datavorous/entrance-exam-dataset huggingface dataset:
What is the number of carbon atoms in alkanes found in diesel?

Options:
A $\mathrm{C}_{29}$ to $\mathrm{C}_{31}$
B $\mathrm{C}_{6}$ to $\mathrm{C}_{8}$
C $\mathrm{C}_{1}$ to $\mathrm{C}_{4}$
D $\mathrm{C}_{15}$ to $\mathrm{C}_{18}$

The correct answer is:
$\mathrm{C}_{15}$ to $\mathrm{C}_{18}$

Solution:
Diesel fuel consists mainly of paraffins, aromatics and naphthenes. The hydrocarbons of gasoline contain typically 4-12 carbon atoms with boiling range between 30 and 210 °C, whereas diesel fuel contains hydrocarbons with approximately 15–18 carbon atoms and the boiling range is between 170 and 360 °C.

Topic: Chemistry, Hydrocarbons, MHT CET, MHT CET 2020 (12 Oct Shift 2)
What is the number of chambers in the heart of Periplaneta americana?

Options:
A 2 chambers
B 4 chambers
C 13 chambers
D 11 chambers

The correct answer is:
13 chambers

Solution:
The circulatory system of cockroach:
- It is an open type. Blood vessels are poorly developed and open into haemocoel or open spaces.
- Body fluid present in hemocoel is also known as hemolymph.
- Blood is colorless as there is no respiratory pigment.
- Blood doesn't perform the function of transporting gases.
- Cockroach possesses 13- chambered tubular, muscular, elongated and neurogenic heart.
- Blood circulation in cockroach is maintained by wing-shaped alary muscles (involuntary).

Topic: Biology, Structural Organisation in Animals, NEET
What is the number of solutions to the equation $x^n-5 x^{n-1}+6 x^{n-2}=0$ ?

Options:
A $n$
B $n-1$
C $2$
D $1$

The correct answer is:
$2$

Solution:
To find the solutions, we can factor out an $x^{n-2}$ from the equation:
$x^{n-2}\left(x^2-5 x+6\right)=0$
Now, we solve the quadratic equation $x^2-5 x+6=0$ to find the roots:
$(x-2)(x-3)=0$
So, the roots of the quadratic equation are $x=2$ and $x=3$.
Thus, the equation $x^n-5 x^{n-1}+6 x^{n-2}=0$ has 2 solutions, one for each root of the quadratic equation.

Topic: Mathematics, Quadratic Equation, JEE Main
What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : N2, N 2+, O2, O2+ ?

Options:
A 0, 1, 2, 1
B 2, 1, 2, 1
C 0, 1, 0, 1
D 2, 1, 0, 1

The correct answer is:
0, 1, 2, 1

Solution:
The orbital in which last electron is present is called the highest occupied molecular orbital(HOMO).

N2 : σ1s2σ*1s2σ2s2σ*2s2π2px2=π2py2σ2pz2HOMO

N2+: σ1s2σ*1s2σ*2s2σ*2s2π2px2=π2py2σ2pz1HOMO

O2: σ1s2σ*1s2σ*2s2σ2pz2π2px1=π*2py1HOMO

O2+: σ1s2σ*1s2σ2s2σ*2s2σ2pz2π2px2=π2py2 π*2px1=π*2py0HOMO

N2⇒0 unpaired e– in HOMO

N2+⇒1 unpaired e- in HOMO

O2⇒2 unpaired e- in HOMO

O2+⇒1 unpaired e- in HOMO

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (24 Jan Shift 2)
What is the number of ways in which an examiner can assign 10 marks to 4 questions, giving not less than 2 marks to any questions?

Options:
A 4
B 6
C 10
D 16

The correct answer is:
10

Solution:
$x_{1}+x_{2}+x_{3}+x_{4}=10 \quad x_{i} \geq 2$
$\Rightarrow\left(x_{1}-2\right)+\left(x_{2}-2\right)+\left(x_{3}-2\right)+\left(x_{4}-2\right)=2$
$\begin{array}{llll}\mathrm{y}_{1} & \mathrm{y}_{2} & \mathrm{y}_{3} & \mathrm{y}_{4}\end{array}$
$\mathrm{y}_{1}+\mathrm{y}_{2}+\mathrm{y}_{3}+\mathrm{y}_{4}=2 ; \mathrm{y}_{i} \geq 0$
$\therefore$ N.O.W $={ }^{2+4-1} \mathrm{C}_{4-1}={ }^{5} \mathrm{C}_{3}=10$

Topic: Mathematics, Permutation Combination, JEE Main
What is the number of words that can be formed from the letters of the word 'UNIVERSAL', the vowels remaining always together?

