datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is $0.9$
Given Freezing point depression constant of water $\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of water $\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of ethanol $\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=328 \mathrm{~mm}$ of $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm}$ of $\mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question:
Water is added to the solution $M$ such that the mole fraction of water in the solution becomes $0.9$ The boiling point of this solutions is

Options:
A
$380.4 \mathrm{~K}$

B
$376.2 \mathrm{~K}$

C
$375.5 \mathrm{~K}$

D
$354.7 \mathrm{~K}$

The correct answer is:
$376.2 \mathrm{~K}$

Solution:
$\chi_{\mathrm{H}_2 \mathrm{O}}=0.9$ (solvent) $\chi_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}=0.1$ (solute)
$$
\Delta T_b=k_b \times m=0.52 \times \frac{1.0 \times 1000}{0.9 \times 18}=3.2 \mathrm{~K}
$$
Boiling point, $T_b=373+3.2=376.2 \mathrm{~K}$

Topic: Chemistry, Solutions, JEE Main
Paragraph:
Read the following passage and answer the questions.
For every function $f(x)$ which is twice differentiable, these will be good approximation of $\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}$. Now, if we take $c=\frac{a+b}{2}$, then using above again, we get $\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}$ and so on.
We get approximation for value of $\int_a^b f(x) d x$.Question:
If $\lim _{t \rightarrow a} \frac{\int_a^t f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^3}=0$, then degree of polynomial function $f(x)$ at most is

Options:
A
0

B
1

C
3

D
2

The correct answer is:
1

Solution:
$$
\begin{aligned}
\lim _{t \rightarrow a} \frac{\int_a^t f(t) d t-\frac{(t-a)}{2}(f(t)+f(a))}{(t-a)^3} & =0 \\
\Rightarrow \quad \lim _{h \rightarrow 0} \frac{\int_a^{a+h} f(t) d t-\frac{h}{2}(f(a+h)+f(a))}{h^3} & =0 \\
\Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(a+h)-\frac{1}{2}(f(a+h)+f(a))-\frac{h}{2}\left(f^{\prime}(a+h)\right)}{3 h^2} & =0
\end{aligned}
$$
$f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)$
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{1}{2} f^{\prime}(a+h)-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0$
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{h}{2} f^{\prime}(a+h)}{6 h}=0$
$\Rightarrow \quad f^{\prime \prime}(a)=0, \forall a \in R$
$\Rightarrow(x)$ must have maximum degree 1.

Topic: Mathematics, Limits, JEE Main
Paragraph:
Read the following passage carefully and answer the questions given below it.
Journalism combines writing with news gathering and interpretation. While the journalist's work obviously varies from newspaper to newspaper and from magazine to magazine, all journalists are as much research workers, as they are writers. They cannot write their news or feature stories, unless they locate it and research them first. They must be able to read the in-between lines of the main source-news and interpret that. A great many stories hunt the journalist, who rejects most of them. He carefully sorts and sifts those, taking only a relatively very small proportion of news, may be, coming from an unexpected source. A successful journalist may gather news to the tune of 100 percent but he can use them for his profession only 3-5 percent. He must be able to see or forecast to himself, the news of tomorrow or the day after, from the news of today; because newspapers want advance news or advance warning to give a good, exclusive and exhaustive coverage to anything of interest. But a real 'scoop' is a very rare event. And bogus scoops also bring disrepute to the newspaper.Question:
What is most important for a journalist?

Options:
A He must be able to interpret the news correctly
B He must know how to sort out the news
C He must be able to write effectively
D He must know how to gather the right news

The correct answer is:
He must be able to interpret the news correctly

Solution:
No Solution Available

Topic: Biology, Breathing and Exchange of Gases, JIPMER, JIPMER 2008
Paragraph:
Read the following passage carefully and answer the questions given below it.
Journalism combines writing with news gathering and interpretation. While the journalist's work obviously varies from newspaper to newspaper and from magazine to magazine, all journalists are as much research workers, as they are writers. They cannot write their news or feature stories, unless they locate it and research them first. They must be able to read the in-between lines of the main source-news and interpret that. A great many stories hunt the journalist, who rejects most of them. He carefully sorts and sifts those, taking only a relatively very small proportion of news, may be, coming from an unexpected source. A successful journalist may gather news to the tune of 100 percent but he can use them for his profession only 3-5 percent. He must be able to see or forecast to himself, the news of tomorrow or the day after, from the news of today; because newspapers want advance news or advance warning to give a good, exclusive and exhaustive coverage to anything of interest. But a real 'scoop' is a very rare event. And bogus scoops also bring disrepute to the newspaper.Question:
What does the passage say about successful journalists?

