datavorous/entrance-exam-dataset huggingface dataset:
Statement I If there is no external torque on a body about its centre of mass, then the velocity of the centre of mass remains constant.
Statement II The linear momentum of an isolated system remains constant.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:
If a force is applied at centre of mass of a rigid body, its torque about centre of mass will be zero, but acceleration will be non-zero. Hence, velocity will change.
$\therefore$ Correct option is $(\mathrm{d})$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision.
Statement II In an elastic collision, the linear momentum of the system is conserved.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

Solution:
In case of elastic collision, coefficient of restitution $e=1$ or relative speed of separation.
$\therefore$ Option (b) is correct.

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I In water, orthoboric acid behaves as a weak monobasic acid.
Statement II In water, orthoboric acid acts as proton donor.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is false

Solution:

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I Micelles are formed by surfactant molecules above the critical micellar concentration (CMC).
Statement II The conductivity of a solution having surfactant molecules deceases sharply at the CMC.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

Solution:
At a certain concentration surfactant molecules start to aggregate and form micelle, the concentration is called critical micellisation concentration (CMC).
Aggregation of surfactant molecules (ions; $\mathrm{RCOO}^{-}$) micelle formation cause effective fall in number of free ions to conduct electricity, thus, conductivity decreases at CMC.

Topic: Chemistry, Surface Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I Molecules that are not superimposable on their mirror images are chiral.
Statement II All chiral molecules have chiral centres.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is false

Solution:
Molecules having perpendicular dissymmetric plane are supposed to have chirality without having chiral centre. For e.g.,

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I p-hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid.
Statement II o-hydroxybenzoic acid has intramolecular hydrogen bonding.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Statement I p-hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid.
Statement II o-hydroxybenzoic acid has intramolecular hydrogen bonding.

Options:
A
Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I The curve $y=-\frac{x^2}{2}+x+1$ is symmetric with respect to the line $x=1$
Statement II A parabola is symmetric about its axis.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

Solution:
$$
\begin{array}{rlrl}
y & =-\frac{x^2}{2}+x+1 \\
\Rightarrow & y-\frac{3}{2} & =-\frac{1}{2}(x-1)^2
\end{array}
$$
It is symmetric about $x=1$.

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2007 (Paper 2)
Statement I The formula connecting $u, v$ and $f$ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.
Statement II Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is false

Solution:
Laws of reflection can be applied to any type of surface.
$\therefore$ Correct option is (c).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2007 (Paper 1)
Statement I The total translational kinetic energy of all the molecules of a given mass of an ideal gas is $1.5$ times the product of its pressure and its volume.
Statement II The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I

Solution:
Total translational kinetic energy $=\frac{3}{2} n R T=\frac{3}{2} P V=1.5 P \mathrm{~V}$
$\therefore$ Correct option is (b).

Topic: Physics, Kinetic Theory of Gases, JEE Advanced, JEE Advanced 2007 (Paper 2)
Steel wire of length $L$ at $40^{\circ} \mathrm{C}$ is suspended from the ceiling and then a mass $m$ is hung from its free end. The wire is cooled down from $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$ to regain its original length $L$. The coefficient of linear thermal expansion of the steel is $10^{-5} /{ }^{\circ} \mathrm{C}$, Young's modulus of steel is $10^{11} \mathrm{~N} / \mathrm{m}^2$ and radius of the wire is $1 \mathrm{~mm}$. Assume that $L>$ diameter of the wire. Then, the value of $m$ in $\mathrm{kg}$ is nearly

Options:

The correct answer is:
3

Solution:
$\Delta l_1=\frac{F L}{A Y}=\frac{m g L}{\pi r^2 Y}=$ Increase in length $\Delta l_2=L \alpha \Delta \theta=$ Decrease in length To regain its original length,
$$
\begin{aligned}
& \Delta l_1 & =\Delta l_2 \\
\therefore & \frac{m g L}{\pi r^2 Y} & =L \alpha \Delta \theta \\
\therefore & m & =\left(\frac{r^2 Y \alpha \Delta \theta}{g}\right)
\end{aligned}
$$
Substituting the values, we get
$$
m \simeq 3 \mathrm{~kg}
$$
$\therefore$ Answer is 3 .
Analysis of Question
Question is very simple. In my opinion any student who have gone through the syllabus once can solve this problem easily.

