datavorous/entrance-exam-dataset huggingface dataset:
In the electrochemical cell
Zn|ZnSO4(0.01M)CuSO4(1.0M)|Cu, the emf of this Daniel cell is E1 . When the concentration of ZnSO4 is changed to 1.0M and that of CuSO4 changed to 0.01M, the emf changes to E2 . From the followings, which one is the relationship between E1 and E2 ?
(Given, RTF=0.059 )

Options:
A E1<E2
B E1>E2
C E2=0≠E1
D E1=E2

The correct answer is:
E1>E2

Solution:
E  =E o − 0.059 2 log 10 Zn 2+ Cu 2+
E 1 =EÂ°â 0.059 2 log 10 0.01 (1) 2
E 2 =EÂ°â 0.059 2 log 10 1 (0.01) 2
Hence E1>E2

Topic: Chemistry, Electrochemistry, JEE Main
In the electrolysis of alumina, cryolite is added to

Options:
A lower the melting point of alumina.
B Minimise the anodic effect
C Remove impurities from alumina
D Increase the electrical conductivity

The correct answer is:
lower the melting point of alumina.

Solution:
The electrolysis of pure alumina faces some difficulties. Pure alumina is a bad conductor of electricity. The fusion temperature of pure alumina is about $2000^{\circ} \mathrm{C}$ and at this temperature when electrolysis is carried out on the fused mass, the metal formed vaporizes, as the boiling point of aluminium is $1800^{\circ} \mathrm{C}$. These difficulties are overcome by using a mixture containing alumina, cryolite $\left(\mathrm{Na}_3, \mathrm{AlF}_6\right)$, and fluorspar $\left(\mathrm{CaF}_2\right.$, ). But the addition of cryolite do not minimize the anode effect and do not remove the impurities from alumina.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the electrolysis of alumina to obtain aluminium metal, cryolite is added mainly to

Options:
A
lower the melting point of alumina

B
dissolve alumina in molten cryolite

C
remove the impurities of alumina

D
increase the electrical conductivity

The correct answer is:
lower the melting point of alumina

Solution:
Fused alumina $\left(\mathrm{Al}_2 \mathrm{O}_3\right)$ is a bad conductor of electricity. Therefore, cryolite $\left(\mathrm{Na}_3 \mathrm{AlF}_6\right)$ and fluorspar $\left(\mathrm{CaF}_2\right)$ are added to purified alumina which not only make alumina a good conductor of electricity but also reduce the melting point of the mixture to around $1140 \mathrm{~K}$.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2012 (12 May Online)
In the electrolytic extraction of aluminium, cryolite is used

Options:
A To obtain more aluminium
B To decrease temperature to dissolve bauxite
C To protect the anode
D As reducing agent

The correct answer is:
To decrease temperature to dissolve bauxite

Solution:
Cryolite $N a_3 A l F_6$ is added
(1)To decrease the melting temp
(2) To increase the electrical conductivity of solution

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the electrolytic refining of blister copper, the total number of main impurities, from the following, removed as anode mud is ___________ .
Pb, Sb, Se, Te, Ru, Ag, Au and Pt

Options:

The correct answer is:
6

Solution:
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (31 Aug Shift 2)
In the electrolytic refining of zinc

Options:
A Graphite is at the anode.
B The impure metal is at the cathode.
C The metal ion get reduced at the anode.
D Acidified zinc sulphate is the electrolyte.

The correct answer is:
Acidified zinc sulphate is the electrolyte.

Solution:
In the electrolytic refining of zinc, anode is made up of impure zinc while a strip of pure zinc acts as cathode. An acidified solution of zinc sulphate acts as electrolyte. When electricity is passed, following reactions occur.
At cathode
Zn 2+ +2e â â Zn (pure)
At anode
Zn (impure) â Zn 2+ â+â 2e â

Topic: Chemistry, d and f Block Elements, JEE Main
In the electron transport system present in the inner mitochondrial membrane, complexes I and IV are respectively ___________.

