datavorous/entrance-exam-dataset huggingface dataset:
$2.5 \mathrm{~mL}$ of $\frac{2}{5} \mathrm{M}$ weak monoacidic base $\left(K_b=1 \times 10^{-12}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$ is titrated with $\frac{2}{15}$ $\mathrm{M} \mathrm{HCl}$ in water at $25^{\circ} \mathrm{C}$. The concentration of $\mathrm{H}^{+}$at equivalence point is $\left(K_w=1 \times 10^{-14}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$

Options:
A
$3.7 \times 10^{-13} \mathrm{M}$

B
$3.2 \times 10^{-7} \mathrm{M}$

C
$3.2 \times 10^{-2} \mathrm{M}$

D
$2.7 \times 10^{-2} \mathrm{M}$

The correct answer is:
$2.7 \times 10^{-2} \mathrm{M}$

Solution:
Weak monoacidic base, e.g. $\mathrm{BOH}$ is neutralised.
$$
\mathrm{BOH}+\mathrm{HCl} \longrightarrow \mathrm{BCl}+\mathrm{H}_2 \mathrm{O}
$$
At equivalence point all $\mathrm{BOH}$ gets converted into salt and remember! the concentration of $\mathrm{H}^{+}$(or $\mathrm{pH}$ of solution) is due to hydrolysis of resultant salt ( $\mathrm{BCl}$, cationic hydrolysis here)
$$
\underset{\mathrm{C}(1-h)}{\mathrm{B}^{+}}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\mathrm{Ch}}{\rightleftharpoons} \mathrm{OH}+\underset{\mathrm{Ch}}{\mathrm{H}^{+}}
$$
Volume of $\mathrm{HCl}$ used up,
$$
V_a=\frac{N_b V_b}{N_a}=\frac{2.5 \times 2 \times 15}{2 \times 5}=7.5 \mathrm{~mL}
$$
Concentration of salt,
$$
\begin{aligned}
{[\mathrm{BCl}] } & =\frac{\text { Concentration of base }}{\text { Total volume }}=\frac{2 \times 25}{5(7.5+2.5)}=\frac{1}{10}=0.1 \\
K_h & =\frac{C h^2}{1-h}=\frac{K_w}{K_b}
\end{aligned}
$$
( $h$ should be estimated whether that can be neglected or not) on calculating $h=0.27$ (significant, not negligible)
$$
\left[\mathrm{H}^{+}\right]=C h=0.1 \times 0.27=2.7 \times 10^{-2} \mathrm{M}
$$
When $\left[\mathrm{H}^{+}\right]$is asked to calculate in connection with neutralisation, it
Should be calculated:
Before neutralisation Using Oslwald's dilution law
During neutralisation $\quad$ Considering buffer solution
Half neutralisation $\mathrm{pH}=\mathrm{p} K_a$ and $\mathrm{p} K_b$
At the end of neutralisation $\quad$ Considering hydrolysis of salt

Topic: Chemistry, Ionic Equilibrium, JEE Advanced, JEE Advanced 2008 (Paper 1)
$5.6 \mathrm{~L}$ of helium gas at STP is adiabatically compressed to $0.7 \mathrm{~L}$. Taking the initial temperature to be $T_1$, the work done in the process is

Options:
A
$\frac{9}{8} R T_1$

B
$\frac{3}{2} R T_1$

C
$\frac{15}{8} R T_1$

D
$\frac{9}{2} R T_1$

The correct answer is:
$\frac{9}{8} R T_1$

Solution:
At STP, 22.4 L of any gas is 1 mole.
$$
\therefore \quad 5.6 \mathrm{~L}=\frac{5.6}{22.4}=\frac{1}{4} \text { moles }=n
$$
In adiabatic process,
$$
\begin{array}{rlrl}
& T V^{\gamma-1} & =\text { constant } \\
& \therefore & T_2 V_2^{\gamma-1} & =T_1 V_1^{\gamma-1} \\
& \text { or } & T_2 & =T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}
\end{array}
$$
$$
\begin{aligned}
& \gamma=\frac{C_p}{C_V}=\frac{5}{3} \text { for monoatomic He gas. } \\
& \therefore \quad T_2=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T_1 \\
& \because \text { Further in adiabatic process, } \\
& Q=0 \\
& \therefore \quad W+\Delta U=0 \\
& \text { or } \quad W=-\Delta U \\
& =-n C_V \Delta T \\
& =-n\left(\frac{R}{\gamma-1}\right)\left(T_2-T_1\right) \\
& =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T_1-T_1\right) \\
& =-\frac{9}{8} R T_1 \\
&
\end{aligned}
$$
$\therefore$ Correct option is (a).
Analysis of Question
(i) From calculation point of view question is moderately tough.
(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2011 (Paper 1)
$29.2 \%(\mathrm{w} / \mathrm{w}) \mathrm{HCl}$ stock solution has a density of $1.25 \mathrm{gmL}^{-1}$. The molecular weight of $\mathrm{HCl}$ is $36.5 \mathrm{~g} \mathrm{~mol}^{-1}$. The volume $(\mathrm{mL})$ of stock solution required to prepare a 200 $\mathrm{mL}$ solution of $0.4 \mathrm{M} \mathrm{HCl}$ is :