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2011 (Paper 1)
Students I, II and III perform an experiment for measuring the acceleration due to gravity $(g)$ using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.
Least count for length $=0.1 \mathrm{~cm}$
Least count for time $=0.1 \mathrm{~s}$
If $E_{\mathrm{I}}, E_{\mathrm{II}}$ and $E_{\mathrm{III}}$ are the percentage errors in $g$, i.e., $\left(\frac{\Delta g}{g} \times 100\right)$ for students I, II and III, respectively.

Options:
A
$E_{\mathrm{I}}=0$

B
$E_{\mathrm{I}}$ is minimum

C
$E_{\mathrm{I}}=E_{\mathrm{II}}$

D
$E_{\mathrm{II}}$ is maximum

The correct answer is:
$E_{\mathrm{I}}$ is minimum

Solution:
$$
\begin{aligned}
& T & =2 \pi \sqrt{\frac{l}{g}} \quad \text { or } \quad \frac{t}{n}=2 \pi \sqrt{\frac{l}{g}} \\
\therefore \quad & g & =\frac{\left(4 \pi^2\right)\left(n^2\right) l}{t^2}
\end{aligned}
$$
$\%$ error in $g=\frac{\Delta g}{g} \times 100=\left(\frac{\Delta l}{l}+\frac{2 \cdot \Delta t}{t}\right) \times 100$
$$
\begin{aligned}
& E_{\mathrm{I}}=\left(\frac{0.1}{64}+\frac{2 \times 0.1}{128}\right) \times 100=0.3125 \% \\
& E_{\mathrm{II}}=\left(\frac{0.1}{64}+\frac{2 \times 0.1}{64}\right) \times 100=0.46875 \% \quad E_{\mathrm{III}}=\left(\frac{0.1}{20}+\frac{2 \times 0.1}{36}\right) \times 100=1.055 \%
\end{aligned}
$$
Hence, $E_{\mathrm{I}}$ is minimum.
$\therefore$ correct option is (b).

Topic: Physics, Experimental Physics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Suppose four distinct positive numbers $a_1, a_2, a_3$ and $a_4$ are in GP. Let $b_1=a_1$, $b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.
Statement 1 The numbers $b_1, b_2, b_3$ and $b_4$ are neither in AP nor in GP.
Statement 2 The numbers $b_1, b_2, b_3$ and $b_4$ are in HP.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
Let $a_1=1, a_2=2, a_3=4, a_4=8$
$$
\therefore b_1=1, b_2=3, b_3=7, b_4=15
$$
Clearly, $b_1, b_2, b_3$ and $b_4$ are not in HP.
Statement 2 is false.

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2008 (Paper 2)
System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass $2 m$ and $m$ just after the string is cut will be

Options:
A
$g / 2$ upwards, $g$ downwards

B
$g$ upwards, $g / 2$ downwards

C
$g$ upwards, $2 g$ downwards

D
$2 g$ upwards, $g$ downwards

The correct answer is:
$g / 2$ upwards, $g$ downwards

Solution:
Initially under equilibrium of mass ' $m$ '
$$
T=m g
$$
Now, the string is cut. Therefore, $T=m g$ force is decreased on mass $m$ upwards and downwards on mass $2 m$.
$$
a_m=\frac{m g}{m}=g \text { (downwards) and }
$$
$$
a_{2 m}=\frac{m g}{2 m}=\frac{g}{2} \text { (upwards) }
$$

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2006
Tangent is drawn at any point $P$ of a curve which passes through $(1,1)$ cutting $X$-axis and $Y$-axis at $A$ and $B$, respectively. If $A P: B P=3: 1$, then

Options:
A
differential equation of the curve is $3 x \frac{d y}{d x}+y=0$

B
differential equation of the curve is $3 x \frac{d y}{d x}-y=0$

C
curve is passing through $\left(\frac{1}{8}, 2\right)$

D
normal at $(1,1)$ is $x+3 y=4$

The correct answers are:
differential equation of the curve is $3 x \frac{d y}{d x}+y=0$
,
curve is passing through $\left(\frac{1}{8}, 2\right)$

