datavorous/entrance-exam-dataset huggingface dataset:
The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L −1 ) at 750 torr partial pressure is:

Options:
A 0.0075
B 0.005
C 0.02
D 0.015

The correct answer is:
0.015

Solution:
p = KH X
where p→ Partial pressure of gas
X2→ Mole fraction of gas in solution
KH = Henry's law constant
So, p∝ solubility
p1p2 = s1s2 ⇒ 5000.01 = 750x
∴ x = 0.015 g/L

Topic: Chemistry, Solutions, JEE Main, JEE Main 2016 (09 Apr Online)
The solubility of $\mathrm{PbCl}_2$ is

Options:
A $\sqrt{K_{s p}}$
B $\sqrt[3]{K_{s p}}$
C $\sqrt[3]{\frac{K_{s p}}{4}}$
D $\sqrt{8 K_{s p}}$

The correct answer is:
$\sqrt[3]{\frac{K_{s p}}{4}}$

Solution:
$\mathrm{PbCl}_2 \rightleftharpoons \mathrm{Pb}^{2+}+\underset{S}{2 \mathrm{Cl}^{-}}$
$\begin{gathered}K_{s p} \text { of } \mathrm{PbCl}_2=\left[\mathrm{Pb}^{2+}\right] \times\left[\mathrm{Cl}^{-}\right]^2 ; K_{s p}=S \times(2 S)^2 \\ K_{s p}=S \times 4 S^2=4 S^3 ; S^3=\frac{K_{s p}}{4} ; S=\sqrt[3]{\frac{K_{s p}}{4}}\end{gathered}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of $\mathrm{PbI}_2$ at $25^{\circ} \mathrm{C}$ is $0.7 \mathrm{~g} \mathrm{~L}^{-1}$. The solubility product of $\mathrm{PbI}_2$ at this temperature is (molar mass of $\mathrm{PbI}_2=461.2 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Options:
A
$1.40 \times 10^{-9}$

B
$0.14 \times 10^{-9}$

C
$140 \times 10^{-9}$

D
$14.0 \times 10^{-9}$

The correct answer is:
$14.0 \times 10^{-9}$

Solution:
$\mathrm{PbI}_2 \leftrightharpoons \mathrm{Pb}^{++}+2 \mathrm{I}^{-}$
$$
\begin{aligned}
K_{\text {sp }}= & s \times(2 s)^2=4 s^3 \\
& =4 \times\left(\frac{0.7}{461.2}\right)^3=14.0 \times 10^{-9}
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2012 (26 May Online)
The solubility of silver chromate in $0.01 \mathrm{M} \mathrm{K}_2 \mathrm{CrO}_4$ is $2 \times 10^{-8} \mathrm{~mol} \mathrm{dm}^{-3}$. The solubility product of silver chromate will be

Options:
A $8 \times 10^{-24}$
B $16 \times 10^{-24}$
C $1.6 \times 10^{-18}$
D $16 \times 10^{-18}$

The correct answer is:
$16 \times 10^{-18}$

Solution:
$K_{s p}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right]=[2 S]^2[0.01]$
$=4 S^2[0.01]=4\left[2 \times 10^{-8}\right]^2 \times 0.01=16 \times 10^{-18}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility of sulphates of alkaline earth metals in water decreases down the group because

Options:
A The lattice energy of sulphates of group II decreases down the group
B The lattice energy of sulphates of group II increases down the group
C Both hydration and lattice energies decreases down the group
D The decrease in hydration energy is more than the decreases in lattice energy

The correct answer is:
The decrease in hydration energy is more than the decreases in lattice energy

Solution:
Both lattice energy and hydration energy decreases down the group of alkaline earth sulphates but hydration energy decreases more rapidly than lattice energy.

