datavorous/entrance-exam-dataset huggingface dataset:
If dydx+ytanx=sin2x and y0=1, then yπ is equal to

Options:
A -1
B 5
C 1
D -5

The correct answer is:
-5

Solution:
The given differential equation dydx+ytanx=sin2x is linear differential equation of the form

dydx+Pxy=Qx with P(x)=tanx, Q(x)=sin2x

Now, I.F.=e∫Pxdx=e∫tanxdx=elnsecx=secx

And, the solution is yI.F.=∫QxI.F.dx+c

⇒ysecx=∫secxsin2xdx+c

⇒ysecx=∫1cosx2sinxcosxdx+c

⇒ysecx=∫2sinxdx+c

⇒ysecx=-2cosx+c

Given, y0=1

⇒1·sec0=-2cos0+c

⇒1=-2+c

⇒c=3

⇒ysecx=-2cosx+3

Now, at x=π

yπsecπ=-2cosπ+3

⇒yπ-1=-2-1+3

⇒yπ=-5.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2014 (19 Apr Online)
If decomposers are removed what will happen to the ecosystem?

Options:
A Energy cycle is stopped
B Mineral cycle is stopped
C Consumers cannot absorb solar energy
D Rate of decompostion of mineral increases

The correct answer is:
Mineral cycle is stopped

Solution:
Decomposers are heterotropic organisms mostly bacteria and fungi which live on dead organic matter or detritus. They release different enzymes from their bodies into the dead and decaying plant and animal remains. The extra cellular digestion of the dead remains, lead to the release of simple inorganic substances.

Topic: Biology, Ecosystem, JIPMER, JIPMER 2005
If denote microwaves, \( X \)-rays, infrared, gamma rays, ultraviolet, radio waves and visible parts of the electromagnetic
spectrum by \( M, X, I, G, U, R \) and \( V \) the following is the arrangement in ascending order of the wavelength

Options:
A \( I, M, R, U, V, X \) and \( G \)
B \( R, M, I, V, U, X \) and \( G \)
C \( M, R, V, X, U, G \) and \( I \)
D \( G, X, U, V, I, M \) and \( R \)

The correct answer is:
\( G, X, U, V, I, M \) and \( R \)

Solution:
The electromagnetic (EM) spectrum is the range of all types of EM radiation. The γ‐ray has a maximum frequency or minimum wavelength; since the wavelength is inversely proportional to frequency, while radio wave has a minimum frequency and maximum wavelength. The order of wavelength will be; Gamma rays G < X-rays X < Ultra violet U < Visible rays V < Infrared rays I < Microwaves M < Radio waves R.

Topic: Physics, Electromagnetic Waves, JEE Main
If $\phi$ denotes reduction potential, then which is true ?

Options:
A
$E_{\text {cell }}^0=\phi_{\text {right }}-\phi_{\text {left }}$

B
$\mathrm{E}_{\text {cell }}^0=\phi_{\text {left }}+\phi_{\text {right }}$

C
$\mathrm{E}_{\text {cell }}^0=\phi_{\text {left }}-\phi_{\text {right }}$

D
$\mathrm{E}_{\text {cell }}^0=-\left(\phi_{\text {left }}+\phi_{\text {right }}\right)$

The correct answer is:
$E_{\text {cell }}^0=\phi_{\text {right }}-\phi_{\text {left }}$

Solution:
$E_{\text {cell }}=E_{\text {right (cathode) }}-E_{\text {left (anode) }}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2002
If θ denotes the acute angle between the curves, y=10-x2 and y=2+x2 at a point of their intersection, then tan⁡θ is equal to:

Options:
A 49
B 817
C 717
D 815

The correct answer is:
815

Solution:
Intersection points of given curves are ±2, 6, for 1st curve dydx=-2x
∴m1=4 or -4
For 2nd curve, dydx=2x
∴m2=-4 or 4
∴ |tanθ|=4--41+4-4 or -4-41+-44
=815 (In both cases)

