datavorous/entrance-exam-dataset huggingface dataset:
The value of $\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \log (1+t)}{t^4+4} d t$ is

Options:
A
0

B
$\frac{1}{12}$

C
$\frac{1}{24}$

D
$\frac{1}{64}$

The correct answer is:
$\frac{1}{12}$

Solution:
$\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^{x t} \frac{\log (1+t)}{4+t^4} d t$
Using L' Hospital's rule,
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\frac{x \log (1+x)}{4+x^4}}{3 x^2} \\
= & \lim _{x \rightarrow 0} \frac{\log (1+x)}{3 x} \cdot \frac{1}{4+x^4} \\
= & \frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}\left[\text { using, } \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]
\end{aligned}
$$

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2010 (Paper 1)
The value of $\log _{10} \mathrm{~K}$ for a reaction $A \rightleftharpoons B$ is (Given: $\Delta_{\mathrm{r}} H_{298 \mathrm{~K}}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{r}} S_{298 \mathrm{~K}}^{\circ}=10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ and $R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} ; 2.303 \times 8.314 \times 298=5705$ )

Options:
A
5

B
10

C
95

D
100

The correct answer is:
10

Solution:
$$
\begin{aligned}
A & \rightleftharpoons B \\
\Delta G^{\circ} & =\Delta H^{\circ}-T \Delta S^{\circ} \\
\Delta G^{\circ} & =-2303 R T \log K \\
\log K & =\frac{\Delta H^{\circ}-T \Delta S^{\circ}}{-2303 R T} \\
& =\frac{-54.07 \times 10^3-298 \times 10}{-2303 \times 8.314 \times 298} \\
& =\frac{-57050}{-5705}=10
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2007 (Paper 1)
The value of $n$ in the molecular formula $\mathrm{Be}_n \mathrm{Al}_2 \mathrm{Si}_6 \mathrm{O}_{18}$ is

Options:

The correct answer is:
3

Solution:
For $\mathrm{Be}_n \mathrm{Al}_2 \mathrm{Si}_6 \mathrm{O}_{18} \quad n=3$
$$
2 n+6+24-36=0 \quad n=3
$$
Chemical bonding
Basic application
II

Topic: Chemistry, Redox Reactions, JEE Advanced, JEE Advanced 2010 (Paper 1)
The value of the integral $\int_{-\pi / 2}^{\pi / 2}\left(x^{2}+\ln \frac{\pi+x}{\pi-x}\right) \cos x d x$ is

Options:
A 0
B $\frac{\pi^{2}}{2}-4$
C $\frac{\pi^{2}}{2}+4$
D $\frac{\pi^{2}}{2}$

The correct answer is:
$\frac{\pi^{2}}{2}-4$

Solution:
$\int_{-\pi / 2}^{\pi / 2}\left[x^{2}+\ln \left(\frac{\pi+x}{\pi-x}\right)\right] \cos x d x$ $=\int_{-\pi / 2}^{\pi / 2} x^{2} \cos x d x+\int_{-\pi / 2}^{\pi / 2} \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x d x$ $=2 \int_{0}^{\pi / 2} x^{2} \cos x d x+0 \quad\left[\because x^{2} \cos x\right.$ is an even function and $\ln \left(\frac{\pi+x}{\pi-x}\right) \cos x$ is an odd function] $=2\left[x^{2} \sin x+2 x \cos x-2 \sin x\right]_{0}^{\pi / 2}=\frac{\pi^{2}}{2}-4$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2012 (Paper 2)
The value of $\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}$ is

Options:
A
$2 \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$

B
$2 \sqrt{2+\frac{2}{x^2}+\frac{1}{x^4}}+C$

C
$\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$

D
None of these

The correct answer is:
$\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C$

Solution:
Let $\quad I=\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}$,
On dividing numerater and denominator by $x^5$, we get
$$
=\int \frac{\left(\frac{1}{x^3}-\frac{1}{x^5}\right) d x}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}}
$$
put $2-\frac{2}{x^2}+\frac{1}{x^4}=t \Rightarrow\left(\frac{4}{x^3}-\frac{4}{x^5}\right) d x=d t$
$$
\therefore \quad I=\frac{1}{4} \int \frac{d t}{\sqrt{t}}=\frac{1}{4} \cdot \frac{t^{1 / 2}}{1 / 2}=\frac{1}{2} \sqrt{t}+c=\frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C
$$