Topic: Chemistry, s Block Elements, JEE Main
The solubility order for alkali metal fluoride in water is :

Options:
A
$\mathrm{LiF} < \mathrm{RbF} < \mathrm{KF} < \mathrm{NaF}$
B
$\mathrm{RbF} < \mathrm{KF} < \mathrm{NaF} < \mathrm{LiF}$
C
$\mathrm{LiF}>\mathrm{NaF}>\mathrm{KF}>\mathrm{RbF}$

D
$\mathrm{LiF} < \mathrm{NaF} < \mathrm{KF} < \mathrm{RbF}$

The correct answer is:
$\mathrm{LiF} < \mathrm{NaF} < \mathrm{KF} < \mathrm{RbF}$

Solution:
Higher the lattice enthalpy lower will be solubilityi.e., lattice enthalpy $\propto \frac{1}{\text { Solubility }}$
Since the lattice enthalpy of alkali metals follow the order $\mathrm{Li}>\mathrm{Na}>\mathrm{K}>\mathrm{Rb}$ Hence the correct order of solubility is $\mathrm{LiF} < \mathrm{NaF} < \mathrm{KF} < \mathrm{RbF}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (22 Apr Online)
The solubility order for alkali metal fluoride in water is

Options:
A LiF<RbF<KF<NaF
B RbF<KF<NaF<LiF
C LiF>NaF>KF>RbF
D LiF<NaF<KF<RbF

The correct answer is:
LiF<NaF<KF<RbF

Solution:
Higher the lattice enthalpy lower will be solubility
i.e., lattice enthalpy ∝1Solubility
Since the lattice enthalpy of alkali metals follow the order
Li>Na>K>Rb
Hence the correct order of solubility is
LiF<NaF<KF<RbF

Topic: Chemistry, s Block Elements, JEE Main
The solubility product of AgCl is 1.8 Ã 10â10. Precipitation of AgCl will occur only when equal volumes of which of the following solutions are mixed?

Options:
A 10â4 M Ag+ and 10â4 M Clâ
B 10â7Â MÂ Ag+ and 10â7Â MÂ Clâ
C 10â5Â MÂ Ag+ and 10â5Â MÂ Clâ
D 10â10Â MÂ Ag+ and 10â10Â MÂ Clâ

The correct answer is:
10â4 M Ag+ and 10â4 M Clâ

Solution:
After mixing,
Ag+=12 Ã 10â4=5 Ã 10â5 M
Clâ =12Ã 10â4=5 Ã 10â5 M
Q sp = Ag + Cl â = 5 Ã 10 â5 2 =2.5 Ã 10 â9
Since, Q sp > K sp â precipition takes place

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility product of a salt having general formula $M X_2$, in water is: $4 \times 10^{-12}$. The concentration of $\mathrm{M}^{2+}$ ions in the aqueous solution of the salt is

Options:
A
$2.0 \times 10^{-6} \mathrm{M}$

B
$1.0 \times 10^{-4} \mathrm{M}$

C
$1.6 \times 10^{-4} \mathrm{M}$

D
$4.0 \times 10^{-10} \mathrm{M}$

The correct answer is:
$1.0 \times 10^{-4} \mathrm{M}$

Solution:
$\mathrm{~K}_{\mathrm{sp}}=4 \mathrm{~s}^3, \mathrm{~S}=\sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=1 \times 10^{-4}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility product of a salt having general formula $M X_2$, in water is: $4 \times 10^{-12}$. The concentration of $\mathrm{M}^{2+}$ ions in the aqueous solution of the salt is

Options:
A
$2.0 \times 10^{-6} \mathrm{M}$

B
$1.0 \times 10^{-4} \mathrm{M}$

C
$1.6 \times 10^{-4} \mathrm{M}$

D
$4.0 \times 10^{-10} \mathrm{M}$

The correct answer is:
$1.0 \times 10^{-4} \mathrm{M}$

Solution:
$\mathrm{~K}_{\mathrm{sp}}=4 \mathrm{~s}^3, \mathrm{~S}=\sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=1 \times 10^{-4}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2005
The solubility product of $\mathrm{Ag}_{2} \mathrm{CrO}_{4}$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_{4}^{2-}$ ions in that
solution?