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (09 Jan Shift 1)
If $[\cdot]$ denotes the greatest integer function, then the domain and range of the function $f(x)=\frac{\sin [x] \pi+\tan [x] \pi}{1+[x]^2+[x]^4}$ are respectively

Options:
A $\mathbb{R}-\{0\}, \mathbb{R}-\{0\}$
B $\mathbb{R}^{+},\{0\}$
C $\mathbb{R}^{+}, \mathbb{R}$
D $\mathbb{R},\{0\}$

The correct answer is:
$\mathbb{R},\{0\}$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 2)
If [ ] denotes the greatest integer function, then the integral $\int_0^\pi[\cos x d x$ is equal to:

Options:
A
$\frac{\pi}{2}$

B
0

C
$-1$

D
$-\frac{\pi}{2}$

The correct answer is:
$-\frac{\pi}{2}$

Solution:
Let $\mathrm{I}=\int_0^\pi[\cos x] d x$
$$
\mathrm{I}=\int_0^\pi[\cos (\pi-x)] d x=\int_0^\pi[-\cos x] d x
$$
On adding (1) and (2), we get
$$
\begin{aligned}
&2 \mathrm{I}=\int_0^\pi[\cos x] d x+\int_0^\pi[-\cos x] d x \\
&2 \mathrm{I}=\int_0^\pi[\cos x]+[-\cos x] d x \\
&2 \mathrm{I} \\
&=\int_0^\pi-1 d x \quad(\because[x]+[-x]=-1 \text { if } x \notin Z) \\
&2 \mathrm{I}=-\left.x\right|_0 ^\pi=-\pi \\
&\Rightarrow \mathrm{I}=\frac{-\pi}{2}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2014 (12 Apr Online)
If density of vapours of a substance of molar mass 18âgm/mole at 1âatm pressure and 500âK is 0.36âkgâ m â3 , then value of Z for the vapours is (Take R=0.082 Lâ atm mole â1 K â 1 )

Options:
A 4150
B 5041
C 1.8
D 0.9

The correct answer is:
5041

Solution:
Vreal=molar massdensity=180.36
Videal=nRTP=1Ã0.082Ã5001
So, Z=VrealVideal=500.082Ã500=5041

Topic: Chemistry, States of Matter, JEE Main
If Dichloromethane (DCM) and water H2O are used for differential extraction, which one of the following statements is correct?

Options:
A DCM and H2O would stay as upper and lower layer respectively in the separating funnel (S.F.)
B DCM and H2O will be miscible clearly
C DCM and H2O would stay as lower and upper layer respectively in the S.F
D DCM and H2O will make turbid/colloidal mixture

The correct answer is:
DCM and H2O would stay as lower and upper layer respectively in the S.F

Solution:
Dichloromethane and water are immiscible in nature. Also, Dichloromethane has higher density than water so forms bottom layer (Layer II) in the separating funnel.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If dielectric is inserted in charged capacitor (battery removed), then quantity that remains constant is

Options:
A capacitance
B potential
C intensity
D charge

The correct answer is:
charge

Solution:
Variation of different variables (Q, $C, V, E$ and $U$ ) of parallel plate capacitor when dielectric $(K)$ is introduced when battery is removed is
$$
\begin{array}{ll}
C^{\prime}=K C & E^{\prime}=E / K \\
Q^{\prime}=Q & U^{\prime}=U / K \\
V^{\prime}=V / K &
\end{array}
$$

Topic: Physics, Capacitance, JEE Main
If dielectric is inserted in charged capacitor (battery removed), then quantity that remains constant is

Options:
A capacitance
B potential
C intensity
D charge

The correct answer is:
charge

Solution:
Variation of different variables (Q, $C, V, E$ and $U$ ) of parallel plate capacitor when dielectric $(K)$ is introduced when battery is removed is
$$
\begin{array}{ll}
C^{\prime}=K C & E^{\prime}=E / K \\
Q^{\prime}=Q & U^{\prime}=U / K \\
V^{\prime}=V / K &
\end{array}
$$

