datavorous/entrance-exam-dataset huggingface dataset:
The mean of the numbers $a, b, 8,5$ and 10 is 6 and the variance is 6.80 , then the possible values of $a$ and $b$ are

Options:
A $a=2, b=3$
B $a=4, b=5$
C $a=1, b=3$
D $a=3, b=4$

The correct answer is:
$a=3, b=4$

Solution:
Given data, $a, b, 8,5,10$
$\therefore \quad$ Mean $=\frac{a+b+8+5+10}{5}$
$6=\frac{a+b+23}{5}$
$\Rightarrow \quad a+b=7$ ...(i)
Variance
$=\frac{(a-6)^2+(b-6)^2+(8-6)^2+(5-6)^2+(10-6)^2}{5}$
$\Rightarrow 6.80=\frac{a^2+36-12 a+b^2+36-12 b+4+1+16}{5}$
$\begin{array}{ll}\Rightarrow & 34=a^2+b^2-12 a-12 b+93 \\ \Rightarrow & a^2+b^2-12(a+b)=-59\end{array}$
$\begin{array}{rrr}\Rightarrow & a^2+b^2-12(7) & =-59 \\ \Rightarrow & a^2+b^2 =25\end{array}$
$\Rightarrow \quad a^2+(7-a)^2=25$ [from Eq. (i), $b=7-a$ ]
$\begin{array}{ll}\Rightarrow & 2 a^2+49-14 a=25 \\ \Rightarrow & 2 a^2-14 a+24=0\end{array}$
$\Rightarrow \quad a^2-7 a+12=0$
$(a-4)(a-3)=0$
$\Rightarrow \quad a=4$ or $a=3$
If $a=3 \Rightarrow b=7-3=4$

Topic: Mathematics, Statistics, JEE Main
The mean square deviation of a set of observations x1,x2......xn about a point m is defined as 1nâi=1nxi-m2. If the mean square deviations about â1 and 1 of a set of observations are 7 and 3 respectively. The standard deviation of those observations is

Options:
A 2
B 2
C 5
D 3

The correct answer is:
3

Solution:
∑i=1nxi+12n=7(given)
⇒∑xi2+2∑xi+n=7n
⇒∑xi2+2∑xi=6n … (1)
Also, ∑xi-12n=2 (given)
⇒∑xi2-2∑xi+n=3n
⇒∑xi2-2∑xi=2n … (2)
From (1) and (2)
∑xi2=4n, ∑xi=n
Standard deviation=∑xi2n-∑xin2
=4nn-1=3

Topic: Mathematics, Statistics, JEE Main
The means of 5 observations is 4.4 and variance is 8.24 . If three of the five observations are 1,2 and 6 , then the values of other two observations are

Options:
A 5,7
B 4,9
C 3,9
D 4,8

The correct answer is:
4,9

Solution:
Let the two observation be $\mathrm{x}$ and $\mathrm{y}$
$$
\begin{aligned}
& 1+2+6+x+y=5 \times 4.4=22 \\
& \Rightarrow x+y=13
\end{aligned}
$$
Also $\frac{1^2+2^2+6^2+x^2+y^2}{5}-(4.4)^2=8.24$
$$
\begin{aligned}
& \Rightarrow \frac{41+x^2+y^2}{5}=8.24+19.36=27.60 \\
& \Rightarrow x^2+y^2=138-41 \\
& \Rightarrow x^2+(13-x)^2=97 \\
& \Rightarrow x^2-13 x+36=0 \\
& \Rightarrow x=4,9
\end{aligned}
$$
$\Rightarrow y=9,4$

Topic: Mathematics, Statistics, JEE Main
The means of two samples of size 40 and 50 were found to be 54 and 63 respectively. Their standard deviations were 6 and 9 respectively. The variance of the combined sample of size 90 is

Options:
A 90
B 7
C 9
D 81

The correct answer is:
81

Solution:
n1=40, n2=50
x¯1=54, x¯2=63
σ1=6, σ2=9
x¯=n1x¯1+n2x¯2n1+n2=40×54+50×6390=59
d1=x¯-x¯1=5
d2=x¯2-x¯=4
Combined S.D=n1d12+n2d22+n1σ12+n2σ22n1+n2
=40×25+50×16+40×36+50×8190
=9
Variance =81

Topic: Mathematics, Statistics, JEE Main
The _______ means the number of births per ________ in a particular area for a particular time.