Options:

The correct answer is:
8

Solution:
Molarity of stock solution of $\mathrm{HCl}$

$=\frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}$

Let the volume of stock solution required $=V \mathrm{~mL}$

Thus, $V \times \frac{29.2 \times 1000 \times 1.25}{100 \times 36.5}=200 \times 0.4$ $\Rightarrow V=8 \mathrm{~mL}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Advanced, JEE Advanced 2012 (Paper 1)
$75.2 \mathrm{~g}$ of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$ (phenol) is dissolved in a solvent of $K_f=14$. If the depression in freezing point is $7 \mathrm{~K}$, then find the $\%$ of phenol that dimerises.

Options:

The correct answer is:
35

Solution:
Phenol is dimerised as follows
At $t=0$
$$
2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right)_2
$$
At equilibrium
$$
(1-\alpha) \quad \alpha / 2
$$
Total moles after dimerisation $=1-\alpha+\alpha / 2=1-\alpha / 2$
$$
\begin{aligned}
\text { Van't Hoff factor (i) } & =\frac{\text { Normal molecular wt. of phenol }}{\text { Abnormal molecular wt. of phenol }}=\frac{94}{m_{a b}} \\
m_{a b} & =\frac{1000 \times K_f \times w}{\Delta T_f \times W}
\end{aligned}
$$
Given that
$$
\begin{aligned}
& K_f=14 \\
& w=72.5 \mathrm{gm} \\
& W=1000 \mathrm{gm} \\
& \therefore \quad m_{a b}=\frac{1000 \times 14 \times 72.5}{7 \times 1000}=145 \\
&
\end{aligned}
$$
$$
\therefore \quad i=\frac{94}{145}
$$
$$
\text { (i) }=\frac{\text { No. of particles after dimerisation }}{\text { No. of particles before dimerisation }}=\frac{1-\alpha / 2}{1}
$$
From eq. (i) and (ii)
$$
\begin{aligned}
1-\alpha / 2 & =\frac{94}{145}=0.65 \\
\frac{\alpha}{2} & =1-0.65=0.35
\end{aligned}
$$
Hence, $35 \%$ phenol is present in dimeric form.

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2006
A $0.1 \mathrm{~kg}$ mass is suspended from a wire of negligible mass. The length of the wire is $1 \mathrm{~m}$ and its cross-sectional area is $4.9 \times 10^{-7} \mathrm{~m}^2$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 \mathrm{rad} \mathrm{s}^{-1}$. If the Young's modulus of the material of the wire is $n \times 10^9 \mathrm{Nm}^{-2}$, the value of $n$ is

Options:

The correct answer is:
4

Solution:
$\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^9\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}$
Putting, $\omega=140 \operatorname{rad} \mathrm{s}^{-1}$ in above equation we get,
$$
n=4
$$
$\therefore$ Answer is 4 .

Topic: Physics, Mechanical Properties of Solids, JEE Advanced, JEE Advanced 2010 (Paper 1)
A $2 \mu \mathrm{F}$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position 2 , is

Options:
A
$0 \%$

B
$20 \%$

C
$75 \%$

D
$80 \%$

The correct answer is:
$80 \%$

Solution:
$q_i=C_i V=2 V=q$ (say)
This charge will remain constant after switch is shifted from position 1 to position 2.
$$
\begin{aligned}
U_i & =\frac{1}{2} \frac{q^2}{C_i}=\frac{q^2}{2 \times 2}=\frac{q^2}{4} \\
U_f & =\frac{1}{2} \frac{q^2}{C_f}=\frac{q^2}{2 \times 10}=\frac{q^2}{20}
\end{aligned}
$$
$\therefore$ Energy dissipated $=U_i-U_f=\frac{q^2}{5}$
This energy dissipated $\left(=\frac{q^2}{5}\right)$ is $80 \%$ of the initial stored energy $\left(=\frac{q^2}{4}\right)$.
Analysis of Question
(i) This question is moderately tough.
(ii) In a capacitor circuit, redistribution of charge takes place under following three conditions.
(a) A switch is closed.
(b) A closed switch is opened.
(c) A switch is shifted from one position to another position.
In the redistribution of charge, energy is dissipated.