Topic: Physics, Capacitance, MHT CET, MHT CET 2009
If dimensions of critical velocity, vc of a liquid flowing through a tube are expressed as ηxρyrz, where, η, ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube, respectively, then, the values of x, y and z are given by

Options:
A -1, -1, 1
B -1, -1, -1
C 1, 1, 1
D 1, -1, -1

The correct answer is:
1, -1, -1

Solution:
Dimensions of velocity, coefficient of viscosity and density are,

v=LT-1

η=FAdvdx=MLT-2L2T-1=ML-1T-1

ρ=ML-3

Now, v=ηxρyrz

⇒LT-1=ML-1T-1xML-3yLz

Now, by using dimensional homogeneity, we will get,

M0=Mx+y

⇒y=-x

For T,

x=1

⇒y=-x=-1

For L,

1=-x-3y+z

⇒z=-1

Topic: Physics, Units and Dimensions, JEE Main
If domain of the function loge6x2+5x+12x-1+cos-12x2-3x+43x-5 is α,β∪γ,δ, then 18α2+β2+γ2+δ2 is equal to

Options:

The correct answer is:
20

Solution:
We need to find the domain of the function loge6x2+5x+12x-1+cos-12x2-3x+43x-5.

⇒6x2+5x+12x-1>0 .....(1) and

⇒ -1≤2x2-3x+43x-5≤1......(2)
Now From 1,
⇒6x2+5x+12x-1>0

⇒(3x+1)(2x+1)(2x-1)>0

Now we get the common region here as
⇒x∈-12,-13∪12,∞.......(a)

From 2 we get that
⇒-1≤2x2-3x+43x-5≤1

⇒2x2-3x+43x-5≥-1......3

and 2x2-3x+43x-5≤1.....4

From 3 we get
⇒2x2-3x+43x-5+1≥0

⇒x∈-12,12∪53,∞.....b

From 4 we get
⇒2x2-3x+43x-5-1≤0

⇒2x2-6x+93x-5≤0

As D=-62-429<0 and 2>0, we observe that 2x2-6x+9>0 ∀x∈ℝ
⇒13x-5≤0 ∵2x2-6x+9>0∀x∈R

⇒ x∈-∞,53 ......c

⇒ Intersection of (a), (b) and (c) gives us
x∈-12,-13∪12,12.

On comparing this with α,β∪γ,δ,
⇒α2+β2+γ2+δ2=109
⇒18α2+β2+γ2+δ2=20.

Hence this is the required answer.

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (08 Apr Shift 2)
If ∫ dx x - 1 3 / 4 x + 2 5 / 4 = k x - 1 x + 2 1 / 4 + c , then the number of divisors of 30k is

Options:

The correct answer is:
8

Solution:
I = ∫ dx x - 1 x + 2 4 x + 2 2 put x - 1 x + 2 = t

On solving, we get,

I=43x-1x+214+c
k = 4 3 ⇒ 3 0 k = 4 0 = 2 3 × 5 1
Hence, total divisors are 4×2 = 8

Topic: Mathematics, Indefinite Integration, JEE Main
If $E_0$ represents the peak value of the voltage in an ac circuit, the r.m.s. value of the voltage will be

Options:
A $\frac{E_0}{\pi}$
B $\frac{E_0}{2}$
C $\frac{E_0}{\sqrt{\pi}}$
D $\frac{E_0}{\sqrt{2}}$

The correct answer is:
$\frac{E_0}{\sqrt{2}}$

Solution:
If $\mathrm{E}_{\mathrm{o}}$ represents the peak value of the voltage in an ac circuit then, we know,
$\mathrm{E}_{\mathrm{rms}}=\frac{\mathrm{E}_{\mathrm{o}}}{\sqrt{2}}$

Topic: Physics, Alternating Current, JEE Main
If e1 and e2 are the eccentricities of a hyperbola 3x2-3y2=25 and its conjugate, then

Options:
A e12+e22=2
B e12+e22=4
C e1+e2=4
D e1+e2=2

The correct answer is:
e12+e22=4

Solution:
Given equation can be written as

x2-y2=253

∴ e1=1+b2a2

=1+1=2

The equation of conjugate hyperbola is

-x2+y2=253

∴ e2=1+b2a2=1+1=2

∴ e12+e22=22+22=4

Hence, e12+e22=4.