Options:
A $2 \times 10^{-4}$
B $16 \times 10^{-4}$
C $8 \times 10^{-4}$
D $12 \times 10^{-4}$

The correct answer is:
$2 \times 10^{-4}$

Solution:
$\begin{aligned} \mathrm{Ag}_{2} \mathrm{CrO}_{4} \longrightarrow 2 \mathrm{Ag}^{+}+\mathrm{CrO}_{4}^{2-} \\ & 2 s \\ K_{\mathrm{sp}} &=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right] \\ &=(2 s)^{2}(s) \\ K_{\mathrm{sp}} &=4 s^{3} \\ s &=\left(\frac{K_{\mathrm{sp}}}{4}\right)^{1 / 3} \\ &=\left(\frac{32 \times 10^{-12}}{4}\right)^{1 / 3} \\ &=2 \times 10^{-4} \mathrm{M} \end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility product of $\mathrm{AgCl}$ is $4.0 \times 10^{-10}$ at $298 \mathrm{~K}$. The solubility of $\mathrm{AgCl}$ in $0.04 \mathrm{~m} \mathrm{CaCl}{ }_2$ will be

Options:
A $2.0 \times 10^{-5} \mathrm{~m}$
B $1.0 \times 10^{-4} \mathrm{~m}$
C $5.0 \times 10^{-9} \mathrm{~m}$
D $2.2 \times 10^{-4} \mathrm{~m}$

The correct answer is:
$5.0 \times 10^{-9} \mathrm{~m}$

Solution:
Given:
\(\mathrm{Ksp}\) of \(\mathrm{AgCl}=4.0 \times 10^{-10}\)
Concentration of \(\mathrm{CaCl}_2=0.04 \mathrm{M}\)
When \(\mathrm{CaCl}_2\) dissolves, it dissociates completely into \(\mathrm{Ca}^{2+}\) and \(2 \mathrm{Cl}^{-}\):
\(\mathrm{CaCl}_2 \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-}\)
So, the concentration of \(\mathrm{Cl}^{-}\)from \(\mathrm{CaCl}_2\) will be:
\(\left[\mathrm{Cl}^{-}\right]=2 \times 0.04 \mathrm{M}=0.08 \mathrm{M}\)
Let the solubility of \(\mathrm{AgCl}\) in the solution be \(s\). At equilibrium, the concentration of \(\mathrm{Ag}^{+}\)ions will be \(s\) and the concentration of \(\mathrm{Cl}^{-}\)ions will be \(0.08 \mathrm{M}\) plus \(s\).
However, since \(s\) will be very small compared to \(0.08 \mathrm{M}\), we can approximate the concentration of \(\mathrm{Cl}^{-}\)ions as \(0.08 \mathrm{M}\).
The Ksp expression for \(\mathrm{AgCl}\) is:
\(\mathrm{Ksp}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\)
Substitute the known values into the equation:
\(4.0 \times 10^{-10}=s \times 0.08\)
Solve for \(s\) :
\(\begin{aligned}
& s=\frac{4.0 \times 10^{-10}}{0.08} \\
& s=5.0 \times 10^{-9} \mathrm{M}
\end{aligned}\)
Therefore, the solubility of \(\mathrm{AgCl}\) in \(0.04 \mathrm{M} \mathrm{CaCl}_2\) is \(5.0 \times 10^{-9} \mathrm{M}\), which corresponds to option (3).

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility product of BaCrO4 is 2.4Ã10-10M2 . The maximum concentration of Ba(NO3)2 possible without precipitation in a 6Ã10-4M K2CrO4 solution is

Options:
A 4×10-7M
B 1.2×1010M
C 6×10-4M
D 3×10-4M

The correct answer is:
4×10-7M

Solution:
Precipitation occurs, if Qc>Ksp . The maximum concentration of BaNO33 or Ba2+ without precipitation is obtained if
Qc=Ksp=Ba2+CrO42-
2.4×10-10M2=Ba2+6×10-4M
or Ba2+=2.4×10-10M26×10-4M
=0.4×10-6M
=4×10-7M .

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility product of BaSO4 is 1×10-10 at 298 K. The solubility of BaSO4 in 0.1M K2SO4(aq) solution is ------×10-9 g L-1(nearest integer).
Given: Molar mass of BaSO4 is 233 g mol-1

Options:

The correct answer is:
233

Solution:
For the insoluble salt barium sulphate, sulphate ion added is the common ion. Due to the common ion effect the solubility of barium sulphate decreases. Let us assume solubility of barium sulphate is S.