Options:
A NADH dehydrogenase and FADH2
B FADH2 and NADH dehydrogenase
C NADH dehydrogenase and cytochrome oxidase complex
D NADH dehydrogenase and ATP synthase

The correct answer is:
NADH dehydrogenase and cytochrome oxidase complex

Solution:
The series of redox reactions taking place between compounds involving electron transfer from donors to acceptors is known as the electron transport chain. In complex I (NADH dehydrogenase) which is the main site of production of superoxide where the NADH loses two electrons which are transferred to ubiquinone. In complex IV (cytochrome oxidase) two molecules of water are produced due to the transfer of electrons from four molecules of cytochrome C to the molecule oxygen.

Topic: Biology, Respiration in Plants, NEET
In the ellipse, minor axis is 8 and eccentricity is $\frac{\sqrt{5}}{3}$. Then major axis is

Options:
A 6
B 12
C 10
D 16

The correct answer is:
12

Solution:
Given, minor axis of ellipse $(2 b)=8$ or $b=4$ and eccentricity $(e)=\sqrt{5} / 3$. We know that in an ellipse,
$e^2=1-\frac{b^2}{a^2} \quad$ or $\quad \frac{5}{9}=1-\frac{16}{a^2} \quad$ or $\quad \frac{16}{a^2}=1-\frac{5}{9} \quad$ or $a^2=\frac{16 \times 9}{4}=36$ or $a=6$. We also know that major axis of the ellipse $=2 a=2 \times 6=12$.

Topic: Mathematics, Ellipse, JEE Main
In the ellipse $x^{\wedge} 2 \div a^{\wedge} 2+y^{\wedge} 2 \div b^{\wedge} 2=1 y=x$ and $3 y+2 x=0$ are a pair of conjugate diameters then the eccentricity of the ellipse is

Options:
A $1 / \sqrt{ 3}$
B $\sqrt{ 3}$
C $1 / \sqrt{ 5}$
D $\sqrt{ 5}$

The correct answer is:
$1 / \sqrt{ 3}$

Solution:
$S: \frac{x^4}{a^2}+\frac{y^2}{b^2}=1$
Since, $y=x$ and $3 y=-2 x$ is a pair of conjugate diameters therefore:
$\begin{aligned} & m_1 m_2=-\frac{b^2}{a^2} \\ & (1)\left(-\frac{2}{3}\right)=-\frac{b^2}{a^2} \\ & \frac{b^2}{a^2}=\frac{2}{3} \\ & e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
In the equation, GPP-R=NPP, R represents:

Options:
A Radiant energy
B Retardation factor
C Environment factor
D Respiration losses

The correct answer is:
Respiration losses

Solution:
The gross primary productivity (GPP) refers to the rate of energy stored by photosynthesis in plants. The R refers to the maintenance and reproduction of plants from the energy expended. GPP minus respiration losses (R) is the net primary productivity (NPP). GPP -R= NPP.

Topic: Biology, Ecosystem, NEET, NEET 2021
In the equation
$\mathrm{H}_2 \mathrm{~S}+2 \mathrm{HNO}_3 \rightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}_2+\mathrm{S}$
The equivalent weight of hydrogen sulphide is

Options:
A $16$
B $68$
C $34$
D $17$

The correct answer is:
$17$

Solution:
$\begin{aligned} & \mathrm{H}_2 \mathrm{~S} \rightarrow \stackrel{0}{\mathrm{~S}}+2 e \\ & \text { Equivalent wt. }=\frac{\text { Mol. wt. }}{2}=\frac{34}{2}=17\end{aligned}$

Topic: Chemistry, Redox Reactions, JEE Main
In the equation of state of an ideal gas $P V=n R T$, the value of the universal gas constant would depend only on

Options:
A The nature of the gas
B The pressure of the gas
C The units of the measurement
D None of these

The correct answer is:
The units of the measurement

Solution:
The units of the measurement

Topic: Chemistry, States of Matter, JEE Main
In the equation $P V=n R T$, which one cannot be the numerical value of $R$

Options:
A $8.31 \times 10^7 \mathrm{erg} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$
B $8.31 \times 10^7$ dyne $\mathrm{cmK}^{-1} \mathrm{~mol}^{-1}$
C $8.31 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
D $8.31 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

The correct answer is:
$8.31 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Solution:
$\left(\mathrm{atm} . \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$ is not a unit of $R$

Topic: Chemistry, States of Matter, JEE Main
In the equation X+aY2[Y-b]=RT, X is pressure, Y is volume, R is universal gas constant and T is temperature. The physical quantity equivalent to the ratio ab is:

Options:
A Pressure gradient
B Energy
C Impulse
D Coefficient of viscosity

The correct answer is:
Energy

Solution:
X & aY2 have the same dimensions. Y & b also have the same dimensions.