Topic: Physics, Capacitance, JEE Advanced, JEE Advanced 2011 (Paper 1)
A $20 \mathrm{~cm}$ long string, having a mass of $1.0 \mathrm{~g}$, is fixed at both the ends. The tension in the string is $0.5 \mathrm{~N}$. The string is set into vibration using an external vibrator of frequency 100 $\mathrm{Hz}$. Find the separation (in $\mathrm{cm}$ ) between the successive nodes on the string.

Options:

The correct answer is:
5

Solution:
Distance between the successive nodes,
$$
\begin{aligned}
d & =\frac{\lambda}{2}=\frac{v}{2 f} \\
& =\frac{\sqrt{T / \mu}}{2 f}
\end{aligned}
$$
Substituting the values we get
$$
d=5 \mathrm{~cm}
$$

Topic: Physics, Waves and Sound, JEE Advanced, JEE Advanced 2009 (Paper 2)
A ball is dropped from a height of $20 \mathrm{~m}$ above the surface of water in a lake. The refractive index of water is $\frac{4}{3}$. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is $12.8 \mathrm{~m}$ above the water surface, the fish sees the speed of ball as

Options:
A
$9 \mathrm{~ms}^{-1}$

B
$12 \mathrm{~ms}^{-1}$

C
$16 \mathrm{~ms}^{-1}$

D
$21.33 \mathrm{~ms}^{-1}$

The correct answer is:
$16 \mathrm{~ms}^{-1}$

Solution:
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 7}=12 \mathrm{~ms}^{-1}$
In this case when eye is inside water, $x_{\text {app. }}=\mu x \Rightarrow \frac{d x_{\text {app. }}}{d t}=\mu \cdot \frac{d x}{d t}$ or $\quad v_{\text {app. }}=\mu v=\frac{4}{3} \times 12=16 \mathrm{~ms}^{-1}$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2009 (Paper 1)
A ball moves over a fixed track as shown in the figure. From $A$ to $B$ the ball rolls without slipping. If surface $B C$ is frictionless and $K_A, K_B$ and $K_C$ are kinetic energies of the ball at $A, B$ and $C$ respectively, then

Options:
A
$h_A>h_C ; K_B>K_c$

B
$h_A>h_c ; K_c>K_A$

C
$h_A=h_c ; K_B=K_c$

D
$h_A < h_c ; K_B>K_c$

The correct answers are:
$h_A>h_C ; K_B>K_c$

Solution:
In smooth part $B C$, due to zero torque, angular velocity and hence the rotational kinetic energy remains constant. While moving from $B$ to $C$ translational kinetic energy converts into gravitational energy.

Topic: Physics, Work Power Energy, JEE Main
A ball moves over a fixed track as shown in the figure. From $A$ to $B$ the ball rolls without slipping. If surface $B C$ is frictionless and $K_A, K_B$ and $K_C$ are kinetic energies of the ball at $A, B$ and $C$ respectively, then

Options:
A
$h_A>h_C ; K_B>K_c$

B
$h_A>h_c ; K_c>K_A$

C
$h_A=h_c ; K_B=K_c$

D
$h_A < h_c ; K_B>K_c$

The correct answers are:
$h_A>h_C ; K_B>K_c$

Solution:
In smooth part $B C$, due to zero torque, angular velocity and hence the rotational kinetic energy remains constant. While moving from $B$ to $C$ translational kinetic energy converts into gravitational energy.