Options:
A productivity, 100
B mortality, 1000
C natality, 1000
D natality, 100

The correct answer is:
natality, 1000

Solution:
Natality means the number of births per 1000 in a particular area for a particular time and mortality is the number of deaths per 1000 in a particular area for a particular time. In short, natality is birth rate and mortality is death rate.
If the natality and mortality remains the same then the population is neither growing nor reducing and it remains stable.
Declining population is the one where the old individuals outnumber young individuals and having more of death rate than birth rate.
Growing population is where the birth rate is higher than death rate.

Topic: Biology, Reproductive Health, NEET
The measure of dispersion is

Options:
A Mean deviation
B S.D.
C Quartile deviation
D All of these

The correct answer is:
All of these

Solution:
Measures of dispersion include:
1)Sample standard deviation
2)Interquartile range (IQR) or Interdecile range
3)Range
4)Mean difference
5)Median absolute deviation (MAD)
6)Average absolute deviation (or simply called average deviation)
7)Distance standard deviation

Topic: Mathematics, Statistics, JEE Main
The measured freezing point depression for a $0.1 \mathrm{m}$ aqueous $\mathrm{CH}_{3} \mathrm{COOH}$ solution is $0.19^{\circ} \mathrm{C}$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $=1.86 \mathrm{K}$ $\mathrm{kg} \mathrm{mol}^{-1}$ )

Options:
A $4.76 \times 10^{-5}$
B $4 \times 10^{-5}$
C $8 \times 10^{-5}$
D $2 \times 10^{-5}$

The correct answer is:
$4 \times 10^{-5}$

Solution:
$\begin{aligned} \Delta T_{f} &=i \times K_{t} \times m \\ \therefore & i=\frac{\Delta T_{f}}{K_{t} \times m}=\frac{0.19}{1.86 \times 0.1}=1.02 \end{aligned}$
Again from, $\alpha=\frac{i-1}{n-1}=\frac{1.02-1}{2-1}$ $=0.02=2.0 \times 10^{-2}$
$$
\begin{aligned}
\text { for } \mathrm{CH}_{3} \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+} \\
K \mathrm{a} &=\mathrm{C} \alpha^{2} \\
&=0.1 \times\left(2 \times 10^{-2}\right)^{2} \\
&=4 \times 10^{-5}
\end{aligned}
$$

Topic: Chemistry, Solutions, JEE Main
The measured voltage of the cell, Pt (s)  H 2 1.0 atm   H + (q)| Ag + 1.0 M  Ag (s) is 1.02 V at.298 K and Ecello is 0.80 V, Find the pH of the solution.
Report your answer by rounding it upto nearest whole number.

Options:

The correct answer is:
4

Solution:
E cell  =  E cell o − 0.059 n  log  [H + ] 2 [Ag + ] 2
=  E cell o − 0.059 × 2 2  log  H + Ag +
E cell o −0.059 × 2 log  H + 1
= E cell o +0.059  (− log[H + ]) 1
= E cell o +0.059 pH
So pH =  E cell − E cell o 0.059  =  1.02−0.80 0.059  = 3.728

Topic: Chemistry, Electrochemistry, JEE Main
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1}$. The uncertainty in electron velocity is, (mass of an electron is $9 \times 10^{-28} \mathrm{~g}$ )

Options:
A $1 \times 10^{11} \mathrm{~cm} \mathrm{~s}^{-1}$
B $1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}$
C $1 \times 10^6 \mathrm{~cm} \mathrm{~s}^{-1}$
D $1 \times 10^5 \mathrm{~cm} \mathrm{~s}^{-1}$

The correct answer is:
$1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}$

Solution:
$$
\begin{aligned}
& \Delta p=1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1} \\
& \text { or } \Delta p=\mathrm{m} \Delta \mathrm{v} \\
& \text { or } \Delta v=\frac{\Delta p}{\mathrm{~m}}=\frac{1 \times 10^{-18}}{9 \times 10^{-28}} \simeq 1 \times 10^9 \mathrm{~cm} / \text { second }
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1}$. The uncertainty in electron velocity is, (mass of an electron is $9 \times 10^{-28} \mathrm{~g}$ )