BaSO4s⇌Ba2+Saq+SO42-S+0.1aq
K2SO40.1-0.1⟶ K+0.1aq+SO42-0.1aq
KSP=Ba2+SO42-
1×10-10=S(S+0.1)
≃0.1 S
S=10-9M=233×10-9gL-1

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The solubility product of Cr(OH)3 at 298K is 6.0×10-31. The concentration of hydroxide ions in a saturated solution of CrOH3 will be

Options:
A 2.22×10-3114
B 18×10-3114
C 18×10-3112
D 4.86×10-2914

The correct answer is:
18×10-3114

Solution:
CrOH3âCr+3+3O-
Ksp=s.3S3
⇒6Ã10-31=27.3s4
⇒s=627Ã10-311/4
OH-=3s
=3Ã627Ã10-3114=18Ã10-311/4M

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The solubility product of PbI2 is 8.0×10-9. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x×10-6 mol/L. The value of x is _________ (Rounded off to the nearest integer)

[Given 2=1.41]

Options:

The correct answer is:
141

Solution:
Given: KspPbI2=8×10-9

To calculate : solubility of PbI2 in 0.1M solution of PbNO22

I PbNO32→Pb(aq)+2+2NO3-aq

0.1 M--- 0.1 M 0.2 M

II PbI2s⇌Pb+2aq+2I-aq

s 2s

[Pb+2]=s+0.1

≃0.1

Now: Ksp=8×10-9=Pb+2I-2

⇒8×10-9=0.1×2 s2

⇒8×10-8=4 s2⇒s=2×10-4

⇒S=141×10-6M

⇒x=141

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The solubility product values of barium sulphate and barium carbonate are 1.0 × 10–10 and 5.0 × 10–9 respectively When the two salts are simultaneously equilibrated in pure water the ratio of SO 4 2− CO 3 2− is

Options:
A 0.02
B 0.10
C 0.05
D 0.12

The correct answer is:
0.02

Solution:
K sp 1 = Ba 2 + SO 4 2 -
K sp 2 = Ba 2 + CO 3 2 -
K sp 1 K sp 2 = SO 4 2 - CO 3 2 - = 1 × 1 0 - 1 0 5 × 1 0 - 9 = 0.02

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solubility (S) of a solute in water varies with temperature as given by S=Ae-ΔH/RT,ΔH being the enthalpy of solution. For a given solute, variation of lnS with temperature is as shown in figure. The solute is expected to be

Options:
A CaO
B MgSO4
C CuSO4
D CuSO4.5H2O

The correct answer is:
CuSO4.5H2O

Solution:
lnS = lnA− ΔH RT
Slope =  −ΔH RT
lnS vs  1 T graph with negative slope.
So, ΔH>0
Salts with water of crystallisation.
i.e., CuSO4.5H2O .

Topic: Chemistry, Ionic Equilibrium, JEE Main
The solution curve of the differential equation, 1+e-x1+y2dydx=y2 which passes through the point 0, 1, is

Options:
A y2+1=yloge1+e-x2+2
B y2+1=yloge1+ex2+2
C y2=1+yloge1+ex2
D y2=1+yloge1+e-x2

The correct answer is:
y2=1+yloge1+ex2

Solution:
1+e-x1+y2dydx=y2

Separate variables

⇒1+y2dyy2=dx1+e-x

Integrate both side

⇒∫1+y2dyy2=∫dx1+e-x

⇒∫1+1y2dy=∫exdxex+1

⇒y-1y=lnex+1+C

Since curve pass through 0, 1.