Therefore,

aY2=ML-1T-2⇒a=ML-1T-2×L6=ML5T-2

and b=L3.

So, the dimensions of ab is,

ab=ML5T-2L3=ML2T-2.

This is the dimensional form of energy. Hence, this is the right option.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2023 (13 Apr Shift 2)
In the equilateral triangle ABC, the equation of the side BC is x+y-2=0 and the centroid of âABC is 0,0. If the points A,B,C are in anticlockwise order, then the midpoint of the line segment joining A and C is

Options:
A -3+12,3+12
B -3-12,3-12
C -3-12,3+12
D -3+12,3-12

The correct answer is:
-3-12,3-12

Solution:
Foot of 0,0 on BC is D1,1âOD=2
â Slope of OD is 1
âOE makes an angle 45o+120o=165o with the positive x-axis
âE=2cosâ¡165o,2sinâ¡165o=-3-12,3-12

Topic: Mathematics, Straight Lines, JEE Main
In the equilibrium, $A B \rightleftharpoons A+B$, if the equilibrium concentration of $A$ is double, then equilibrium concentration of $B$ will be

Options:
A half
B twice
C $\frac{1}{4}$ th
D $\frac{1}{8}$ th

The correct answer is:
half

Solution:
$A B \rightleftharpoons A+B$
or $K=\frac{[A][B]}{[A B]}$
if concentration of $A$ is doubled, the equilibrium concentration of $B$ becomes half to maintain $K$ constant.

Topic: Chemistry, Chemical Equilibrium, JEE Main
In the equilibrium, $A B \rightleftharpoons A+B$, if the equilibrium concentration of $A$ is double, then equilibrium concentration of $B$ will be

Options:
A half
B twice
C $\frac{1}{4}$ th
D $\frac{1}{8}$ th

The correct answer is:
half

Solution:
$A B \rightleftharpoons A+B$
or $K=\frac{[A][B]}{[A B]}$
if concentration of $A$ is doubled, the equilibrium concentration of $B$ becomes half to maintain $K$ constant.

Topic: Chemistry, Chemical Equilibrium, COMEDK, COMEDK 2021
In the equilibrium, $\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI},$ if at a given temperature the concentrations of the reactants are increased, the value of the cquilibrium constant, $K_{C}$, will

Options:
A increase
B decrease
C remain the same
D cannot be predicted with certainty

The correct answer is:
remain the same

Solution:
In the equilibrium $\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}$, if at a
given temperature, the concentrations of the reactants are increased, the value of the equilibrium constant $K_{C}$ will remain the same because equilibrium constant does not depend on the molar concentration of reactants.
The option (c) is correct answer.

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2019
In the esterification C2H5OHl+CH3COOH lâCH3COOC2H5l+H2Ol an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction = 0.333. The equilibrium constant is.

Options:

The correct answer is:
4

Solution:
As alcohol and acid are equimolar so the initial concentration would also be same. Suppose the initial concentration is 'C'.
xC-x+C-x+x+x=0.33=13
3x=2C
x=2C3
Kc=2C3×2C3C-2C3×C-2C3=4

Topic: Chemistry, Chemical Equilibrium, JEE Main
In the esterification reaction of alcohols

Options:
A $\mathrm{OH}^{-}$is replaced by $\mathrm{C}_6 \mathrm{H}_3 \mathrm{OH}$
B $\mathrm{H}^{+}$is replaced by sodium metal
C $\mathrm{OH}^{-}$is replaced by chlorine
D $\mathrm{OH}^{-}$is replaced by $\mathrm{CH}_3 \mathrm{COO}^{-}$group

The correct answer is:
$\mathrm{OH}^{-}$is replaced by $\mathrm{CH}_3 \mathrm{COO}^{-}$group

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the estimation of bromine, 0.5 g of an organic compound gave 0.40 g of silver bromide. The percentage of bromine in the given compound is % (nearest integer)
(Relative atomic masses of Ag and Br are 108u and 80u, respectively).

