Options:
A Lanthanoid metals
B actinoid and transition metals
C lanthanoid and actinoid metals
D actinoid metals

The correct answer is:
Lanthanoid metals

Solution:
Misch metal consists of Lanthanide metal ≈95%and iron ≈5% and traces of S, C, Ca and Al.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Mitosis is characterised by

Options:
A
reduction division

B
equal division

C
Both (a) and (b)

D
pairing of homologous chromosomes

The correct answer is:
equal division

Solution:
Mitosis is characterised by equal division because the chromosome numbers in the daughter cells remains same as that of the parent cell. Reduction division is the characteristic of meiosis.

Topic: Biology, Cell Cycle and Cell Division
Mitotic anaphase differs from metaphase in possessing

Options:
A Same number of chromosomes and same number of chromatids
B Half number of chromosomes and half number of chromatids
C Half number of chromosomes and same number of chromatids
D Same number of chromosomes and half number of chromatids

The correct answer is:
Same number of chromosomes and half number of chromatids

Solution:
In mitotic anaphase, the sister chromatids separate from each other and begin to move towards the opposite poles. Hence, the number of chromosomes remains the same but the amount of DNA or chromatids becomes half.

Topic: Biology, Cell Cycle and Cell Division, NEET
Mitotic spindle have main protein

Options:
A tubulin
B myosin
C tropomyocin
D dynein

The correct answer is:
tubulin

Solution:
Microtubules are unbranched, hollow, submicroscopic tubules, which form the structural material of cilia, flagella, basal bodies, centrioles and spindle apparatus. Microtubules are built up of protofilaments formed of two related proteins, $\alpha$ and $\beta$-tubulins.

Topic: Biology, Morphology of Flowering Plants, JIPMER, JIPMER 2008
Mitotic spindle is mainly composed of which protein?

Options:
A Actin
B Tubulin
C Actomyosin
D Myoglobin

The correct answer is:
Tubulin

Solution:
Spindle fibres are composed of a large number of microtubules which is made up of a single layer of globulin protein. These microtubules are made by polymerisation of globular tubulin proteins.
Related Theory
The two centrosomes give rise to the mitotic spindle which regulates the movement of chromosomes during mitosis. At the centre of each animal cell, the centrosomes of animal cells are associated with a pair of rod-like objects, the centrioles, which are at right angles to each other.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2002
Mixed He+ and O2+ ions (mass of He+=4 amu and that of O2+=16 amu)
beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same, then

Options:
A He+ ions will be deflected more than those of O2+
B He+ ions will be deflected less than that of O2+
C All the ions will be deflected equally
D No ions will be deflected

The correct answer is:
All the ions will be deflected equally

Solution:
E=12mv2
or v=2Em
q v B=mv2r
or r=mvqB=mqBÃ2Em=2EmqB
or râmq
Hence,
r1=r2

Topic: Physics, Magnetic Effects of Current, JEE Main
Mixing of non-reacting gases is generally accompanied by

Options:
A Decrease in entropy
B Increase in entropy
C Change in enthalpy
D Change in free energy

The correct answer is:
Increase in entropy

Solution:
Mixing of non-reacting gases increases randomness and so increase entropy .

Topic: Chemistry, Thermodynamics (C), JEE Main
Mixture of 1 g each of Na2CO3 and NaHCO3 is reacted with 0.1 N HCl. The quantity of 0.1 M HCl required to react completely with the above mixture is

Options:
A 15.78 mL
B 157.8 mL
C 198.4 mL
D 308 mL

The correct answer is:
308 mL

Solution:
Na 2 CO 3 + 2 HCl â 2 NaCl + H 2 O + CO 2 1 mol 2 mol
Na 2 CO 3 = 1 g= 1 106 mol.N a 2 C o 3 ≡ 2 106 mol.HCl
NaHCO 3 + HCl â NaCl + H 2 O + CO 2
NaHCO 3 = 1 g = 1 8 4 mol â¡ 1 8 4 mol HCl
Total HCl required = 1 5 3 + 1 8 4 mol HCl
Let volume of 0.1 N HCl = V mL
then, Moles = 0.1 Ã V 1 0 0 0 = V 1 0 0 0 0
â´ V 1 0 0 00 = 1 5 3 + 1 8 4
= 0.0308 mol
V = 0.0308 Ã 1 0 0 0 0
= 308 mL