Options:
A $1 \times 10^{11} \mathrm{~cm} \mathrm{~s}^{-1}$
B $1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}$
C $1 \times 10^6 \mathrm{~cm} \mathrm{~s}^{-1}$
D $1 \times 10^5 \mathrm{~cm} \mathrm{~s}^{-1}$

The correct answer is:
$1 \times 10^9 \mathrm{~cm} \mathrm{~s}^{-1}$

Solution:
$$
\begin{aligned}
& \Delta p=1 \times 10^{-18} \mathrm{~g} \mathrm{~cm} \mathrm{~s}^{-1} \\
& \text { or } \Delta p=\mathrm{m} \Delta \mathrm{v} \\
& \text { or } \Delta v=\frac{\Delta p}{\mathrm{~m}}=\frac{1 \times 10^{-18}}{9 \times 10^{-28}} \simeq 1 \times 10^9 \mathrm{~cm} / \text { second }
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, NEET, NEET 2008 (Mains)
The mechanism of action of "Terfenadine" (Seldane) is :

Options:
A Activates the histamine receptor
B Inhibits the secretion of histamine
C Helps in the secretion of histamine
D Inhibits the action of histamine receptor

The correct answer is:
Inhibits the action of histamine receptor

Solution:
Seldane act as antihistamines and interfere with the natural action of histamine by competing with histamine for binding sites of receptor.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The mechanism of Sn1 reaction is given as:

R-X⟶R⊕X⊖⟶R⊕ ‖ X⊖⟶Y⊖R-Y+X⊖ lon pair Solvent Speparated ion pair

A student writes general characteristics based on the given mechanism as :
(a) The reaction is favoured by weak nucleophiles.
(b) R⊕ would be easily formed if the substituents are bulky y
(c) The reaction is accompanied by recemization
(d) The reaction is favoured by non-polar solvents.

Which observations are correct?

Options:
A (a) and (b)
B (a) and (c)
C (a), (b) and (c)
D (b) and (d)

The correct answer is:
(a), (b) and (c)

Solution:
Above reaction isSN1 reaction as it proceed via formation of carbocation. Polar protic solvent is more suitable for SN1 and recemisation takes place.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The median of 100 observations grouped in classes of equal width is 25 . If the median class interval is 20-30 and the number of observations less than 20 is 45 , then the frequency of median class is

Options:
A
10

B
20

C
15

D
12

The correct answer is:
10

Solution:
Median is given as
$$
M=l+\frac{\frac{N}{2}-F}{f} \times C
$$
where
$l=$ lower limit of the median - class
$f=$ frequency of the median class
$N=$ total frequency
$F=$ cumulative frequency of the class just before the median class
$C=$ length of median class
Now, given, $M=25, N=100, F=45$, $C=20-30=10, l=20$.
$\therefore$ By using formula, we have
$$
\begin{aligned}
& 25=20+\frac{50-45}{f} \times 10 \\
& 25-20=\frac{50}{f} \Rightarrow 5=\frac{50}{f} \Rightarrow f=10
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main, JEE Main 2012 (19 May Online)
The median of a set of 9 distinct observations is 20.5 . If each of the largest 4 observation of the set is increased by 2 , then the median of the new set

Options:
A Is increased by 2
B Is decreased by 2
C Is two times the original median
D Remains the same as that of the original set

The correct answer is:
Remains the same as that of the original set

Solution:
Since $n=9$, then median term $=\left(\frac{9+1}{2}\right)^{\text {th }}=5^{\text {th }}$ term .
Now, last four observations are increased by 2 .
$\because$ The median is $5^{\text {th }}$ observation, which is remaining unchanged.
$\therefore$ There will be no change in median.