Options:

The correct answer is:
34

Solution:
The estimation of halogens in organic compounds is usually done by carius method.

Percentage of Bromine =Atomic mass of BrMolecular mass of AgBr×Mass of AgBr×100Mass of the organic compound

=80×0.4×100188×0.5

=34%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (28 Jun Shift 1)
In the expansion of $\left(1+3 x+2 x^2\right)^6$ the coefficient of $x^{11}$ is

Options:
A $144$
B $288$
C $216$
D $576$

The correct answer is:
$576$

Solution:
$\left(1+3 x+2 x^2\right)^6=[1+x(3+2 x)]^6$
$=1+{ }^6 C_1 x(3+2 x)+{ }^6 C_2 x^2(3+2 x)^2$ $+{ }^6 C_3 x^3(3+2 x)^3+{ }^6 C_4 x^4(3+2 x)^4$ $+{ }^6 C_5 x^5(3+2 x)^5+{ }^6 C_6 x^6(3+2 x)^6$
Only $x^{11}$ gets from ${ }^6 C_6 x^6(3+2 x)^6$
${ }^6 C_6 x^6(3+2 x)^6=x^6(3+2 x)^6$
$\therefore \text { Coefficient of } x^{11}={ }^6 C_5 3.2^5=576$

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $\left(1+3 x+3 x^2+x^3\right)^{2 n}$, the term which has greatest binomial coefficient, is

Options:
A $(3 n)$ th term
B $(3 n+1)$ th term
C $(3 n-1)$ th term
D $(3 n+2)$ th term

The correct answer is:
$(3 n+1)$ th term

Solution:
$\because$ Middle term has greatest binomial coefficient.
In the expansion of $\left(1+3 x+3 x^2+x^3\right)^{2 n}$
$=\left((1+x)^3\right)^{2 n}=(1+x)^{6 n}$
$\because 6 n$ is even
So, middle term of $(1+x)^{6 n}=T\left(\frac{6 n}{2}+1\right)$ $=T_{(3 n+1)}=(3 n+1)$ th term.

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $(1+x)^5$, the sum of the coefficient of the terms is

Options:
A $80$
B $16$
C $32$
D $64$

The correct answer is:
$32$

Solution:
Sum of the coefficients $=(1+1)^5=2^5=32$.

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $(1+x)^{50}$, the sum of the coefficient of odd powers of $x$ is

Options:
A $0$
B $2^{49}$
C $2^{50}$
D $2^{51}$

The correct answer is:
$2^{49}$

Solution:
We have $(1+x)^{50}=\sum_{r=0}^{50}{ }^{50} C_r x^r$
Therefore, sum of coefficients of odd power of
$\begin{aligned} x & ={ }^{50} C_1+{ }^{50} C_3+\ldots+{ }^{50} C_{49} \\
& =\frac{1}{2}\left[{ }^{50} C_0+{ }^{50} C_1+\ldots+{ }^{50} C_{50}\right]=\frac{1}{2}\left[2^{50}\right]=2^{49}\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $\left(1+x+x^3+x^4\right)^{10}$, the coefficient of $x^4$ is

Options:
A ${ }^{40} \mathrm{C}_4$
B ${ }^{10} C_4$
C $210$
D $310$

The correct answer is:
$310$

Solution:
$\begin{aligned} & \left(1+x+x^3+x^4\right)^{10}=(1+x)^{10}\left(1+x^3\right)^{10} \\ & =\left(1+{ }^{10} C_1 \cdot x+{ }^{10} C_2 \cdot x^2+\ldots . .\right)\left(1+{ }^{10} C_1 \cdot x^3+{ }^{10} C_2 \cdot x^6+\ldots . .\right) \\ & \therefore \text { Coefficient of } x^4={ }^{10} C_1 \cdot{ }^{10} C_1+{ }^{10} C_4=310 .\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of a+bn, first three terms are 243,810 and 1080 respectively, then the fourth term of the expansion is n∈N