Options:
A 720
B 1440
C 17280
D 21540

The correct answer is:
17280

Solution:
Consider the word UNIVERSAL Total no. of vowels $=\mathrm{U}, \mathrm{I}, \mathrm{E}, \mathrm{A}=4$
Let us consider these as a single letter. UIEA N VRSL Then, total no. of letters $=6$
Then, number of ways to arrange them $=6 !=720$ But vowels can also arranged in $4 !$ or 24 ways. Hence, total number of ways $=720 \times 24=17280$

Topic: Mathematics, Permutation Combination, NDA, NDA 2010 (Phase 2)
What is the objective of plant breeding in animal husbandry practices?

Options:
A To reduce the yield of the food
B To increase the protein content of the food
C To maintain the yield of the food
D To decrease the quality of food

The correct answer is:
To increase the protein content of the food

Solution:
Plant breeding may be used to create varieties, which are resistant to pathogens and insect pests, and also to increase the protein content of the plant foods and thereby enhance the quality of food.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
What is the OH â concentration of a 0.04âM solution of CH 3 COONa?
[K a ofâ CH 3 COONaâ=â2âÃâ 10 â5 , âlog 2â=â0.30, âlog 5â=â0.69,â 10 0.7 â5]

Options:
A 5Ã 10-6
B 6Ã10-6
C 2Ã10-9
D 3Ã10-9

The correct answer is:
5Ã 10-6

Solution:
pK a â=ââlogâ K a â=ââlogâ(2âÃâ 10 â5 )
=5-log 2=5-0.3=4.7
pHâ=â 1 2 pK w â+â pK a â+âlogâC
pH=12[14+4.7+logâ¡0.04]
pH=8.63
pOH = 14−8.63 ≅ 5.3
OH-=5Ã10-6

Topic: Chemistry, Ionic Equilibrium, JEE Main
What is the $\left[\mathrm{OH}^{-}\right]$in the final solution prepared by mixing $20.0 \mathrm{~mL}$ of $0.050 \mathrm{M} \mathrm{HCl}$ with 30.0 $\mathrm{mL}$ of $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$ ? -

Options:
A $0.12 \mathrm{M}$
B $0.10 \mathrm{M}$
C $0.40 \mathrm{M}$
D $0.0050 \mathrm{M}$

The correct answer is:
$0.10 \mathrm{M}$

Solution:
$20 \mathrm{~mL}$ of $0.50 \mathrm{M} \mathrm{HCl}=20 \times 0.050 \mathrm{~m} \mathrm{~mol}$ $=1.0 \mathrm{~m} \mathrm{~mol}=1.0$ meq. of $\mathrm{HCl}$
$30 \mathrm{~mL}$ of $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$
$$
\begin{aligned}
& =30 \times 0.1 \mathrm{~m} \mathrm{~mol} \\
& =3 \mathrm{~m} \mathrm{~mol}=3 \times 2 \mathrm{meq} \\
& =6 \mathrm{meq} \mathrm{Ba}(\mathrm{OH})_2
\end{aligned}
$$
1 meq of $\mathrm{HCl}$ will neutralize 1 meq of $\mathrm{Ba}(\mathrm{OH})_2$
$$
\mathrm{Ba}(\mathrm{OH})_2 \text { left }=5 \text { meq. }
$$
Tatal volume $=50 \mathrm{~mL}$ $\mathrm{Ba}(\mathrm{OH})_2$ conc. in final solution
$$
=\frac{5}{50} \mathrm{~N}=0.1 \mathrm{~N}=0.05 \mathrm{M}
$$
$$
\left[\mathrm{OH}^{-}\right]=2 \times 0.05 \mathrm{M}=0.10 \mathrm{M}
$$
Alternatively,
$$
\mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+2 \mathrm{H}_2 \mathrm{O}
$$
$2 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{HCl}$ neutralize $1 \mathrm{~m}$ mole of $\mathrm{Ba}(\mathrm{OH})_2$
$1 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{HCl}$ neutralize $0.5 \mathrm{~m} \mathrm{~mol} \mathrm{of}$ $\mathrm{Ba}(\mathrm{OH})_2$
$\mathrm{Ba}(\mathrm{OH})_2 1 \mathrm{eft}=3-0.5 \mathrm{~m} \mathrm{~mol}=2.5 \mathrm{mmol}$
$$
\left[\mathrm{Ba}(\mathrm{OH})_2\right]=\frac{2.5}{50} \mathrm{M}=0.05 \mathrm{M}
$$
or $[\mathrm{OH}]^{-}=2 \times 0.05=0.1 \mathrm{M}$