⇒1-11=lne0+1+C ⇒C=-ln2

⇒y-1y=lnex+1-ln2

⇒y2-1y=lnex+12

⇒y2=1+yloge1+ex2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The solution curve of the differential equation ydxdy=xlogex-logey+1, x>0, y>0 passing through the point (e, 1) is

Options:
A logeyx=x
B logeyx=y2
C logexy=y
D 2logexy=y+1

The correct answer is:
logexy=y

Solution:
Given: ydxdy=xlogex-logey+1, x>0, y>0

⇒dxdy=xylogexy+1

Putting, x=vy

⇒dxdy=v+ydvdx

⇒v+ydvdy=vyylogevyy+1

⇒v+ydvdy=vlogev+1

⇒v+ydvdy=vlogev+v

⇒ydvdy=vlogev

⇒dvvlogev=dyy

⇒∫dvvlogev=∫dyy

Putting, logev=t

⇒dvv=dt

⇒∫dtt=∫dyy

⇒logt=logy+c

⇒loglogexy=logy+c

Using point e,1

⇒loglogee=log1+c

⇒c=0

⇒loglogexy=logy

⇒logexy=y

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The solution for the differential equation \( \frac{d y}{y}+\frac{d x}{x}=0 \)

Options:
A \( \frac{1}{y}+\frac{1}{x}=C \)
B \( \log x \cdot \log y=c \)
C \( x y=c \)
D \( x+y=c \)

The correct answer is:
\( x y=c \)

Solution:
Given differential equation is,
\( \frac{d y}{y}+\frac{d x}{x}=0 \)
Integrating both the sides
\[
\begin{array}{l}
\int \frac{d y}{y}+\int \frac{d x}{x}=\int \operatorname{Od} x \\
\Rightarrow \log y+\log x=\log c \\
\Rightarrow \log (x y)=\log c \\
\Rightarrow x y=c
\end{array}
\]

Topic: Mathematics, Differential Equations, JEE Main
The solution for $x$ of the equation
$\int_{\sqrt{2}}^x \frac{d t}{t \sqrt{t^2-1}}=\frac{\pi}{2}$ is

Options:
A
$2$

B
$\pi$

C
$\frac{\sqrt{3}}{2}$

D
None of these

The correct answer is:
None of these

Solution:
$\int_{\sqrt{2}}^x \frac{d t}{t \sqrt{t^2-1}}=\frac{\pi}{2}$
$\left[\sec ^{-1} t\right]_{\sqrt{2}}^x=\frac{\pi}{2}$
$\sec ^{-1} x-\frac{\pi}{4}=\frac{\pi}{2}$
$\sec ^{-1} x=\frac{3 \pi}{4}$
$x=-\sqrt{2}$.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2007
The solution from the following with highest depression in freezing point/lowest freezing point is

Options:
A 180 g of acetic acid dissolved in 1 L of aqueous solution.
B 180 g of acetic acid dissolved in benzene
C 180 g of benzoic acid dissolved in benzene
D 180 g of glucose dissolved in water

The correct answer is:
180 g of acetic acid dissolved in 1 L of aqueous solution.

Solution:
Depression in freezing point or elevation in melting point depends upon the number of solute particles rather than the kind of solute particles.

Depression in freezing point (ΔTf) is directly proportional to number of solute particles.

Also, ΔTf is directly proportional to molality.

ΔTf=m×kf

Amongst the given options acetic acid has lowest molecular mass.

The molar mass of acetic acid is 60.05 g mol-1.

Hence if the molecular mass is less, molality will be more, so the lowest freezing point will be shown by 180 gm of acetic acid in 1L aqueous solution.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The solution of $\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0$ is :

Options:
A $3 x\left(1+y^2\right)=4 y^3+c$
B $3 y\left(1+x^2\right)=4 x^3+c$
C $3 x\left(1-y^2\right)=4 y^3+c$
D $3 y\left(1+y^2\right)=4 x^3+c$

The correct answer is:
$3 y\left(1+x^2\right)=4 x^3+c$

Solution:
$\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0$
$\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}$
On comparing with $\frac{d y}{d x}+P y=Q$, we get
$P=\frac{2 x}{1+x^2}, Q=\frac{4 x^2}{1+x^2}$
I.F. $=e^{\int P d x}=e^{\log \left(1+x^2\right)}$
$=1+x^2$
The solution is
$y \text { (I.F.) }=\int Q \text { (I. F. ) } d x+c_1$
$\Rightarrow \quad y\left(1+x^2\right)=\int \frac{4 x^2}{\left(1+x^2\right)}\left(1+x^2\right) d x+C_1$
$\Rightarrow \quad y\left(1+x^2\right)=\frac{4 x^3}{3}+c_1$
$\Rightarrow \quad 3 y\left(1+x^2\right)=4 x^3+c$

