datavorous/entrance-exam-dataset huggingface dataset:
1.86 g of aniline completely reacts to form acetanilide. 10% of the product is lost during purification. Amount of acetanilide obtained after purification (in g) is ___________ ×10-2.

Options:

The correct answer is:
243

Solution:
Given 1.86 g

⇒1 mol C6H5NH2 give 1 mol

∴ moles of C6H5NH2= moles of

⇒1.8693=Wacetanilide 135

⇒Wacelanilide =1.86×13593 g=2.70 g

But efficiency of reaction is 90% only

∴ Mass of acetanilide produced=2.70×90100 g

=2.43 g

=243×10-2 g

⇒x=243

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (24 Feb Shift 2)
$\frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}+\ldots .+\frac{C_n}{n+1}=$

Options:
A $\frac{2^n}{n+1}$
B $\frac{2^n-1}{n+1}$
C $\frac{2^{n+1}-1}{n+1}$
D None of these

The correct answer is:
$\frac{2^{n+1}-1}{n+1}$

Solution:
Proceeding as above and putting $n+1=N$.
So given term can be written as
$\begin{aligned}
& \frac{1}{N}\left\{{ }^N C_1+{ }^N C_2+{ }^N C_3+\ldots .\right\} \\
= & \frac{1}{N}\left\{2^N-1\right\}=\frac{1}{n+1}\left(2^{n+1}-1\right) \quad(N=n+1)\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main
$\frac{\sin 2 A}{1+\cos 2 A} \cdot \frac{\cos A}{1+\cos A}=$

Options:
A $\tan \frac{A}{2}$
B $\cot \frac{A}{2}$
C $\sec \frac{A}{2}$
D $\operatorname{cosec} \frac{A}{2}$

The correct answer is:
$\tan \frac{A}{2}$

Solution:
$\begin{aligned}
& \left(\frac{\sin 2 A}{1+\cos 2 A}\right)\left(\frac{\cos A}{1+\cos A}\right) \\
& =\frac{2 \sin A \cos A}{2 \cos ^2 A} \frac{\cos A}{1+\cos A}=\frac{\sin A}{1+\cos A}=\tan \frac{A}{2}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$ is

Options:
A
$\frac{\pi^2}{4}$

B
$\pi^2$

C
zero

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

The correct answer is:
$\pi^2$

Solution:
$$
\begin{aligned}
& \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\int_{-\pi}^\pi \frac{2 x}{1+\cos ^2 x}+2 \int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} \\
& =0+4 \int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x} I=4 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} \\
& I=4 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} \Rightarrow I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-4 \pi \int \frac{x \sin x}{1+\cos ^2 x} \Rightarrow 2 I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x
\end{aligned}
$$
put $\cos x=t$ and solve it.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2002
$\left(\frac{1+\cos \phi+i \sin \phi}{1+\cos \phi-i \sin \phi}\right)^n=$

Options:
A $\cos n \phi-i \sin n \phi$
B $\cos n \phi+i \sin n \phi$
C $\sin n \phi+i \cos n \phi$
D $\sin n \phi-i \cos n \phi$

The correct answer is:
$\cos n \phi+i \sin n \phi$

Solution:
L.H.S. $=\left[\frac{2 \cos ^2(\phi / 2)+2 i \sin (\phi / 2) \cos (\phi / 2)}{2 \cos ^2(\phi / 2)-2 i \sin (\phi / 2) \cos (\phi / 2)}\right]^n$
$\begin{aligned}=\left[\frac{\cos (\phi / 2)+i \sin (\phi / 2)}{\cos (\phi / 2)-i \sin (\phi / 2)}\right]^n= & {\left[\frac{e^{i(\phi / 2)}}{e^{-i(\phi / 2)}}\right]^n=\left(e^{i \phi}\right)^n } \\ & =\cos n \phi+i \sin n \phi .\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
$\int \frac{x+\sin x}{1+\cos x} d x=$

Options:
A $x \tan \frac{x}{2}+C$
B $x \sin \frac{x}{2}+\cos \frac{x}{2}+C$
C $x \tan \frac{x}{2}+\sec \frac{x}{2}+C$
D $x \sec \frac{x}{2}+\tan \frac{x}{2}+C$