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2006
A ball of mass $0.2 \mathrm{~kg}$ rests on a vertical post of height $5 \mathrm{~m}$. A bullet of mass $0.01 \mathrm{~kg}$, travelling with a velocity $v \mathrm{~m} / \mathrm{s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20 \mathrm{~m}$ and the bullet at a distance of $100 \mathrm{~m}$ from the foot of the post. The initial velocity $v$ of the bullet is

Options:
A
$250 \mathrm{~m} / \mathrm{s}$

B
$250 \sqrt{2} \mathrm{~m} / \mathrm{s}$

C
$400 \mathrm{~m} / \mathrm{s}$

D
$500 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$500 \mathrm{~m} / \mathrm{s}$

Solution:
Time taken by the bullet and ball to strike the ground is
$$
t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{~s}
$$
Let $v_1$ and $v_2$ are the velocities of ball and bullet after collision.
Then applying
We have, $20=v_1 \times 1$
or $v_1=20 \mathrm{~ms}^{-1}$
$$
100=v_2 \times 1 \text { or } v_2=100 \mathrm{~m} / \mathrm{s}^{-1}
$$
Now, from conservation of linear momentum before and after collision we have,
$0.01 v=(0.2 \times 20)+(0.01 \times 100)$
On solving, we get
$$
v=500 \mathrm{~ms}^{-1}
$$
$\therefore$ Correct answer is (d).
Analysis of Question
Question is moderately lengthy from calculation point of view, otherwise it is simple.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball of mass $0.2 \mathrm{~kg}$ rests on a vertical post of height $5 \mathrm{~m}$. A bullet of mass $0.01 \mathrm{~kg}$, travelling with a velocity $v \mathrm{~m} / \mathrm{s}$ in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of $20 \mathrm{~m}$ and the bullet at a distance of $100 \mathrm{~m}$ from the foot of the post. The initial velocity $v$ of the bullet is

Options:
A
$250 \mathrm{~m} / \mathrm{s}$

B
$250 \sqrt{2} \mathrm{~m} / \mathrm{s}$

C
$400 \mathrm{~m} / \mathrm{s}$

D
$500 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$500 \mathrm{~m} / \mathrm{s}$

Solution:
Time taken by the bullet and ball to strike the ground is
$$
t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{~s}
$$
Let $v_1$ and $v_2$ are the velocities of ball and bullet after collision.
Then applying
We have, $20=v_1 \times 1$
or $v_1=20 \mathrm{~ms}^{-1}$
$$
100=v_2 \times 1 \text { or } v_2=100 \mathrm{~m} / \mathrm{s}^{-1}
$$
Now, from conservation of linear momentum before and after collision we have,
$0.01 v=(0.2 \times 20)+(0.01 \times 100)$
On solving, we get
$$
v=500 \mathrm{~ms}^{-1}
$$
$\therefore$ Correct answer is (d).
Analysis of Question
Question is moderately lengthy from calculation point of view, otherwise it is simple.

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2011 (Paper 2)
A ball of mass $(m) 0.5 \mathrm{~kg}$ is attached to the end of a string having length $(L) 0.5 \mathrm{~m}$. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is $324 \mathrm{~N}$. The maximum possible value of angular velocity of ball (in rad/s) is

Options:
A
9

B
18

C
27

D
36

The correct answer is:
36

Solution:
$T \cos \theta$ component will cancel $\mathrm{mg}$.
$T \sin \theta$ component will provide necessary centripetal force to the ball towards centre $C$.
$$
\begin{array}{rlrl}
& \therefore & T \sin \theta & =m r \omega^2=m(l \sin \theta) \omega^2 \\
& \text { or } & T & =m l \omega^2 \\
& \therefore & \omega & =\sqrt{\frac{T}{m l}} \\
& \text { or } & \omega_{\max } & =\sqrt{\frac{T_{\max }}{m l}}=\sqrt{\frac{324}{0.5 \times 0.5}} \\
& = & 36 \mathrm{rad} / \mathrm{s}
\end{array}
$$
$\therefore$ Correct option is (d).
Analysis of Question
(i) Question is simple.
(ii) This is called the conical pendulum.
(iii) The interesting fact in this problem is that $\omega$ or $T$ is independent of $\theta$.
$\omega \propto \sqrt{T}$
If $\omega$ is increased, $T$ will also increase.