Topic: Mathematics, Hyperbola, JEE Main
If e1 and e2 are the eccentricities of the ellipse x218+y24=1 and the hyperbola x29-y24=1 respectively and e1,e2 is a point on the ellipse 15x2+3y2=k , then the value of k is equal to

Options:
A 16
B 17
C 15
D 14

The correct answer is:
16

Solution:
e1=1-418=79=73
e1=1+49=139=133

Also,
15e12+3e22=k⇒k=1579+3139
∴k=16

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If e1 and e2 are the eccentricities of the ellipse x218+y24=1 and the hyperbola x29-y24=1 respectively and e1,e2 is a point on the ellipse 15x2+3y2=k , then the value of k is equal to

Options:
A 16
B 17
C 15
D 14

The correct answer is:
16

Solution:
e1=1-418=79=73
e1=1+49=139=133

Also,
15e12+3e22=k⇒k=1579+3139
∴k=16

Topic: Mathematics, Hyperbola, JEE Main
If $e_{1}$ and $e_{2}$ represent the eccentricity of the curves $16 x^{2}-9 y^{2}=144$ and $9 x^{2}-16 y^{2}=144$
respectively. Then, $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}$ is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
Given curves are $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
$\therefore$
$e_{1}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$
and $e_{2}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
Now, $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}=\frac{9}{25}+\frac{16}{25}=1$

Topic: Mathematics, Hyperbola, JEE Main
If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ and $e_2$ is the eccentricity of a hyperbola passing through the foci of the given ellipse and $e_1 e_2=1$, then the equation of such a hyperbola among the following is

Options:
A $\frac{x^2}{9}-\frac{y^2}{16}=1$
B $\frac{y^2}{9}-\frac{x^2}{16}=1$
C $\frac{x^2}{9}-\frac{y^2}{25}=1$
D $\frac{x^2}{25}-\frac{y^2}{9}=1$

The correct answer is:
$\frac{y^2}{9}-\frac{x^2}{16}=1$

Solution:
Equation of ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$
Foci $\quad=(0, \pm 3)$
$\Rightarrow \quad e_2=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ given $e_2 e_2=1$
(where $e_2$ is eccentricity of hyperbola)
Let equation of hyperbola $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
its passes through $(0, \pm 3)$
$$
\begin{array}{ll}
\therefore & b^2=9 \\
\Rightarrow & e_2=\sqrt{1+\frac{a^2}{b^2}} \\
\Rightarrow & e_2^2=1+\frac{a^2}{b^2} \\
\Rightarrow & \frac{1}{e_2^2}=1+\frac{a^2}{9} \Rightarrow \frac{25}{9}=1+\frac{a^2}{9} \\
\Rightarrow & a^2=16
\end{array}
$$
Hence, the equation of hyperbola is $\frac{y^2}{9}-\frac{x^2}{16}=1$

Topic: Mathematics, Hyperbola, JEE Main
If E and B denote electric and magnetic fields respectively, which of the following is dimensionless?

Options:
A Î¼0 Îµ0EB
B Î¼0Îµ0EB
C Î¼0Îµ0Â BE2
D μ0EBε0

The correct answer is:
Î¼0 Îµ0EB

Solution:
EB=c (Velocity of light if we consider free space) and 1Î¼0 Îµ0=c
So Î¼0 Îµ0EB is dimensionless.

