datavorous/entrance-exam-dataset huggingface dataset:
The deposition of X and Y on ground surfaces is referred as wet and dry depositions, respectively. X and Y are :

Options:
A X=CO2 , Y=SO2
B

X= Ammonium salts

Y=SO2

C X= Ammonium saltsY=CO2
D X=SO2 Y= Ammonium salts

The correct answer is:

X= Ammonium salts

Y=SO2

Solution:
Wet deposition is the acidic water received through rain, fog, snow, dew, mist, etc. Dry deposition is deposition of wind blown acidic gases and particles on the ground. Ammonium salt in rain drop result in wet deposition. Oxides of Nitrogen and sulphur settle down on ground as dry deposition.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The depression in freezing point observed for a formic acid solution of concentration 0.5 mL L-1 is 0.0405 °C. Density of formic acid is 1.05 g mL-1. The Van't Hoff factor of the formic acid solution is nearly: (Given for water kf=1.86 K kg mol-1)

Options:
A 0.8
B 1.1
C 1.9
D 2.4

The correct answer is:
1.9

Solution:
HCOOH=0.5ml L-1

⇒0.5ml×1.05 g ml-1 HCOOH in 1 L

⇒0.525 g HCOOH in 1 L

Weight of water will be = 1 kg since density of water is 1 kg/L

m=0.525461 kg mol [Assuming dilute solution]

∴ ΔTf=iKfm⇒i=ΔTfkfm=0.0405×461.86×0.525=1.9

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The depth at which acceleration due to gravity becomes $\frac{\mathrm{g}}{\mathrm{n}}$ is [ $\mathrm{R}$ = radius of earth, $\mathrm{g}$ = acceleration due to gravity, $\mathrm{n}=$ integer]

Options:
A $\frac{R(n-1)}{n}$
B $\frac{(\mathrm{n}-1)}{\mathrm{nR}}$
C $\frac{\mathrm{Rn}}{(\mathrm{n}-1)}$
D $\frac{\mathrm{n}}{\mathrm{R}(\mathrm{n}-1)}$

The correct answer is:
$\frac{R(n-1)}{n}$

Solution:
$\begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \\ & \frac{\mathrm{g}}{\mathrm{n}}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \\ & \mathrm{d}=\frac{\mathrm{R}(\mathrm{n}-1)}{\mathrm{n}}\end{aligned}$

Topic: Physics, Gravitation, JEE Main
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02% is _____ m.

(Take density of sea water =103 kg m-3, Bulk modulus of rubber =9×108 N m-2, and g=10 m s-2)

Options:

The correct answer is:
18

Solution:
Bulk modulus is given by, B=-∆P∆VV

⇒∆P=-B∆VV

⇒ρgh=-B∆VV

⇒103×10×h=-9×108×-0.02100

⇒h=18 m

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [ $R=$ radius of the earth]

Options:
A $\frac{R}{n}$
B $R\left(\frac{n-1}{n}\right)$
C $\frac{R}{n^2}$
D $R\left(\frac{n}{n+1}\right)$

The correct answer is:
$R\left(\frac{n-1}{n}\right)$

Solution:
$g^{\prime}=g\left(1-\frac{d}{R}\right) \Rightarrow \frac{g}{n}=g\left(1-\frac{d}{R}\right) \Rightarrow d=\left(\frac{n-1}{n}\right) R$

Topic: Physics, Gravitation, JEE Main
The derivate of (logx)sinx with respect to cosx at x=π2 is

Options:
A -4π
B -π2
C -2π
D -π4

The correct answer is:
-2π

Solution:
Let fx=logxsinx, gx=cosx

Note: Assume base of logarithm as e

Now, logfx=sinx loglogx

Differentiating function with respect to x

f'xfx=cosx·loglogx+sinx·logx·1x

⇒f'x=sinxlogxsinx-1·1x+logxsinx·loglogx·cosx

g'x=-sinx

Now, derivative of fx w.r.t. gx is

=sinxlogxsinx-1·1x+logxsinx·loglogx·cosx-sinx

Put x=π2

=1·1·2π+0-1=-2π

Topic: Mathematics, Differentiation, JEE Main
The derivative of $a^{\sec x}$ w.r.t. $a^{\tan x}(a>0)$ is