Topic: Physics, Motion In Two Dimensions, JEE Advanced, JEE Advanced 2011 (Paper 1)
A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surface are of the same radius of curvature $R=14$ $\mathrm{cm}$. For this bi-convex lens, for an object distance of 40 $\mathrm{cm}$, the image distance will be

Options:
A $-280.0 \mathrm{~cm}$
B $40.0 \mathrm{~cm}$
C $21.5 \mathrm{~cm}$
D $13.3 \mathrm{~cm}$

The correct answer is:
$40.0 \mathrm{~cm}$

Solution:
The focal length $\left(f_{1}\right)$ of the plano-cÃ¶neex Tens with $n=1.5$ using lens-maker formula

$\frac{1}{f_{1}}=\left(n_{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.5-1)\left[\frac{1}{14}-\frac{1}{\infty}\right]=\frac{1}{28}$

The focal length $\left(f_{2}\right)$ of the plano-convex lens with $n=1.2$ $\frac{1}{f_{2}}=\left(n_{2}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.2-1)\left[\frac{1}{\infty}-\frac{1}{-14}\right]=\frac{1}{70}$ Focal length F of the combination

$\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{20}$

Now, applying lens formula for the combination of lens

$\begin{array}{ll}

& \frac{1}{V}-\frac{1}{U}=\frac{1}{F} \Rightarrow \frac{1}{V}-\frac{1}{-40}=\frac{1}{20} \quad[\text { Given } \mu=40 \mathrm{~cm}] \\

\therefore \quad & V=40 \mathrm{~cm}

\end{array}$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2012 (Paper 1)
A biconvex lens of focal length $15 \mathrm{~cm}$ is in front of a plane mirror. The distance between the lens and the mirror is 10 $\mathrm{cm}$. A small object is kept at a distance of $30 \mathrm{~cm}$ from the lens. The final image is

Options:
A
virtual and at a distance of $16 \mathrm{~cm}$ from the mirror

B
real and at a distance of $16 \mathrm{~cm}$ from the mirror

C
virtual and at a distance of $20 \mathrm{~cm}$ from the mirror

D
real and at a distance of $20 \mathrm{~cm}$ from the mirror

The correct answer is:
real and at a distance of $16 \mathrm{~cm}$ from the mirror

Solution:
Object is placed at distance $2 f$ from the lens. So first image $I_1$ will be formed at distance $2 f$ on other side. This image $I_1$ will behave like a virtual object for mirror. The second image $I_2$ will be formed at distance $20 \mathrm{~cm}$ in front of the mirror, or at distance $10 \mathrm{~cm}$ to the left hand side of the lens.
Now applying lens formula
$$
\begin{array}{rlr}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
& \frac{1}{v}-\frac{1}{+10}=\frac{1}{+15} \\
\text { or } & v=6 \mathrm{~cm}
\end{array}
$$
Therefore, the final image is at distance $16 \mathrm{~cm}$ from the mirror. But, this image will be real.
This is because ray of light is travelling from right to left.
$\therefore$ The correct option is (b).

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2010 (Paper 2)
A biconvex lens of focal length $f$ forms a circular image of radius $r$ of sun in focal plane. Then, which option is correct?

Options:
A
$\pi r^2 \propto f$

B
$\pi r^2 \propto f^2$

C
If lower half part is covered by black sheet, then area of the image is equal to $\pi r^2 / 2$

D
If $f$ is doubled, intensity will increase

The correct answer is:
$\pi r^2 \propto f^2$

Solution:
$$
r=f \tan \theta
$$
or
$$
\begin{array}{lc}
\text { or } & r \propto f \\
\therefore & \pi r^2 \propto f^2
\end{array}
$$

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2006
A binary star consists of two stars $A$ (mass $2.2 M_S$ ) and $B$ (mass $11 M_s$ ), where $M_S$ is the mass of the sun. They are separated by distance $d$ and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of $\operatorname{star} B$ about the centre of mass is

Options:

The correct answer is:
6

Solution:
$\frac{L_{\text {Total }}}{L_B}=\frac{\left(I_A+I_B\right) \omega}{I_B \cdot \omega}$ (as $\omega$ will be same in both cases)
$$
\begin{aligned}
& =\frac{I_A}{I_B}+1=\frac{m_A r_A^2}{m_B r_B^2}+1 \\
& \left.=\frac{r_A}{r_B}+1 \quad \quad \text { (as } m_A r_A=m_B r_B\right) \\
& =\frac{11}{2.2}+1 \quad \quad\left(\text { as } r \propto \frac{1}{m}\right) \\
& =6
\end{aligned}
$$
$\therefore$ The correct answer is 6 .

Topic: Physics, Gravitation, JEE Advanced, JEE Advanced 2010 (Paper 1)
A black body of temperature $T$ is inside chamber of temperature $T_0$. Now, the closed chamber is slightly opened to sun such that temperature of black body $(T)$ and chamber $\left(T_0\right)$ remains constant.