Topic: Mathematics, Statistics, JEE Main
The median of a set of 9 distinct observations is $20.5$. If each of the largest 4 observations of the set is increased by 2 , then median of the new set

Options:
A
remains the same as that of the original set

B
is increased by 2

C
is decreased by 2

D
is two times the original median

The correct answer is:
remains the same as that of the original set

Solution:
$\mathrm{n}=9$ then median term $=\left(\frac{9+1}{2}\right)^{\text {th }}=5^{\text {th }}$ term. Last four observations are increased by 2 . The median is 5 th observation which is remaining unchanged. $\therefore$ There will be no change in median.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2003
The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 4, then the median of the new set

Options:
A is increased by 4
B is decreased by 4
C is two times the original median
D remains the same as that of the original set

The correct answer is:
remains the same as that of the original set

Solution:
If x1,x2,x3â¦,x9 are the observations such that, xi+1â¥xi i â1,2,â¦8
Median =x5
âµx6,x7,x8,x9 are increased by 4 hence, median x5 remains the same.

Topic: Mathematics, Statistics, JEE Main
The median of all 4- digit numbers that are divisible by 7 is

Options:
A 5797
B 5498.5
C 5499.5
D 5490

The correct answer is:
5498.5

Solution:
Four digits number which is

divisible by 7 are 1001,1008,1015,… 9996

Hence, total number of such numbers
=1286

N2thobservation

Median =+N2+1th observation 2

[∵N is even ]

12862th observation

Median =+12862+1th observation 2

=643th+644th2

=(1001+(642)7)+(1001)+(643)7)2

=2(1001)+7(642+643)2

=2(1001)+7(1285)2

=1001+4497.5=5498.5

Topic: Mathematics, Statistics, JEE Main
The melting of ice

Options:
A is the phase transformation
B takes place at constant pressure and temperature
C takes place at $373 \mathrm{~K}$
D Both (a) and (b)

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

Solution:
Phase transformation involve the energy changes e.g., ice requires heat for melting. Normally this meltir: trkes place at constant pressure and during phase change, temperature remains constant (at $273 \mathrm{~K}$ ).

Topic: Chemistry, Thermodynamics (C), COMEDK, COMEDK 2021
The metal atom present in the complex MABXL (where A, B, X and L are unidentate ligands and \(\mathrm{M}\) is metal) involves \(\mathrm{sp}^3\) hybridization. The number of geometrical isomers exhibited by the complex is:

Options:
A 2
B 0
C 4
D 3

The correct answer is:
0

Solution:
Tetrahedral complex does not show geometrical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The metal extracted by leaching with a cyanide is

Options:
A
Mg

B
Ag

C
Cu

D
Na

The correct answer is:
Ag

Solution:
Silver ore forms a soluble complex with NaCN from which silver is precipitated using scrap zinc.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2002
The metal M crystallizes in a body cantered lattice with cell edge 400 pm . The atomic radius of M is

Options:

The correct answer is:
173

Solution:
a = 400 pm
In body central lattice 3a=4r
r=3a4
r=34Ã400
r=3Ã100=173 pm

Topic: Chemistry, Solid State, JEE Main
The metal mainly used in devising photoelectric cells is :

Options:
A Na
B Li
C Rb
D Cs

The correct answer is:
Cs

Solution:
Cesium has lowest ionisation enthalpy and hence it can show photoelectric effect to the maximum extent hence it is used in photo electric cell.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The metal showing a resemblance with lithium is:

Options:
A \( \mathrm{Be} \)
B \( \mathrm{Mg} \)
C \( \mathrm{Al} \)
D \( \mathrm{Na} \)

The correct answer is:
\( \mathrm{Mg} \)

Solution:
The diagonally adjacent elements of period 2 and 3 show similarity in their properties, which is known as diagonal relationship. This is accounted to the similarity in the ionic radii of these elements or their charge to size ratio. Examples of such elements are: Li and Mg; Be and Al.

In Mg, there is an increase in the number of shells, along with the completely filled valence shell, due to which its size is comparable to that of Li. Some of the similarities between Li and Mg are:

Both Li and Mg form covalent organometallic compounds.

Their hardness is higher than the other elements in their respective groups.

They react with nitrogen to form their respective nitrides.

On heating their carbonates, they yield their respective oxides along with the liberation of carbon dioxide.