The correct answer is:
$x \tan \frac{x}{2}+C$

Solution:
$\begin{aligned} & \text { Let } I=\int \frac{x+\sin x}{1+\cos x} d x \\ & =\int \frac{x}{1+\cos x} d x+\int \frac{\sin x}{1+\cos x} d x \\ & =\int \frac{x}{2 \cos ^2 \frac{x}{2}} d x+\int \frac{\sin x}{2 \cos ^2 \frac{x}{2}} d x \\ & =\frac{1}{2} \int x \sec ^2 \frac{x}{2} d x+\int \frac{2 \sin x / 2 \cos x / 2}{2 \cos ^2 x / 2} d x \\ & =\frac{1}{2} \int x \sec ^2 \frac{x}{2} d x+\int \tan \frac{x}{2} d x \\ & =\frac{1}{2}\left[x \frac{\tan \frac{x}{2}}{1 / 2}-\int \frac{\tan x / 2}{\frac{1}{2}} d x\right]+\int \tan \frac{x}{2} d x \\ & =x \tan \frac{x}{2}-\int \tan \frac{x}{2} d x+\int \tan \frac{x}{2} d x=x \tan \frac{x}{2}+C\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
(1) Centre of gravity (CG) of a body is the point at which the weight of the body acts.
(2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius.
(3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its CG.
(4) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.
Which one of the following pairs of statements is correct?

Options:
A (4) and (1)
B (1) and (2)
C (2) and (3)
D (3) and (4)

The correct answer is:
(4) and (1)

Solution:
Centre of gravity of a body is the point at which the weight of the body acts and the radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
(1) Centre of gravity (CG) of a body is the point at which the weight of the body acts.
(2) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius.
(3) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its CG.
(4) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.
Which one of the following pairs of statements is correct?

Options:
A (4) and (1)
B (1) and (2)
C (2) and (3)
D (3) and (4)

The correct answer is:
(4) and (1)

Solution:
Centre of gravity of a body is the point at which the weight of the body acts and the radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the CG of the body to the axis.

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2010 (Mains)
$\int \sqrt{1+\cos x} d x$ equals

Options:
A $2 \sqrt{2} \sin \frac{x}{2}+c$
B $-2 \sqrt{2} \sin \frac{x}{2}+c$
C $-2 \sqrt{2} \cos \frac{x}{2}+c$
D $2 \sqrt{2} \cos \frac{x}{2}+c$

The correct answer is:
$2 \sqrt{2} \sin \frac{x}{2}+c$

Solution:
$\begin{aligned} & I=\int \sqrt{1+\cos x} d x=\int \sqrt{2 \cos ^2(x / 2)} d x \\ & I=\sqrt{2} \int \cos (x / 2) d x=2 \sqrt{2} \sin (x / 2)+c\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \sqrt{1+\cos x} d x$ is equal to

Options:
A $2 \sqrt{2} \cos \frac{\mathrm{x}}{2}+\mathrm{C}$
B $2 \sqrt{2} \sin \frac{\mathrm{x}}{2}+\mathrm{C}$
C $\sqrt{2} \cos \frac{x}{2}+C$
D $\sqrt{2} \sin \frac{x}{2}+C$

The correct answer is:
$2 \sqrt{2} \sin \frac{\mathrm{x}}{2}+\mathrm{C}$

Solution:
Hints: $\int \sqrt{1+\cos x} d x=\sqrt{2} \int \cos \left(\frac{x}{2}\right) d x=2 \sqrt{2} \sin \left(\frac{x}{2}\right)+c$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x=$

Options:
A $\log\left[\left(1+e^x\right)\left(2+e^x\right)\right]+c$
B $\log \left[\frac{1+e^x}{2+e^x}\right]+c$
C $\log \left[\left(1+e^x\right) \sqrt{2+e^x}\right]+c$
D None of these

The correct answer is:
$\log \left[\frac{1+e^x}{2+e^x}\right]+c$

Solution:
$\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)} d x=\int\left\{\frac{e^x}{1+e^x}-\frac{e^x}{2+e^x}\right\} d x$
Now put $1+e^x=t$ and $2+e^x=t$, then the required integral
$=\log \left(1+e^x\right)-\log \left(2+e^x\right)=\log \left(\frac{1+e^x}{2+e^x}\right)+c .$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{d x}{1+e^x}=$