Options:
A $a^{\sec x-\tan x}$
B $\sin x a^{\sec x-\tan x}$
C $\sin x a^{\tan x-\sec x}$
D $\sec x a^{\sec x-\tan x}$

The correct answer is:
$\sin x a^{\sec x-\tan x}$

Solution:
Let $u=a^{\sec x} \Rightarrow v=a^{\tan x}$
$\log u=\sec x \log a$
$\frac{d u}{d x}=u \log a \sec x \tan x$
$=a^{\tan x} \log a \sec x \tan x$
and $\quad \begin{aligned} \log v &=\tan x \log a \\ \frac{d v}{d x} &=v \log a \sec ^{2} x \\ &=a^{\tan x} \log a \sec ^{2} x \end{aligned}$
$\begin{aligned} \frac{d u}{d v} &=\frac{\left(\frac{d u}{d x}\right)}{\frac{d v}{d x}}=\frac{a^{\sec x} \log a \sec x \tan x}{a^{\tan x} \log a \sec ^{2} x} \\ &=\frac{a^{\sec x} \sin x}{a^{\tan x}}=\sin x a^{\sec x-\tan x} \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
The derivative of cosh-1x with respect to logx at x=5 is

Options:
A 526
B 126
C 126
D 526

The correct answer is:
526

Solution:
Let u=cosh-1x=logx+x2-1.

Differentiating w.r.t. x, we get

dudx=1+xx2-1x+x2-1=1x2-1

Again let v=logx

Differentiating w.r.t. x, we get

dvdx=1x

Hence, dudv=dudxdvdx=xx2-1

at x=5

dudv=526

Topic: Mathematics, Differentiation, JEE Main
The derivative of $e^{x^3}$ with respect to $\log x$ is

Options:
A $\mathrm{e}^{\mathrm{x}^3}$
B $3 \mathrm{x}^2 2 \mathrm{e}^{\mathrm{x}^3}$
C $3 \mathrm{x}^3 \mathrm{e}^{\mathrm{x}^3}$
D $3 \mathrm{x}^3 \mathrm{e}^{\mathrm{x}^3}+3 \mathrm{x}^2$

The correct answer is:
$3 \mathrm{x}^3 \mathrm{e}^{\mathrm{x}^3}$

Solution:
Let $y=\mathrm{e}^{\mathrm{x}^3}, \mathrm{z}=\log \mathrm{x}$
On differentiating w.r.t.x, we get
$\frac{d y}{d x}=e^{x^3}\left(3 x^2\right)=3 x^2 e^{x^3} \text { and } \frac{d z}{d x}=\frac{1}{x}$
$\therefore \frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{3 x^2 e^{x^3}}{\left(\frac{1}{x}\right)}=3 x^3 e^{x^3}$

Topic: Mathematics, Differentiation, JEE Main
The derivative of $f(x)=x^{\tan ^{-1} x}$ with respect to $g(x)=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ is

Options:
A $\frac{1}{2} \sqrt{1-x^2} x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]$
B $-\frac{1}{2} \sqrt{1-x^2} x^{x^{-a^{-1}}}\left[\log \left(\tan ^{-1} x\right)+x\left(1+x^2\right) \tan ^{-1} x\right]$
C $\frac{\left.-2 \tan ^{-1} \frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]}{\sqrt{1-x^2}}$
D $-\frac{1}{2} \sqrt{1-x^2} x^{\tan x^{-1}}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]$