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Mixture of \( 1 \mathrm{~g} \) each of \( \mathrm{Na}_{2} \mathrm{CO}_{3} \) and \( \mathrm{NaHCO}_{3} \) is reacted with \( 0.1 \mathrm{NHCl} \). The quantity of \( 0.1 \mathrm{M} \mathrm{HCl} \) required to react completely with the above mixture is

Options:
A \( 15.78 \mathrm{~mL} \)
B \( 157.8 \mathrm{~mL} \)
C \( 198.4 \mathrm{~mL} \)
D \( 308 \mathrm{~mL} \)

The correct answer is:
\( 308 \mathrm{~mL} \)

Solution:
Na 2 CO 3 + 2 HCl â 2 NaCl + H 2 O + CO 2 1 mol 2 mol
Na 2 CO 3 = 1 g= 1 106 mol.N a 2 C o 3 ≡ 2 106 mol.HCl
NaHCO 3 + HCl â NaCl + H 2 O + CO 2
NaHCO 3 = 1 g = 1 8 4 mol â¡ 1 8 4 mol HCl
Total HCl required = 1 5 3 + 1 8 4 mol HCl
Let volume of 0.1 N HCl = V mL
then, Moles = 0.1 Ã V 1 0 0 0 = V 1 0 0 0 0
â´ V 1 0 0 00 = 1 5 3 + 1 8 4
= 0.0308 mol
V = 0.0308 Ã 1 0 0 0 0
= 308 mL

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Mixture of voltaic component A and B has total vapure pressure (in torr)

P=254-119XA .

Where XA is mole fraction of A in the mixture. Hence, PA0 & PB0 (in torr) are:

Options:
A 254,Â 199
B 119,Â 254
C 135, 254
D 119,Â 373

The correct answer is:
135, 254

Solution:
For X A =0, P B 0 =254
For X A =1, P A 0 =254–119 =135

Topic: Chemistry, Solutions, JEE Main
$\mathrm{MnO}_{2}$ when fused with $\mathrm{KOH}$ and oxidized in air gives a dark green compound $\mathrm{X}$. In acidic $\mathrm{X}$. In acidic solution, $\mathrm{X}$ undergoes disproportion to give an intense purple compound $\mathrm{Y}$ and $\mathrm{MnO}_{2}$. The compounds $\mathrm{X}$ and $\mathrm{Y}$. respectively, are

Options:
A $\mathrm{K}_{2} \mathrm{MnO}_{4}$ and $\mathrm{KMnO}_{4}$
B $\mathrm{Mn}_{2} \mathrm{O}_{7}$ and $\mathrm{KMnO}_{4}$
C $\mathrm{K}_{2} \mathrm{MnO}_{4}$ and $\mathrm{Mn}_{2} \mathrm{O}_{7}$
D $\mathrm{KMnO}_{4}$ and $\mathrm{K}_{2} \mathrm{MnO}_{4}$

The correct answer is:
$\mathrm{K}_{2} \mathrm{MnO}_{4}$ and $\mathrm{KMnO}_{4}$

Solution:
$2 \mathrm{MnO}_{2}+4 \mathrm{KOH}+\mathrm{O}_{2} \rightarrow 2 \mathrm{~K}_{2} \mathrm{MnO}_{4}+2 \mathrm{H}_{2} \mathrm{O}$
$3 \mathrm{~K}_{2} \mathrm{MnO}_{4}+4 \mathrm{HCl} \rightarrow 2 \mathrm{KMnO}_{4}+\mathrm{MnO}_{2}+4 \mathrm{KCl}+2 \mathrm{H}_{2} \mathrm{O}$

Topic: Chemistry, Redox Reactions, JEE Main
$\mathrm{MnO}_4^{2-}$
( 1 mole) in neutral aqueous medium is disproportionate to