Options:
A They reject a major portion of the gathered news
B They can use a relatively small portion of the news
C Their news come from unexpected sources
D They present advance news

The correct answer is:
They can use a relatively small portion of the news

Solution:
No Solution Available

Topic: Biology, Biomolecules (B), JIPMER, JIPMER 2008
Paragraph:
Read the following passage carefully and answer the questions given below it.
Journalism combines writing with news gathering and interpretation. While the journalist's work obviously varies from newspaper to newspaper and from magazine to magazine, all journalists are as much research workers, as they are writers. They cannot write their news or feature stories, unless they locate it and research them first. They must be able to read the in-between lines of the main source-news and interpret that. A great many stories hunt the journalist, who rejects most of them. He carefully sorts and sifts those, taking only a relatively very small proportion of news, may be, coming from an unexpected source. A successful journalist may gather news to the tune of 100 percent but he can use them for his profession only 3-5 percent. He must be able to see or forecast to himself, the news of tomorrow or the day after, from the news of today; because newspapers want advance news or advance warning to give a good, exclusive and exhaustive coverage to anything of interest. But a real 'scoop' is a very rare event. And bogus scoops also bring disrepute to the newspaper.Question:
What is the basic requisite for exclusive and exhaustive coverage of newspapers?

Options:
A Forecast of tomorrow's news from today's news
B Collecting the news of a 'real scoop'
C Selection of a small portion of the gathered news
D Effective style of writing

The correct answer is:
Forecast of tomorrow's news from today's news

Solution:
No Solution Available

Topic: Biology, Evolution, JIPMER, JIPMER 2008
Paragraph:
Read the following passage carefully and answer the questions given below it.
Journalism combines writing with news gathering and interpretation. While the journalist's work obviously varies from newspaper to newspaper and from magazine to magazine, all journalists are as much research workers, as they are writers. They cannot write their news or feature stories, unless they locate it and research them first. They must be able to read the in-between lines of the main source-news and interpret that. A great many stories hunt the journalist, who rejects most of them. He carefully sorts and sifts those, taking only a relatively very small proportion of news, may be, coming from an unexpected source. A successful journalist may gather news to the tune of 100 percent but he can use them for his profession only 3-5 percent. He must be able to see or forecast to himself, the news of tomorrow or the day after, from the news of today; because newspapers want advance news or advance warning to give a good, exclusive and exhaustive coverage to anything of interest. But a real 'scoop' is a very rare event. And bogus scoops also bring disrepute to the newspaper.Question:
What, according to the passage, is the common characteristic of all journalists?

Options:
A They all write for the newspapers
B They all have to first probe into the news they gather
C They all make an adventure to find a worthy news
D They all can forecast future news

The correct answer is:
They all have to first probe into the news they gather

Solution:
No Solution Available

Topic: Biology, Human Reproduction, JIPMER, JIPMER 2008
Paragraph:
Read the following passage carefully and answer the questions given below it.
Journalism combines writing with news gathering and interpretation. While the journalist's work obviously varies from newspaper to newspaper and from magazine to magazine, all journalists are as much research workers, as they are writers. They cannot write their news or feature stories, unless they locate it and research them first. They must be able to read the in-between lines of the main source-news and interpret that. A great many stories hunt the journalist, who rejects most of them. He carefully sorts and sifts those, taking only a relatively very small proportion of news, may be, coming from an unexpected source. A successful journalist may gather news to the tune of 100 percent but he can use them for his profession only 3-5 percent. He must be able to see or forecast to himself, the news of tomorrow or the day after, from the news of today; because newspapers want advance news or advance warning to give a good, exclusive and exhaustive coverage to anything of interest. But a real 'scoop' is a very rare event. And bogus scoops also bring disrepute to the newspaper.Question:
According to the passage, a journalist's work includes

