datavorous/entrance-exam-dataset huggingface dataset:
One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat (in at constant volume) is

Options:
A $18.7$
B $18.9$
C $19.2$
D None of the above

The correct answer is:
$18.7$

Solution:
$C_{V_{\text {mix }}}=\frac{\mu_1 C_{V_1}+\mu_2 C_{V_2}}{\mu_1+\mu_2}=\frac{1 \times \frac{3}{2} R+3 \times \frac{5}{2} R}{1+3}$ $=\frac{3}{4} R=\frac{3}{4} \times 8.3=18.7$

Topic: Physics, Kinetic Theory of Gases, JEE Main
One mole of monoatomic gas and three moles of diatomic gas are put together in a container. The molar specific heat \( \left(\right. \) in \( \left.\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right) \) at constant volume is \( \left(R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \)

Options:
A \( 18.7 \)
B \( 18.9 \)
C \( 19.2 \)
D None of these

The correct answer is:
\( 18.7 \)

Solution:
Ratio of specific heat for a monoatomic gas is 53 and for diatomic gas is 75.
Given, n1=1, n2=3, n=4
∴ nγ-1=n1γ1-1+n2γ2-1
4γ-1=153-1+375-1
⇒ 4γ-1=32+152=9
∴ 4=9γ-9
⇒ 9γ=13 ⇒γ=139
Now, CVγ-1=R
or CV=Rγ-1=8.3139-1=8.3×94
⇒ CV=18.7 J mol-1-K-1

Topic: Physics, Kinetic Theory of Gases, JEE Main
One mole of N2H4 looses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y? (there is no change in the oxidation state of hydrogen)

Options:
A +2
B -2
C +3
D +4

The correct answer is:
+3

Solution:
N2H4→Y+10e-
Oxidation state of N in N2H4 :
2X+4=0
X=-2
The two, nitrogen atoms will balance the charge of 10e-1 .
So, oxidation state of N will increase by +5, that is., from -2 to +3.

Topic: Chemistry, Redox Reactions, JEE Main
One mole of $\mathrm{N}_2 \mathrm{H}_4$ loses 10 moles of electrons to form a new compound $Z$. Assuming that all the nitrogens appear in the new compound, what is the oxidation state of nitrogen in $Z$ ? (There is no change in the oxidation state of hydrogen.)

Options:
A $-1$
B $-3$
C $+3$
D $+5$

The correct answer is:
$+3$

Solution:
$\mathrm{N}_2 \mathrm{H}_4-10 e^{-} \longrightarrow X^{10+}$ (with two $\mathrm{N}$ atoms)
Total oxidation number of two $\mathrm{N}$ atoms in $\mathrm{N}_2 \mathrm{H}_4$,
$\begin{aligned} \Rightarrow \quad 2 x+4 & =0 \\ 2 x & =-4\end{aligned}$
$\mathrm{N}_2 \mathrm{H}_4$ loses 10 electrons, so total oxidation number of two $\mathrm{N}$-atoms increases by 10 , i.e., the total oxidation number of two $\mathrm{N}$-atoms in
$Y=-4+10=+6$
$\therefore$ Oxidation number of each $\mathrm{N}$ atom in
$X^{10+}=+3$

Topic: Chemistry, Redox Reactions, JEE Main
One mole of N2O4 (g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4 (g) decomposes to NO2 (g). The resultant pressure (in atm) is

Options:

The correct answer is:
2.4

Solution:
N2O411-0.2↔2NO200.4
=0.8
Number of moles at 300 K = 1
No. of moles at 600 K = 0.8 + 0.4 = 1.2
V is constant
P2P1=n2T2n1T1
P21=1.2×6001×300=2.4atm

Topic: Chemistry, States of Matter, JEE Main
One mole of $\mathrm{O}_{2(\mathrm{~g})}$ and two moles of $\mathrm{SO}_{2(\mathrm{~g})}$ were heated in a closed vessel of one-litre capacity at $1098 \mathrm{~K}$. At equilibrium $1.6$ moles of $\mathrm{SO}_{3(\mathrm{~g})}$ were found. The equilibrium constant $K_c$ of the reaction would be