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

Solution:
We have,
$$
\begin{aligned}
f(x) & =x^{\tan ^{-1} x} \\
\Rightarrow \quad \log f(x) & =\tan ^{-1} x \log x
\end{aligned}
$$
$\begin{aligned} & \therefore \quad \frac{1}{f(x)} \cdot \frac{d}{d x} f(x)=\frac{1}{1+x^2} \log x+\frac{\tan ^{-1} x}{x} \\ & \Rightarrow \quad \frac{d f(x)}{d x}=x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right] \\ & \text { Also, } \quad g(x)=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right) \\ & =\cos ^{-1}\left(2 x^2-1\right)=2 \cos ^{-1} x \\ & \therefore \quad \frac{d}{d x} g(x)=\frac{-2}{\sqrt{1-x^2}} \\ & \therefore \quad \frac{d f(x)}{d \cdot g(x)}=\frac{\frac{d f(x)}{d x}}{\frac{d g(x)}{d x}} \\ & =\frac{x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right]}{\frac{-2}{\sqrt{1-x^2}}} \\ & =\frac{-1}{2} \sqrt{1-x^2} \cdot x^{\tan ^{-1} x}\left[\frac{\log x}{1+x^2}+\frac{\tan ^{-1} x}{x}\right] \\ & \end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
The derivative of $\log |x|$ is

Options:
A $\frac{1}{x}, x>0$
B $\frac{1}{|x|}, x \neq 0$
C $\frac{1}{x}, x \neq 0$
D None of these

The correct answer is:
$\frac{1}{x}, x \neq 0$

Solution:
We have, $y=\log |x|=\left\{\begin{array}{ll}\log x & , x>0 \\ \log (-x), & x < 0\end{array}\right.$
$$
\begin{array}{l}
\therefore \quad \frac{d y}{d x}=\left\{\begin{array}{ll}
\frac{1}{x}, & x>0 \\
\frac{1}{-x}(-1)=\frac{1}{x}, & x < 0
\end{array}\right. \\
\Rightarrow \quad \frac{d y}{d x}=\frac{1}{x}, x \neq 0
\end{array}
$$

Topic: Mathematics, Differentiation, JEE Main
The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ equals

Options:
A $2$
B $\frac{1}{2}$
C $\frac{1}{4}$
D $4$

The correct answer is:
$4$

Solution:
Let $u=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ and $v=\sqrt{1-x^2}$
Put $x=\cos \theta$
$\therefore \quad u=\sec ^{-1}\left(\frac{1}{2 \cos ^2 \theta-1}\right)$ and $v=\sqrt{1-\cos ^2 \theta}$
$\begin{aligned} & \Rightarrow u=\sec ^{-1}(\sec 2 \theta) \text { and } v=\sin \theta \\ & \Rightarrow u=2 \theta \text { and } v=\sin \theta \\ & \Rightarrow \frac{d u}{d \theta}=2 \text { and } \frac{d v}{d \theta}=\cos \theta \Rightarrow \frac{d u}{d v}=\frac{d u / d \theta}{d v / d \theta}=\frac{2}{\cos \theta}\end{aligned}$
$\Rightarrow \frac{d u}{d v}=\frac{2}{x}$ $[\because \cos \theta=x]$
At $\quad x=\frac{1}{2} \Rightarrow\left(\frac{d u}{d v}\right)_{x=1 / 2}=\frac{2}{1 / 2}=4$

Topic: Mathematics, Differentiation, JEE Main
The derivative of $\sin \left(x^{3}\right)$ w.r.t. $\cos \left(x^{3}\right)$ is

Options:
A $-\tan \left(x^{3}\right)$
B $\tan \left(x^{3}\right)$
C $-\cot \left(x^{3}\right)$
D $\cot \left(x^{3}\right)$

The correct answer is:
$-\cot \left(x^{3}\right)$

Solution:
Let $u=\sin \left(x^{3}\right)$
and $v=\cos \left(x^{3}\right)$
Then, $\quad \frac{d u}{d x}=\cos \left(x^{3}\right) \cdot\left(3 x^{2}\right)$
and $\quad \frac{d v}{d x}=-\sin \left(x^{3}\right)\left(3 x^{2}\right)$
Now, $\quad \frac{d u}{d v}=\frac{\left(\frac{d u}{d x}\right)}{\left(\frac{d v}{d x}\right)}$
$$
=\frac{3 x^{2} \cos \left(x^{3}\right)}{-3 x^{2} \sin \left(x^{3}\right)}=-\cot \left(x^{3}\right)
$$

Topic: Mathematics, Differentiation, JEE Main
The derivative of tan-1sinx-cosxsinx+cosx with respect to x2, where x∈0,π2, is