Topic: Chemistry, Structure of Atom, JEE Main
What is the $\left[\mathrm{OH}^{-}\right]$in the final solution prepared by mixing $20.0 \mathrm{~mL}$ of $0.050 \mathrm{M} \mathrm{HCl}$ with 30.0 $\mathrm{mL}$ of $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$ ? -

Options:
A $0.12 \mathrm{M}$
B $0.10 \mathrm{M}$
C $0.40 \mathrm{M}$
D $0.0050 \mathrm{M}$

The correct answer is:
$0.10 \mathrm{M}$

Solution:
$20 \mathrm{~mL}$ of $0.50 \mathrm{M} \mathrm{HCl}=20 \times 0.050 \mathrm{~m} \mathrm{~mol}$ $=1.0 \mathrm{~m} \mathrm{~mol}=1.0$ meq. of $\mathrm{HCl}$
$30 \mathrm{~mL}$ of $0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2$
$$
\begin{aligned}
& =30 \times 0.1 \mathrm{~m} \mathrm{~mol} \\
& =3 \mathrm{~m} \mathrm{~mol}=3 \times 2 \mathrm{meq} \\
& =6 \mathrm{meq} \mathrm{Ba}(\mathrm{OH})_2
\end{aligned}
$$
1 meq of $\mathrm{HCl}$ will neutralize 1 meq of $\mathrm{Ba}(\mathrm{OH})_2$
$$
\mathrm{Ba}(\mathrm{OH})_2 \text { left }=5 \text { meq. }
$$
Tatal volume $=50 \mathrm{~mL}$ $\mathrm{Ba}(\mathrm{OH})_2$ conc. in final solution
$$
=\frac{5}{50} \mathrm{~N}=0.1 \mathrm{~N}=0.05 \mathrm{M}
$$
$$
\left[\mathrm{OH}^{-}\right]=2 \times 0.05 \mathrm{M}=0.10 \mathrm{M}
$$
Alternatively,
$$
\mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_2+2 \mathrm{H}_2 \mathrm{O}
$$
$2 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{HCl}$ neutralize $1 \mathrm{~m}$ mole of $\mathrm{Ba}(\mathrm{OH})_2$
$1 \mathrm{~m} \mathrm{~mol}$ of $\mathrm{HCl}$ neutralize $0.5 \mathrm{~m} \mathrm{~mol} \mathrm{of}$ $\mathrm{Ba}(\mathrm{OH})_2$
$\mathrm{Ba}(\mathrm{OH})_2 1 \mathrm{eft}=3-0.5 \mathrm{~m} \mathrm{~mol}=2.5 \mathrm{mmol}$
$$
\left[\mathrm{Ba}(\mathrm{OH})_2\right]=\frac{2.5}{50} \mathrm{M}=0.05 \mathrm{M}
$$
or $[\mathrm{OH}]^{-}=2 \times 0.05=0.1 \mathrm{M}$

Topic: Chemistry, Structure of Atom, NEET, NEET 2009 (Mains)
What is the order of reactivity of these alkenes (CH3)2C=CH2(I),CH3CH=CH2(II) and CH2=CH2(III) when subject to acid-catalysed hydration?

Options:
A I > II > III
B I > III > II
C III > II > I
D II > I > III

The correct answer is:
I > II > III

Solution:
The order of acid catalysed hydration can be given on the basis of stability of carbocations formed as follows

Topic: Chemistry, Hydrocarbons, JEE Main
What is the outcome of bioprocess engineering?