Options:
A
30

B
40

C
80

D
60

The correct answer is:
80

Solution:
$\quad \mathrm{O}_2+2 \mathrm{SO}_2 \rightleftharpoons 2 \mathrm{SO}_3$ at $\mathrm{t}=0 \quad 1 \quad 2 \quad 0$ at equil. $(1-\alpha) 2(1-\alpha) \quad 2 \alpha$ Given at equilibrium
$$
\begin{aligned}
& 2 \alpha=1.6 \\
& K_{\mathrm{C}}=\frac{(2 \alpha)^2}{(1-\alpha)(2-2 \alpha)^2}=\frac{\begin{array}{c}
\alpha=0.8 \\
(0.8)^2
\end{array}}{(1-0.8)(1-0.8)^2} \\
& =\frac{0.64}{0.002} \\
& K_{\mathrm{C}}=80 \\
&
\end{aligned}
$$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2012 (26 May Online)
One mole of $\mathrm{O}_{2}$ gas having a volume equal to 22.4 Litres at $0^{\circ} \mathrm{C}$ and 1 atmospheric pressure in compressed isothermally so that its volume reduces to 11.2 litres. The work done in this process is-

Options:
A $1672.5 \mathrm{~J}$
B $1728 \mathrm{~J}$
C $-1728 \mathrm{~J}$
D $-1572.5 \mathrm{~J}$

The correct answer is:
$-1572.5 \mathrm{~J}$

Solution:
Work done in adiabatic process, \(\mathrm{W}=\mu \mathrm{RT} \log _{\mathrm{e}}\left(\frac{\mathrm{V}_{\mathrm{f}}}{\mathrm{V}_{\mathrm{i}}}\right)\)

\(\mathrm{W}=1 \times 8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{11.2}{22.4}\right)\)

\(\mathrm{W}=8.314 \times 273 \times \log _{\mathrm{e}}\left(\frac{1}{2}\right)\)

\(\mathrm{W}=8.314 \times 273 \times\left[\log _{\mathrm{e}}(1)-\log _{\mathrm{e}}(2)\right]\)

\(\mathrm{W}=8.314 \times 273 \times(0-0.69)\)

\(\mathrm{W}=-1566.1 \mathrm{~J}\)

Topic: Physics, Thermodynamics, JEE Main
One mole of oxygen gas is made to undergo a process in which its molar heat capacity C depends on its absolute temperature T as C=αT, where α is a positive constant. Work done by the gas when heated from an initial temperature T0 to a final temperature 2T0 will be

Options:
A 4αT02
B αT0-R3T02
C 3 Î± T 0 - 5 R T 0 2
D none of these

The correct answer is:
3 α T 0 - 5 R T 0 2

Solution:
C=Cv+W'
⇒W'=dWndT (where W' is the work done by the gas per unit mole per unit rise in temperature)
⇒W'=αT-Cv=αT-5R/2
ΔW=∫ W'dT=∫T02T0αT-5R2dT
=3αT0-5RT02

Topic: Physics, Thermodynamics, JEE Main
One mole of P4 reacts with 8 moles of SOCl2 to give 4 moles of A, x mole of SO2 and 2 moles of B. A, B and x respectively are

Options:
A POCl3, S2Cl2 and 2
B PCl3, S2Cl2 and 4
C PCl3, S2Cl2 and 2
D POCl3, S2Cl2 and 4

The correct answer is:
PCl3, S2Cl2 and 4

Solution:
Thionyl chloride reacts with white phosphorus to produce phosphorus trichloride. The other side products in this reaction are sulphur dioxide and Sulphur monochloride.

