datavorous/entrance-exam-dataset huggingface dataset:
Solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass precipitation of $\mathrm{AgBr}$ is

Options:
A
$1.2 \times 10^{-10} \mathrm{~g}$

B
$1.2 \times 10^{-9} \mathrm{~g}$

C
$6.2 \times 10^{-5} \mathrm{~g}$

D
$5.0 \times 10^{-8} \mathrm{~g}$

The correct answer is:
$1.2 \times 10^{-9} \mathrm{~g}$

Solution:
$$
\mathrm{Ag}^{+}+\mathrm{Br}^{-} \rightleftharpoons \mathrm{AgBr}
$$
Precipitation starts when ionic product just exceeds solubility product
$$
\begin{aligned}
& \mathrm{K}_{\text {sp }}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right] \\
& {\left[\mathrm{Br}^{-}\right]=\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{Ag}^{+}\right]}=\frac{5 \times 10^{-13}}{0.05}=10^{-11}}
\end{aligned}
$$
i.e., precipitation just starts when $10^{-11}$ moles of $\mathrm{KBr}$ is added to $1 \mathrm{~L}$ of $\mathrm{AgNO}_3$ solution.
No. of moles of $\mathrm{KBr}$ to be added
$$
\begin{aligned}
\therefore \text { weight of } \mathrm{KBr} \text { to be added } & =10^{-11} \times 120 \\
& =1.2 \times 10^{-9} \mathrm{~g}
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2010
Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2 °C. The percentage association of solute A in water, is

[Given: Molar mass of A=93 g mol-1. Molal depression constant of water is 1.86 K kg mol-1]

Options:
A 50%
B 60%
C 70%
D 80%

The correct answer is:
80%

Solution:
Depression in freezing point,ΔTf=iKf×m

2.0=ix1.86×7×100093×42

2.0=i3.33

i=0.6

i=1+1n-1β

0.6=1+12-1β

β=0.8

%β=80

Hence, the percentage association of solute A in water is 80.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Solution of 12 g of non-electrolyte (A) prepared by dissolving it in 1000 mL of water exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The empirical formula of A is CH2O. The molecular mass of A is ________g. (Nearest integer)

Options:

The correct answer is:
240

Solution:
The pressure which applied externally in the solution side to prevent the osmosis process is known as osmotic pressure. It can be denoted by π.

Since, (π1)non electrolyte = (π2)glucose
C1RT = C2RT
W1M1=W2M2

Empirical mass of
12M1 = 0.05
M1 = 120.05 = 240

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Solution of differential equation $2 x y \frac{d y}{d x}=x^2+3 y^2$ is

Options:
A $x^3+y^2=p x^2$
B $\frac{x^2}{2}+\frac{y^3}{x}=y^2+p$
C $x^2+y^3=p x^2$
D $x^2+y^2=p x^3$

The correct answer is:
$x^2+y^2=p x^3$

Solution:
It is homogeneous equation $\frac{d y}{d x}=\frac{x^2+3 y^2}{2 x y}$
Put $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$
So, we get $x \frac{d v}{d x}=\frac{1+v^2}{2 v}$
$\Rightarrow \frac{2 v d v}{1+v^2}=\frac{d x}{x}$
On integrating, we get $x^2+y^2=p x^3$.

Topic: Mathematics, Differential Equations, JEE Main
Solution of differential equation $\frac{d y}{d x}=\frac{y-x}{y+x}$ is

Options:
A $\log _e\left(x^2+y^2\right)+2 \tan ^{-1} \frac{y}{x}+c=0$
B $\frac{y^2}{2}+x y=x y-\frac{x^2}{2}+c$
C $\left(1+\frac{x}{y}\right) y=\left(1-\frac{x}{y}\right) x+c$
D $y=x-2 \log _e y+c$

The correct answer is:
$\log _e\left(x^2+y^2\right)+2 \tan ^{-1} \frac{y}{x}+c=0$