Options:
A 2
B 12
C 23
D 1

The correct answer is:
2

Solution:
Given function can be written as

y=tan-1sinx-cosxsinx+cosx

⇒ y=tan-1tanx-1tanx+1

⇒ y=tan-1tanx-π4 given x∈0,π2
⇒ y=x-π4
Now dydx2=2dydx=2

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The derivative of tan-11+x2-1x with respect to tan-12x1-x21-2x2 at x=12 is :

Options:
A 235
B 312
C 233
D 310

The correct answer is:
310

Solution:
Let x=tanθ
y1=tan-1secθ-1tanθ=tan-1tanθ2=θ2=12tan-1x
x=sinϕ, y2=tan-12sinϕcosϕcos2ϕ=tan-1(tan2ϕ)=2ϕ=2sin-1x

dy1dy2=dy1/dxdy2/dx=11+x2·122·11-x2
=1-x241+x2=1-1441+14=310

Topic: Mathematics, Differentiation, JEE Main
The derivative of tan-11+x2-1x with respect to tan-12x1-x21-2x2 at x=12 is :

Options:
A 235
B 312
C 233
D 310

The correct answer is:
310

Solution:
Let x=tanθ
y1=tan-1secθ-1tanθ=tan-1tanθ2=θ2=12tan-1x
x=sinϕ, y2=tan-12sinϕcosϕcos2ϕ=tan-1(tan2ϕ)=2ϕ=2sin-1x

dy1dy2=dy1/dxdy2/dx=11+x2·122·11-x2
=1-x241+x2=1-1441+14=310

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The descending order of acidity for the following carboxylic acid is-
(A) CH3COOH
(B) F3C-COOH
(C) ClCH2-COOH
(D) FCH2-COOH
(E) BrCH2-COOH
Choose the correct answer from the options given below:

Options:
A B>C>D>E>A
B E>D>B>A>C
C B>D>C>E>A
D D>B>A>E>C

The correct answer is:
B>D>C>E>A

Solution:
The correct order of acidic strength of the given carboxylic acids will be decided by the stability of their conjugate bases. The +I effect of CH3 group destabilises the carboxylate anion. But the -I effect of halogen substituted methyl group stabilises the carboxylate anion. Higher the -I effect of the halogen substituted methyl group, higher the acidic strength of the corresponding acid.
∴ Correct order of acidic strength is
F3C-COOH (B) >FCH2-COOH (D) >CICH2-COOHC>BrCH2-COOHE>CH3-COOHA

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The detergent used in hair conditioner is

Options:
A $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3 \mathrm{Cl}^{-}$
B $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{NH}_2$
C $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{NHCH}_3$
D $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{~N}\left(\mathrm{CH}_3\right)_2$

The correct answer is:
$\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3 \mathrm{Cl}^{-}$

Solution:
Cationic part possesses a long hydrocarbon chain and a positive charge on nitrogen atom. Hence, these are called cationic detergents.Cety|trimethy|ammonium chloride $\mathrm{CH}_3\left(\mathrm{CH}_2\right)_{15} \mathrm{~N}^{+}\left(\mathrm{CH}_3\right)_3 \mathrm{Cl}^{-}$is a popular cationic detergent and is used in hair conditioners.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The determinant $\left|\begin{array}{ccc}1 & (x-3) & (x-3)^{2} \\ 1 & (x-4) & (x-4)^{2} \\ 1 & (x-5) & (x-5)^{2}\end{array}\right|$

Options:
A 3 values of $x$
B 2 values of $x$
C 1 values of $x$
D No value of $x$

The correct answer is:
No value of $x$

Solution:
The given determinant vanishes, i.e.,

$$

\left|\begin{array}{ccc}

1 & x-3 & (x-3)^{2} \\

1 & x-4 & (x-4)^{2} \\

1 & x-5 & (x-5)^{2}

\end{array}\right|=0

$$

Expanding along $\mathrm{C}_{1},$ we get

$$

\begin{array}{l}

(x-4)(x-5)^{2}-(x-5)(x-4)^{2}-\left\{(x-3)(x-5)^{2}\right. \\

\left.-(x-5)(x-3)^{2}\right\}+(x-3)(x-4)^{2} \\

-(x-4)(x-3)^{2}=0 \\

\Rightarrow(x-4)(x-5)(x-5-x+4) \\

-(x-3)(x-5)(x-5-x+3) \\

+(x-3)(x-4)(x-4-x+3)=0

\end{array}

$$

$$

\Rightarrow-(x-4)(x-5)+2(x-3)(x-5)-(x-3)