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

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

Solution:
$\int \frac{d x}{1+e^x}=\int \frac{e^{-x}}{1+e^{-x}} dx$
Put $1+e^{-x}=t \Rightarrow e^{-x} d x=-d t$, then it reduces to $-\int \frac{d t}{t}=-\log t=-\log \left(1+e^{-x}\right)$

Topic: Mathematics, Indefinite Integration, JEE Main
1 g of 79 Au 198 t 1/2 =65 h give stable mercury by Î²- emission. What amount of mercury will left after 260 h?

Options:
A 0.9375 g
B 0.3758 g
C 0.7586 g
D 0.9000 g

The correct answer is:
0.9375 g

Solution:
79Au198â -B 80Hg198
k =  0.693 t 1/2  =  0.693 65
After 260 hr,
k =  2.303 260  log  a a−x
0.693 65  =  2.303 260  log  a a−x
a a−x  = 16
1 1−x  = 16
x =  15 16  g = 0.9375 g

Topic: Chemistry, Chemical Kinetics, JEE Main
1 g of a carbonate M2CO3 on treatment with excess HCl produces 0.01 mol of CO2. The molar mass of M2CO3 is gmol-1. (Nearest integer)

Options:

The correct answer is:
100

Solution:
From Balanced chemical reaction,

M2CO3+2HCl→2MCl+H2O+CO2

Here from this, we get:

0.01 mol CO2≡0.01 mol M2CO3≡1 g M2CO3

We know that,

Number of moles = Given massMolar mass

⇒Molar mass of M2CO3=Given mass of M2CO3Number of Moles of M2CO3=10.01=100 g/mol

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (13 Apr Shift 2)
1 g of a complex [Cr(H 2 O) 5 Cl]Cl 2 ⋅ H 2 O (mol. Wt. 266.5) was passed through a cation exchanger to produce HCl . The acid liberated was diluted to 1 litre. The normality of this acid solution is

Options:
A 5×10–3 N
B 7.5×10–3 N
C 7.5×10–2 N
D 7.5×10–1 N

The correct answer is:
7.5×10–3 N

Solution:
Mole of complex =Wmol.wt.=1266.5
1 mole of complex produces 2 mole HCl
[Cr(H 2 O) 5 Cl]Cl 2 → Cr(H 2 O)Cl] 2+ 2Cl –
Mole of HCl formed =1266.5×2
N HCl  =  n V  × n−factor =  2 266.5×1
= 7.5 ×  10 –3  N

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
1 g of a liquid is converted to vapour at 3×105 Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm3 during this phase change, then the increase in internal energy in the process will be :

Options:
A 4320 J
B 432000 J
C 4800 J
D 4.32×108 J

The correct answer is:
4320 J

Solution:
As pressure against which expansion is happening is constant, work done can be written as =PΔV

=3×105×1600×10-6

=480 J
As only 10% of heat is used in work done.
Hence ΔQ=W0.1=4800 J
The rest goes in increase in internal energy, which is 90% of heat.
Change in internal energy =0.9×4800=4320 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (24 Jan Shift 1)
1 g of a monobasic acid (HA) in 100 g water lowers the freezing point by 0.1 K. If 0.75 g of same acid requires 15 ml of 15 N NaOH solution for complete neutralization, then % degree of ionization of acid in water is Kf of H2O=1.86 Kkgmol-1

Options:
A 0.22
B 0.25
C 0.34
D 0.50

The correct answer is:
0.34

Solution:
Normal molecular weight of acid can be found as
Meq of acid = Meq of NaOH
N1V1=N2V2; N1V1=M2V2
0.75MacidÃ1000=15Ã15; Macid=250
Observed molecular weight of acid can be formed as
âTf=KfWacidMacidÃ1000WsolventÃi ; 0.1=1.86Ã1250Ã1000100Ã1+Î±
1.34=1+Î±
Î±=0.34

Topic: Chemistry, Solutions, JEE Main
1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1:5. The ratio of the elevation in their boiling points, ∆TbA∆TbB, is: (assuming they have the same molar mass)