Solution:
$\begin{aligned} & \text { Put } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \therefore v+x \frac{d v}{d x}=\frac{v-1}{v+1} \Rightarrow x \frac{d v}{d x}=\frac{v-1}{v+1}-v \\ & \Rightarrow \quad x \frac{d v}{d x}=-\frac{v^2+1}{v+1} \Rightarrow \int \frac{d x}{x}=-\int \frac{v+1}{v^2+1} d v \\ & \Rightarrow \quad-\log _e x=\frac{1}{2} \int \frac{2 v}{v^2+1} d v+\int \frac{1}{v^2+1} d v \\ & \Rightarrow \quad-\log _e x=\frac{1}{2} \log \left(v^2+1\right)+\tan ^{-1} v+c\end{aligned}$
$\begin{aligned} & \Rightarrow \quad-2 \log _e x=\log \left(\frac{x^2+y^2}{x^2}\right)+2 \tan ^{-1}\left(\frac{y}{x}\right)+c \\ & \Rightarrow \log _e\left(x^2+y^2\right)+2 \tan ^{-1} \frac{y}{x}+c=0 .\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
Solution of differential equation

$$

x^{2}=1+\left(\frac{x}{y}\right)^{-1} \frac{d y}{d x}+\frac{\left(\frac{x}{y}\right)^{-2}\left(\frac{d y}{d x}\right)^{2}}{2 !}+\frac{\left(\frac{x}{y}\right)^{-3}\left(\frac{d y}{d x}\right)^{3}}{3 !}+\ldots \text { is }

$$

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

The correct answer is:
$y^{2}=x^{2}\left(\ln x^{2}-1\right)+C$

Solution:
$x^{2}=e^{\left(\frac{x}{y}\right)^{-1}\left(\frac{d y}{d x}\right)} \Rightarrow x^{2}=e^{\left(\frac{y}{x}\right)\left(\frac{d y}{d x}\right)}$

$\Rightarrow \ln x^{2}=\frac{y}{x} \frac{d y}{d x}$ or $\int x \ln x^{2} d x=\int y d y$

Put $x^{2}=t \Rightarrow 2 x d x=d t \therefore \frac{1}{2} \int \ln t d t=\frac{y^{2}}{2}$

$C+t \ln t-t=y^{2}$ or $y^{2}=x^{2}\left(\ln x^{2}-1\right)+C$

Topic: Mathematics, Differential Equations, JEE Main
Solution of $e^{d y / d x}=x$ when $x=1$ and $y=0$ is

Options:
A $y=x(\log x-1)+4$
B $y=x(\log x-1)+3$
C $y=x(\log x+1)+1$
D $y=x(\log x-1)+1$

The correct answer is:
$y=x(\log x-1)+1$

Solution:
Given, $e^{d y / d x}=x$
Taking logarithm on both sides, we get
$$
\begin{aligned}
\log e^{d y} / d x &=\log x \\
\frac{d y}{d x} &=\log x \\
d y &=\log x d x
\end{aligned}
$$
On integrating, we get
$$
\begin{aligned}
&\int \begin{aligned}
\int \mathrm{dy} &=\int \log \mathrm{xdx} \\
&=\log \mathrm{x} \int 1 \mathrm{dx}-\int\left[\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x} \int 1 \mathrm{dx}\right] \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\int \frac{1}{\mathrm{x}} \times \mathrm{x} \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\int \mathrm{dx} \\
&=\mathrm{x} \log \mathrm{x}-\mathrm{x} \\
\mathrm{y} &=\mathrm{x}(\log \mathrm{x}-1)+\mathrm{C} \\
\text { when } \mathrm{x} &=1 \text { and } \mathrm{y}=0 \\
\Rightarrow \quad 0=1(\log 1-1)+\mathrm{C}
\end{aligned}
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & 0=(0-1)+C \\
\Rightarrow & C=1
\end{array}
$$
$\therefore \mathrm{Eq}$. (i) becomes
$$
y=x(\log x-1)+1
$$