$$

$(x-4)=0$

$$

\Rightarrow-x^{2}+9 x-20+2 x^{2}-16 x+30-x^{2}+7 x-12=0

$$

$\Rightarrow-32+30=0 \Rightarrow-2=0$

Which is not possible, hence no value of $\mathrm{x}$ satisfies the given condition.

Topic: Mathematics, Determinants, JEE Main
The determinant \( \left|\begin{array}{lll}\cos (\theta+\phi) & -\sin (\theta+\phi) & \cos 2 \phi \\ \sin \theta & \cos \theta & \sin \phi \\ -\cos \theta & \sin \theta & \cos \phi\end{array}\right| \) is -

Options:
A \( 0 \)
B independent of \( \theta \)
C independent of \( \phi \)
D independent of both \( \theta \) and \( \phi \)

The correct answer is:
independent of \( \theta \)

Solution:
Δ =cos⁡θ+ϕ. cos⁡θcos⁡ϕ-sin⁡θsin⁡ϕ
+sin⁡θ+ϕ. sin⁡θcos⁡ϕ+cos⁡θsin⁡ϕ
+cos⁡2ϕ. sin2⁡θ +cos2⁡θ
Δ=cos2⁡θ+ϕ+sin2⁡θ+ϕ+cos⁡2ϕ
Δ=1 +cos⁡2ϕ
∴Δ is independent of θ

Topic: Mathematics, Determinants, JEE Main
The device which can remove particulate matter present in the exhaust from a thermal power plant is:

Options:
A Incinerator
B Electrostatic Precipitator
C Catalytic Convertor
D STP

The correct answer is:
Electrostatic Precipitator

Solution:
There are several ways of removing particulate matter; the most widely used of which is the electrostatic precipitator, which can remove over 99% of particulate matter present in the exhaust from a thermal power plant.

It has electrode wires that are maintained at several thousand volts, which produce a corona that releases electrons. These electrons attach to dust particles giving them a net negative charge. The collecting plates are grounded and attract the charged dust particles.

Topic: Biology, Environmental Issues, NEET, NEET 2022 (Phase 1)
The diagonal relationship in Be and Al is due to

Options:
A Similar ionisation enthalpy
B Similar metallic character
C Similar ionic size and charge/radius ratio
D Anthraquinone

The correct answer is:
Similar ionic size and charge/radius ratio

Solution:
The diagonal relationship in Be and Al is due to similar ionic size and charge/radius ratio. Generally, on moving from left to right across a period the ionic charge/radius ratio. Period, the ionic charge increase to maximum and then decreases, while the ionic size decreases causing an increase in its polarising power (Fajan’s). on the other hand, on moving down a group the ionic charge remains the same while ionic size increases. Therefore, polarising power decreases. On moving diagonally, therefore, there is no marked change in their properties.

Topic: Chemistry, s Block Elements, MHT CET, MHT CET 2019 (Shift 1)
The diagram below shows how things get to and from the liver. They are labeled as A, B, C, D, E and F. Which one of the following labellings is the correct one?

Options:
A A is the hepatic portal vein and E is the hepatic vein
B C is the intestine and F is the hepatic portal vein
C D is the hepatic portal vein and F is hepatic vein
D B is the pancreatic artery and E is the hepatic artery

The correct answer is:
D is the hepatic portal vein and F is hepatic vein

Solution:
The hepatic portal vein transports blood high on glucose concentration and amino acids from the small intestine to the liver while the hepatic vein transports deoxygenated blood from the liver to the heart and remaining glucose and amino acids to the rest of the body.

Topic: Biology, Body Fluids and Circulation, NEET
The diagram below shows three plants with identical leaf surface areas:

Assuming all other conditions were identical for all three plants, which of the plants would likely to photosynthesize slowest and fastest.