Options:
A $2 / 3$ mole of $\mathrm{MnO}_4^{-}$and $1 / 3$ mole of $\mathrm{MnO}_2$
B $1 / 3$ mole of $\mathrm{MnO}_4^{-}$and $2 / 3$ mole of $\mathrm{MnO}_2$
C $1 / 3$ mole of $\mathrm{Mn}_2 \mathrm{O}_7$ and $1 / 3$ mole of $\mathrm{MnO}_2$
D $2 / 3$ mole of $\mathrm{Mn}_2 \mathrm{O}_7$ and $1 / 3$ mole of $\mathrm{MnO}_2$

The correct answer is:
$2 / 3$ mole of $\mathrm{MnO}_4^{-}$and $1 / 3$ mole of $\mathrm{MnO}_2$

Solution:
$\mathrm{MnO}_{42}$ in neutral aqueous medium is
disproportionate to $2 / 3$ mole of $\mathrm{MnO}_4$ and $1 / 3$ mole of $\mathrm{MnO}_2$

Topic: Chemistry, Redox Reactions, JEE Main
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If for an $n$-type semiconductor, the density of electrons is $10^{19} \mathrm{~m}^{-3}$ and their mobility is $1.6 \mathrm{~m}^2 /(\mathrm{V}-\mathrm{s})$ then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to

Options:
A $0.2 \mathrm{~m}$
B $4 \mathrm{~m}$
C $2 \mathrm{~m}$
D $0.4 \mathrm{~m}$

The correct answer is:
$0.4 \mathrm{~m}$

Solution:
$J=$ neV $_{\mathrm{d}}$
Resistivity, $\rho=E / j=E / n e V_d=1 / n e\left(v_d / E\right)=1 / n e \mu_e$
Resistivity, $1 /\left(10^{19} \times 1.6 \times 10^{-19} \times 1.6\right)=0.39 \Omega \mathrm{m}=0.4 \Omega \mathrm{m}$

Topic: Physics, Semiconductors, JEE Main
Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an N-type semiconductor, the density of electrons is 1019 m-3 and their mobility is 1.6 m2 V-1 s-1, then the resistivity of the semiconductor (since it is an N-type semiconductor contribution of holes is ignored) is close to:

Options:
A 0.2 Ω m
B 4 Ω m
C 2 Ω m
D 0.4 Ω m

The correct answer is:
0.4 Ω m

Solution:
Resistivity ρ=EJ=Eneυd=1neμ
μ= mobility
ρ=11019×1.6×10-19×1.6=12.56 Ω m
=0.4 Ω m

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Model a torch battery of length l to be made up of a thin cylindrical bar of radius a and a concentric thin cylindrical shell of radius b filled in between with an electrolyte of resistivity ρ (see figure). If the battery is connected to a resistance of value R, the maximum Joule heating in R will take place for :

Options:
A R=ρ2πℓba
B R=ρ2πℓlnba
C R=ρπℓInba
D R=2ρπℓlnba

The correct answer is:
R=ρ2πℓlnba

Solution:
Let a small elementary cylinder in the shell of thickness dr.

The resistance of a wire is given by R=ρLA.

Here ρ is the resistivity of the material.

The resistance of small elementdR=ρdr2πrl

Integrating both sides,

⇒R=ρ2πl∫abdrr

⇒R=ρ2πllnba

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Modern detergents contain enzyme preparations of

Options:
A Thermophiles
B Acidophiles
C Alkaliphiles
D Thermoacidophiles

The correct answer is:
Acidophiles

Solution:
Enzymes used in detergents are proteases and lipases, which are obtained from fungi. Fungi grows properly in acidic medium.