Options:
A Altering the phenotype of host organisms
B Maintaining a sterile environment for microbe growth
C Large quantities of biotechnological product
D Correcting a defective gene

The correct answer is:
Large quantities of biotechnological product

Solution:
The outcome of bioprocess engineering is the large quantities of biotechnological products like antibiotics, vaccines, enzymes, etc.

Topic: Biology, Biotechnology - Principles and Processes, NEET
What is the outcome of meiosis in terms of chromosome number?

Options:
A It remains the same
B It increases
C It decreases
D It becomes zero

The correct answer is:
It decreases

Solution:
Meiosis results in reduction of chromosome number by half.

Topic: Biology, Cell Cycle and Cell Division, NEET
What is the outcome of reactions in which only small amounts of products are formed at equilibrium?

Options:
A The reactants are completely consumed.
B Most of the reactants remain unchanged.
C The concentrations of reactants and products are equal.
D The reaction is very fast.

The correct answer is:
Most of the reactants remain unchanged.

Solution:
In reactions where only small amounts of products are formed, most of the reactants remain unchanged at equilibrium.

Topic: Chemistry, Chemical Equilibrium, NEET
What is the oxidation number of $\mathrm{Co}$ in $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{C} / \mathrm{NO}_2\right]$

Options:
A $+2$
B $+3$
C $+4$
D $+5$

The correct answer is:
$+2$

Solution:
$\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{C} / \mathrm{NO}_2\right]$
$\begin{aligned} & x+4(0)+1(-1)+1(-1)=0 \\ & x-2=0 ; x=2\end{aligned}$

Topic: Chemistry, Redox Reactions, JEE Main
What is the pOH of 0.1MKB (salt of weak acid and strong base) at 25oC ? (Given: pKb of B-=7)

Options:

The correct answer is:
4

Solution:
For salt of weak acid and strong base;
pH=7+12pKa+12logC at 25o
=7+12×7+12log10-1
=7+3.5-0.5
∴pH=10
So, pOH=4 at 25o C

Topic: Chemistry, Ionic Equilibrium, JEE Main
What is the packet of energy called

Options:
A Electron
B Photon
C Positron
D Proton

The correct answer is:
Photon

Solution:
According to quantum theory of radiation, a hot body emits radiant energy not continuously but discontinuously in the form of small packets of energy called quanta or photons.

Topic: Chemistry, Structure of Atom, JEE Main
What is the particle $\mathrm{x}$ in the following nuclear reaction :
${ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+\mathrm{x}$

Options:
A electron
B proton
C Photon
D Neutron

The correct answer is:
Neutron

Solution:
${ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 \mathrm{X}$
Hence X represents neutron $\left({ }_0^1 n\right)$

Topic: Physics, Nuclear Physics, JEE Main
What is the path difference of destructive interference

Options:
A $n \lambda$
B $n(\lambda+1)$
C $\frac{(n+1) \lambda}{2}$
D $\frac{(2 n+1) \lambda}{2}$

The correct answer is:
$\frac{(2 n+1) \lambda}{2}$

Solution:
For destructive interference path difference is odd multiple of $\frac{\lambda}{2}$.

Topic: Physics, Wave Optics, JEE Main
What is the percentage error in the physical quantity A If it is related to four other physical quantities a, b, c and d as A=a2b3dc? The percentage error in measurement of a, b, c and d are1%, 3%, 2% and 1% respectively.

Options:

The correct answer is:
13.00

Solution:
Percentage error in A

=2Δaa+3Δbb+12Δcc+Δdd

=2×1+3×3+12×2+1×1%=13%

Topic: Physics, Experimental Physics, JEE Main
What is the percentage of nitrogen in the compound, if \( 1.4 \mathrm{~g} \) of an organic compound was digested according to Kjeldahl's method and the ammonia evolved was absorbed in \( 60 \mathrm{~mL} \) of \( \mathrm{M} / 10 \mathrm{H}_{2} \mathrm{SO}_{4} \) solution. The excess sulphuric acid required \( 20 \mathrm{~mL} \) of \( \mathrm{M} / 10 \mathrm{NaOH} \) solution for neutralization?