Options:
A
Black body will absorb more radiation

B
Black body will absorb less radiation

C
Black body emit more energy

D
None of these

The correct answers are:
None of these

Solution:
Since, the temperature of black body is constant, total heat absorbed $=$ total heat radiated.

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2006
A block $(B)$ is attached to two unstretched $S_1$ and $S_2$ with spring constants $k$ and $4 k$, respectively (see figure I). The other ends are attached to identical supports $M_1$ and $M_2$ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block $B$ is displaced towards wall 1 by a small distance $x$ (figure II) and released. The block returns and moves a maximum distance $y$ towards wall 2 . Displacements $x$ and $y$ are measured with respect to the equilibrium position of the block $B$. The ratio $\frac{y}{-1}$ is

Options:
A
4

B
2

C
$\frac{1}{2}$

D
$\frac{1}{4}$

The correct answer is:
$\frac{1}{2}$

Solution:
From energy conservation,
$$
\begin{aligned}
\frac{1}{2} k x^2 & =\frac{1}{2}(4 k) y^2 \\
\frac{y}{x} & =\frac{1}{2}
\end{aligned}
$$
$\therefore$ correct option is (c).

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2008 (Paper 2)
A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define $N=10 \mu$, then $N$ is

Options:

The correct answer is:
5

Solution:
$$
\begin{aligned}
& F_1=m g \sin \theta+\mu m g \cos \theta \\
& F_2=m g \sin \theta-\mu m g \cos \theta
\end{aligned}
$$
Given that, $F_1=3 F_2$
or $\left(\sin 45^{\circ}+\mu \cos 45^{\circ}\right)$
$$
=3\left(\sin 45^{\circ}-\mu \cos 45^{\circ}\right)
$$
On solving, we get $\mu=0.5$
$$
\therefore \quad N=10 \mu=5
$$
$\therefore$ Answer is 5 .
Analysis of Question
Question is simple. Only one has to take care of direction and magnitude of friction.

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2011 (Paper 1)
A block of base $10 \mathrm{~cm} \times 10 \mathrm{~cm}$ and height $15 \mathrm{~cm}$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then,

Options:
A
at $\theta=30^{\circ}$, the block will start sliding down the plane

B
the block will remain at rest on the plane up to certain $\theta$ and then it will topple

C
at $\theta=60^{\circ}$, the block will start sliding down the plane and continue to do so at higher angles

D
at $\theta=60^{\circ}$, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$

The correct answer is:
the block will remain at rest on the plane up to certain $\theta$ and then it will topple

Solution:
Condition of sliding is $m g \sin \theta>\mu m g \cos \theta$
or $\tan \theta>\mu$
or $\quad \tan \theta>\sqrt{3}$
Condition of toppling is
Torque of $m g \sin \theta$ about $0>$ torque of $m g \cos \theta$ about
$\therefore \quad(m g \sin \theta)\left(\frac{15}{2}\right)>(m g \cos \theta)\left(\frac{10}{2}\right)$
or $\quad \tan \theta>\frac{2}{3}$
With increase in value of $\theta$, condition of sliding is satisfied first.

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2009 (Paper 1)
A block of mass $2 \mathrm{~kg}$ is free to move along the $x$-axis. It is at rest and from $t$ $=0$ onwards it is subjected to a time-dependent force $F(t)$ in the $x$ direction. The force $F(t)$ varies with $t$ as shown in the figure. The kinetic energy of the block after $4.5 \mathrm{~s}$ is

Options:
A
$4.50 \mathrm{~J}$

B
$7.50 \mathrm{~J}$

C
$5.06 \mathrm{~J}$

D
$14.06 \mathrm{~J}$

The correct answer is:
$5.06 \mathrm{~J}$

Solution:
Area under $F$ - $t$ graph $=$ momentum
$$
\begin{aligned}
&=P=\sqrt{2 k m} \\
& \therefore \quad k=\frac{A^2}{2 m} \\
&(A=\text { net area of } F-t \text { graph) } \\
&=\frac{\left\{\left(\frac{4 \times 3}{2}\right)-\left(\frac{1.5 \times 2}{2}\right)\right\}^2}{2 \times 2}=5.0625 \mathrm{~J}
\end{aligned}
$$
$\therefore$ The correct option is (c).

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2010 (Paper 2)