Options:
A writing
B news gathering
C sorting out the news
D All of the above

The correct answer is:
All of the above

Solution:
No Solution Available

Topic: Biology, Evolution, JIPMER, JIPMER 2008
Paragraph:
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential $\left(E^{\circ}\right)$ of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their $E^{\circ}(V$ with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
$$
\begin{array}{cl}
\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} & E^{\circ}=0.54 \\
\mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} & E^{\circ}=1.36 \\
\mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} & E^{\circ}=1.50 \\
\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} & E^{\circ}=0.77 \\
\mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} & E^{\circ}=1.23
\end{array}
$$Question:
While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because

Options:
A
$\mathrm{O}_2$ oxidises $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$

B
$\mathrm{O}_2$ oxidises both $\mathrm{Mn}^{2+}$ and $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$

C
$\mathrm{Fe}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

D
$\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

The correct answer is:
$\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$

Solution:
$$
\begin{aligned}
& {\left[\mathrm{Mn}^{3+}+e^{-}\right.}\left.\longrightarrow \mathrm{Mn}^{2+}\right] \times 4 \\
& 2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+\mathrm{O}_2+4 e^{-} \\
& 4 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+}+\mathrm{O}_2 \\
& E_{\text {cell }}^{\circ}= 1.50-1.23=0.27 \mathrm{~V} \\
& E_{\text {cell }}^{\circ} \text { is + ve }
\end{aligned}
$$

Topic: Chemistry, Redox Reactions, JEE Main
Paragraph:
Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $A$ and $B$.Question:
The equation of the locus of the point whose distance from the point $P$ and the line $A B$ are equal, is

Options:
A
$9 x^2+y^2-6 x y-54 x-62 y$
$$
+241=0
$$

B
$x^2+9 y^2+6 x y-54 x+62 y$
$$
-241=0
$$

C
$9 x^2+9 y^2-6 x y-54 x-62 y$
$$
-241=0
$$

D
$x^2+y^2-2 x y+27 x+31 y$
$$
-120=0
$$

The correct answer is:
$9 x^2+y^2-6 x y-54 x-62 y$
$$
+241=0
$$

Solution:
Equation of $A B$ is $y-0=-\frac{1}{3}(x-3)$
$$
\begin{gathered}
x+3 y-3=0 \\
|x+3 y-3|^2=10\left[(x-3)^2+(y-4)^2\right]
\end{gathered}
$$
(Look at coefficient of $x^2$ and $y^2$ in the
answers)

Topic: Mathematics, Ellipse, JEE Main
Paragraph:
The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question:
Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Options:
A
$2 x-\sqrt{5} y-20=0$

B
$2 x-\sqrt{5} y+4=0$

C
$3 x-4 y+8=0$

D
$4 x-3 y+4=0$

The correct answer is:
$2 x-\sqrt{5} y+4=0$

Solution:
Equation of tangent to hyperbola having slope $m$ is
$$
y=m x+\sqrt{9 m^2-4}
$$
Equation of tangent to circle is
$$
y=m(x-4)+\sqrt{16 m^2+16}
$$
Eqs. (i) and (ii) will be identical for $m=\frac{2}{\sqrt{5}}$ satisfy.
$\therefore$ Equation of common tangent is $2 x-\sqrt{5} y+4=0$

Topic: Mathematics, Hyperbola, JEE Main
Paragraph:
The coordination number of $\mathrm{Ni}^{2+}$ is 4 .
$\mathrm{NiCl}_2+\mathrm{KCN}$ (excess) $\rightarrow A$ (cyano complex)
$\mathrm{NiCl}_2+$ conc. $\mathrm{HCl}$ (excess) $\rightarrow B$ (chloro complex)Question:
The hybridization of $A$ and $B$ are

Options:
A
$d s p^2, s p^3$

B
$s p^3, s p^3$

C
$d s p^2, d s p^2$

D
$s p^3 d^2, d^2 s p^3$

The correct answer is:
$d s p^2, s p^3$

Solution:
Hybridisation of $\mathrm{Ni}$ in $\mathrm{K}_2\left[\mathrm{Ni}(\mathrm{CN})_4\right]$ is $d s p^2$ and in $\mathrm{K}_2\left[\mathrm{Ni}^2\left(\mathrm{Cl}_4\right)\right]$ is $s p^3$ (Detail see in question no. 25).