Topic: Biology, Biomolecules (B), NEET, NEET 2008 (Mains)
Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 × 1 0 - 1 5 atm. at room temperature (300 K). Taking R = 8.3 JK-1 mole-1, 1 atm = 105 Pa and N Avogadro = 6 × 1 0 2 3 mole - 1 , the mean distance between molecules of gas in an evacuated vessel will be of the order of :

Options:
A 0.2 μm
B 0.2 mm
C 0.2 cm
D 0.2 nm

The correct answer is:
0.2 mm

Solution:
Let the inter molecular distance be D
Then in a volume of 4 3 Ï D 3 there is only one molecule.
4 3 Ï D 3 P=( 1 N A )RT
D= ( 3RT 4Ï N A P ) 1 3
Substituting P=4Ã 10 â10 Pa,ââR=8.3,ââ N A =6Ã 10 23 ,ââT=300K
D=0.2âmm

Topic: Physics, Kinetic Theory of Gases, JEE Main
Modern vacuum pumps can evacuate a vessel down to a pressure of 4.0 × 1 0 - 1 5 atm. at room temperature (300 K). Taking R = 8.3 JK-1 mole-1, 1 atm = 105 Pa and N Avogadro = 6 × 1 0 2 3 mole - 1 , the mean distance between molecules of gas in an evacuated vessel will be of the order of :

Options:
A 0.2 μm
B 0.2 mm
C 0.2 cm
D 0.2 nm

The correct answer is:
0.2 mm

Solution:
Let the inter molecular distance be D
Then in a volume of 4 3 Ï D 3 there is only one molecule.
4 3 Ï D 3 P=( 1 N A )RT
D= ( 3RT 4Ï N A P ) 1 3
Substituting P=4Ã 10 â10 Pa,ââR=8.3,ââ N A =6Ã 10 23 ,ââT=300K
D=0.2âmm

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2014 (09 Apr Online)
Modulus of $\left(\frac{3+2 i}{3-2 i}\right)$ is

Options:
A $1$
B $1 / 2$
C $2$
D $\sqrt{2}$

The correct answer is:
$1$

Solution:
$\left(\frac{3+2 i}{3-2 i}\right)=\left(\frac{3+2 i}{3-2 i}\right)\left(\frac{3+2 i}{3+2 i}\right)=\frac{9-4+12 i}{13}=\frac{5}{13}+i\left(\frac{12}{13}\right)$
Modulus $=\sqrt{\left(\frac{5}{13}\right)^2+\left(\frac{12}{13}\right)^2}=1$

Topic: Mathematics, Complex Number, JEE Main
Molality ( \(\mathrm{m}\) ) of \(3 \mathrm{M}\) aqueous solution of \(\mathrm{NaCl}\) is :
(Given : Density of solution \(=1.25 \mathrm{~g} \mathrm{~mL}^{-1}\), Molar mass in \(\mathrm{g} \mathrm{mol}^{-1}: \mathrm{Na}-23, \mathrm{Cl}-35.5\))

Options:
A \(1.9 \mathrm{~m}\)
B \(3.85 \mathrm{~m}\)
C \(2.79 \mathrm{~m}\)
D \(2.90 \mathrm{~m}\)

The correct answer is:
\(2.79 \mathrm{~m}\)

Solution:
3 moles are present in 1 litre solution
$\text {molality }=\frac{3 \times 1000}{1.25 \times 1000-[3 \times 58.5]}=2.79 \mathrm{~m}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Molality of 0.8M H2SO4 solution (density 1.06 g cm-3 ) is _______×10-3 m.

Round off your answer to the nearest integer.

Options:

The correct answer is:
815

Solution:
1000 ml solution contains 0.8 mol of H2SO4

Mass of solution = 100cm3×1.06 gcm3= 1060 g

Mass of solute (H2SO4) =0.8 × 98 =78.4g

Mass of solvent = 1060 -78.4 =981.6 g

Molality = moles of H2SO4mass of solvent (in kg)

=0.8981.6×10−3 = 0.815m

= 815×10-3 m

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Molality of an aqueous solution of urea is \(4.44 \mathrm{~m}\). Mole fraction of urea in solution is \(x \times 10^{-3}\). Value of \(x\) is \(\qquad\) - (Integer answer)

Options:

The correct answer is:
74

Solution:
Molality of urea is $4.44 \mathrm{~m}$, that means 4.44 moles of urea present in $1000 \mathrm{gm}$ of water.
$\therefore \mathrm{X}_{\text {ura }}=\frac{4.44}{4.44+\frac{1000}{18}}$
$=0.0740$
OR
$\begin{aligned} & 74 \times 10^{-3} \\ & X=74\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Molar conductivities $\left(\Lambda_m^{\circ}\right)$ at infinite dilution of $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{CH}_3 \mathrm{COONa}$ are 126.4, 425.9 and $91.0 \mathrm{~S} \mathrm{~cm} \mathrm{cmol}^{-1}$ respectively. $\Lambda^{\circ}{ }_m$ for $\mathrm{CH}_3 \mathrm{COOH}$ will be

Options:
A $425.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
B $180.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
C $290.8 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
D $390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

The correct answer is:
$390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution:
$$
\begin{aligned}
& \text { } \mathrm{CH}_3 \mathrm{COONa}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CH}_3 \mathrm{COOH} \\
& 91+425.9=126.4+x \\
& \therefore \quad x=516.9-126.4 \\
& =390.5 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \\
&
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2012 (Mains)
Molar conductivity of a solution of an electrolyte AB3 is 150 Scm2mol-1 . If it ionizes as AB 3 → A 3+ +3 B − , its equivalent conductivity will be

Options:
A 150(in Scm2eq-1)
B 75(in Scm2eq-1)
C 50(in Scm2eq-1)
D 80(in Scm2eq-1 )

The correct answer is:
50(in Scm2eq-1)

Solution:
∧eq=1n+×Z+∧m
Where, ∧m= molar conductance
n+= number of cations
Z+= charge of cation
∧eq=11×3×150=50

Topic: Chemistry, Electrochemistry, JEE Main
Molar entropy and temperature are:

Options:
A Intensive properties
B Extensive properties
C Intensive and extensive properties respectively
D Extensive and intensive properties respectively

The correct answer is:
Intensive properties

Solution:
Thermodynamics properties which are independent of mass are known as intensive properties
∵ Molar entropy and temperature are mass independent
∴ they are intensive properties.

Topic: Chemistry, Thermodynamics (C), JEE Main
Molar ionic conductivities of divalent cation and anion are \(57 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) and \(73 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) respectively. The molar conductivity of solution of an electrolyte with the above cation and anion will be :

Options:
A \(187 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
B \(260 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
C \(130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
D \(65 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

The correct answer is:
\(130 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Solution:
$\begin{aligned} & \Lambda_{\mathrm{C}}^{+2}=57 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\ & \Lambda_{\mathrm{A}}^{+2}=73 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\ & \Lambda_{\text {Solution }}=\lambda_{\mathrm{C}}^{+2}+\Lambda_{\mathrm{A}}^{-2} \\ & =57+73=130\end{aligned}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Molar mass of the hydrocarbon (X) which on ozonolysis consumes one mole of O3 per mole of (X) and gives one mole each of ethanal and propanone is ______ g mol–1 (Molar mass of C : 12 g mol–1 , H : 1 g mol–1 )

Options:

The correct answer is:
70

Solution:
When 2-methyl butene-2 is subjected to ozonolysis and then reduced in presence of dimethyl sulphide or Zn dust, it yields acetone and acetaldehyde. Cycloaddition of a molecule of ozone occurs first and then it is cleaved reductively to give aldehyde and ketone.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Molar mass of the salt from NaBr, NaNO3, KI and CaF2 which does not evolve coloured vapours on heating with concentrated H2SO4 is ____ g mol–1, (Molar mass in g mol–1 : Na : 23, N : 14, K : 39, O : 16, Br : 80, I : 127, F : 19, Ca : 40)

Options:

The correct answer is:
78

Solution:
On heating NaBr with concentrated sulphuric acid gives brown colour bromine vapours.

On heating NaNO3 with concentrated sulphuric acid will release a brown colour NO2 gas

On heating I- with concentrated sulphuric acid gives a dark purple colour.