Topic: Mathematics, Differential Equations, JEE Main
The solution of $\frac{d y}{d x}-1=e^{x-y}$ is

Options:
A $e^{x-y}+x=c$
B $e^{-(x-y)}+x=c$
C $e^{-(x-y)}=x+c$
D $e^{x-y}=x+c$

The correct answer is:
$e^{-(x-y)}=x+c$

Solution:
Given, differential equation is
$\frac{d y}{d x}-1=e^{x-y}$
$\begin{array}{lrl}
\text { Put } & & x-y=t \\
\Rightarrow & 1-\frac{d y}{d x}=\frac{d t}{d x}
\end{array}$
So, given equation becomes,
$$
\begin{aligned}
-\frac{d t}{d x} &=e^{t} \\
-e^{-t} d t &=d x
\end{aligned}
$$
Integrating both sides, we get
$$
\begin{aligned}
&e^{-t}=x+c \\
&e^{-(x-y)}=x+c
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main
The solution of $\frac{d y}{d x}=e^x(\sin x+\cos x)$

Options:
A $y=e^x(\sin x-\cos x)+c$
B $y=e^x(\cos x-\sin x)+c$
C $y=\mathrm{e}^x \sin x+c$
D $y=e^x \cos x+c$

The correct answer is:
$y=\mathrm{e}^x \sin x+c$

Solution:
Given equation $\frac{d y}{d x}=e^x(\sin x+\cos x)$
$\Rightarrow d y=e^x(\sin x+\cos x) d x$
On integrating, we get $y=e^x \sin x+c$

Topic: Mathematics, Differential Equations, JEE Main
The solution of $\frac{d y}{d x}=x \log x$ is

Options:
A $y=x^2 \log x-\frac{x^2}{2}+c$
B $y=\frac{x^2}{2} \log x-x^2+c$
C $y=\frac{1}{2} x^2+\frac{1}{2} x^2 \log x+c$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & \frac{d y}{d x}=x \log x \Rightarrow d y=x \log x d x \\ & \Rightarrow \int d y=\int x \log x d x \Rightarrow y=\frac{x^2}{2} \log x-\frac{x^2}{4}+c\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of dy=cosx2-y cosec xdx, where y=2 when x=Ï/4, is

Options:
A y=sinâ¡x+12 cosec x
B y=tanâ¡(x/2)+cotâ¡x/2
C y=1/2secâ¡x/2+2cosâ¡x/2
D None of the above

The correct answer is:
y=sinâ¡x+12 cosec x

Solution:
Given, dydx=2cosâ¡x-ycosâ¡x cosec x
â dydx+ycotâ¡x=2cosâ¡x
â´ IF=eâ«cotâ¡x dx=elnâ¡sinâ¡x=sinâ¡x
â´ Solution is ysinâ¡x=â«2cosâ¡xsinâ¡x dx+c
â ysinâ¡x=â«sinâ¡2x dx+c
â ysinâ¡x=-cosâ¡2x2+c
At x=Ï4,y=2
â´ 2sinâ¡Ï4=-cosâ¡2(Ï/4)2+c
â c=1
â´ ysinâ¡x=-12cosâ¡2x+1
â y=-12. cosâ¡2xsinâ¡x+cosec x
â y=-12sinâ¡x1-2sin2â¡x+cosec x
â y=12 cosec x+sinâ¡x

Topic: Mathematics, Differential Equations, JEE Main
The solution of dydx=cosx+y+sinx+y, is

Options:
A log1+tanx+y2+C=0
B log1+tanx+y2=x+C
C log1-tanx+y2=x+C
D None of the above