Topic: Mathematics, Differential Equations, JEE Main
Solution of II A group precipitate in dil. $\mathrm{HNO}_3$ when treated with $\mathrm{NH}_4 \mathrm{OH}$ becomes blue because of the presence of

Options:
A $M g$
B $\mathrm{Cd}$
C $B i$
D $\mathrm{Cu}$

The correct answer is:
$\mathrm{Cu}$

Solution:
$\mathrm{CuCl}+2 \mathrm{HNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{HCl}$
$\mathrm{Cu}\left(\mathrm{NO}_3\right)_2$ reacts with $\mathrm{NH}_4 \mathrm{OH}$ and gives $\mathrm{Cu}\left(\mathrm{NH}_3\right)_4$ which has deep blue colour.

Topic: Chemistry, Practical Chemistry, JEE Main
Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0 < x < \frac{\pi}{2}$ is

Options:
A
$y \sec x=\tan x+c$

B
$y \tan x=\sec x+c$

C
$\tan x=(\sec x+c) y$

D
$\sec x=(\tan x+c) y$

The correct answer is:
$\sec x=(\tan x+c) y$

Solution:
$$
\begin{aligned}
& \cos x d y=y(\sin x-y) d x \\
& \frac{d y}{d x}=y \tan x-y^2 \sec x \\
& \frac{1}{y^2} \frac{d y}{d x}-\frac{1}{y} \tan x=-\sec x \\
& \text { Let } \frac{1}{y}=t \\
& -\frac{1}{y^2} \frac{d y}{d x}=\frac{d t}{d x} \\
& -\frac{d y}{d x}-t \tan x=-\sec x \Rightarrow \frac{d t}{d x}+(\tan x) t=\sec x . \\
& \text { I.F. }=e^{f \tan x d x}=\sec x
\end{aligned}
$$
Solution is $t\left(\right.$ I.F) $=\int$ (I.F) $\sec x d x$
$\frac{1}{y} \sec x=\tan x+c$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2010
Solution of the differential equation \( \frac{x+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !} \cdots}{1+\frac{x^{2}}{2}+\frac{x^{4}}{4} \cdots}=\frac{d x-d y}{d x+d y} \) is

Options:
A \( 2 y e^{2 x}=c e^{2 x}+1 \)
B \( 2 y e^{2 x}=2 c e^{2 x}-1 \)
C \( y e^{2 x}=c e^{2 x}+2 \)
D None of these

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

Solution:
ex=1+x1!+x22!+x33!+...............∞e-x=1-x1!+x22!-x33!+...............∞x+x33!+x55!…1+x22!+x44!…=dx-dydx+dyex-e-x2ex+e-x2=dx-dydx+dy.

Apply componendo and dividendo

dydx=e-2x

∫dydx=∫e-2x⇒y=-12e-2x+c

2ye2x=2ce2x-1.

Topic: Mathematics, Differential Equations, JEE Main
Solution of the differential equation, $y d x-x d y+x y^2 d x=0$ can be

Options:
A $2 x+x^2 y=\lambda y$
B $2 y+y^2 x=\lambda y$
C $2 y-y^2 x=\lambda y$
D None of these

The correct answer is:
$2 x+x^2 y=\lambda y$

Solution:
$\frac{y d x-x d y}{y^2}=-x d x \Rightarrow d\left(\frac{x}{y}\right)=-x d x$
Integrating both side, we get $\frac{x}{y}=-\frac{x^2}{2}+c$
$\Rightarrow 2 x+x^2 y=2 c y \Rightarrow 2 x+x^2 y=\lambda y \quad[\lambda=2 c]$

Topic: Mathematics, Differential Equations, JEE Main
Solution of the equation dydx=ex-yex-ey is