Topic: Physics, Units and Dimensions, JEE Main
If E and E0 denote energies at time t and t0 respectively, and L and L0 distance from some point at t and t0 respectively, then which of the following equations can be declared to be incorrect on dimensional grounds

(A) E=2E0LL0

(B) E=E0e-2LL0

(C) E=2Le-LE0

(D) E=2E0L0×e-LL0

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

The correct answer is:
C,D only

Solution:
Dimensions of energy =E=E0=ML2T-2

Dimensions of length

=L=L0=M0L1T0

For A:

LHS =ML2T-2

RHS =ML2T-2LL=ML2T-2

In this case, LHS=RHS

For B:

Exponent should be dimensionless. As we can see

-2LL0=LL=M0L0T0

and LHS =ML2T-2

and RHS =ML2T-2

So in this case, LHS=RHS

For C:

Exponent should be dimensionless but here

-L0E0=M0L1T0ML2T-2=M-1L-1T2

is not dimensionless therefore it is wrong.

For D:

Exponent should be dimensionless. Here

-LL0=M0L0T0M0L0T0=M0L0T0

Now, LHS =E=ML2T-2

and RHS =ML2T-2L=ML1T-2

As in this case, LHS ≠RHS, hence it is wrong.

Topic: Physics, Units and Dimensions, JEE Main
If $e$ and $e^{\prime}$ are eccentricities of hyperbola and its conjugate respectively, then

Options:
A $\left(\frac{1}{e}\right)^2+\left(\frac{1}{e^{\prime}}\right)^2=1$
B $\frac{1}{e}+\frac{1}{e^{\prime}}=1$
C $\left(\frac{1}{e}\right)^2+\left(\frac{1}{e^{\prime}}\right)^2=0$
D $\frac{1}{e}+\frac{1}{e^{\prime}}=2$

The correct answer is:
$\left(\frac{1}{e}\right)^2+\left(\frac{1}{e^{\prime}}\right)^2=1$

Solution:
Let hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ....(i)
Then its conjugate will be, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ ....(ii)
If $e$ is eccentricity of hyperbola (i), then $b^2=a^2\left(e^2-1\right)$
or $\frac{1}{e^2}=\frac{a^2}{\left(a^2+b^2\right)}$ ....(iii)
Similarly if $e^{\prime}$ is eccentricity of conjugate (ii), then $a^2=b^2\left(e^{\prime 2}-1\right)$ or $\frac{1}{e^{\prime 2}}=\frac{b^2}{\left(a^2+b^2\right)}$ .....(iv)
Adding (iii) and (iv),
\(\left(\frac{1}{e}\right)^2+\left(\frac{1}{e^1}\right)^2=1\)

Topic: Mathematics, Hyperbola, JEE Main
If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5 x^2+9 y^2=45$ and the hyperbola $5 x^2-4 y^2=45$ respectively, then $e e^{\prime}$ is equal to

Options:
A $1$
B $4$
C $5$
D $9$

The correct answer is:
$1$

Solution:
Equation of ellipse is
or
$$
\begin{array}{r}
5 x^2+9 y^2=45 \\
\frac{x^2}{9}+\frac{y^2}{5}=1 \\
e=\sqrt{\frac{1-b^2}{a^2}}=\sqrt{\frac{1-5}{9}}=\frac{2}{3}
\end{array}
$$
and equation of hyperbola is
$$
5 x^2-4 y^2=45
$$
or
$$
\frac{x^2}{9}-\frac{y^2}{\left(\frac{45}{4}\right)}=1
$$
$$
e^{\prime}=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{45}{36}}=\sqrt{\frac{81}{36}}=\frac{9}{6}=\frac{3}{2}
$$
Then, $e e^{\prime}=\frac{2}{3} \cdot \frac{3}{2}=1$

Topic: Mathematics, Ellipse, JEE Main
If $E$ and $F$ are independent events such that $0 \lt P(E) \lt 1$ and $0 \lt P(F) \lt 1$,then

Options:
A $E$ and $F^{\circ}$ (the complement of the event $F$ ) are independent
B $E^c$ and $F^c$ are independent
C $P\left(\frac{E}{F}\right)+P\left(\frac{E^c}{F^c}\right)=1$
D All of the above