The correct answer is:
log1+tanx+y2=x+C

Solution:
Put x+y=v and 1+dydx=dvdx

Therefore, the differential equation reduces to

dvdx=1+cosv+sinv=2cos2v2+2sinv2cosv2

=2cos2v21+tanv2

⇒ ∫sec2V221+tanv2dv=∫dx

∴ log1+tanx+y2=x+C

Topic: Mathematics, Differential Equations, JEE Main
The solution of differential equation \(2 x \frac{d y}{d x}-y=3\) represents a family of

Options:
A circles
B straight lines
C ellipses
D parabola

The correct answer is:
parabola

Solution:
We have, \(2 x \frac{d y}{d x}=y+3 \Rightarrow \frac{2}{y+3} d y=\frac{d x}{x}\) integrating, \(2 \ln (y+3)=\ln x+\ln c=\ln c x\)
\(\Rightarrow \ln (y+3)^2=\ln \mathrm{cx} \Rightarrow(y+3)^2=\mathrm{cx}\)
which is a family of parabolas.

Topic: Mathematics, Vector Algebra, JEE Main
The solution of differential equation $\frac{d y}{d x}+\sin ^2 y=0$

Options:
A $y+2 \cos y=c$
B $y-2 \sin y=c$
C $x=\cot y+c$
D $y=\cot x+c$

The correct answer is:
$x=\cot y+c$

Solution:
$\frac{d y}{d x}+\sin ^2 y=0 \Rightarrow-\frac{d y}{\sin ^2 y}=d x$
On integrating, we get $x-\cot y+c$

Topic: Mathematics, Differential Equations, JEE Main
The solution of differential equation \( \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y} f^{\prime}(\mathrm{x})-\mathrm{y}^{2}}{f(\mathrm{x})} \) is

Options:
A \( f(x)=y(x-c) \)
B \( f(x)=y(c-x) \)
C \( f(x)=y(x+c) \)
D None of these

The correct answer is:
\( f(x)=y(x+c) \)

Solution:
The given equation is
dydx=yÆ′x-y2Æx
âÂ yÆ′Â xdxÂ âÂ Æxdy=y2dx
âyÆ′xdx-Æxdyy2=dxâÂ dÂ Æxy=dx
On integration, we get
Æxy=x+cÂ âÆx=yx+c

Topic: Mathematics, Differential Equations, JEE Main
The solution of differential equation \( \left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0 \) is

Options:
A \( 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k \)
B \( 2 x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}+k \)
C \( x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}+k \)
D \( x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}-k \)

The correct answer is:
\( 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k \)

Solution:
Given equation can be rewritten as

dxdy+11+y2x=etan-1y1+y2

∴ If =e∫11+y2 dy=etan1y

∴ Required solution is

xetan1y=∫etan-1yetan-1y1+y2 dy

Put etan-1y=t⇒etan-1y 11+y2dy=dt

∴ xetan-1y=∫t dt=t22+c

⇒ 2xetan-1y=e2tan-1y+k

Topic: Mathematics, Differential Equations, JEE Main
The solution of differential equation $x^{2} \frac{d y}{d x}=y^{2}+x y$ is

Options:
A $\frac{x}{y}+\log |x|=c$
B $\frac{y}{x}+\log |x|=c$
C $\frac{x}{y}-\log |x|=c$
D $\frac{y}{x}-\log |x|=c$

The correct answer is:
$\frac{x}{y}+\log |x|=c$

Solution:
We have $x^{2} \frac{d y}{d x}=y^{2}+x y$
$\begin{array}{l}
\therefore \frac{d y}{d x}=\frac{y^{2}+x y}{x^{2}} ...(1) \\
\therefore \text { Put } y=u x \\
\therefore \frac{d y}{d x}=u+x \frac{d u}{d x}
\end{array}$
$\therefore$ Equation (1) becomes
$u+x \frac{d u}{d x}=\frac{u^{2} x^{2}+x(u x)}{x^{2}} \Rightarrow u+x \frac{d u}{d x}=u^{2}+u$
$\begin{array}{l}
\therefore \mathrm{x} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}^{2} \\
\therefore \int \frac{\mathrm{du}}{\mathrm{u}^{2}}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\
\therefore \frac{-1}{\mathrm{u}}=\log |\mathrm{x}|+\mathrm{c}_{1} \Rightarrow \quad \frac{-\mathrm{x}}{\mathrm{y}}=\log |\mathrm{x}|+\mathrm{c}_{1} \\
\therefore \frac{\mathrm{x}}{\mathrm{y}}+\log |\mathrm{x}|=\mathrm{c}
\end{array}$