Topic: Mathematics, Indefinite Integration, JEE Advanced, JEE Advanced 2006
The value(s) of $\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$ is (are)

Options:
A
$\frac{22}{7}-\pi$

B
$\frac{2}{105}$

C
0

D
$\frac{71}{15}-\frac{3 \pi}{2}$

The correct answers are:
$\frac{22}{7}-\pi$

Solution:
Let $I=\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x$
$$
\begin{aligned}
& =\int_0^1 \frac{\left(x^4-1\right)(1-x)^4+(1-x)^4}{\left(1+x^2\right)} d x \\
& =\int_0^1\left(x^2-1\right)(1-x)^4 d x \\
& +\int_0^1 \frac{\left(1+x^2-2 x\right)^2}{\left(1+x^2\right)} d x \\
& =\int_0^1\left\{\left(x^2-1\right)(1-x)^4+\left(1+x^2\right)-4 x\right. \\
& \left.+\frac{4 x^2}{\left(1+x^2\right)}\right\} d x \\
& =\int_0^1\left(\left(x^2-1\right)(1-x)^4+\left(1+x^2\right)-4 x\right. \\
& \left.+4-\frac{4}{1-x^2}\right) d x \\
& =\int_0^1\left(x^6-4 x^5+5 x^4-4 x^2\right. \\
& \left.+4-\frac{4}{1+x^2}\right) d x \\
& =\left[\frac{x^7}{7}-\frac{4 x^6}{6}+\frac{5 x^5}{5}\right. \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \left.\quad-\frac{4 x^3}{3}+4 x-4 \tan ^{-1} x\right]_0^1 \\
= & \frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\left(\frac{\pi}{4}-0\right) \\
= & \frac{22}{7}-\pi
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2010 (Paper 1)
The vector(s) which is/are coplanar with vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, are perpendicular to the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is/are

Options:
A
$\hat{\mathbf{j}}-\hat{\mathbf{k}}$

B
$-\hat{\mathbf{i}}+\hat{\mathbf{j}}$

C
$\hat{\mathbf{i}}-\hat{\mathbf{j}}$

D
$-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

The correct answers are:
$\hat{\mathbf{j}}-\hat{\mathbf{k}}$
,
$-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

Solution:
Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\therefore$ A vector coplanar to $\mathbf{a}$ and $\mathbf{b}$, and perpendicular to $\mathbf{c}$.
Now, $\quad \lambda(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$
$\Rightarrow \quad \lambda\{(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}\}$
$\Rightarrow \quad \lambda\{(1+1+4)(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
$-(1+2+1)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})\}$
$\Rightarrow \lambda\{6 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}-6 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}\}$
$\Rightarrow \lambda\{6 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\} \Rightarrow 6 \lambda(\hat{\mathbf{j}}-\hat{\mathbf{k}})$
For $\lambda=\frac{1}{6} \Rightarrow$ Option (a) is correct.
For $\lambda=-\frac{1}{6} \Rightarrow$ Option (d) is correct.

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2011 (Paper 1)
The volume (in $\mathrm{mL}$ ) of $0.1 \mathrm{M} \mathrm{AgNO}_3$ required for complete precipitation of chloride ions present in $30 \mathrm{~mL}$ of $0.01 \mathrm{M}$ solution of $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}_{)_5} \mathrm{Cl}\right] \mathrm{Cl}_2\right.$, as silver chloride is close to

Options:

The correct answer is:
6

Solution:
$\mathrm{mmol}$ of complex $=30 \times 0.01=0.3$ Also, 1 mole of complex $\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$ gives only two moles of chloride ion when dissolved in solution
$$
\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}^{2+}+2 \mathrm{Cl}^{-}\right.
$$
$\Rightarrow \mathrm{mmol}$ of $\mathrm{Cl}^{-}$ion produced from its $0.3 \mathrm{mmol}=0.6$
Hence, $0.6 \mathrm{mmol}$ of $\mathrm{Ag}^{+}$would be required for precipitation.
$$
\begin{aligned}
& \Rightarrow 0.60 \mathrm{mmol} \text { of } \mathrm{Ag}^{+}=0.1 \mathrm{M} \times \mathrm{V}(\text { in } \mathrm{mL}) \\
& \Rightarrow \quad V=6 \mathrm{~mL} .
\end{aligned}
$$