Topic: Chemistry, s Block Elements, JEE Main
The metal that can be purified economically by fractional distillation method is:

Options:
A Fe
B Zn
C Cu
D Ni

The correct answer is:
Zn

Solution:
Zinc can be purified economically by fractional distillation because of low melting point.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The metal that cannot be obtained by the electrolysis of an aqueous solution of its salt is

Options:
A Ag
B Ca
C Cu
D Cr

The correct answer is:
Ca

Solution:
For the reduction of H2O at cathode,
2H2O+2e-→H2g+2OH-; Ered∘=-1.23V
For Ca2++2e-→Ca; Ered∘=-2.87V
i.e., water will be reduced at the cathode, in preference to the liberation of Ca.
However, in the case of Ag, Cu and Cr, their standard reduction potentials are higher than -1.23 V and will be liberated/ deposited at the cathode and H2O will not be reduced.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2014 (06 Apr)
The metal that does not give the borax-bead test is

Options:
A Chromium
B Nickel
C Lead
D Manganese

The correct answer is:
Lead

Solution:
Borax bead test is generally given by transition elements

Topic: Chemistry, Practical Chemistry, JEE Main
The metal that forms nitride by reacting directly with N2 of air is:

Options:
A Li
B Rb
C Cs
D K

The correct answer is:
Li

Solution:
Among the alkali metals, only lithium is able to form a stable nitride.
6Li+N2→2Li3N
Lithium nitride

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The metal that gives hydrogen gas upon treatment with both the acid as well as the base is

Options:
A Zinc
B Magnesium
C Iron
D Mercury

The correct answer is:
Zinc

Solution:
(i) Mg+HCl→MgCl2+H2↑ Mg+NaOH→OH2+Na
(ii) Fe+HCl→FeCl3+H2↑ Fe+2OH-→FeOH2+2Na
(iii) Zn is amphoteric in nature, so it reacts with acid as well as base.
Zn+HCl→ZnCl2+12H2g↑Zn+NaOH→ZnO2-2+H2g↑
(iv)
Hg+NaOH→Hg+HgO+H2OHg+2HCl→Hg2Cl2+H2(g)

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The metal that has very low melting point and its periodic position is closer to a metalloid is

Options:
A Al
B Ga
C Se
D In

The correct answer is:
Ga

Solution:
The expected order of melting points of group 13 elements is B>Al>Ga>In>Tl.

But due to structural changes, melting point instead of decreasing, increases from Ga to Tl and Ga has the lowest melting point. Hence, the actual order is B>Al>Tl>In>Ga

Melting point

Al→933 K

Ga→303 K

In→430 K

Se→490 K

Ga has low meting point and it is closer to metalloid in periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The metal that is not extracted from its sulphide ore is

Options:
A Aluminium
B Iron
C Lead
D Zinc

The correct answer is:
Aluminium

Solution:
The sulphide ore of iron is iron pyrite.

Iron can be extracted from iron pyrites

Galena is the sulphide ore of lead (PbS)

The sulphide containing ore of zinc is zinc blende (ZnS).

Al is extracted from Al2O3·2H2O i.e., Bauxite ore

Hence, Iron, lead and zinc can be extracted from their sulphide ores but not aluminium.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The metal used for making X-ray tube window is:

Options:
A Mg
B Na
C Be
D Ca

The correct answer is:
Be

Solution:
Many metals can be used to make the anode of a X-ray tube, but only one metal is most suitable for the windows. It is beryllium because beryllium has lowest electron density outside its nucleus, and will absorb least X-ray.

Beryllium is used in making windows in X-ray tubes.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The metal used for making X-ray tube window is

Options:
A \( M g \)
B \( N a \)
C \( B e \)
D \( \mathrm{Ca} \)

The correct answer is:
\( B e \)

Solution:
Many metals can be used to make the anode of a X-ray tube, but only one metal is most suitable for the windows. It is beryllium because beryllium has lowest electron density outside its nucleus, and will absorb least X-ray.

Beryllium is used in making windows in X-ray tubes.