Topic: Mathematics, Differential Equations, JEE Main
The solution of differential equation $x \frac{d y}{d x}+\boldsymbol{y}=\boldsymbol{y}^2$ is

Options:
A $y=1+c x y$
B $y=\log \{c x y\}$
C $y+1=c x y$
D $y=c+x y$

The correct answer is:
$y=1+c x y$

Solution:
$\begin{aligned} & x \frac{d y}{d x}+y=y^2 \Rightarrow x \frac{d y}{d x}=y^2-y \\ & \Rightarrow \frac{d y}{y^2-y}=\frac{d x}{x} \Rightarrow\left[\frac{1}{y-1}-\frac{1}{y}\right] d y=\frac{d x}{x}\end{aligned}$
On integrating, we get $\log (y-1)-\log y-\log x+\log c$
$\Rightarrow \frac{y-1}{y}=x c \Rightarrow y=1+c x y$.

Topic: Mathematics, Differential Equations, JEE Main
The solution of \( \frac{d z}{d x}+\frac{z}{x} \log z=\frac{z}{x^{2}}(\log z)^{2} \) is

Options:
A \( \left(\frac{1}{\log z}\right) x=2-x^{2} c \)
B \( \left(\frac{1}{\log z}\right) x=2+x^{2} c \)
C \( \left(\frac{1}{\log z}\right) x=x^{2} c \)
D \( \left(\frac{1}{\log z}\right) x=\frac{1}{2}+c x^{2} \)

The correct answer is:
\( \left(\frac{1}{\log z}\right) x=\frac{1}{2}+c x^{2} \)

Solution:
Given, dzdx+zxlogz=zx2 logz2

Dividing given equation by zlogz2,

⇒1zlogz2dzdx+1x1logz=1x2
Put 1logz=t⇒-1logz2·1z·dzdx=dtdx
∴ -dtdx+tx=1x2
⇒dtdx-tx=-1x2 ...i

Now, integrating factorI.F=e∫-dxx=e-lnx=eln1x=1x
⇒tx=-∫1x3dx=12x2+c, where, c is the constant of integration.

⇒1xlogz=12x2+c
⇒1logzx=12+cx2.

Topic: Mathematics, Differential Equations, JEE Main
The solution of \( \left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x \) is

Options:
A \( \log x=\log (x-y)+\frac{y}{x}+C \)
B \( \log x=2 \log (x-y)+\frac{y}{x}+C \)
C \( \log x=\log (x-y)+\frac{x}{y}+C \)
D none of these

The correct answer is:
\( \log x=2 \log (x-y)+\frac{y}{x}+C \)

Solution:
x2dy+xydy=x2dx+y2dx

dividing both sides by x2

dy+yxdy=dx+yx2dx

1+yxdydx=1+yx2 ....i

As it is of type dydx=f(yx)

Put yx=t⇒dydx=t+xdtdx ....ii

from equations (i) and (ii), we get

(1+t)t+xdtdx=1+t2

xdtdx=1-t1+t⇒-1+tt-1dt=dxx ....iiiintegrating both sides,

-∫2t-1+t-1t-1dt=∫dxx

-2ln|t-1|-t=lnx+C

-2lnyx-1-yx=lnx+C

lnx=2ln|y-x|+yx+C

or, lnx=2ln|x-y|+yx+C.

Topic: Mathematics, Differential Equations, JEE Main
The solution of Na2CO3 has pH

Options:
A greater than 7
B less than 7
C equal to 7
D equal to zero

The correct answer is:
greater than 7

Solution:
Na2CO3 is the salt of strong base and weak acid.
Na 2 CO 3 + 2 H 2 O → 2 NaOH + H 2 CO 3 Strong base Weak acid
Hence, the solution of Na2CO3 has pH greater than 7.

Topic: Chemistry, Ionic Equilibrium, JEE Main