Options:
A Y : slowest, Z : fastest
B Z : slowest, X : fastest
C X : slowest, Z : fastest
D Z : slowest, Y : fastest

The correct answer is:
Z : slowest, Y : fastest

Solution:
Photosynthesis is a process of using light energy to make glucose. The main component helping in making use of light energy is the pigment chlorophyll. This pigment is green in colour, which means it will reflect the green wavelength of the visible spectrum. It will not be active in green light as it cannot use it. Chlorophyll is more active in red and blue light and rate of photosynthesis increases when exposed to such light. In the given arrangement, Z plant will show least speed for photosynthesis due to green light and Y will show the highest speed due to red light. At X, chlorophyll activity will be low due to very low temperature.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is

Options:
A

B

C

D

The correct answer is:

Solution:
$V_{\text {in }}=\frac{-G m}{2 R}\left[3-\left(\frac{r}{R}\right)^2\right], V_{\text {surface }}=\frac{-G M}{R}, V_{\text {out }}=\frac{-G M}{r}$

Topic: Physics, Gravitation, JEE Main
The diagram shows a horizontal cylindrical container of length 30 cm, which is partitioned by a tight-fitting separator. The separator is diathermic but conducts heat very slowly. Initially, the separator is in the state shown in the diagram. The temperature of the left part of the cylinder is 100 K and that on the right part is 400 K. Initially the separator is in equilibrium. As heat is conducted from the right to left part, separator displaces to the right. Find the displacement of separator (in cm) after a long when gases on the two parts of the cylinder are in thermal equilibrium.

Options:

The correct answer is:
10

Solution:
It is given that initially, the separator is in equilibrium; thus pressure on both sides of the gas is equal say, it is Pi. If A be the area of cross-section of cylinder, number of moles of gas in the left and right part, n1 and n2, can be given as
n1=Pi10AR100
and n2=Pi20AR400
Finally, if the separator is displaced to right by a distance x,
we have
n1=Pfâ¡i10+xARTfâ¡
and n2=Pfâ¡20-xARTfâ¡
Here P fâ¡ and T fâ¡ are the final pressure and temperature on both sides after a long time.
Now if we equate the ratio of moles n1/n2 in initial and final states, we get
n1n2=10A/10020A/400=10+xA20-xA
220-x=10+x
x=10 cm
Thus in the final state when gases in both parts are in thermal equilibrium, the piston is displaced 10 cm rightward from its initial position.

Topic: Physics, Thermodynamics, JEE Main
The diagram shows a spherical surface which separates two media of refractive indices, Î¼1 and Î¼2, respectively. Now, a point object is placed on the principal axis as shown in the figure. Then

Options:
A Real image will form if μ1>μ2 and for all values of u
B Real image for some values of u if μ1>μ2
C Virtual image will form if μ1>μ2
D Virtual image will form if μ1<μ2

The correct answer is:
Virtual image will form if μ1>μ2

Solution:
μ2v-μ1-u=μ2-μ1R

⇒v=uRμ2uμ2-μ1-Rμ1

⇒v=uRμ2uμ2-μ1u+R

We can see that μ1>μ2 then image distance is negative which means virtual image

Topic: Physics, Ray Optics, JEE Main
The diagram shows different transitions across the energy levels for an electron in a certain atom. Among these, which transition represents the emission of a photon with the most energy?

Options:
A (II)
B (I)
C (IV)
D (III)

The correct answer is:
(III)

Solution:
Since, energy, $E=13.6 Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$
where, $Z$ is atomic number and $n$ is orbital number.
From given diagram,
$\begin{aligned}
& E_{\mathrm{I}}=13.6 Z^2\left(\frac{1}{3^2}-\frac{1}{1}\right)=-13.6 Z^2(0.88) \\
& E_{\mathrm{II}}=13.6 Z^2\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=13.6 Z^2(0.049) \\
& E_{\mathrm{III}}=13.6 Z^2\left(\frac{1}{1^2}-\frac{1}{2}\right)=13.6 Z^2(0.75) \\
& E_{\mathrm{IV}}=13.6 Z^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=13.6 Z^2(0.187)
\end{aligned}$
Here, $E_{\mathrm{III}}$ is highest.