P4+8SOCl2→4PCl3+4SO2+2S2Cl2
Hence A=PCl3, x=4, B=S2Cl2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (11 Apr Shift 2)
One mole of the complex compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3$, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $\mathrm{AgNO}_3$ solution to yield two moles of $\mathrm{AgCl}$ (s). The structure of the complex is

Options:
A
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right] \cdot 2 \mathrm{NH}_3$

B
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} . \mathrm{NH}_3$

C
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\right] \mathrm{Cl}_2 . \mathrm{NH}_3$

D
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

Solution:
$\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{+2}+2 \mathrm{Cl}^{-}$
$\therefore$ Structure is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$.

Topic: Chemistry, Coordination Compounds, JEE Main
One mole of the complex compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3$, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $\mathrm{AgNO}_3$ solution to yield two moles of $\mathrm{AgCl}$ (s). The structure of the complex is

Options:
A
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right] \cdot 2 \mathrm{NH}_3$

B
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl} . \mathrm{NH}_3$

C
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}\right] \mathrm{Cl}_2 . \mathrm{NH}_3$

D
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

The correct answer is:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$

Solution:
$\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}_3\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right]^{+2}+2 \mathrm{Cl}^{-}$
$\therefore$ Structure is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2003
One mole of the ideal gas goes through the process $p=p_0\left[1-\alpha\left(\frac{V}{V_0}\right)^3\right]$, where $p$ and $V$ are pressure and volume, $p_0, V_0$ and $\alpha$ are constants. If the maximum attainable temperature of the gas is $\left(\frac{3}{4}\right) \frac{p_0 V_0}{R}$, then the value of $\alpha$ is

Options:
A 2
B $\frac{1}{2}$
C $\frac{1}{4}$
D 4

The correct answer is:
$\frac{1}{4}$

Solution:
For 1 mole of an ideal gas,
$$
p=\frac{R T}{V}
$$
But,
$$
p=p_0\left(1-\frac{\alpha V^3}{V_0^3}\right)
$$
(given)
So,
$$
\begin{gathered}
\frac{R T}{V}=p_0-\frac{\alpha p_0}{V_0^3} V^3 \\
T=\frac{p_0 V}{R}-\frac{\alpha p_0 V^4}{V_0^3 R}
\end{gathered}
$$
For maximum $T$,
$$
=\frac{d T}{d V}=0
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{d T}{d V}=\frac{p_0}{R}-\frac{4 \alpha p_0}{R V_0^3} V^3 \quad \Rightarrow \quad \frac{p_0}{R}=\frac{4 \alpha p_0}{R V_0^3} V^3 \\
& \Rightarrow \quad V^3=\frac{V_0^3}{4 \alpha} \quad \Rightarrow \quad V=\frac{V_0}{(4 \alpha)^{1 / 3}} \\
&
\end{aligned}
$$
Substituting in Eq. (i), we get
$$
T_{\max }=\frac{p_0 V_0}{R(4 \alpha)^{1 / 3}}-\frac{\alpha p_0 V_0^4}{V_0^3 R(4 \alpha)^{4 / 3}}=\frac{p_0 V_0}{R(4 \alpha)^{1 / 3}}-\frac{p_0 V_0}{R 4^{4 / 3} \alpha^{1 / 3}}
$$
Substituting, $T_{\max }=\frac{3 p_0 V_0}{4 R}$, we get
$$
\frac{3}{4}=\frac{1}{4^{1 / 3} \alpha^{1 / 3}}-\frac{1}{4^{4 / 3} \alpha^{1 / 3}}
$$
Now from options, we can check that only $\alpha=\frac{1}{4}$, balances the equation.
So, $\quad \alpha=\frac{1}{4}$

Topic: Physics, Thermodynamics, JEE Main
One of the by-products formed during the recovery of NH3 from Solvay process is:

Options:
A Ca(OH)2
B NaHCO3
C CaCl2
D NH4Cl

The correct answer is:
CaCl2

Solution:
The Solvay process is also called ammonia recovery.

The calcium hydroxide is reacted with the ammonium chloride separated out of the carbonating tower by filtration.

Ca(OH)2(aq)+2NH4Cl(aq)→CaCl2(aq)+2H2O(l)+2NH3(g)

The ammonia is recycled back into the process to form ammoniated brine.