Options:
A 32
B 720
C 510
D 420

The correct answer is:
720

Solution:
T1=an,T2=nan-1b,T3=nn-12an-2⋅b2
Now, T22T1T3=2nn-1=810×810243×1080=52

⇒n=5
So, a=3 and b=2,

Hence, T4=C53⋅32⋅23=720

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of ax+b2020, if the coefficients of x2 and x3 are equal, then the value of 9100ba is equal to

Options:

The correct answer is:
60.54

Solution:
ax+b2020=b+ax2020
Given, coefficient of x2=coefficient of x3
â2020C2 b2018a2=2020C3 b2017a3
âba=2020C32020C2=2020!3!2017!2!2018!2020!=20183
â9100ba=60.54

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $\frac{a+b x}{e^{x}}$, the coefficient of $x^{r}$ is

Options:
A $\frac{a-b}{r !}$
B $\frac{a-b r}{r !}$
C $(-1)^{r} \frac{a-b r}{r !}$
D None of these

The correct answer is:
$(-1)^{r} \frac{a-b r}{r !}$

Solution:
$\begin{array}{l}
(a+b x) e^{-x} \\
=(a+b x) \\
\quad\left\{1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots+(-1)^{n} \frac{x^{n}}{n !}+\ldots\right\}
\end{array}$
$\therefore$ The coefficient of $x^{\mathrm{r}}=\mathrm{a}$.
$\frac{(-1)^{\mathrm{r}}}{\mathrm{r} !}+\mathrm{b} \frac{(-1)^{\mathrm{r}-1}}{(\mathrm{r}-1) !}=\frac{(-1)^{\mathrm{r}}}{\mathrm{r} !}(\mathrm{a}-\mathrm{br})$

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of \( \left(2-3 x^{3}\right)^{20} \), if the ratio of \( 10 \) th term to \( 11 \) th term is \( \frac{45}{22} \), then \( x \) is equal to

Options:
A \( -\frac{2}{3} \)
B \( \frac{-3}{2} \)
C \( -\sqrt[3]{\frac{2}{3}} \)
D \( -\sqrt[3]{\frac{3}{2}} \)

The correct answer is:
\( -\frac{2}{3} \)

Solution:
The given expression in the question is 2-3x320

10th term is 20C9-1921139x27 and 11th term is 20C10210310x30.

Ratio is given, so
20C9-1921139x27 20C10⋅210⋅310⋅x30=4522
⇒-1011⋅23x3=4522
⇒x3=-827⇒x=-23

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$ where n is a positive integer, the sum of the coefficients of $\mathrm{x}^{5}$ and $\mathrm{x}^{10}$ is $0 .$
What is the sum of the coefficients of the two middle terms?