Topic: Physics, Atomic Physics, JEE Main
The diagram shows part of a carbohydrate molecule formed by glucose monomer. What is the name of the molecule?

Options:
A Amylose
B Cellulose
C Starch
D Glycogen

The correct answer is:
Glycogen

Solution:
Glycogen is a branched polymer of glucose monomer. Glucose units are linked together linearly by α1→4 glycosidic bonds from one glucose to the next whereas the branches are linked to the chains by α1→6 glycosidic bonds between the first glucose of the branch and a glucose on the main chain.

Topic: Chemistry, Biomolecules, NEET
The diagram shows the arrangement of three small uniformly charged spheres \( A, B \) and \( C \). The arrows indicate the direction of the electrostatic forces acting between the spheres(for example, the left arrow on sphere \( A \) indicates the electrostatic force on sphere \( A \) due to sphere \( B \) ). At least two of the spheres are positively charged. Which sphere, if any, could be negatively charged?

Options:
A Sphere \( A \)
B Sphere \( B \)
C Sphere \( C \)
D No sphere

The correct answer is:
Sphere \( A \)

Solution:
It can be seen from the diagram that only the sphere B and sphere C repel. Hence, they both must be of same type. According to the fact that at least two spheres are positively charged, therefore both spheres should be positively charged. Since attraction occurs for two remaining pairs it can be concluded that the sphere A is negatively charged.

Topic: Physics, Electrostatics, JEE Main
The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy ?

Options:
A
III

B
IV

C
I

D
II

The correct answer is:
III

Solution:
$\Delta \mathrm{E} \propto\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2005
The diamagnetic species is

Options:
A NO
B \( \mathrm{NO}_{2} \)
C \( \mathrm{O}_{2} \)
D \( \mathrm{CO}_{2} \)

The correct answer is:
\( \mathrm{CO}_{2} \)

Solution:
NO & O2 are paramagnetic, which can be understood by the molecular orbital theory, as they have electronic configuration of NO=σ1s2σ1s2*σ2s2σ2s2*σ2p2π2p2=π2p2π2p1*=π2p0*O2=σ1s2σ1s2*σ2s2σ2s2*σ2p2π2p2=π2p2π2p1*=π2p1*

As we can see, both have unpaired electrons. So, it will show paramagnetic in nature.

Lewis structure of NO2:

So, we can see one unpaired electron over nitrogen, which makes it paramagnetic in nature.

Structure of CO2:

O=C=O

So, no unpaired electrons, and it is diamagnetic in nature.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The diameter and height of a cylinder are measured by a meter scale to be 12.6±0.1 cm and 34.2±0.1 cm , respectively. What will be the value of its volume in appropriate significant figures?

Options:
A 4264±81cm3
B 4264.4±81.0 cm3
C 4260±80 cm3
D 4300±80 cm3

The correct answer is:
4260±80 cm3

Solution:
V=πd2h4=π412.6234.2 cm3
dVV=2ddd+dhh
Converting to errors
∆VV=2∆dd+∆hh
∆V4260=20.112.6+0.134.2
⇒∆V=80 cm3
V=4260±80 cm3

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The diameter of $16 x^2-9 y^2=144$, which is conjugate to $x=2 y$, is

Options:
A $y=\frac{16 x}{9}$
B $y=\frac{32 x}{9}$
C $x=\frac{16 y}{9}$
D $x=\frac{32 y}{9}$

The correct answer is:
$y=\frac{32 x}{9}$

Solution:
Two diameters of a hyperbola $y=m x$ and $y=m_1 x$ are said to be conjugate if they follow the relation
$\mathrm{mm}_1=\frac{\mathrm{b}^2}{\mathrm{a}^2}$
So here,
$\begin{aligned}
& \mathrm{m}=\frac{1}{2} \\
& \Rightarrow \frac{1}{2} \mathrm{~m}_1=\frac{16}{9} \\
& \Rightarrow \mathrm{m}_1=\frac{32}{9}
\end{aligned}$
Hence the equation of other diameter is
$y=\frac{32}{9} x$

Topic: Mathematics, Hyperbola, JEE Main