Options:
A \( 24 \)
B \( 3 \)
C \( 5 \)
D \( 10 \)

The correct answer is:
\( 10 \)

Solution:
60Ã110=6 mm H2SO4 used
Excess H2SO4â¡20Ã110Ã12=1 mm H2SO4
H2SO4 used =6-1=5 mm
2NH3+H2SO4â¶NH42SO4
mm of NH3=10 mm
Mass of N=10Ã10-3Ã14gmole=0.140g
% N2=0.1401.4Ã100=10%

Topic: Chemistry, General Organic Chemistry, JEE Main
What is the period for which pollen grains remain viable?

Options:
A 30 minutes
B Months
C Highly variable
D Years

The correct answer is:
Highly variable

Solution:
The period for which pollen grains remain viable is highly variable and depends on the prevailing temperature and humidity. Some cereal pollens lose viability within 30 minutes, while some members of other plant families maintain viability for months.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
What is the $\mathrm{pH}$ of $0.02 \mathrm{M} \mathrm{NaOH}$ solution?

Options:
A 10.3
B 11.3
C 11.7
D 12.3

The correct answer is:
12.3

Solution:
$$
[\mathrm{NaOH}]=\left[\mathrm{OH}^{-}\right]=2 \times 10^{-2} \mathrm{M}
$$
$\mathrm{NaOH}$ is strong Base.
$$
\begin{aligned}
& \mathrm{P}^{\mathrm{OH}}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(2 \times 10^{-2}\right) \\
& =2-\log 2 \\
& =2-0.3=1.7 \\
& \mathrm{P}^{\mathrm{H}}+\mathrm{P}^{\mathrm{OH}}=14 \\
& \mathrm{p}^{\mathrm{H}}=14-\mathrm{P}^{\mathrm{OH}}=14-1.7=12.3
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
What is the $\mathrm{pH}$ of a $10^{-4} \mathrm{M} \mathrm{OH}^{-}$solution at $330 \mathrm{~K}$, if $\mathrm{K}_{\mathrm{w}}$ at $330 \mathrm{~K}$ is $10^{-13.6}$ ?

Options:
A
4

B
$9.0$

C
10

D
$9.6$

The correct answer is:
$9.6$

Solution:
Given at $330 \mathrm{~K}$
$$
\begin{aligned}
\mathrm{K}_w & =10^{-13.6} \\
\text { i.e. } \quad \mathrm{pK}_w & =\mathrm{pH}+\mathrm{pOH} \\
\because \quad \mathrm{pOH}^2 & =-\log \left[\mathrm{OH}^{-}\right] \\
13.6 & =\mathrm{pH}+\mathrm{pOH} \\
\mathrm{pOH} & =-\log 10^{-4} \\
13.6 & =\mathrm{pH}+4 \\
\therefore \quad \mathrm{pH} & =13.6-4 \\
& =9.6
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2013 (23 Apr Online)
What is the $\mathrm{pH}$ of $\mathrm{Ba}(\mathrm{OH})_2$ if normality is 10

Options:
A 4
B 10
C 7
D 9

The correct answer is:
10

Solution:
$\mathrm{Ba}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}$
One molecule on dissociation furnishes $2 \mathrm{OH}^{-}$ions
$\begin{aligned} & \text { So, }\left[\mathrm{OH}^{-}\right]=2 \times 10^{-4} \mathrm{~N} \\ & N=M \times 2 ; M=\frac{N}{2}=\frac{2 \times 10^{-4}}{2}=10^{-4} \\ & p \mathrm{OH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(1 \times 10^{-4}\right)=-4 \\ & p H+p O H=14 ; p H=14-4=10 .\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
What is the pH of the intestinal juice?

Options:
A 1.8
B 6.8
C 7.8
D 9.8

The correct answer is:
7.8

Solution:
The intestinal juice or succus entericus has a pH of 7.8, which is neutral. This pH is maintained by the bicarbonates from the pancreas and the mucus that is secreted by the goblet cells, which protects the mucosa from acid. This pH is optimal for enzymatic activities that take place in the small intestine.

Topic: Biology, Digestion and Absorption, NEET
What is the phase difference between two successive crests in the wave

Options:
A $\pi$
B $\pi / 2$
C $2 \pi$
D $4 \pi$

The correct answer is:
$2 \pi$

Solution:
Since distance between two consecutive crests is $\lambda$, so
$\phi=\frac{2 \pi}{\lambda} \times \lambda=2 \pi$.