Calcium chloride is formed as a by-product of the Solvay process.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (17 Mar Shift 2)
One of the commonly used electrode is calomel electrode. Under which of the following categories, calomel electrode comes?

Options:
A Oxidation - Reduction electrodes
B Metal ion - Metal electrodes
C Gas - Ion electrodes
D Metal - Insoluble Salt - Anion electrodes

The correct answer is:
Metal - Insoluble Salt - Anion electrodes

Solution:
Theory based

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (04 Apr Shift 1)
One of the commonly used plant growth hormone in tea plantations is

Options:
A ethylene
B abscissic acid
C zeatin
D indole-3-acetic acid

The correct answer is:
indole-3-acetic acid

Solution:
Tea plants are grown by stem cuttings. If we dip the lower cut end of a cutting in dilute solution of auxins (IAA, IBA, etc), very soon large number of roots are developed on the cut ends, due to which these cutting develop into successful plants.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2010 (Mains)
One of the complex roots of the equation $x^{11}-x^6-x^5+1=0$ is

Options:
A $\operatorname{cis} \frac{3 \pi}{5}$
B $\operatorname{cis} \frac{\pi}{3}$
C $\operatorname{cis} \frac{5 \pi}{6}$
D cis $\frac{7 \pi}{5}$

The correct answer is:
$\operatorname{cis} \frac{\pi}{3}$

Solution:
$\begin{aligned} & \quad x^{11}-x^6-x^5+1=0 \\ & \Rightarrow \quad\left(x^6-1\right)\left(x^5-1\right)=0 \\ & \Rightarrow \quad x^6=1 \text { or } x^5=1 \\ & \Rightarrow \quad x=(1)^{\frac{1}{6}} \text { or } x=(1)^{\frac{1}{5}} \\ & x=\cos \frac{2 k \pi}{6}+i \sin \frac{2 k \pi}{6} \text { or } x=\cos \frac{2 r \pi}{5}+i \sin \frac{2 r \pi}{5} \\ & \end{aligned}$
where,
$$
\begin{aligned}
k & =0,1,2,3,4,5 \text { where } r=0,1,2,3,4 \\
x & =\cos \frac{k \pi}{3}+i \sin \frac{k \pi}{3}
\end{aligned}
$$
When $k=1$,
$$
\cos \frac{\pi}{3}+i \sin \pi / 3=C \text { is } \frac{\pi}{3} \text {. }
$$

Topic: Mathematics, Complex Number, JEE Main
One of the constituents of the pancreatic juice while poured into the duodenum in humans, is

Options:
A trypsinogen
B chymotrypsin
C trypsin
D enterokinase

The correct answer is:
trypsinogen

Solution:
Trypsinogen is the precursor form or zymogen of the pancreatic enzyme trypsin. It is found in pancreatic juice, along with amylase, lipase and chymotrypsinogen. It is activated by enteropeptidase which is found in the intestinal mucosa, to form trypsin.

Topic: Biology, Digestion and Absorption, NEET, NEET 2011 (Mains)
One of the cube roots of unity is

Options:
A $\frac{-1+i \sqrt{3}}{2}$
B $\frac{1+i \sqrt{3}}{2}$
C $\frac{1-i \sqrt{3}}{2}$
D $\frac{\sqrt{3}-i}{2}$

The correct answer is:
$\frac{-1+i \sqrt{3}}{2}$

Solution:
$\frac{-1+i \sqrt{3}}{2}$

Topic: Mathematics, Complex Number, JEE Main
One of the ex-situ conservation methods for endangered species is

Options:
A Wildlife Sanctuaries
B Biosphere Reserves
C Cryopreservation
D National parks

The correct answer is:
Cryopreservation

Solution:
Preservation at -196o liquid N2 can maintain tissue culture, embryos, animal cells/tissues, spermatozoa, etc., which is called cryopreservation. It is an ex-situ conservation method, i.e., conservation of organisms outside their natural homes. Other ex-situ conservation methods are off-site collection and gene banks.