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2011 (Paper 2)
The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561 Å$. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

Options:
A
$1215 Å$

B
$1640 Å$

C
$2430 Å$

D
$4687 Å$

The correct answer is:
$1215 Å$

Solution:
For hydrogen or hydrogen type atoms
$$
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
$$
In the transition from $n_i \longrightarrow n_f$
$$
\begin{aligned}
& \therefore \quad \lambda \propto \frac{1}{Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)} \\
& \therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{Z_1^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_1}{Z_2^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_2} \\
&
\end{aligned}
$$
$$
\lambda_2=\frac{\lambda_1 Z_1^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_1}{Z_2^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)_2}
$$
Substituting the values, we have
$$
=\frac{(6561 Å)(1)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{(2)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}=1215 Å
$$
$\therefore$ Correct option is (a).
Analysis of Question
(i) Question is simple.
(ii) In modern physics, mostly questions are asked on the emission of photon by the transition of electron from some higher energy state to some lower energy state.
(iii) For hydrogen and hydrogen like atoms which have only single electron, we can use the formula
$$
\frac{1}{\lambda}=R Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)
$$
(iv) Further, the student should also remember different series, as shown in figure below.
$\mathrm{I} \rightarrow$ Lymen series
II $\rightarrow$ Balmer series
III $\rightarrow$ Paschen series
IV $\rightarrow$ Brackett series
$\mathrm{V} \rightarrow$ Pfund series

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2011 (Paper 1)
The work function ( $\Phi$ ) of some metals is listed below. The number of metals which will show photoelectric effect when light of $300 \mathrm{~nm}$ wavelength falls on the metal is

Options:

The correct answer is:
4

Solution:
Energy of photon $=\frac{h c}{\lambda} \mathrm{J}=\frac{h c}{e \lambda} \mathrm{eV}=\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9} \times 1.602 \times 10^{-19}}=4.14 \mathrm{eV}$
For photoelectric effect to occur, energy of incident photons must be greater than work function of metal. Hence, only $\mathrm{Li}, \mathrm{Na}, \mathrm{K}$ and $\mathrm{Mg}$ have work functions less than $4.14 \mathrm{~V}$.

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2011 (Paper 1)
The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t=\frac{4}{3} \mathrm{~s}$ is

Options:
A
$\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}$

B
$\frac{-\pi^2}{32} \mathrm{cms}^{-2}$

C
$\frac{\pi^2}{32} \mathrm{cms}^{-2}$

D
$-\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}$

The correct answer is:
$-\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}$

Solution:
$T=8 \mathrm{~s}, \omega=\frac{2 \pi}{T}=\left(\frac{\pi}{4}\right) \operatorname{rads}^{-1}$
$$
\begin{aligned}
x & =A \sin \omega t \\
\therefore \quad a=-\omega^2 x & =-\left(\frac{\pi^2}{16}\right) \sin \left(\frac{\pi}{4} t\right)
\end{aligned}
$$
Substituting $t=\frac{4}{3} \mathrm{~s}$, we get
$$
a=-\left(\frac{\sqrt{3}}{32} \pi^2\right) \mathrm{cms}^{-2}
$$

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2009 (Paper 1)
There is a rectangular plate of mass $M \mathrm{~kg}$ of dimensions $(a \times b)$. The plate is held in horizontal position by striking $n$ small balls each of mass $m$ per unit area per unit time. These are striking in the shaded half region of the plate. The balls are colliding elastically with velocity $v$. What is $v$ ? It is given $n=100, M=3 \mathrm{~kg}, m=0.01 \mathrm{~kg}$; $b=2 \mathrm{~m} ; a=1 \mathrm{~m} ; g=10 \mathrm{~m} / \mathrm{s}^2$.

Options:

The correct answer is:
10

Solution:
$F=\frac{\Delta p}{\Delta t}=n \times\left(a \times \frac{b}{2}\right) \times(2 m v)$
Equating the torque about hinge side, we have
$$
n \times\left(a \times \frac{b}{2}\right) \times(2 m v) \times \frac{3 b}{4}=M g \times \frac{b}{2}
$$
Substituting the given values we get,
$$
v=10 \mathrm{~m} / \mathrm{s}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
There is a rectangular plate of mass $M \mathrm{~kg}$ of dimensions $(a \times b)$. The plate is held in horizontal position by striking $n$ small balls each of mass $m$ per unit area per unit time. These are striking in the shaded half region of the plate. The balls are colliding elastically with velocity $v$. What is $v$ ? It is given $n=100, M=3 \mathrm{~kg}, m=0.01 \mathrm{~kg}$; $b=2 \mathrm{~m} ; a=1 \mathrm{~m} ; g=10 \mathrm{~m} / \mathrm{s}^2$.