Topic: Physics, Waves and Sound, JEE Main
What is the possible product of the reaction in which 1 mole of HNO3 is reduced by absorption of 4 moles electrons?

Options:
A 0.5 mole of N2
B 0.5 mole of N2O
C 1 mole of NO3
D 1 mole of NH3

The correct answer is:
0.5 mole of N2O

Solution:
Here, when nitric acid is converted into any of the nitrogen compound, the oxidation state of nitrogen is changed and for oxygen and hydrogen, it remains unchanged.

Let the oxidation state of Nitrogen in HNO3 be x.

Then, +1+x+3-2=0⇒x-5=0⇒x=+5.

Now, after absorption of 4 moles of electrons, the oxidation state of nitrogen will become

N5++4e-→N+

Out of the given compounds, the compound having the oxidation state of nitrogen as +1 is N2O.

Thus, here nitric acid is getting converted into dinitrogen oxide.

1 mole of HNO3→givesx moles of N2O

Applying principle of atom conservation for nitrogen,

1×moles of HNO3=2×moles of N2O1×1=2xx=12=0.5 mol

Topic: Chemistry, Redox Reactions, NEET
What is the potential of a half-cell consisting of zinc electrode in $0.01 \mathrm{~m} \mathrm{ZnSO}_4$ solution at $258^{\circ} \mathrm{C}$
$\left(E^{\circ}=0.763 \mathrm{~V}\right)$

Options:
A $0.8221 \mathrm{~V}$
B $8.221 \mathrm{~V}$
C $0.5282 \mathrm{~V}$
D $9.232 \mathrm{~V}$

The correct answer is:
$0.8221 \mathrm{~V}$

Solution:
The half cell reaction is $\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$
The Nernst equation for the oxidation half- cell reaction is
$\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059 \mathrm{H}}{\mathrm{n}} \log \left[\mathrm{Zn}^{2+}\right]$
The number of electrons transferred $\mathrm{n}=2$ and $\mathrm{E}_0=0.763 \mathrm{~V}$
substituting these values in the Nernst equation
we have, $\mathrm{E}=0.763-\frac{0.059 \mathrm{H}}{2} \operatorname{og}[-2]$
$\mathrm{E}=0.793+0.0591=0.8221 \mathrm{~V}(0.8221 \mathrm{~V})$

Topic: Chemistry, Electrochemistry, JEE Main
What is the potential of an electrode which originally contained 0.1 M NO 3 - and 0.4 MH+ and which has been treated by 60% of the cadmium necessary to reduce all the NO 3 - to NO(g) at 1 atm.

Given, NO 3 - + 4 H + + 3 e - â NO + 2 H 2 O , E â = 0.95 V and log2 = 0.3010

Options:
A 0.52 V
B 0.44 V
C 0.86 V
D 0.78 V

The correct answer is:
0.86 V

Solution:
After addition of Cd it oxidises into Cd2+
NO 3 - aq + 4 H + aq + 3 e - → NO g + 2 H 2 O l 0.1 - x 0.4 - 4 x
X = 0.06
NO 3 - remaining = 0.1 - 0.06 ≈ 0.04 M
H + remaining = 0.4 - 4 × 0.06 = 0.4 - 0.24 = 0.16 M
E NO 3 - / NO = E NO 3 - / NO â - 0.591 3 log 1 NO 3 - H + 4
= 0.95 - 0.0591 3 log 1 0.04 0.16 4 = 0.95 - 0.0591 3 log 1 4 × 1 0 - 2 × 1 0 - 8 × 1 6 4
= 0.95 - 0.0591 3 log 1 4 × 1 0 - 2 × 1 0 - 8 × 2 + 1 6
= 0.95 - 0.0591 3 log 1 0 1 0 2 2 × 2 1 6
= 0.95 - 0.0591 3 1 0 - 1 8 × 0.3010 = 0.95 - 0.0591 3 × 4.582
= 0.95 - 0.09 = 0.86 V

Topic: Chemistry, Electrochemistry, NEET
What is the practical solution for the slow reaction rate at very low temperatures in ammonia production?