Options:
A ey=ex-1+ce-ex
B ey-x=-1+ce-ex
C ex+ey=ce-ex
D None of these

The correct answer is:
ey=ex-1+ce-ex

Solution:
We have,

dydx=ex-yex-ey

⇒ eydydx+eyex=e2x

Put ey=v

∴ eydydx=dvdx

∴ dvdx+vex=e2x

This is linear equation

I.F==e∫exdx=eex

∴ Required solution is

v·eex=∫eex·e2xdx+C

v·eex=eexex-1+C

⇒ ey·eex=eexex-1+C

⇒ ey=ex-1+ce-ex

Topic: Mathematics, Differential Equations, JEE Main
Solution of $(x+y)^{2} \frac{d y}{d x}=a^{2}$ ('a' being a constant) is

Options:
A $\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C$ is an arbitrary constant
B $x y=a \tan C x, C$ is an arbitrary constant
C $\frac{x}{a}=\tan \frac{y}{C}, C$ is an arbitrary constant
D $x y=\tan (x+C), C$ is an arbitrary constant

The correct answer is:
$\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C$ is an arbitrary constant

Solution:
We have, $(x+y)^{2} \frac{d y}{d x}=a^{2}$
Let $x+y=v$
$\therefore$
$1+\frac{d y}{d x}=\frac{d v}{d x}$
$\frac{d y}{d x}=\frac{d y}{d x}-1$
$v^{2}\left(\frac{d v}{d x}-1\right)=a^{2}$
$\Rightarrow$
$v^{2} \frac{d v}{d x}=v^{2}+a^{2}$
$\quad \frac{d v}{d x}=\frac{v^{2}+a^{2}}{v^{2}}$
$\Rightarrow \quad \frac{v^{2}}{v^{2}+d^{2}} d v=d x$
On integrating both sides, we get $\int \frac{v^{2}}{v^{2}+a^{2}} d v=\int d x$
$\int\left(1-\frac{a^{2}}{v^{2}+a^{2}}\right) \mathrm{d} v=x+C^{\prime}$
$\Rightarrow \quad v-\frac{a^{2}}{a} \tan ^{-1} \frac{v}{a}=x+C^{\prime}$
$\Rightarrow \quad x+y-a \tan ^{-1} \frac{x+y}{a}=x+C^{\prime}$
$\Rightarrow \quad y=a \tan ^{-1} \frac{x+y}{a}+C^{\prime}$
$=\frac{y-C^{\prime}}{a}=\tan ^{-1} \frac{x+y}{a}$
$\Rightarrow \quad\left(\frac{x+y}{a}\right)=\tan \left(\frac{y+C}{a}\right)$
where $-C^{\prime}=C$.

Topic: Mathematics, Differential Equations, JEE Main
Solution set of [sin-1⁡x]>[cos-1⁡x], where [.] denotes the greatest integer function, is

Options:
A 12,1
B (cosâ¡1, sinâ¡1)
C [sin 1, 1]
D None of these

The correct answer is:
[sin 1, 1]

Solution:
∵[ sin −1 x ]>[ cos −1 x ]
⇒ x>0
Here, cos-1 x=0,x∈(cos 1, 1]1,x∈(0, cos 1]
and sin-1 x=0, x∈(0, sin 1 )1,x∈[sin 1, 1]
∴ x∈[sin 1, 1]

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Solution " $X$ " contains $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$. $20 \mathrm{~mL}$ of $X$ when titrated using methyl orange indicator consumed $60 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{HCl}$ solution. In another experiment, $20 \mathrm{~mL}$ of $X$ solution when titrated using phenolphthalein consumed $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{HCl}$ solution. The concentrations (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$ in $X$ are respectively