Options:
A 10:1
B 1:5
C 1:0.2
D 5:1

The correct answer is:
1:5

Solution:
Elevation in Boiling point ∆Tb=i×kb×m
For non-electrolyte, i=1

m= Molar mass of solute

m1=molality of solution 1m2= molality of solution 2
∆Tb1=k1×1×m1, m1=1100×m×1000=10m
∆Tb2=k2×1×m2, m2=1100×m×1000=10m
m1=m2
∆Tb1∆Tb2=k1k2=1:5

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (10 Apr Shift 2)
1 g of Au 79 198 t 1 2 = 65 h decays by Beta emission and produce stable mercury. How much mercury will be present after 260 h.

Options:
A 0.93 g
B 0.85 g
C 1 g
D 0.79 g

The correct answer is:
0.93 g

Solution:
Au19198âHg80108+Î¸-10
t12=65h
T=260h
T=t12Ãâ²nâ²
Number of Lay live (n) =26065=4
Amount of undecayed Au=N02n=124=116g
Amount of decayed Au=1-116=1516=0.93 g

Topic: Chemistry, Chemical Kinetics, JEE Main
1 g of non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1:5. The ratio of the elevation in their boiling point, ΔTbAΔTbB, is

Options:
A 5 : 1
B 10 : 1
C 1 : 5
D 1 : 3

The correct answer is:
1 : 5

Solution:
ΔTb=Kb×m⇒ΔTb∝Kb
As mA=mB (given)
So ΔTbAΔTbB=KbAKbB
As KbAKbB=15
So ΔTbAΔTbB=15

Topic: Chemistry, Solutions, JEE Main
1 gram of a carbonate M2CO3 on treatment with excess HCl produces 0.01186 moles of CO2 . The molar mass of M2CO3 in gmol-1 is:

Options:
A 84.3
B 118.6
C 11.86
D 1186

The correct answer is:
84.3

Solution:
M2CO3+2HCl→2MCl+H2O+CO2
From reaction,
Moles of M2CO3=Moles of CO2

Weight of M2CO3Molar Mass of M2CO3 =Weight of CO2Molar mass of CO2

1Molar Mass of M2CO3 =0.01186 ×4444
1Molar Mass of M2CO3=0.01186
Molar Mass =10.01186=84.3 g/mol

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2017 (02 Apr)
$\left|\begin{array}{ccc}1+i & 1-i & i \\ 1-i & i & 1+i \\ i & 1+i & 1-i\end{array}\right|=$

Options:
A $-4-7 i$
B $4+7 i$
C $3+7 i$
D $7+4 i$

The correct answer is:
$4+7 i$

Solution:
$\begin{aligned} & \Delta=(2+i)\left|\begin{array}{ccc}1 & 1 & i \\ 1 & 1+2 i & 1+i \\ 1 & 2 & 1-i\end{array}\right| \\ & =(2+i)\left|\begin{array}{ccc}0 & -2 i & -1 \\ 0 & -1+2 i & 2 i \\ 1 & 2 & 1-i\end{array}\right| \text { by } \begin{array}{l}R_1 \rightarrow R_1-R_2 \\ R_2 \rightarrow R_2-R_3\end{array} \\ & =(2+i)\left\{-4 i^2+(-1+2 i)\right\}=(2+i)(4-1+2 i) \\ & =(2+i)(3+2 i)=4+7 i .\end{aligned}$

Topic: Mathematics, Determinants, JEE Main
$\frac{1-i}{1+i}$ is equal to

Options:
A $\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}$
B $\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}$
C $\sin \frac{\pi}{2}+i \cos \frac{\pi}{2}$
D None of these

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

Solution:
$\frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{1+(i)^2-2 i}{1+1}=-i$
which can be written as $\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}$

Topic: Mathematics, Complex Number, JEE Main
$\left\{\frac{2 i}{1+i}\right\}^2=$

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

The correct answer is:
$2 i$

Solution:
$\left[\frac{2 i}{1+i}\right]^2=\left[\left(\frac{2 i}{1+i}\right)\left(\frac{1-i}{1-i}\right)\right]^2=(i+1)^2=i^2+1+2 i=2 i$

Topic: Mathematics, Complex Number, JEE Main
$1+i^2+i^4+i^6+\ldots .+i^{2 n}$ is

Options:
A Positive
B Negative
C Zero
D Cannot be determined

The correct answer is:
Cannot be determined

Solution:
$S=1+i^2+i^4+\ldots . .+i^{2 n}=1-1+1-1+\ldots \ldots+(-1)^n$
Obviously it depends on $n$.
Hence cannot be determined unless $n$ is known.