Options:
A Increasing the pressure
B Using a catalyst
C Reducing the pressure
D Adding more reactants

The correct answer is:
Using a catalyst

Solution:
To overcome the slow reaction rate at very low temperatures in ammonia production, a catalyst is used to speed up the reaction while maintaining a low temperature for a high yield of ammonia.

Topic: Chemistry, Chemical Equilibrium, NEET
What is the predominant intermolecular force of attraction between the adjacent chains of polymer molecules in natural
rubber ?

Options:
A H-bonds
B Dipole-Dipole attraction
C Van der Waal's force
D Ionic attraction

The correct answer is:
Van der Waal's force

Solution:
Natural rubber is a pure hydrocarbons therefore the only force of attraction between the adjacent chain is van der Waal's force.

Topic: Chemistry, Polymers, JEE Main
What is the primary characteristic feature of ice and water in equilibrium?

Options:
A Changing mass
B Changing temperature
C No activity at the boundary
D Constant mass and temperature

The correct answer is:
Constant mass and temperature

Solution:
In equilibrium, the mass of ice and water remains constant, and the temperature remains constant.

Topic: Chemistry, Chemical Equilibrium, NEET
What is the primary characteristic of a Schottky defect in ionic solids?

Options:
A It increases the density of the substance.
B It creates vacancies at interstitial sites.
C It changes the electrical conductivity of the crystal.
D It involves an equal number of missing cations and anions.

The correct answer is:
It involves an equal number of missing cations and anions.

Solution:
A Schottky defect in ionic solids involves an equal number of missing cations and anions to maintain electrical neutrality.

Topic: Chemistry, Solid State, NEET
What is the primary function of buffer solutions in clinical chemistry tests?

Options:
A To perform radiological imaging
B To regulate body temperature
C To stabilize pH during experiments
D To maintain constant pH for accurate test results

The correct answer is:
To maintain constant pH for accurate test results

Solution:
Although the paragraph doesn't explicitly mention clinical chemistry tests, the role of buffer solutions in maintaining a constant pH is essential for accurate test results in healthcare.

Topic: Chemistry, Ionic Equilibrium, NEET
What is the primary function of the corpus luteum during the menstrual cycle?

Options:
A Regulating the growth of primary follicles
B Releasing estrogens
C Secreting large amounts of progesterone
D Inducing ovulation

The correct answer is:
Secreting large amounts of progesterone

Solution:
During the luteal phase of the menstrual cycle, the remaining parts of the Graafian follicle transform into the corpus luteum which secretes large amounts of progesterone. This progesterone is essential for maintenance of the endometrium necessary for implantation of the fertilised ovum and other events of pregnancy.

Topic: Biology, Human Reproduction, NEET
What is the primary function of the duodenum?

Options:
A To absorb remaining nutrients from food
B To mix food with stomach acid and enzymes
C To neutralise stomach acid before it enters the small intestine
D To digest and absorb carbohydrates

The correct answer is:
To neutralise stomach acid before it enters the small intestine

Solution:
The duodenum is the first part of the small intestine. Its primary function is to neutralise the acidic chyme that enters from the stomach, by secreting bicarbonate and other alkaline substances, to create a neutral environment for enzymes to work efficiently.

Topic: Biology, Digestion and Absorption, NEET
What is the primary function of the excretory organs in animals?

Options:
A Elimination of nitrogenous wastes
B Maintenance of ionic balance of body fluids
C Both A and B
D None of the above

The correct answer is:
Both A and B

Solution:
The excretory organs in animals not only eliminate nitrogenous wastes but also help in the maintenance of ionic and acid-base balance of body fluids.

Topic: Biology, Excretory Products and their Elimination, NEET
What is the primary function of the jejunum?

Options:
A To absorb remaining nutrients from food
B To mix food with stomach acid and enzymes
C To neutralize stomach acid before it enters the small intestine
D To digest and absorb carbohydrates

The correct answer is:
To absorb remaining nutrients from food

Solution:
The jejunum is the middle section of the small intestine. Its primary function is to absorb remaining nutrients from food, such as carbohydrates, proteins, and fats, that have not been absorbed in the duodenum. This is accomplished by the presence of villi and microvilli, which increase the surface area for absorption.

Topic: Biology, Digestion and Absorption, NEET