Topic: Biology, Biodiversity and its Conservation, NEET
One of the following conversion results in the change of hybridization and geometry :

Options:
A \( \mathrm{CH}_{4} \) to \( \mathrm{C}_{2} \mathrm{H}_{6} \)
B \( N H_{3} \) to \( N H \)
C \( B F_{3} \) to \( B F_{4} \)
D \( \mathrm{H}_{2} \mathrm{O} \) to \( \mathrm{H}_{3} \mathrm{O} \)

The correct answer is:
\( B F_{3} \) to \( B F_{4} \)

Solution:
For the conversion, \(\mathrm{CH}_{4} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}\),
There is no change in the hybridisation of the carbon atom. Each carbon loses an atom of hydrogen and the two carbon hydrogen bonds are replaced by a carbon-carbon bond.
Both \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) have \(\mathrm{sp}^{3}\) hybridisation and tetrahedral shape.
For the conversion, \(\mathrm{NH}_{3} \rightarrow \mathrm{NH}_{4}^{+}\)
Nitrogen in \(\mathrm{NH}_{3}\) has 3 bonds and one lone pair. It has \(\mathrm{sp}^{3}\) hybridisation, thus has a tetrahedral geometry but due to lone pair of electrons, its shape is pyramidal. On conversion to \(\mathrm{NH}_{4}^{+}\), a proton attaches to the lone pair of electrons. The hybridisation remains the same, that is \(\mathrm{pp}^{3}\) but the shape of the molecule becomes tetrahedral.
For the conversion, \(\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}\),
Oxygen in \(\mathrm{H}_{2}\) Oforms two bonds with two hydrogen atoms and has two lone pairs of electrons.
Thus, it has a hybridisation of \({s p^{3}}^{3}\), a tetrahedral geometry but a bent shape. On conversion to
\(\mathrm{H}_{3} \mathrm{O}^{+}\), a proton attaches to the lone pair of electrons. The hybridisation remains the same,
that is \(\operatorname{sp}^{3}\) but the shape of the molecule now becomes pyramidal.
For the conversion, \(\mathrm{BF}_{3} \rightarrow \mathrm{BF}_{4}^{-}\)
Boron in \(\mathrm{BF}_{3}\) forms three bonds with three fluorine atoms and has no lone pair of electrons.
Thus, it has the hybridisation \(\mathbf{s p}^{2}\) and trigonal planar shape. On conversion to \(\mathrm{BF}_{4}^{-}\), a fluoride
ion puts its lone pair in the empty orbital of boron atom changing the hybridisation to \(\mathrm{sp}^{3}\) and shape to tetrahedral.

Topic: Chemistry, Chemical Bonding and Molecular Structure, KCET, KCET 2015
One of the following statements is incorrect with reference to biodiversity. Identify it.

Options:
A The areas with very few plant and animal species (low species richness) with no threatened species are called biodiversity hotspots.
B Biodiversity increases from higher altitudes to lower altitudes.
C Biodiversity decreases from the equator to polar regions.
D Depletion in genetic diversity of crop plants is mainly due to the introduction of better varieties with high yield, disease resistance, etc.

The correct answer is:
The areas with very few plant and animal species (low species richness) with no threatened species are called biodiversity hotspots.

Solution:
Biodiversity hotspots are areas of high endemism and high level of species richness.
They are determined mainly by three factors:
(i) Number of species/species diversity
(ii) Degree of endemism
(iii) Degree of threat

Topic: Biology, Biodiversity and its Conservation, AIIMS, AIIMS 2016
One of the hydrolysed product of the following compound does not react with silica of glass vessel

Options:
A BF3
B ClF5
C XeF2
D SF4

The correct answer is:
BF3

Solution:
In case of hydrolysis of BF3 , HF is not formed so no reaction with Silica glass vessel.
4BF3+3H2OâRoom TemperatureH3BO3+3H[BF4]âdoesnotreactwithsilica(SiO2)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
One of the limit point of the coaxial system of circles containing $x^2+y^2-6 x-6 y+4=0$, $x^2+y^2-2 x-4 y+3=0$ is