Topic: Chemistry, Coordination Compounds, JEE Main
Paragraph:
The hydrogen-like species $\mathrm{Li}^{2+}$ is in a spherically symmetric state $S_1$ with one radial node. Upon absorbing light the ion undergoes transition to a state $S_2$. The state $S_2$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.Question:
Energy of the state $S_1$ in units of the hydrogen atom ground state energy is

Options:
A
$0.75$

B
$1.50$

C
$2.25$

D
$4.50$

The correct answer is:
$2.25$

Solution:
Energy of state, $S_1=-13.6\left(\frac{3^2}{2^2}\right) \mathrm{eV} /$ atom $=\frac{9}{4}$ of energy of $\mathrm{H}$-atom in ground state $=2.25$ of energy of $\mathrm{H}$-atom in ground state

Topic: Chemistry, Structure of Atom, JEE Main
Paragraph:
The hydrogen-like species $\mathrm{Li}^{2+}$ is in a spherically symmetric state $S_1$ with one radial node. Upon absorbing light the ion undergoes transition to a state $S_2$. The state $S_2$ has one radial node and its energy is equal to the ground state energy of the hydrogen atom.Question:
The state $S_1$ is

Options:
A
$1 s$

B
$2 \mathrm{~s}$

C
$2 p$

D
$3 \mathrm{~s}$

The correct answer is:
$2 \mathrm{~s}$

Solution:
Given : Hydrogen like species, $\mathrm{Li}^{2+}$
Spherically symmetric state, $S_1 \Rightarrow s$-orbital
$S_1$ has radial node $1 \Rightarrow 2 s \quad$ (No. of radial node $=1=n-l-1$, so $n=2$ ) Species undergoes state $S_2$
$S_2$ has energy equal to the ground state energy of $\mathrm{H}$-atom $=-13.6 \mathrm{eV}$
So, $\quad-13.6=-13.6\left(\frac{3^2}{n^2}\right)$
$n=3$
Again radial node $=1=n-1-1$
or $1=3-l-1 \quad$ so $\quad l=1$
It suggests $S_2$ is $3 p$

Topic: Chemistry, Structure of Atom, JEE Main
Paragraph:
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantization condition.Question:
A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the $n$th level ( $n=0$ is not allowed) is

Options:
A
$\frac{1}{n^2}\left(\frac{h^2}{8 \pi^2 I}\right)$

B
$\frac{1}{n}\left(\frac{h^2}{8 \pi^2 I}\right)$

C
$n\left(\frac{h^2}{8 \pi^2 I}\right)$

D
$n^2\left(\frac{h^2}{8 \pi^2 I}\right)$

The correct answer is:
$n^2\left(\frac{h^2}{8 \pi^2 I}\right)$

Solution:
$$
\begin{aligned}
L=I \omega & =\frac{n h}{2 \pi} \therefore \omega=\frac{n h}{2 \pi I} \\
K & =\frac{1}{2} I \omega^2=\frac{1}{2} I\left(\frac{n h}{2 \pi I}\right)^2 \\
& =\frac{n^2 h^2}{8 \pi^2 I}
\end{aligned}
$$
$\therefore$ The correct answer is (d).

Topic: Physics, Atomic Physics, JEE Main
Paragraph:
There are some deposits of nitrates and phosphates in earth's crust. Nitrates are more soluble in water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily. Ammonia forms large number of complexes with transition metal ions. Hybridisation easily explains the ease of sigma donation capability of $\mathrm{NH}_3$ and $\mathrm{PH}_3$. Phosphine is a flammable gas and is prepared from white phosphorus.Question:
Among the following, the correct statement is

Options:
A
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is a better electron donor because the lone pair of electrons occupies spherical 's' orbital and is less directional

B
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{PH}_3$ is a better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional

C
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is a better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional

D
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{PH}_3$ is a better electron donor because the lone pair of electrons occupies spherical ' $s$ ' orbital and is less directional