Topic: Mathematics, Complex Number, JEE Main
$(-1+i \sqrt{3})^{20}$ is equal to

Options:
A $2^{20}(-1+i \sqrt{3})^{20}$
B $2^{20}(1-i \sqrt{3})^{20}$
C $2^{20}(-1-i \sqrt{3})^{20}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & \text { Let } z=-1+i \sqrt{3}, r=\sqrt{1+3}=2 \\ & \theta=\tan ^{-1}\left(\frac{\sqrt{3}}{-1}\right)=\frac{2 \pi}{3} \\ & \therefore z=2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right) \\ & \therefore(z)^{20}=\left[2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)\right]^{20} \\ & =2^{20}\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)^{20}=2^{20}\left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{20}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
$1 \mathrm{Kcal}$ of heat flowing through a rod of iron. When the rod is cut down to 4 pieces then what will be the heat flowing through each piece having same differential temperature?

Options:
A $1 / 2 \mathrm{Kcal}$
B $1 / 4 \mathrm{Kcal}$
C $1 \mathrm{Kcal}$
D $1 / 15 \mathrm{Kcal}$.

The correct answer is:
$1 \mathrm{Kcal}$

Solution:
We know that the heat flowing through a conductor is given by
$$
\begin{aligned}
\frac{Q}{t} & =-K A\left(\frac{d \theta}{d x}\right) \\
\Rightarrow Q & =-K A\left(\frac{d \theta}{d x}\right) t
\end{aligned}
$$
Where
$$
\begin{aligned}
K & =\text { thermal conductivity of the material } \\
A & =\text { area of cross-section } \\
\frac{d \theta}{d x} & =\text { temperature gradient } \\
t & =\text { time }
\end{aligned}
$$
Now if we cut the rod into 4 pieces, $A$ remains same $K$ remains same, $\frac{d \theta}{d x}$ is said to be same, so for same time $t, Q$ is going to be same as earlier.

Topic: Physics, Thermal Properties of Matter, AIIMS, AIIMS 2001
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to -4°C before freezing. The amount of ice (in g) that will be separated out is ___________. (Nearest integer) [Given : KfH2O=1.86 K kg mol-1]

Options:

The correct answer is:
518

Solution:
Let mass of water initially present =xg

Mass of sucrose =(1000-x)g

Number of moles sucrose =1000-x342

Molality M=1000-x342x/1000=0.75

2565.5x=106-1000x

x=795.86 g

no. of moles of sucrose=0.5969

ΔTf=ikf m

4=1×1.86×0.5969(795.86-a)g×1000

Ice separated=518.3 g

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (27 Aug Shift 1)
1 kg of water, at 20 °C is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 Ω. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully is close to
[ Specific heat of water =4200 J kg-1 °C-1 Latent heat of water =2260 kJ kg-1 ]

Options:
A 3 min
B 16 min
C 22 min
D 10 min

The correct answer is:
22 min

Solution:
Given specific heat of the water 4200 J kg-1 °C-1
Latent heat of the water =2260  kJ kg-1
Q=P×t
Q=mcΔT+mL
⇒P×t=mcΔT+mL
⇒Vrms2R×t=mcΔT+mL
⇒200220×t=4200×80+2260×103
⇒t=1298 s≃22 min

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (12 Apr Shift 2)
1 kg of water, at 20 Â°C is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20âÎ©. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully is close to
[ Specific heat of water =4200 J kg-1 Â°C-1 Latent heat of water =2260 kJ kg-1 ]

Options:
A 3 min
B 16 min
C 22 min
D 10 min

The correct answer is:
22 min

Solution:
Given specific heat of the water 4200 J kg-1 °C-1
Latent heat of the water =2260 kJ kg-1
Q=P×t
Q=mcΔT+mL
⇒P×t=mcΔT+mL
⇒Vrms2R×t=mcΔT+mL
⇒200220×t=4200×80+2260×103
⇒t=1298 s≃22 min

Topic: Physics, Thermal Properties of Matter, NEET
1 L aqueous solution of H2SO4 contains 0.02 m mol H2SO4.50% of this solution is diluted with deionized water to give 1 L solution A. In solution A, 0.01 m mol of H2SO4 are added. Total m mols of H2SO4 in the final solution is____×10-3 m moles.