The correct answer is:
All of the above

Solution:
$P(E \cap F)=P(E) \cdot P(F)$
Now, $P\left(E \cap F^c\right)=P(E)-P(E \cap F)=P(E)[1-P(F)]=P(E) \cdot P\left(F^c\right)$
and $P\left(E^c \cap F^c\right)=1-P(E \cup F)=1-[P(E)+P(F)-P(E \cap F)$
$=[1-P(E)][1-P(F)]=P\left(E^c\right) P\left(F^c\right)$
Also $P(E / F)=P(E)$ and $P\left(E^c / F^c\right)=P\left(E^c\right)$
$\Rightarrow P(E / F)+P\left(E^c / F^c\right)=1$

Topic: Mathematics, Probability, JEE Main
If E and G respectively denote energy and gravitational constant, then EG has the dimensions of:

Options:
A [M]L0T0
B [M]L-1T-1
C M2L-1T0
D M2L-2T-1

The correct answer is:
M2L-1T0

Solution:
Dimensional Formulas,

Energy, [E]=M1L2T-2

Gravitational constant

[G]=Fr2M2=M1L1T-2×L2M2

[G]=M-1L3T-2

Now,

EG=ML2T-2M-1L3T-2=M2L-1T0

Correct option (3)

Topic: Physics, Units and Dimensions, JEE Main
If E and H represents the intensity of electric field and magnetizing field respectively, then the unit of EH will be:

Options:
A joule
B ohm
C newton
D mho

The correct answer is:
ohm

Solution:
The unit of, EH=FqIr=FrI2t, here the units are, F=N, r=m, I=A and t=s then the net unit of EH,is J s-1 A-2, which is equivalent to Ω.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{K}}$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by: ( $\omega$ - angular frequency) :

Options:
A $\frac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$
B $\omega(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{K}})$
C $\omega(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$
D $\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}}$

The correct answer is:
$\frac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$

Solution:
For an EM wave
$\overrightarrow{\mathrm{B}}=\frac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$

Topic: Physics, Electromagnetic Waves, JEE Main
If E ClO 3 - / ClO 4 - 0 = - 0 . 3 6 V & E ClO 3 - / ClO 2 - 0 = 0 . 3 3 V at 300 K. The equilibrium concentration of perchlorate ion ClO 4 - which was initially 1.0 M in ClO 3 - when the reaction starts to attain the equilibrium,

2 ClO 3 - â ClO 2 - + ClO 4 -

Given : Anti log(0.509) = 3.329

Options:
A 0.0236 M
B 0.0190 M
C 0.123 M
D 0.191 M

The correct answer is:
0.191 M

Solution:
The redox reaction can be split as
Cl + 5 O 3 -1 â¶ Cl + 7 O 4 -1 + 2 e OHR - anode Cl + 5 O 3 -1 + 2 e â¶ Cl + 3 O 2 -1 RHR - cathode
-------------------------------------------------------
2 ClO 3 - 1 â n = 2e - ClO 4 - 1 + ClO 2 - 1
E cell o = E red cath o - E red anode o
= 0.33 - 0.36 = -0.03 V
as E red Anode o = E ClO 4 - 1 / ClO 3 - 1 o = + 0 . 3 6 V
= - E ClO 3 - 1 / ClO 4 - 1 o
at equilibrium Ecell = 0
E cell o = 0 . 0 5 9 n log K eq = 0 . 0 5 9 2 log K eq
Writing concentration of speces at equilibrium
2 ClO 3 - 1 â ClO 4 - 1 + ClO 2 - 1 1 - 2 x x x Molar Molar Molar
E cell o = - 0 . 0 3 = 0 . 0 5 9 2 log x 2 1 - 2 x 2 = 0 . 0 5 9 log x 1 - 2 x
- 0 . 0 3 0 . 0 5 9 = log x 1 - 2 x ⇒ log 1 - 2 x x = 0 . 0 3 0 . 0 5 9
log 1 - 2 x x = 0 . 5 0 9
1 - 2 x x = 3 . 2 2 9
1 - 2x = 3.229 x ⇒ 5.229x = 1
x = 1 5 . 2 2 9 = 0 . 1 9 1 M .