Options:
A $0.01,0.02$
B $0.1,0.1$
C $0.01,0.01$
D $0.1,0.01$

The correct answer is:
$0.1,0.1$

Solution:
For titration of a basic solution of $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$ against $\mathrm{HCl}$, if phenolphthalein is used as indicator, the end point is indicated only for half neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$, i.e., (upto $\mathrm{NaHCO}_3$ ).
$$
\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaHCO}_3+\mathrm{NaCl}
$$
The remaining solution then contains the unreacted $\mathrm{NaHCO}_3$ from this reaction plus the unreacted $\mathrm{NaHCO}_3$ originally in the solution. At the phenolphthalein end point, there is no reaction between $\mathrm{HCl}$ and $\mathrm{NaHCO}_3$.
From the equations
$$
\begin{aligned}
\text { Mol of } \mathrm{HCl} \text { consumed } & =\mathrm{mol} \text { of } \mathrm{Na}_2 \mathrm{CO}_3 \\
20 \mathrm{~mL} \text { of } 0.1 \mathrm{M} & =20 \mathrm{~mL} \text { of } 0.1 \mathrm{M}
\end{aligned}
$$
$\therefore$ The concentration of $\mathrm{Na}_2 \mathrm{CO}_3$ in solution $X=0.1 \mathrm{M}$.
Note that for a quantity of $\mathrm{Na}_2 \mathrm{CO}_3$, exactly half volume of the $\mathrm{HCl}$ is used at the phenolphthalein end point and the second half volume of the $\mathrm{HCl}$ is required for complete neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$ at methyl orange end point.
$$
\mathrm{NaHCO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$$
$\therefore$ Volume of $\mathrm{HCl}$ required to neutralize
$$
\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 \text { in original sample } & =2 \times 20 \mathrm{~mL} \\
& =40 \mathrm{~mL}
\end{aligned}
$$
If methyl orange is used, the end point is indicated when all the alkali is neutralized.
$$
\mathrm{NaHCO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$$
As $40 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ is consumed in complete neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$ at methyl orange end point, so the volume of $\mathrm{HCl}$ used to neutralized $\mathrm{NaHCO}_3$ from the original sample would be
Remaining $\mathrm{HCl}=60-40=20 \mathrm{~mL}$ of $0.1 \mathrm{M}$
As per equation $=1 \mathrm{~mol}$ of $\mathrm{NaHCO}_3=1 \mathrm{~mol}$ of $\mathrm{HCl}$
$\therefore 0.1 \mathrm{~mol}$ of $\mathrm{NaHCO}_3=0.1 \mathrm{~mol}$ of $\mathrm{HCl}$,

Topic: Chemistry, Hydrogen, AP EAMCET, AP EAMCET 2013
Solution " $X$ " contains $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$. $20 \mathrm{~mL}$ of $X$ when titrated using methyl orange indicator consumed $60 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{HCl}$ solution. In another experiment, $20 \mathrm{~mL}$ of $X$ solution when titrated using phenolphthalein consumed $20 \mathrm{~mL}$ of $0.1 \mathrm{M}$ $\mathrm{HCl}$ solution. The concentrations (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$ in $X$ are respectively