Options:

The correct answer is:
10

Solution:
$F=\frac{\Delta p}{\Delta t}=n \times\left(a \times \frac{b}{2}\right) \times(2 m v)$
Equating the torque about hinge side, we have
$$
n \times\left(a \times \frac{b}{2}\right) \times(2 m v) \times \frac{3 b}{4}=M g \times \frac{b}{2}
$$
Substituting the given values we get,
$$
v=10 \mathrm{~m} / \mathrm{s}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2006
Thermococcus, Methanococcus and Methanobacterium are groups of

Options:
A bacteria containing a cytoskeleton and all membrane bound organelles
B archaebacteria with peptidoglycan in their cell wall
C archaebacteria that consists of protein homologous to eukaryotic core histones
D most advanced type of bacteria

The correct answer is:
archaebacteria that consists of protein homologous to eukaryotic core histones

Solution:
Archaebacteria are the most primitive and ancient bacterial species.
Key characteristic features are
(i) Lack of peptidoglycan in cell wall instead pseudomurein is present.
(ii) Lipids in cell membrane have different structure than in all other organisms.
(iii) They lack membrane bound organelles.

Topic: Biology, Biological Classification, JIPMER, JIPMER 2015
Three concentric metallic spherical shells of radii $R, 2 R, 3 R$ are given charges $Q_1, Q_2, Q_3$, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, $Q_1: Q_2: Q_3$, is

Options:
A
$1: 2: 3$

B
$1: 3: 5$

C
$1: 4: 9$

D
$1: 8: 18$

The correct answer is:
$1: 3: 5$

Solution:
$Q_1=\sigma\left(4 \pi R^2\right)=4 \pi \sigma R^2$
$$
\begin{aligned}
& Q_2=16 \pi \sigma R^2-Q_1=12 \pi \sigma R^2 \\
& Q_3=36 \pi \sigma R^2-16 \pi \sigma R^2=20 \pi \sigma R^2 \\
& Q_1: Q_2: Q_3=1: 3: 5
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2009 (Paper 1)
Three objects $A, B$ and $C$ are kept in a straight line on a frictionless horizontal surface. These have masses $m, 2 m$ and $m$, respectively. The object $A$ moves towards $B$ with a speed $9 \mathrm{~ms}^{-1}$ and makes an elastic collision with it. Thereafter, $B$ makes completely inelastic collision with $C$. All motions occur on the same straight line. Find the final speed (in $\mathrm{ms}^{-1}$ ) of the object $C$.

Options:

The correct answer is:
4

Solution:
After elastic collision,
$$
\begin{aligned}
v_A^{\prime} & =\left(\frac{m-2 m}{m+2 m}\right)(9)+\frac{2(2 m)}{m+2 m}(0) \\
& =-3 \mathrm{~ms}^{-1}
\end{aligned}
$$
Now from conservation of linear momentum after all collisions are complete,
$$
\begin{array}{rlrl}
m\left(+9 \mathrm{~ms}^{-1}\right) & =m\left(-3 \mathrm{~ms}^{-1}\right)+3 m\left(v_C\right) \\
& \text { or } \quad v_C & =4 \mathrm{~ms}^{-1}
\end{array}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2009 (Paper 2)
Three objects $A, B$ and $C$ are kept in a straight line on a frictionless horizontal surface. These have masses $m, 2 m$ and $m$, respectively. The object $A$ moves towards $B$ with a speed $9 \mathrm{~ms}^{-1}$ and makes an elastic collision with it. Thereafter, $B$ makes completely inelastic collision with $C$. All motions occur on the same straight line. Find the final speed (in $\mathrm{ms}^{-1}$ ) of the object $C$.