Options:

The correct answer is:
20

Solution:
Initial moles of H2SO4 (in/Lit.) =0.02

In 50% solution moles of H2SO4=0.01 m moles

Added moles of H2SO4=0.01 m moles

Total moles of H2SO4 in resulting solution =0.02 m moles

=20×10-3 m moles

=20×10-3 milimoles

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
$\int\left\{\frac{(\log x-1)}{\left(1+(\log x)^2\right.}\right\}^2 d x$ is equal to

Options:
A
$\frac{\log x}{(\log x)^2+1}+C$

B
$\frac{x}{x^2+1}+C$

C
$\frac{x e^x}{1+x^2}+C$

D
$\frac{x}{(\log x)^2+1}+C$

The correct answer is:
$\frac{x}{(\log x)^2+1}+C$

Solution:
$$
\begin{aligned}
& \int \frac{(\log x-1)^2}{\left(1+(\log x)^2\right)^2} d x \\
& =\int\left[\frac{1}{\left(1+(\log x)^2\right)}-\frac{2 \log x}{\left(1+(\log x)^2\right)^2}\right] d x \\
& =\int\left[\frac{e^t}{1+t^2}-\frac{2 t e^t}{\left(1+t^2\right)^2}\right] d t \text { put } \log x=t \Rightarrow d x=e^t d t \\
& \int e^t\left[\frac{1}{1+t^2}-\frac{2 t}{\left(1+t^2\right)^2}\right] d t \\
& =\frac{e^t}{1+t^2}+c=\frac{x}{1+(\log x)^2}+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int\left\{\frac{(\log x-1)}{\left(1+(\log x)^2\right.}\right\}^2 d x$ is equal to

Options:
A
$\frac{\log x}{(\log x)^2+1}+C$

B
$\frac{x}{x^2+1}+C$

C
$\frac{x e^x}{1+x^2}+C$

D
$\frac{x}{(\log x)^2+1}+C$

The correct answer is:
$\frac{x}{(\log x)^2+1}+C$

Solution:
$$
\begin{aligned}
& \int \frac{(\log x-1)^2}{\left(1+(\log x)^2\right)^2} d x \\
& =\int\left[\frac{1}{\left(1+(\log x)^2\right)}-\frac{2 \log x}{\left(1+(\log x)^2\right)^2}\right] d x \\
& =\int\left[\frac{e^t}{1+t^2}-\frac{2 t e^t}{\left(1+t^2\right)^2}\right] d t \text { put } \log x=t \Rightarrow d x=e^t d t \\
& \int e^t\left[\frac{1}{1+t^2}-\frac{2 t}{\left(1+t^2\right)^2}\right] d t \\
& =\frac{e^t}{1+t^2}+c=\frac{x}{1+(\log x)^2}+c
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2005
\( 1 \) litre of a hydrocarbon weighs as much as one litre of \( \mathrm{CO}_{2} \) at same \( \mathrm{P} \) and \( \mathrm{T} \). Then the molecular formula of the
hydrocarbon is:

Options:
A \( \mathrm{C}_{3} \mathrm{H}_{8} \)
B \( \mathrm{C}_{2} \mathrm{H}_{6} \)
C \( \mathrm{C}_{2} \mathrm{H}_{4} \)
D \( \mathrm{C}_{3} \mathrm{H}_{6} \)

The correct answer is:
\( \mathrm{C}_{3} \mathrm{H}_{8} \)

Solution:
1 litre of a hydrocarbon weighs as much as one litre of CO2.

Hence, 22.4 L of hydrocarbon will weigh as much as 22.4 L of CO2.