Options:
A 0
B 1
C $-1$
D None of these

The correct answer is:
0

Solution:
$\left(x^{3}-\frac{1}{x^{2}}\right)^{n}$
$\quad$ General term, $T_{r+1}={ }^{n} C_{r}\left(x^{3}\right)^{n-r} \cdot\left(-\frac{1}{x^{2}}\right)^{r}$
$\quad={ }^{n} C_{r} \cdot x^{(3 n-3 r)} \cdot(-1)^{r} \cdot x^{-2 r}$
$={ }^{n} C_{r} \cdot(-1)^{r} \cdot x^{(3 n-5 r)}$ ...(i)
For the coefficient $\mathrm{x}^{5}$
Put $3 n-5 r=5$
$5 r=3 n-5$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-1$
$\therefore$ Coefficient of $x^{5}={ }^{n} C\left(\frac{3 n}{5}-1\right)^{(-1)}\left(\frac{3 n}{5}-1\right)$
For the coefficient of $\mathrm{x}^{10}$
Put $3 n-5 r=10$
$5 r=3 n-10$
$\therefore \mathrm{r}=\frac{3 \mathrm{n}}{5}-2$
$\therefore$ Coefficient of $x^{10}={ }^{n} C_{\left(\frac{3 n}{5}-2\right)^{(-1)}}\left(\frac{3 n}{5}-2\right)$
The sum of the coefficient of $x^{5}$ and $x^{10}=0$
$\Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-1\right)^{(-1)}\left(\frac{3 \mathrm{n}}{5}-1\right)}+{ }^{\mathrm{n}} \mathrm{C}_{\left(\frac{3 \mathrm{n}}{5}-2\right)}(-1)^{\left(\frac{3 \mathrm{n}}{5}-2\right)}=0$
$\Rightarrow$
$\left.(-1)^{\frac{3 n}{5}}\left[{ }^{n} C_{\left(\frac{3 n}{5}-1\right)} \cdot(-1)^{-1}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right.}\right) \cdot(-1)^{(-2)}\right]=0$
$\Rightarrow-{ }^{n} C_{\left(\frac{3 n}{5}-1\right)}+{ }^{n} C_{\left(\frac{3 n}{5}-2\right)}=0$ ...(ii)
$\mathrm{n}=15$
Total term in the expansion of $\left(\mathrm{x}^{3}-\frac{1}{\mathrm{x}^{2}}\right)^{15}$ is 16 . $\therefore$ middle term $=8^{\text {th }}$ term and $9^{\text {th }}$ term $\mathrm{T}_{8}=\mathrm{T}_{(7+1)}={ }^{15} \mathrm{C}_{7} \cdot(-1)^{7} \cdot \mathrm{x}^{(3 \times 15-5 \times 7)}$
$=-{ }^{15} \mathrm{C}_{7} \cdot \mathrm{x}^{10} \quad$ (from eq. (i))
$\mathrm{T}_{9}=\mathrm{T}_{(8+1)}={ }^{15} \mathrm{C}_{8} \cdot(-1)^{8} \cdot \mathrm{x}^{(3 \times 15-5 \times 8)}$
$=-{ }^{15} \mathrm{C}_{8} \cdot \mathrm{x}^{5} \quad$ (from eq. (ii))
The sum of the coefficients of the two middle terms
$=-{ }^{15} \mathrm{C}_{7}+{ }^{15} \mathrm{C}_{8}=-{ }^{15} \mathrm{C}_{7}+{ }^{15} \mathrm{C}_{7} \cdot\left[\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-\mathrm{r}}\right]$
$=0$

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of x45+x-15n, the coeficient of the 8th and 19th terms are equal. The term independent of x is given by

Options:
A C2127
B C2025
C C2125
D C2227

The correct answer is:
C2025

Solution:
Since there is no constant term the coefficient of 8 th and 19 th term are same as the binomial coefficients of 8 th and 19 th term.
C7n=nC18ân=7+18=25
Tn+1=25Crx4525-rr-15r
Cr25 x45(25-r)-r5
To be independent of x,100-5r5=0âr=20
Hence, the required term is C2025

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $(x+a)^n$, the sum of odd terms is $P$ and sum of even terms is $Q$, then the value of $\left(P^2-Q^2\right)$ will be

Options:
A $\left(x^2+a^2\right)^n$
B $\left(x^2-a^2\right)^n$
C $(x-a)^{2 n}$
D $(x+a)^{2 n}$

The correct answer is:
$\left(x^2-a^2\right)^n$

Solution:
$(x+a)^n=x^n+{ }^n C_1 x{ }^{n-1} a+{ }^n C_2 x^{n-2} a^2+{ }^n C_3 x^{n-3} a^3+$
$=\left(x^n+{ }^n C_2 x^{n-2} a^2+\ldots \ldots\right.$ $+\left({ }^n C_1 x^{n-1} a+{ }^n C_3 x^{n-3} a^3+\ldots ..\right)$
$=P+Q$
$\therefore(x-a)^n=P-Q$, As the terms are alter. $+v e$ and $-v e$
$\begin{aligned}\therefore & P^2-Q^2=(P+Q)(P-Q)=(x+a)^n(x-a)^n \\
& P^2-Q^2=\left(x^2-a^2\right)^n\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of xcos⁡θ+1xsin⁡θ16, if l1 is the least value of the term independent of x when π8≤θ≤π4 and l2 is the least value of the term independent of x when π16≤θ≤π8, then the ratio l2:l1 is equal to:

Options:
A 1:8
B 16:1
C 8:1
D 1:16

The correct answer is:
16:1

Solution:
Tr+1=Cr16xcos⁡θ16-r1xsin⁡θr
for r=8 term is free from 'x'
T9=C8161sin8⁡θcos8⁡θ
T9=C81628sin⁡2θ8
In θ∈π8,π4,l1=C81628 (∵ Minimum value of l1 at θ=π4)
In θ∈π16,π8,l2=C81628128=C816·28·24 (∵Minimum value of l2 at θ=π8)
l2l1=C816.2824C816.28=16

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (09 Jan Shift 2)
In the expansion of $\left(x-\frac{1}{x}\right)^6$, the constant term is

Options:
A $-20$
B $20$
C $30$
D $-30$

The correct answer is:
$-20$

Solution:
In the expansion of $\left(x-\frac{1}{x}\right)^6$
the general term is ${ }^6 C_r x^{6-r}\left(-\frac{1}{x}\right)^r={ }^6 C_r(-1)^r x^{6-2 r}$
For term independent of $x, 6-2 r=0 \Rightarrow r=3$
Thus the required coefficient $=(-1)^3 \cdot{ }^6 C_3=-20$

Topic: Mathematics, Binomial Theorem, JEE Main
In the expansion of $\left(x+\frac{2}{x^2}\right)^{15}$ the term independent of $x$ is

Options:
A ${ }^{15} C_6 2^6$
B ${ }^{15} C_5 2^5$
C ${ }^{15} \mathrm{C}_4 2^4$
D ${ }^{15} C_8 2^8$

The correct answer is:
${ }^{15} C_5 2^5$

Solution:
$(15-r)(1)+r(-2)=0 \Rightarrow 15-3 r=0 \Rightarrow r=5$
Thus term independent of x is $={ }^{15} C_5(1)^{10}(2)^5={ }^{15} C_5 2^5$

Topic: Mathematics, Binomial Theorem, JEE Main
In the experiment of calibration of voltmeter, a standard cell of e.m.f. $1.1$ volt is balanced against $440 \mathrm{~cm}$ of potential wire. The potential difference across the ends of resistance is found to balance against $220 \mathrm{~cm}$ of the wire. The corresponding reading of voltmeter is $0.5$ volt. The error in the reading of volmeter will be:

Options:
A
$-0.15$ volt

B
$0.15$ volt

C
$0.5$ volt

D
$-0.05$ volt

The correct answer is:
$-0.05$ volt

Solution:
In a voltmeter
$$
\begin{gathered}
\mathrm{V} \propto l \\
\mathrm{~V}=\mathrm{k} l
\end{gathered}
$$
Now, it is given $\mathrm{E}=1.1$ volt for $l_1=440 \mathrm{~cm}$ and $\mathrm{V}=0.5$ volt for $l_2=220 \mathrm{~cm}$ Let the error in reading of voltmeter be $\Delta \mathrm{V}$ then,
$$
\begin{aligned}
&1.1=400 \mathrm{~K} \text { and }(0.5-\Delta \mathrm{V})=220 \mathrm{~K} . \\
&\Rightarrow \frac{1.1}{440}=\frac{0.5-\Delta \mathrm{V}}{220} \\
&\therefore \Delta \mathrm{V}=-0.05 \text { volt }
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2014 (12 Apr Online)
In the experiment set up for the measurement of EMF of a half cell using a reference electrode and a salt bridge, when the salt bridge is removed, the voltage

Options:
A Does not change
B Decreases to half the value
C Increase to maximum
D Drops to zero

The correct answer is:
Drops to zero

Solution:
It connect two solutions and complete the circuit.

Topic: Chemistry, Electrochemistry, JEE Main