Options:
A $0.01,0.02$
B $0.1,0.1$
C $0.01,0.01$
D $0.1,0.01$

The correct answer is:
$0.1,0.1$

Solution:
For titration of a basic solution of $\mathrm{Na}_2 \mathrm{CO}_3$ and $\mathrm{NaHCO}_3$ against $\mathrm{HCl}$, if phenolphthalein is used as indicator, the end point is indicated only for half neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$, i.e., (upto $\mathrm{NaHCO}_3$ ).
$$
\mathrm{Na}_2 \mathrm{CO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaHCO}_3+\mathrm{NaCl}
$$
The remaining solution then contains the unreacted $\mathrm{NaHCO}_3$ from this reaction plus the unreacted $\mathrm{NaHCO}_3$ originally in the solution. At the phenolphthalein end point, there is no reaction between $\mathrm{HCl}$ and $\mathrm{NaHCO}_3$.
From the equations
$$
\begin{aligned}
\text { Mol of } \mathrm{HCl} \text { consumed } & =\mathrm{mol} \text { of } \mathrm{Na}_2 \mathrm{CO}_3 \\
20 \mathrm{~mL} \text { of } 0.1 \mathrm{M} & =20 \mathrm{~mL} \text { of } 0.1 \mathrm{M}
\end{aligned}
$$
$\therefore$ The concentration of $\mathrm{Na}_2 \mathrm{CO}_3$ in solution $X=0.1 \mathrm{M}$.
Note that for a quantity of $\mathrm{Na}_2 \mathrm{CO}_3$, exactly half volume of the $\mathrm{HCl}$ is used at the phenolphthalein end point and the second half volume of the $\mathrm{HCl}$ is required for complete neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$ at methyl orange end point.
$$
\mathrm{NaHCO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$$
$\therefore$ Volume of $\mathrm{HCl}$ required to neutralize
$$
\begin{aligned}
\mathrm{Na}_2 \mathrm{CO}_3 \text { in original sample } & =2 \times 20 \mathrm{~mL} \\
& =40 \mathrm{~mL}
\end{aligned}
$$
If methyl orange is used, the end point is indicated when all the alkali is neutralized.
$$
\mathrm{NaHCO}_3+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}
$$
As $40 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$ is consumed in complete neutralization of $\mathrm{Na}_2 \mathrm{CO}_3$ at methyl orange end point, so the volume of $\mathrm{HCl}$ used to neutralized $\mathrm{NaHCO}_3$ from the original sample would be
Remaining $\mathrm{HCl}=60-40=20 \mathrm{~mL}$ of $0.1 \mathrm{M}$
As per equation $=1 \mathrm{~mol}$ of $\mathrm{NaHCO}_3=1 \mathrm{~mol}$ of $\mathrm{HCl}$
$\therefore 0.1 \mathrm{~mol}$ of $\mathrm{NaHCO}_3=0.1 \mathrm{~mol}$ of $\mathrm{HCl}$,

Topic: Chemistry, Hydrogen, TS EAMCET, TS EAMCET 2013
Somaclones are obtained by : -

Options:
A Genetic engineering
B Tissue culture
C Plant breeding
D Irradiation Genetic engineering

The correct answer is:
Tissue culture

Solution:
No Solution Available

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2009 (Mains)
Some amount of a radioactive substance (half-life =10 days ) is spread inside a room and consequently, the level of radiation becomes 50 times the permissible level for normal occupancy of the room. After how many days will the room be safe for occupation?

Options:
A 20 days
B 34.8 days
C 56.4 days
D 62.9 days

The correct answer is:
56.4 days

Solution:
since, the initial activity is 50 times the activity for safe occupancy, therefore R0=50R, where R=λN.
since, R∝N

RR0=NN0=12n=12t/T

Or 12t/10=150

2t/10-50⇒t10log102=log1050

⇒ t=10log1050log102=10×1.6990.301=56.4 days

Topic: Physics, Nuclear Physics, JEE Main
Some amount of a radioactive substance (half-life = 10 days) is spread inside a room and consequently, the level of radiation becomes 50 times the permissible level for normal occupancy of the room. After how many days will the room be safe for occupation?

Options:
A 20 days
B 34.8 days
C 56.4 days
D 62.9 days

The correct answer is:
56.4 days

Solution:
Since, the initial activity is 50 times the activity for safe occupancy, therefore, R0=50R, where R=λN.
Since, R ∝N
⇒ RR0=NN0=12n
⇒ 12t10=150
⇒ 2t10=50
⇒ t10log10⁡2=log10⁡50
⇒ t=10log10⁡50log10⁡2=10×16990.301
=56.4 days.

Topic: Physics, Nuclear Physics, JEE Main
Some cells in testis secrete testosterone and the hormone stimulating such cells are released in blood by

Options:
A Seminal vesicles
B Testis itself
C Hypothalamus
D Anterior pituitary

The correct answer is:
Anterior pituitary

Solution:
FSH and LH are secreted from the anterior pituitary. LH is also known as ICSH (Interstitial cell-stimulating hormone).

Gn	In males	In females

FSH	Regulation of spermatogenesis by FSH and androgens.	FSH stimulates growth and development of the ovarian follicles in females.

LH	Synthesis and release of androgens (male hormones) from the testis.	Induces ovulation of Graafian follicles (fully mature follicles). Formation of corpus luteum from the remnants of the Graafian follicles after ovulation. Maintains the corpus luteum.