Options:

The correct answer is:
4

Solution:
After elastic collision,
$$
\begin{aligned}
v_A^{\prime} & =\left(\frac{m-2 m}{m+2 m}\right)(9)+\frac{2(2 m)}{m+2 m}(0) \\
& =-3 \mathrm{~ms}^{-1}
\end{aligned}
$$
Now from conservation of linear momentum after all collisions are complete,
$$
\begin{array}{rlrl}
m\left(+9 \mathrm{~ms}^{-1}\right) & =m\left(-3 \mathrm{~ms}^{-1}\right)+3 m\left(v_C\right) \\
& \text { or } \quad v_C & =4 \mathrm{~ms}^{-1}
\end{array}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three vectors P→, Q→ and R→ are shown in the figure. Let, S be any point on the vector R→ . The distance between the points P and S is bR→. The general relation among vectors P→, Q→ and S→ are

Options:
A Sâ=1-bÂ Pâ+b2Qâ
B Sâ=B-1Pâ+bQâ
C S→=1-bP→+bQ→
D Sâ=1-b2Pâ+bQâ

The correct answer is:
S→=1-bP→+bQ→

Solution:
Let, vector from point P to point S be C→,

⇒ C→=bR→ R^=bR→ R→R→=bR→=bQ→- P→

From triangle rule of vector addition,

P→+ C→= S→

P→+bQ→- P→= S→

⇒ S→=1-bP→+bQ→

Topic: Physics, Motion In Two Dimensions, JEE Advanced, JEE Advanced 2017 (Paper 2)
Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures $2 T$ and $3 T$ respectively. The temperature of the middle (i.e. second) plate under steady state condition is

Options:
A $\left(\frac{65}{2}\right)^{1 / 4} T$
B $\left(\frac{97}{4}\right)^{1 / 4} T$
C $\left(\frac{97}{2}\right)^{1 / 4} T$
D $(97)^{1 / 4} T$

The correct answer is:
$\left(\frac{97}{2}\right)^{1 / 4} T$

Solution:
Under steady conditions, the heat gained per second by a plate is equal to the heat released per second by the plate.

$\begin{array}{l}

\frac{\text { Heat gained }}{\text { Second }}[\text { by }(2) \text { from }(1)]+\frac{\text { Heat gained }}{\text { Second }} \\

{\left[\left(\text { by }((2) \text { from }(3)]=\frac{\text { Heat gained }}{\text { Second }} \text { (by } 2\right)\right.}

\end{array}$

$\begin{array}{ll}\therefore \quad & \sigma A(2 T)^{4}+\sigma A(3 T)^{4} \\ & =\sigma(2 A)(T)^{4} \\ \therefore \quad & T=\left[\frac{97}{2}\right]^{1 / 4} T\end{array}$

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2012 (Paper 1)
To an evacuated vessel with movable piston under external pressure of $1 \mathrm{~atm}$, $0.1$ mole of He and $1.0$ mole of an unknown compound (vapour pressure $0.68 mathrm{~atm}$ at $0^{\circ} \mathrm{C}$ ) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at $0^{\circ} \mathrm{C}$ is close to

Options:

The correct answer is:
7

Solution:
Since, the external pressure is $1.0 \mathrm{~atm}$, the gas pressure is also $1.0 \mathrm{~atm}$ as piston is movable. Out of this $1.0 \mathrm{~atm}$ partial pressure due to unknown compound is $0.68 \mathrm{~atm}$. Therefore, partial pressure of $\mathrm{He}=1.00-0.68=0.32 \mathrm{~atm}$.
$$
\begin{aligned}
& \Rightarrow \quad \text { Volume }=\frac{n(\mathrm{He}) \mathrm{RT}}{r(\mathrm{He})}=\frac{0.1 \times 0.082 \times 273}{0.32}=7 \mathrm{~L} \\
& \Rightarrow \text { Volume of container }=\text { Volume of } \mathrm{He} .
\end{aligned}
$$

Topic: Chemistry, States of Matter, JEE Advanced, JEE Advanced 2011 (Paper 1)
To determine the half-life of a radioactive element, a student plots a graph of $\ln \left|\frac{d N(t)}{d t}\right|$ versust. Here $\frac{d N(t)}{d t}$ is the rate of radioactive decay at time $t$. If the number of radioactive nuclei of this element decreases by a factor of $p$ after $4.16 \mathrm{yr}$, the value of $p$ is

Options:

The correct answer is:
8

Solution:
$\left|\frac{d N}{d t}\right|=\mid$ Activity of radioactive substance|
$$
=\lambda N=\lambda N_0 e^{-\lambda t}
$$
Taking log both sides
$$
\ln \left|\frac{d N}{d t}\right|=\ln \left(\lambda N_0\right)-\lambda t
$$
Hence, $\ln \left|\frac{d N}{d t}\right|$ versus $t$ graph is a straight line with slope $-\lambda$
From the graph we can see that,
$$
\lambda=\frac{1}{2}=0.5 \mathrm{yr}^{-1}
$$
Now applying the equation,
$$
\begin{aligned}
N & =N_0 e^{-\lambda t} \\
& =N_0 e^{-0.5 \times 4.16} \\
& =N_0 e^{-2.08}=0.125 N_0 \\
& =\frac{N_0}{8}
\end{aligned}
$$
ie, nuclei decreases by a factor of 8 . Hence the answer is 8 .

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2010 (Paper 2)
To verify Ohm's law, a student is provided with a test resistor $R_T$, a high resistance $R_1$, a small resistance $R_2$, two identical galvanometers $G_1$ and $G_2$, and a variable voltage source $V$. The correct circuit to carry out the experiment is

Options:
A

B

C

D

The correct answer is:

Solution:
We will require a voltmeter, an ammeter, a test resistor and a variable battery to verify Ohm's law.
Voltmeter which is made by connecting a high resistance with a galvanometer is connected in parallel with the test resistor.
Further, an ammeter which is formed by connecting a low resistance in parallel with galvanometer is required to measure the current through test resistor.
The correct option is (c).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2010 (Paper 1)
To verify Ohm's law, a student is provided with a test resistor $R_T$, a high resistance $R_1$, a small resistance $R_2$, two identical galvanometers $G_1$ and $G_2$, and a variable voltage source $V$. The correct circuit to carry out the experiment is

Options:
A

B

C

D

The correct answer is:

Solution:
We will require a voltmeter, an ammeter, a test resistor and a variable battery to verify Ohm's law.
Voltmeter which is made by connecting a high resistance with a galvanometer is connected in parallel with the test resistor.
Further, an ammeter which is formed by connecting a low resistance in parallel with galvanometer is required to measure the current through test resistor.
The correct option is (c).

Topic: Physics, Current Electricity, JEE Main
Total number of geometrical isomers for the complex $\left[\mathrm{RhCl}(\mathrm{CO})\left(\mathrm{PPh}_3\right)\left(\mathrm{NH}_3\right)\right]$ is

Options:

The correct answer is:
3

Solution:
$[M(a b c d)]$ complex is square planar so will have 3 geometrical isomers.

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2010 (Paper 2)
Two adjacent sides of a parallelogram $A B C D$ are given by $\overrightarrow{\mathbf{A B}}=2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{A D}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$. The side $A D$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $A D$ becomes $A D^{\prime}$. If $A D^{\prime}$ makes a right angle with the side $A B$, then the cosine of the angle $\alpha$ is given by

Options:
A
$\frac{8}{9}$

B
$\frac{\sqrt{17}}{9}$

C
$\frac{1}{9}$

D
$\frac{4 \sqrt{5}}{9}$

The correct answer is:
$\frac{\sqrt{17}}{9}$

Solution:
$$
\overrightarrow{\mathbf{A B}}=2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+11 \hat{\mathbf{k}}
$$
$$
\overrightarrow{\mathbf{A D}}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}
$$
Angle ' $\theta$ ' between $\overrightarrow{\mathbf{A B}}$ and $\overrightarrow{\mathbf{A D}}$ is
$$
\begin{aligned}
& \cos (\theta)=\left|\begin{array}{l}
\overrightarrow{\mathbf{A B}} \cdot \overrightarrow{\mathbf{A D}} \\
\mid \overrightarrow{\mathbf{A B}}\|\overrightarrow{\mathbf{A D}}\|
\end{array}\right| \\
& =\left|\frac{-2+20+22}{(15)(3)}\right|=\frac{8}{9} \\
& \Rightarrow \quad \sin (\theta)=\frac{\sqrt{17}}{9} \\
& \text { Since, } \quad \alpha+\theta=90^{\circ} \\
& \therefore \quad \cos (\alpha)=\cos \left(90^{\circ}-\theta\right) \\
& =\sin (\theta)=\frac{\sqrt{17}}{9} \\
&
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2010 (Paper 2)