Options:
A $(-1,1)$
B $(-1,2)$
C $(-2,1)$
D $(-2,2)$

The correct answer is:
$(-1,1)$

Solution:
Trick : The equation of radical axis is $S_1-S_2=0$ i.e., $4 x+2 y-1=0$
$\therefore$ The equation of circle of co-axial system can be taken as $\left(x^2+y^2-6 x-6 y+4\right)+\lambda(4 x+2 y-1)=0$
or $x^2+y^2-(6-4 \lambda) x-(6-2 \lambda) y+(4-\lambda)=0$
whose centre is $C(3-2 \lambda, 3-\lambda)$ and radius is $r=\sqrt{(3-2 \lambda)^2+(3-\lambda)^2-(4-\lambda)}$
If $r=0$, then we get $\lambda=2$ or $7 / 5$.
Putting the cl-ordinates of $C$, the limit points are $(-1,1)$ and $\left(\frac{1}{5}, \frac{8}{5}\right)$. One of these limit points is given in (a).

Topic: Mathematics, Circle, JEE Main
One of the lines in the emission spectrum of $\mathrm{Li}^{2+}$ has the same wavelength as that of the $2^{\text {nd }}$ line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is $\mathrm{n}=12 \rightarrow \mathrm{n}=\mathrm{x} .$ Find the value of $\mathrm{x}$

Options:
A 8
B 6
C 7
D 5

The correct answer is:
6

Solution:
For $2^{\text {nd line of Balmer series in hydrogen }}$ spectrum

$$

\frac{1}{\lambda}=\mathrm{R}(1)\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3}{16} \mathrm{R}

$$

$$

\text { For } \mathrm{Li}^{2+}\left[\frac{1}{\lambda}=\mathrm{R} \times 9\left(\frac{1}{\mathrm{x}^{2}}-\frac{1}{12^{2}}\right)=\frac{3 \mathrm{R}}{16}\right]

$$

which is satisfied by $n=12 \rightarrow n=6$

Topic: Physics, Atomic Physics, JEE Main
One of the monomer of Nylon 6,6 is an acid. How many -COOH groups are present in this acidic monomer.

Options:

The correct answer is:
2.00

Solution:
Nylon 6,6 is a polymer of adipic acid and hexa methylene diamine.
Adipic acid (C||H2)4COOHCOOH
It has 2-COOH groups.

Topic: Chemistry, Polymers, JEE Main
One of the points of intersection of the curves $y=1+3 x-2 x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(l \sqrt{5}+\mathrm{m})-\mathrm{n} \log _{\mathrm{e}}(1+\sqrt{5})$, where $l, \mathrm{~m}, \mathrm{n} \in \mathbf{N}$. Then $l+\mathrm{m}+\mathrm{n}$ is equal to

Options:
A 29
B 31
C 30
D 32

The correct answer is:
30

Solution:
$\begin{aligned} & A=\int_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}}\left(1+3 x-2 x^2-\frac{1}{x}\right) \mathrm{dx} \\ & A=\left[x+\frac{3 x^2}{2}-\frac{2 x^3}{3}-\ln x\right]_{\frac{1}{2}}^{1+\sqrt{5}} \\ & A=\frac{1+\sqrt{5}}{2}+\frac{3}{2}\left(\frac{1+\sqrt{5}}{2}\right)^2-\frac{2}{3}\left(\frac{1+\sqrt{5}}{2}\right)^3-\ln \left(\frac{1+\sqrt{5}}{2}\right) \\ & -\frac{1}{2}-\frac{3}{2}\left(\frac{1}{4}\right)+\frac{2}{3}\left(\frac{1}{8}\right)+\ln \left(\frac{1}{2}\right) \\ & A=\frac{1}{2}+\frac{\sqrt{5}}{2}+\frac{3}{8}+\frac{3}{4} \sqrt{5}+\frac{15}{8}-\frac{4}{3}-\frac{2}{3} \sqrt{5} \\ & -\frac{1}{2}-\frac{3}{8}+\frac{1}{12}-\ln (1+\sqrt{5}) \\ & =\sqrt{5}\left(\frac{1}{2}+\frac{3}{4}-\frac{2}{3}\right)+\frac{15}{8}-\frac{4}{3}+\frac{1}{12}-\ln (1+\sqrt{5}) \\ & =\frac{14}{24} \sqrt{5}+\frac{15}{24}-\ln (1+\sqrt{5})\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (04 Apr Shift 1)
One of the processes used for concentration of ores is Froth floatation process. This process is generally used for concentration of sulphide ores. Sometimes in this process we add NaCN as a depressant. NaCN is generally added in case of ZnS and PbS minerals. What is the purpose of addition of NaCN during the process of Froth floatation ?