The correct answer is:
between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is a better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional

Solution:
$\mathrm{NH}_3$ is better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$-orbital and is more directional.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Paragraph:
Tollen's reagent is used for the detection of aldehyde when a solution of $\mathrm{AgNO}_3$ is added to glucose with $\mathrm{NH}_4 \mathrm{OH}$ then gluconic acid is formed
$\begin{array}{rr}\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} & E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \\ \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)+2 \mathrm{H}^{+}+2 e^{-} ; & -E_{\mathrm{red}}^{\circ}=-0.05 \mathrm{~V} \\ \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ; & E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\end{array}$
[Use $2.303 \times \frac{R T}{F}=0.0592$ and $\frac{F}{R T}=38.92$ at $298 \mathrm{~K}$ ]Question:
$2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_4+2 \mathrm{H}^{+}$. Find $\ln \mathrm{K}$ of this reaction.

Options:
A
$66.13$

B
$58.38$

C
$28.30$

D
$46.29$

The correct answer is:
$58.38$

Solution:
Cell reaction,
(i) $\mathrm{Ag}^{+}+e^{-} \longrightarrow \mathrm{Ag}(\mathrm{s}), E_{\mathrm{red}}^{\circ}=0.8 \mathrm{~V}$
(ii) $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7+2 \mathrm{H}^{+}+2 e^{-}, E_{\text {oxid }}=-0.05 \mathrm{~V}$
Hence, for reaction
$$
\begin{aligned}
2 \mathrm{Ag}^{+}+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} & \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O} 7+2 \mathrm{H}^{+} \\
E_{\text {cell }}^{\circ} & =(0.8-0.05)=0.75 \mathrm{~V} \\
E_{\text {cell }}^{\circ} & =E_{\text {anode }}^{\circ}-E_{\text {cathode }}^{\circ}=[-0.05-(-0.8)]_{\mathrm{V}}
\end{aligned}
$$
On the basis of oxidised electrode potential $=0.75 \mathrm{~V}$
By Nernst equation, $E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303 R T}{n F} \log _{10} K$
At equilibrium, $E_{\text {cell }}=0$
$$
\begin{aligned}
E_{\text {cell }}^{\circ} & =\frac{2.303 R T}{n F} \log _{10} K \\
0.75 & =\frac{0.0592}{2} \log _{10} K \\
\log _{10} K & =\frac{2 \times 0.75}{0.0592}=25.33
\end{aligned}
$$
or In $K=2.303 \times \log _{10} K=2.303 \times 25.33=58.35$

Topic: Chemistry, Electrochemistry, JEE Main
Paragraph:
Two plane harmonic sound waves are expressed by the equations.
$$
\begin{aligned}
& y_1(x, t)=A \cos (0.5 \pi x-100 \pi t) \\
& y_2(x, t)=A \cos (0.46 \pi x-92 \pi t)
\end{aligned}
$$
(All parameters are in MKS)Question:
At $x=0$ how many times the amplitude of $y_1+y_2$ is zero in one second at $x=0$ ?

Options:
A
192

B
48

C
100

D
96

The correct answer is:
96

Solution:
At $x=0, y=y_1+y_2=2 A \cos 96 \pi t \cos 4 \pi t$
Frequency of $\cos (96 \pi t)$ function is $48 \mathrm{~Hz}$ and that of $\cos (4 \pi t)$ function is $2 \mathrm{~Hz}$. In one second cos function becomes zero at $2 f$ times, where $f$ is the frequency. Therefore, first function will become zero at 96 times and the second at 4 times. But second overlaps with first. Hence, net $y$ will become zero 96 times in 1 second.

Topic: Physics, Waves and Sound, JEE Main
Paragraph:
Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see in figure).
Question:
In an experiment, electrons are made to pass through a narrow slit of width 'd' comparable to their de Broglie wavelength. They are detected on a screen at a distance ' $D$ ' from the slit (see figure).
Which of the following graph can be expected to represent the number of electrons ' $N$ ' detected as a function of the detector position ' $y$ ' $(y=0$ corresponds to the middle of the slit)?

Options:
A

B

C

D

The correct answer is:

Solution:
Diffraction pattern will be wider than the slit.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2008