Solution:
Since, $\quad A P: B P=3: 1$
where, equation of tangent is $y-y_1=f^{\prime}(x)\left(x-x_1\right)$
$$
\begin{aligned}
\therefore & A\left(x_1-\frac{y_1}{f^{\prime}\left(x_1\right)}, 0\right) \text { and } B\left(0, y_1-x_1 f^{\prime}(x)\right) \\
& A P: B P=3: 1 \Rightarrow \frac{d y}{d x}=\frac{y}{-3 x} \\
\therefore \quad & 3 x \frac{d y}{d x}+y=0 \quad \text { or } \quad \frac{d y}{y}=-\frac{1}{3 x} d x,
\end{aligned}
$$
On integrating both sides, we get
$$
\begin{array}{lrlr}
\Rightarrow & x y^3 & =1 \\
& \therefore \text { At } & x & =\frac{1}{8} \quad \text { and } \quad y=2
\end{array}
$$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2006
Tangents are drawn from the point $(17,7)$ to the circle $x^2+y^2=169$.
Statement I The tangents are mutually perpendicular.
Statement II The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is $x^2+y^2=338$

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

Solution:
Since, the tangents are perpendicular.
So, locus of perpendicular tangents to circle $x^2+y^2=169$ is a director circle having equation
$$
x^2+y^2=338 .
$$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2007 (Paper 1)
Tangents are drawn to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$, parallel to the straight line $2 x-y=1$. The points of contact of the tangents on the hyperbola are

Options:
A $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
B $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
C $(3 \sqrt{3},-2 \sqrt{2})$
D $(-3 \sqrt{3}, 2 \sqrt{2})$

The correct answers are:
$\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$, $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

Solution:
If slope of tangents to hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $m$,

then equations of tangent to the hyperbola is

$y=m x \pm \sqrt{a^{2} m^{2}-b^{2}} \quad$ with the points of contact

$\quad\left(\quad \pm a^{2} m\right.$

$\left.\frac{\pm \sqrt{a^{2} m^{2}-b^{2}}}{\sqrt{a^{2} m^{2}-b^{2}}}\right)$

$\therefore$ Tangent to hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$ is parallel to $2 x-y=1$,

$\therefore$ Slope of tangent $=2$

$\therefore$ Points of contact are $\left(\frac{\pm 9 \times 2}{\sqrt{9 \times 4-4}}, \frac{\pm 4}{\sqrt{9 \times 4-4}}\right)$

i.e. $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{-9}{2 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)$

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2012 (Paper 1)
Tangents drawn from the point $P(1,8)$ to the circle $x^2+y^2-6 x-4 y-11=0$ touch the circle at the points $A$ and $B$. The equation of the circumcircle of $\triangle P A B$ is

Options:
A
$x^2+y^2+4 x-6 y+19=0$

B
$x^2+y^2-4 x-10 y+19=0$

C
$x^2+y^2-2 x+6 y-29=0$

D
$x^2+y^2-6 x-4 y+19=0$

The correct answer is:
$x^2+y^2-4 x-10 y+19=0$

Solution:
For required circle, $P(1,8)$ and $O(3,2)$ will be the end point of its diameter.
$$
\begin{aligned}
& \therefore(x-1)(x-3)+(y-8)(y-2)=0 \\
& \Rightarrow \quad x^2+y^2-4 x-10 y+19=0
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2009 (Paper 1)
The activity of a freshly prepared radioactive sample is $10^{10}$ disintegrations per second, whose mean life is $10^{-9} \mathrm{~s}$. The mass of an atom of this radioisotope is $10^{-25} \mathrm{~kg}$. The mass (in $\mathrm{mg}$ ) of the radioactive sample is

Options:

The correct answer is:
1

Solution:
Activity $\left(-\frac{d N}{d t}\right)=\lambda N=\left(\frac{1}{t_{\text {mean }}}\right) \times N$
$$
\begin{aligned}
& \therefore \quad N=\left(-\frac{d N}{d t}\right) \times t_{\text {mean }} \\
& \quad=\text { Total number of atoms } \\
& \text { Mass of one atom is } 10^{-25} \mathrm{~kg}=m \text { (say) } \\
& \therefore \quad \text { Total mass of radioactive substance } \\
& =(\text { number of atoms) } \\
& \quad \times(\text { mass of one atom) } \\
& =\left(-\frac{d N}{d t}\right)\left(t_{\text {mean }}\right)(m)
\end{aligned}
$$
Substituting the values, we get
Total mass of radioactive substance
$$
=1 \mathrm{mg}
$$
$\therefore$ Answer is 1 .
Analysis of Question
Question is very simple. Again in my opinion one can solve it.

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2011 (Paper 1)
The area of the region between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac{\pi}{4}$ is

Options:
A
$\int_0^{\sqrt{2}-1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

B
$\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

C
$\int_0^{\sqrt{2}+1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

D
$\int_0^{\sqrt{2}+1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

The correct answer is:
$\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

Solution:
$$
\begin{aligned}
& \text { Required area }=\int_0^{\pi / 4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right) d x \quad\left[\because \frac{1+\sin x}{\cos x}>\frac{1-\sin x}{\cos x}>0\right] \\
& \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}} \sqrt{\frac{1-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}}\right) d x \\
& \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}}\right) d x \\
&
\end{aligned}
$$
$$
=\int_0^{\pi / 4} \frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x=\int_0^{\pi / 4} \frac{2 \tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x
$$
Put $\tan \frac{x}{2}=t$
$$
\begin{aligned}
\Rightarrow \quad \frac{1}{2} \sec ^2 \frac{x}{2} d x & =d t \\
\therefore \text { Required area } & =\int_0^{\tan \frac{\pi}{8}} \frac{4 t d t}{\left(1+t^2\right) \sqrt{1-t^2}} \\
& =\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t \quad\left[\text { astan } \frac{\pi}{8}=\sqrt{2}-1\right]
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Advanced, JEE Advanced 2008 (Paper 2)
The bond energy (in $\mathrm{kcal} \mathrm{mol}^{-1}$ ) of $\mathrm{C}-\mathrm{C}$ single bond is approximately

Options:
A
1

B
10

C
100

D
1000

The correct answer is:
100

Solution:
The bond energy of a $\mathrm{C}-\mathrm{C}$ single bond is approximately $100 \mathrm{kcal} \mathrm{mol}^{-1}$.
Energetics
Factual (usually it is not expected to be learnt)
III

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2010 (Paper 1)
The carboxyl functional group $(-\mathrm{COOH})$ is present in

Options:
A picricacid
B barbituric acid
C ascorbic acid
D aspirin

The correct answer is:
aspirin

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The carboxyl functional group $(-\mathrm{COOH})$ is present in

Options:
A picricacid
B barbituric acid
C ascorbic acid
D aspirin

The correct answer is:
aspirin

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Advanced, JEE Advanced 2012 (Paper 1)
The centres of two circles $C_1$ and $C_2$ each of unit radius are at a distance of 6 units from each other. Let $P$ be the mid-point of the line segment joining the centres of $C_1$ and $C_2$ and $C$ be a circle touching circles $C_1$ and $C_2$ externally. If a common tangents to $C_1$ and $C$ passing through $P$ is also a common tangent to $C_2$ and $C$, then the radius of the circle $C$ is

Options:

The correct answer is:
8

Solution:
$(r+1)^2=\alpha^2+9$
$$
\begin{array}{lc}
\Rightarrow & r^2+8=\alpha^2 \\
\Rightarrow & r^2+2 r+1=r^2+8+9 \\
\Rightarrow & 2 r=16 \Rightarrow r=8
\end{array}
$$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2009 (Paper 2)
The centres of two circles $C_1$ and $C_2$ each of unit radius are at a distance of 6 units from each other. Let $P$ be the mid-point of the line segment joining the centres of $C_1$ and $C_2$ and $C$ be a circle touching circles $C_1$ and $C_2$ externally. If a common tangents to $C_1$ and $C$ passing through $P$ is also a common tangent to $C_2$ and $C$, then the radius of the circle $C$ is