22.4 L of CO2 weighs 12+162=44 g at NTP.

Out of all hydrocarbons, C3H8 only weighs 123+18=44 g.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$1 \mathrm{M}$ and 2.5 litre $\mathrm{NaOH}$ solution is mixed with another $0.5 \mathrm{M}$ and 3 litre $\mathrm{NaOH}$ solution, then molarity of the resultant solution will be

Options:
A $1.0 \mathrm{M}$
B $0.73 \mathrm{M}$
C $0.80 \mathrm{M}$
D $0.50 \mathrm{M}$

The correct answer is:
$0.73 \mathrm{M}$

Solution:
$\begin{aligned} & (2.5 \times 1+3 \times 0.5)=M_3 \times 5.5 \\ & \text { or } 2.5+1.5=M_3 \times 5.5 \text { or } M_3=\frac{4}{5.5}=0.73 M \text {. } \\ & \end{aligned}$

Topic: Chemistry, Solutions, JEE Main
1M NH4OH and 1M HCl are mixed to make a total volume of 300 mL. If pH of the mixture is 9.26 and pKaNH4+=9.26 then what would be the volume ratio of NH4OH and HCl .

Options:
A 2:1
B 1:2
C 2:3
D 3:2

The correct answer is:
2:1

Solution:
pOH=14-pH=pKb+logâ¡x300-2x
=14-9.26=-logâ¡10-4.74 +logâ¡x300-2x
â x300-2x=1
â300=3xâx=100 mL
V NH 4 OH V HCl â=â2

Topic: Chemistry, Ionic Equilibrium, JEE Main
1 M NaCl and 1 M HCl are present in an aqueous solution. The solution is

Options:
A
not a buffer solution with pH < 7
B
not a buffer solution with pH > 7

C
a buffer solution with pH < 7
D
a buffer solution with pH > 7

The correct answer is:
not a buffer solution with pH < 7

Solution:
No Solution Available

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2002
1 mol each of AgNO3, CuSO4 and AlCl3 is electrolysed, Then ratio of number of faraday's required are.

Options:
A 1 : 1 : 1
B 1 : 2 : 3
C 3 : 2 : 1
D 1 : 3 : 1

The correct answer is:
1 : 2 : 3

Solution:
Ag++e– ⟶Ag
Cu2++2e–⟶Cu
Al3++3e– ⟶Al
Thus, number of faradays required by 1 mol each=1 : 2: 3.

Topic: Chemistry, Electrochemistry, JEE Main
\( 1 \mathrm{M} \mathrm{NaCl} \) and \( 1 \mathrm{M} \mathrm{HCl} \) are present in an aqueous solution. The solution is

Options:
A a buffer solution with \( \mathrm{pH} < 7 \)
B a buffer solution with \( \mathrm{pH}>7 \)
C uffer solution with \( \mathrm{pH} < 7 \)
D uffer solution with \( \mathrm{pH}>7 \)

The correct answer is:
a buffer solution with \( \mathrm{pH} < 7 \)

Solution:
It is an acidic solution and hence would have pH less than 7. It is not a buffer because buffers contain weak acids or weak bases and their salts.

Topic: Chemistry, Ionic Equilibrium, JEE Main
\( 1 \mathrm{M} \mathrm{NaOH} \) solution was slowly added into \( 1000 \mathrm{~mL} \) of \( 183.75 \mathrm{~g} \) impure \( \mathrm{H}_{2} \mathrm{SO}_{4} \) solution and the following plot was obtained. The percentage purity of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) sample and slope of the curve respectively are :

Options:
A \( 75 \%,-\frac{1}{3} \)
B \( 80 \%,-\frac{1}{2} \)
C \( 80 \%,-1 \)
D None of these

The correct answer is:
\( 80 \%,-1 \)

Solution:
2NaOH + H2SO4 → Na2SO4 + 2H2O

2 Mole of NaOH react with 1 mole of H2SO4.

From given graph,

Moles of NaOH consumed when reacted with H + = 3 × 1
N1V1 = N2 V2

3L x 1M NaOH = 2 x 1L x MH2SO4

moles of H2SO4 present in sample = 1.5
wt. of H2SO4 in sample =1.5×98=147 g
% purity = 1 4 7 183.75 × 1 0 0 = 8 0

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main