On heating with concentrated sulphuric acid with CaF2 does not give any colour because it produces colourless hydrogen fluoride.Molar mass of CaF2 is 78.07 g/mol

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Mole fraction of the toluene in the vapour phase which is in equilibrium with a solution of benzene po=120 Torr and toluene po=80 Torr having 2.0 moles of each, is

Options:
A 0.50
B 0.25
C 0.60
D 0.40

The correct answer is:
0.40

Solution:
Mole fraction of toluene vapour =pToXTPToXT+pBoXB
XT= Mole fraction of toluene in solution =2/4 = 0.5
XB= Mole fraction of benzene = 2/4 = 0.5
pTo= Vapour pressure of pure toluene = 80 mm
pTo= Vapour pressure of benzene = 120 mm
=80×45.80+120×0.5=0.40

Topic: Chemistry, Solutions, JEE Main
Molecular formula $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$ is of

Options:
A Glucose
B Fructose
C Both (a) and (b)
D None of these

The correct answer is:
Both (a) and (b)

Solution:
Glucose and Fructose have same molecular formula of $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$

Topic: Chemistry, Biomolecules, JEE Main
Molecular formula of pentahydroxy acid obtained when glucose is oxidised with $\mathrm{Br}_2$ water is

Options:
A $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7$
B $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_8$
C $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
D $\mathrm{C}_6 \mathrm{H}_{10} \mathrm{O}_6$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7$

Solution:
Glucose on oxidation with bromine water yields gluconic acid $(\mathrm{C} 6 \mathrm{H} 12 \mathrm{O} 7)$. This reaction confirms the presence of an aldehyde group. Because for the formation of gluconic acid, free aldehyde group must be present.

Topic: Chemistry, Biomolecules, JEE Main
Molecule AB has a bond length of 1.617 Å and a dipole moment of 0.38 D. The fractional charge on each atom(absolute magnitude) is: (e0=4.802×10-10 esu)

Options:
A 0.5
B 0.05
C 0
D 1.0

The correct answer is:
0.05

Solution:
Assuming complete separation of charges (i.e. +e, -e on two ends)
Î¼=4.802Ã10-10esuÃ1.617Ã10-8 cm
=7.76Ã10-18 esu.cm=7.76 D
Fractional charge on dipole ends =0.387.76â0.05

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2015 (11 Apr Online)
Molecules of an ideal gas are known to have three translational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ=CpCv are given, respectively, by

Options:
A U=52RT and γ=65
B U=5 R T and γ=75
C U=52RT and γ=75
D U=5RT and γ=65

The correct answer is:
U=52RT and γ=75

Solution:
f = 5

γ=1+2f=1+25=75

CpCv=75

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Molten sodium is used in nuclear reactors to

Options:
A Absorb neutrons in order to control the chain reaction
B Slow down the fast neutrons
C Absorb the heat generated by nuclear fission
D Extract radio-isotopes produced in the reactor

The correct answer is:
Absorb the heat generated by nuclear fission

Solution:
Molten sodium is used as a coolant

Topic: Chemistry, s Block Elements, JEE Main
Moment of inertia M.I. of four bodies, having same mass and radius, are reported as;

I1=M.I. of thin circular ring about its diameter,

I2=M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I3=M.I. of solid cylinder about its axis and

I4=M.I. of solid sphere about its diameter.

Then:

Options:
A I1+I2=I3+52I4.
B I1+I3<I2+I4
C I1=I2=I3>I4
D I1=I2=I3<I4

The correct answer is:
I1=I2=I3>I4

Solution:
Ring I1=MR22 about diameter

Disc I2=MR22

Solid cylinder I3=MR22

Solid sphere I4=25MR2

I1=I2=I3>I4

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis for a duration of:

Options:
A 3 s
B 2Â s
C 2.5Â s
D 5Â s

The correct answer is:
2 s

Solution:
We know that,

Rotational kinetic energy, K=12Iω2

⇒12Iω2=1200
⇒12×1.5×ω2=1200
⇒ω=40 rad s-1

We also know that, ω=αt
⇒t=4020=2 s

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Moment of inertia of a circular wire of mass M and radius R about its diameter is

Options:
A
$\mathrm{MR}^2 / 2$

B
$\mathrm{MR}^2$

C
$2 \mathrm{MR}^2$

D
$\mathrm{MR}^2 / 4$

The correct answer is:
$\mathrm{MR}^2$

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2002