Topic: Biology, Chemical Coordination and Integration, NEET
Some cells in the adult animals do not appear to exhibit division (e.g., heart cells) and many other cells divide only occasionally. Such cells are said to be in

Options:
A G1 phase
B G2 phase
C G0 phase
D S-phase

The correct answer is:
G0 phase

Solution:
Some cells in the adult animals do not appear to exhibit division (e.g., heart cells) and many other cells divide only occasionally, as needed to replace cells that have been lost because of injury or cell death. These cells do not divide further exit G1 phase to enter an inactive stage called quiescent stage (G0) of the cell cycle. Cells in this stage remain metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Topic: Biology, Cell Cycle and Cell Division, NEET
Some charge is being given to a conductor. Then, its potential:

Options:
A is maximum at surface
B is maximum at centre
C is same throughout the conductor
D is maximum somewhere between the surface and the centre

The correct answer is:
is same throughout the conductor

Solution:
We know that inside the conductor $F$ $=0$ so potential remains same.

Topic: Physics, Electrostatics, NEET, NEET 2002
Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total numbers of persons, who participated in the tournament, is ________.

Options:

The correct answer is:
16

Solution:
Given,

Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840,

Now Let total number of persons be 2n, so number of couples will be n

Now number of matches played, so that no couple played in a match is given by,

⇒C2n·C2n-2·2=840

{as if we choose couple as A,B and other couple as C,D and taking other two couple as x,y & w,z, so the possible combination of the match where no couple played in a match will be, A,x→C,w & C,x→A,w , hence there are two possible combination after choosing }

⇒nn-1n-2n-3=5·6·7·8

So, n=8

Hence, the number of persons will be, 2n=16

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Some dividing cells exit the cell cycle and enter vegetative inactive stage. This is called quiescent stage (G0). This process occurs at the end of

Options:
A G1 phase
B S phase
C G2 phase
D M phase

The correct answer is:
G1 phase

Solution:
Quiescent Stage (G0): Cells that do not divide further, exit the G1 phase to enter an inactive stage. This stage is called the quiescent stage (G0) of the cell cycle. The cells in this stage remain metabolically active but do not undergo division. But these cells can resume division as and when required. E.g. heart cells, nerve cells, etc.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2020 (Phase 1)
Some drugs bind to a site other than, the active site of an enzyme. This site is known as

Options:
A non-active site
B allosteric site
C competitive site
D therapeutic site

The correct answer is:
allosteric site

Solution:
Some drugs do not bind to the enzyme’s active site. These bind to a different site of enzyme which is called allosteric site. This binding of inhibitor at allosteric site changes the shape of the active site in such a way that substrate cannot recognise it.

Non-competitive inhibitor changes the active site of enzyme after binding at allosteric site

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Some drugs interact with enzymes and make the biologically inactive. Such drugs are called

Options:
A enzyme promoters
B enzyme inhibitors
C allogens
D all of these

The correct answer is:
enzyme inhibitors

Solution:
When some enzymes are lethal for patient they used to suppressed by drugs which suppress their activity are known as enzyme inhibitors.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r=λ1λ2, is given by:

Options:
A r=13
B r=43
C r=23
D r=34

The correct answer is:
r=13

Solution:
The relation between wave length and energy of electron in nth orbit
λ1=-hcE
λ2=-3hcE
λ1λ2=13

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2017 (02 Apr)
Some gases are responsible for heating of atmosphere (green house effect). Identify from the following the gaseous species which does not cause it.

Options:
A H2O vapour
B O3
C N2
D CH4

The correct answer is:
N2

Solution:
About 75% of the solar energy reaching the earth is absorbed by the earth’s surface, which increases its temperature. The rest of the heat radiates back to the atmosphere. Some of the heat is trapped by gases such as carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFCs) and water vapour in the atmosphere. Thus, they add to the heating of the atmosphere. This causes global warming. Hence, Nitrogen gas is not responsible for green house effect.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2022 (24 Jun Shift 2)