Options:
A NaCN causes reduction by precipitation
B A soluble complex is formed by reaction between NaCN and ZnS while PbS forms froth
C A soluble complex is formed by reaction between NaCN and PbS while ZnS forms froth
D A precipitate of Pb(CN)2 is produced while ZnS remain unaffected.

The correct answer is:
A soluble complex is formed by reaction between NaCN and ZnS while PbS forms froth

Solution:
NaCN + ZnS → Na 2 Zn CN 4 Soluble complex
A layer of this zinc complex is formed on the surface of ZnS and due to this ZnS is prevented from the froth formation while PbS form froth. (i.e., NaCN is added as depressant for ZnS).

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
One of the product formed in the reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO 2 .
FeO s â+âCO g â Fe s â+â CO 2 g ; K P â=â0.265âatm at 1050âK. What are the equilibrium partial pressures of CO and CO2 respectively at 1050âK if the initial partial pressures are: Pco=1.4 atm and PCO2=0.80 atm ?

Options:
A PCO=1.739 atm and PCO2=0.461 atm
B PCO=17.39Â atmÂ andÂ PCO2=0.461Â atm
C PCO=1.79Â atmÂ andÂ PCO2=0.46Â atm
D PCO=2.739Â atmÂ andÂ PCO2=1.461Â atm

The correct answer is:
PCO=1.739 atm and PCO2=0.461 atm

Solution:
KP=0.265=0.8-x1.4+x
On solving, xâ=â0.339
â´PCO=1.739 atm and PCO2=0.461 atm

Topic: Chemistry, Chemical Equilibrium, JEE Main
One of the square faces of a metal slab of side 50 cm and thickness 20 cm is rigidly fixed on a horizontal surface. If a tangential force of 30 N is applied to the top face and it is known that the shear modulus of the material is 4×1010 N/m2, then the displacement (in m) of the top face is

Options:
A 4×10-12
B 4×10-10
C 6×10-10
D 8×10-10

The correct answer is:
6×10-10

Solution:
Shear strain, Îµ=tanâ¡Î¸=Î¸=xl
Î·=ÏÎµ
Or x=FâlAâÎ·
Given F=30 N
A=50Ã50Ã10-4
Î·=4Ã1010
x=30Ã20Ã10-250Ã50Ã10-4Ã4Ã1010
=625Ã4Ã10-2Ã1010
=6Ã10-10 m

Topic: Physics, Mechanical Properties of Solids, JEE Main
One of the synthetic auxin is :

Options:
A IBA
B NAA
C IAA
D GA

The correct answer is:
NAA

Solution:
No Solution Available

Topic: Biology, Plant - Growth and Development, NEET, NEET 2009 (Mains)
One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the centre of the loop BL to that at the centre of the coil BC, i.e. BLBC will be

Options:
A 1N2
B 1N
C N
D N2

The correct answer is:
1N2

Solution:
For a single loop, radius
r1=L2π
BL=μ0I2r1=μ0I2L×2π ...(1)
For a circular coil, radius
r2=L2πN
Bc=μ0IN2r2=μ0IN2L2πN ...(2)
BLBC=1N2 (from eqns (1) and (2))

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (09 Jan Shift 2)