Topic: Chemistry, Electrochemistry, JEE Main
If E=cos2â¡71o+cos2â¡49o+cosâ¡71ocosâ¡49o, then the value of 10E is equal to

Options:
A 7.5
B 2.5
C 12.5
D 10.5

The correct answer is:
7.5

Solution:
E=1+cos⁡142o2+1+cos⁡98o2+122cos⁡71ocos⁡49o
=1+12cos⁡142o+cos⁡98o+12cos⁡120o+cos⁡22o
=1+122cos⁡120ocos⁡22o+12-12+cos⁡22o
=1+-12cos⁡22o-14+12cos⁡22o=34
E=34=0.75
⇒10E=7.5

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If ecos2x+cos4x+cos6x+....∞loge2 satisfies the equation t2-9t+8=0, then the value of 2sinxsinx+3cosx, where 0<x<π2, is equal to

Options:
A 32
B 12
C 3
D 23

The correct answer is:
12

Solution:
ecos2x+cos4x+……∞ln 2=2cos2x+cos4x+……∞

==2cos2x1-cos2x=2cot2x

t2-9t+8=0⇒t=1,8

⇒2cot2x=1, 8⇒cot2x=0,3

0<x<π2⇒cot x=3

⇒2 sin xsin x+3 cos x=21+3 cot x=24=12

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If E= energy, G= gravitational constant, I= impulse and M= mass, then dimensions of GIM2E2 are same as that of

Options:
A Time
B Mass
C Length
D Force

The correct answer is:
Time

Solution:
Dimensions of E=ML2T-2
Dimensions of G=M-1L3T-2
Dimensions of I=MLT-1
And dimension of M=M
∴Dimensions ofGIM2E2=M-1L3T-2MLT-1M2ML2T-22
=T
=Dimensions of time

Topic: Physics, Units and Dimensions, JEE Main
If $E_f$ and $E_r$ are the activation energies of forward and reverse reactions and the reaction is known to be exothermic, then

Options:
A $E_f \gt E_r$
B $E_f \lt E_r$
C $E_f=E_r$
D No relation can be given between $E_f$ and $E_r$ as data are not sufficient

The correct answer is:
$E_f \lt E_r$

Solution:
$E_f \lt E_r$

Topic: Chemistry, Chemical Kinetics, JEE Main
If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity 14πε0e2hc has dimensions of :

Options:
A MLT0
B M0 L0 T0
C MLT-1
D LC-1

The correct answer is:
M0 L0 T0

Solution:
F=14πϵ0e2r2

E=hcλ

e24πε0×1hc=Fr2Eλ=M0 L0 T0

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P=EL2M-5G-2 are:

Options:
A M1L1 T-2
B M0 L1 T0
C M-1L-1T2
D M0 L0 T0

The correct answer is:
M0 L0 T0

Solution:
P=EL2M-5G−2

P=ML2T-2ML2T−12M−5M−1L3T-2−2

P=M0L0T0

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2021 (26 Aug Shift 1)
If ∫esecxsecxtanxfx+secxtanx+sec2xdx=esecxfx+C, then a possible choice of fx is:

Options:
A secx-tanx-12
B secx+tanx+12
C xsecx+tanx+12
D secx+xtanx-12

The correct answer is:
secx+tanx+12

Solution:
∫esec⁡x(sec⁡xtan⁡xfx+sec⁡xtan⁡x+sec2⁡xdx=esec⁡xfx+C

Differentiating both sides w.r.t ‘x’ we get

esecxsecxtanxfx+secxtanx+sec2x=esecx⋅secxtanxf(x)+esecxf'x

⇒f'x=sec2x+tanxsecx

⇒fx=∫sec2x+tanxsecxdx

⇒fx=tanx+secx+c, c∈R
Hence, possible choice is fx=secx+tanx+12.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (09 Apr Shift 2)
If ∫esin⁡θsin⁡θ+sec2⁡θdθ is equal to fθ+C (where, C is the constant of integration) and f0=0, then the value of fπ4 is