Options:

The correct answer is:
8

Solution:
$(r+1)^2=\alpha^2+9$
$$
\begin{array}{lc}
\Rightarrow & r^2+8=\alpha^2 \\
\Rightarrow & r^2+2 r+1=r^2+8+9 \\
\Rightarrow & 2 r=16 \Rightarrow r=8
\end{array}
$$

Topic: Mathematics, Circle, JEE Main
The circle passing through the point $(-1,0)$ and touching the $Y$-axis at $(0,2)$, also passes through the point

Options:
A
$\left(-\frac{3}{2}, 0\right)$

B
$\left(-\frac{5}{2}, 2\right)$

C
$\left(-\frac{3}{2}, \frac{5}{2}\right)$

D
$(-1,-4)$

The correct answer is:
$(-1,-4)$

Solution:
Equation of circle passing through a point $\left(x_1, y_1\right)$ and touching the straight line $L$, is given by
$$
\left(x-x_1\right)^2+\left(y-y_1\right)^2=\lambda L=0
$$
$\therefore$ Equation of circle passing through $(0,2)$ and touching $x=0$.
Now, $(x-0)^2+(y-2)^2+\lambda x=0 \ldots$ (i)
Also, it passes through $(-1,0)$.
So, $1+4-\lambda=0 \Rightarrow \lambda=5$
Eq. (i) becomes,
$$
\begin{aligned}
& x^2+y^2-4 y+4+5 x=0 \\
\Rightarrow \quad x^2+y^2+5 x-4 y+4 & =0
\end{aligned}
$$
For $x$-intercept, put $y=0$,
$$
\begin{array}{rlrl}
& x^2+5 x+4 & =0 \\
\Rightarrow & & (x+1)(x+4) & =0 \\
\therefore & & x & =-1,-4
\end{array}
$$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2011 (Paper 2)
The circular scale of a screw gauge has 50 divisions and pitch of $0.5 \mathrm{~mm}$. Find the diameter of sphere. Main scale reading is 2 .

Options:
A
$1.2$

B
$1.25$

C
$2.20$

D
$2.25$

The correct answer is:
$1.2$

Solution:
Least count $\mathrm{LC}=\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}=\frac{0.1}{50}$ $=0.01 \mathrm{~mm}$
Now, diameter of ball $=(2 \times 0.5 \mathrm{~mm})+(25-5)(0.01)=1.2 \mathrm{~mm}$

Topic: Physics, Experimental Physics, JEE Advanced, JEE Advanced 2006
The colour of light absorbed by an aqueous solution of $\mathrm{CuSO}_{4}$ is:

Options:
A orange-red
B blue-green
C yellow
D violet

The correct answer is:
orange-red

Solution:
$\mathrm{CuSO}_{4}$ will be absorbing orange-red colour and hence will be of blue colour.

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2012 (Paper 1)
The complex showing a spin-only magnetic moment of $2.82$ B.M. is

Options:
A
$\mathrm{Ni}(\mathrm{CO})_4$

B
$\left[\mathrm{NiCl}_4\right]^{2-}$

C
$\mathrm{Ni}\left(\mathrm{PPh}_3\right)_4$

D
$\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

The correct answer is:
$\left[\mathrm{NiCl}_4\right]^{2-}$

Solution:
$\mathrm{Ni}: 3 d^8 4 s^2 \quad \mathrm{Ni}^{2+}: 3 d^8$
Since, $\mathrm{Cl}$ is a weak field ligand, it doesn't cause pairing of electron.
Number of unpaired electrons, $n=2$
$$
\mu=\sqrt{n(n+2)}=\sqrt{2(2+2)} \mathrm{BM}=\sqrt{8} \mathrm{BM}=2.82 \mathrm{BM}
$$
Coordination compound
Conceptual
III

Topic: Chemistry, Coordination Compounds, JEE Advanced, JEE Advanced 2010 (Paper 2)