Topic: Physics, Electromagnetic Waves, JEE Main
The metal which is extracted by oxidation and subsequent reduction from its ore is :

Options:
A Al
B Ag
C Cu
D Fe

The correct answer is:
Ag

Solution:
Cyanide process, also called Macarthur-forrest Process, method of extracting silver and gold from their ores by dissolving them in a dilute solution of sodium cyanide or potassium cyanide. The following redox reaction takes place during the extraction process.

4Ag+8CN-1+O2+2H2O→4[Ag(CN)2 ]+4OH-

2AgCN2-1+Zn→2Ag↓+ZnCN4-2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes 5x s. The value of x will be _____ .

Options:

The correct answer is:
5

Solution:
When the bob is immersed in water its effective weight is mg'=mg-FB, where, FB is buoyant force.

Then, g'=mg-FBm =ρBVg-ρWVgρBV

=ρB-ρwρBg =5-15×g=45g

Now, time period of simple pendulum is given by T=2πlg

Thus, the ratio T'T=gg'=g45g=54

⇒T'=T54=1025

Thus, the new time period is T'=55 s.

Hence, value of x=5.

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The metal's d- orbitals that are directly facing the ligands in K3CoCN6 are:

Options:
A dxy,dxz and dyz
B dx2-y2 and dz2
C dxy and dx2-y2
D dxz,dyz and dz2

The correct answer is:
dx2-y2 and dz2

Solution:
Give K3[CoCN6] is inner orbital complex with hybridization d2sp3 and octahedral geometry. Ligands are approaching metal along the axes. Hence dx2-y2,dz2 orbitals are directly in front of the ligands.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The metals that are employed in the battery industries are A. Fe, B. Mn, C. Ni, D. Cr, E. Cd

Choose the correct answer from the options given below:

Options:
A B, C and E only
B A, B, C, D and E
C A, B, C and D only
D B, D and E only

The correct answer is:
B, C and E only

Solution:
A rechargeable nickel-cadmium cell in a jelly roll arrangement and separated by a layer soaked in moist sodium or potassium hydroxide. Another important secondary cell is the nickel-cadmium cell which has longer life than the lead storage cell but more expensive to manufacture. We shall not go into details of working of the cell and the electrode reactions during charging and discharging. The overall reaction during discharge is.

Cds+2NiOH3→CdOs+2NiOH2s+H2Ol

The primary cell Leclanché cell cell is represented as

Zn|Zn2+∥MnO2|Mn2O3|C

Mn, Ni and Cd metals used in battery industries.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The meter bridge shown in the balance position with $\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{l}_1}{\mathrm{l}_2}$. If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, that will be balanced condition?

Options:
A yes, $\frac{P}{Q}=\frac{I_2-I_1}{I_2+I_1}$
B no, no null point
C yes, $\frac{P}{Q}=\frac{I_2}{I_1}$
D yes, $\frac{P}{Q}=\frac{I_1}{I_2}$

The correct answer is:
yes, $\frac{P}{Q}=\frac{I_1}{I_2}$

Solution:
$$
\text { For balanced position in a meterbridge }
$$
$$
\text { Now, if position of } \mathrm{G} \text { and cell is interchanged, }
$$
The balance condition still remains the same if the jockey points as the same point as given in the initial condition, for which there is no deflection in the galvanometer or no current will be drawn from the cell. Thus, the bridge will work as usual and balance condition is same, $P / Q=l_1 / l_2$

Topic: Physics, Current Electricity, NEET, NEET 2019 (Odisha)
The method of zone refining of metals is based on the principle of

Options:
A greater mobility of the pure metal than that of impurity.
B greater solubility of the impurity in the molten state than in the solid.
C higher melting point of the impurity than that of the pure metal.
D All above the correct

The correct answer is:
greater solubility of the impurity in the molten state than in the solid.

Solution:
A movable heater is fitted around a rod of impure metal. As heater is moved from one end to another pure metal crystallises while impurities pass on to adjacent melted zone.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The method used for the purification of Indium is:

Options:
A vapour phase refining
B liquation
C zone refining
D van Arkel method

The correct answer is:
zone refining

Solution:
Zone refining is used for the purification of indium.

Zone refining is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.

This method is very useful for producing semiconductor and other metals of very high purity, e.g., germanium, silicon, boron, gallium.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (25 Feb Shift 2)