Options:
A e2
B e12
C e2
D e12

The correct answer is:
e12

Solution:
∫esin⁡θsin⁡θdθ+∫esin⁡θsec2⁡θdθ
∫esin⁡θsin⁡θdθ+tan⁡θesin⁡θ-∫tan⁡θesin⁡θcos⁡θdθ
=∫esin⁡θsin⁡θdθ+tan⁡θesin⁡θ-∫esin⁡θsin⁡θdθ=esin⁡θtan⁡θ+C
=esin⁡θtan⁡θ+C
fθ=esin⁡θtan⁡θ
fπ4=e12

Topic: Mathematics, Indefinite Integration, JEE Main
If $\int \frac{e^x-1}{e^x+1} d x=f(x)+c$, then $f(x)$ is equal to

Options:
A $2 \log \left(e^x+1\right)$
B $\log \left(e^{2 x}-1\right)$
C $2 \log \left(e^x+1\right)-x$
D $\log \left(e^{2 x}+1\right)$

The correct answer is:
$2 \log \left(e^x+1\right)-x$

Solution:
We have, $\int \frac{e^x-1}{e^x+1} d x$
$=\int\left(\frac{2 e^x}{e^x+1}-1\right) d x$
$=2 \log \left(e^x+1\right)-x+c$
On comparing with $f(x)+c$, we get $f(x)=2 \log \left(e^x+1\right)-x$

Topic: Mathematics, Indefinite Integration, JEE Main
If $\int e^x(1+x) \cdot \sec ^2\left(x e^x\right) d x$ $=f(x)+$ constant, then $f(x)$ is equal to

Options:
A $\cos \left(x e^x\right.$
B $\sin \left(x e^x\right.$
C $2 \tan ^{-1}(x)$
D $\tan \left(x e^x\right)$

The correct answer is:
$\tan \left(x e^x\right)$

Solution:
Given that,
$$
\int e^x(1+x) \cdot \sec ^2\left(x e^x\right) d x=f(x)+\text { constant }
$$
Put
$$
x e^x=t \text { in LHS }
$$
$$
\begin{aligned}
& \Rightarrow & \left(e^x+x e^x\right) d x & =d t \\
& \therefore & \text { LHS } & =\int \sec ^2 t d t \\
& & & =\tan t+\text { constant } \\
& \Rightarrow & \tan \left(x e^x\right)+\text { constant } & =f(x)+\text { constant } \\
& \Rightarrow & f(x) & =\tan \left(x e^x\right)
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
If ∫e-x22dx=fx and the solution of the differential equation dydx=1+xy is y=kex22f(x)+Cex22, then the value of k is equal to (where C is the constant of integration)

Options:

The correct answer is:
1

Solution:
dydx-xy=1
Here, I.F. =e-â«xdx=e-x22
So, solution is ye-x22=â«e-x22dx+C
y=ex22fx+Cex22
Hence, k=1

Topic: Mathematics, Differential Equations, JEE Main
If $\int e^{x^2} \cdot x^3 d x=e^{x^2} f(x)+C$
(where $C$ is a constant of integration)
and $f(1)=0$, then value of $f(2)$ will be

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

The correct answer is:
$\frac{3}{2}$

Solution:
$\int e^{x^2 x^3}=\frac{1}{2} \int x^2 e^{x^2} 2 x d x=\frac{1}{2} \int t e^t d t \quad\left[\operatorname{let} x^2=t\right]$
Integrating by parts
$\begin{aligned} & =\frac{1}{2}\left[t \cdot e^t-e^t\right]+C \\ & =\frac{1}{2}\left[x^2 e^{x^2}-e^{x^2}\right]+C=e^{x^2} \cdot \frac{1}{2}\left(x^2-1\right)+C \\ & \Rightarrow f(x)=\frac{1}{2}\left(x^2-1\right) \Rightarrow f(2)=\frac{3}{2}\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main