datavorous/entrance-exam-dataset huggingface dataset:
A block of mass 2M is attached to a massless spring with spring-constant k. This block is connected to two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the blocks are a1, a2 and a3 as shown in figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x0. Which of the following option(s) is/are correct?
[ g is the acceleration due to gravity. Neglect friction]

Options:
A x0=4Mgk
B When spring achieves an extension of x02 for the first time, the speed of the block connected to the spring is 3gM5k
C a2-a1=a1-a3
D At an extension of x04 of the spring, the magnitude of acceleration of the block connected to the spring is 3g10

The correct answers are:
a2-a1=a1-a3

Solution:
Using string constraint
a1=a2+a32
⇒2a1=a2+a3
a1-a3=a2-a1⇒C is correct.
for other options use m equivalent

Equivalent mass
Or Reduced mass
meq=2m1×m2m1+m2 (for a two-mass system)
=22m×m2m+m=4m3
⇒T=4m3.g
∴2T=2.4mg3=8m3g
⇒meq=8m3

Using mechanical energy conservation
⇒12kx02=8mg3x0
x0=16mg3k⇒A is incorrect
Option B⇒Vx02=Vmax=x02ω=x02k2m+8m3=x023k14m=g3221k
Option D⇒ax04=x04ω2=x043k14m=3kx042m=8g21
For a simple harmonic motion x0⇒ maximum displacement from extreme to extreme
⇒x0=2A⇒A=x02
and ω=km=3k14m
∴vmax=A.ω
and a=ω2.x

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2019 (Paper 2)
A block of mass $0.18 \mathrm{~kg}$ is attached to a spring of force constant $2 \mathrm{~N} / \mathrm{m}$. The coefficient of friction between the block and the floor is 0.1. Initially, the block is at rest and the spring is unstretched. An impulse is given to the block as shown in the figure. The block slides a distance of $0.06 \mathrm{~m}$ and comes to rest for the first time. The initial velocity of the block in $\mathrm{m} / \mathrm{s}$ is $v=\frac{N}{10}$. Then, $N$ is

Options:

The correct answer is:
4

Solution:
Decrease in mechanical energy
$=$ Work done against friction
$\therefore \quad \frac{1}{2} m v^2-\frac{1}{2} k x^2=\mu m g x$
or $v=\sqrt{\frac{2 \mu m g x+k x^2}{m}}$
Substituting the values, we get
$$
v=0.4 \mathrm{~ms}^{-1}=\left(\frac{4}{10}\right) \mathrm{ms}^{-1}
$$
$\therefore$ Answer is 4 .
Analysis of Question
(i) Question is simple.
(ii) If $\mu_s$ and $\mu_k$ two values of coefficient of friction are given, then we will take $\mu_k$.

Topic: Physics, Work Power Energy, JEE Advanced, JEE Advanced 2011 (Paper 2)
A block of mass $m$ is on an inclined plane of angle $\theta$. The coefficient of friction between the block and the plane is $\mu$ and $\tan \theta>\mu$. The block is held stationary by applying a force $P$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $P$ is varied from $P_1=m g(\sin \theta-\mu \cos \theta) \quad$ to $P_2=m g(\sin \theta+\mu \cos \theta)$, the frictional force $f$ versus $P$ graph will look like

Options:
A

B

C

D

The correct answer is:

Solution:
When
$$
\begin{aligned}
P & =m g(\sin \theta-\mu \cos \theta) \\
f & =\mu m g \cos \theta \quad \text { (upwards) } \\
\text { when } \quad P & =m g \sin \theta \\
f & =0
\end{aligned}
$$
and when $P=m g(\sin \theta+\mu \cos \theta)$
$$
f=\mu m g \cos \theta \text { (downwards) }
$$
Hence friction is first positive, then zero and then negative.
$\therefore$ correct option is (a).

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2010 (Paper 1)
A bob of mass $M$ is suspended by a massless string of length $L$. The horizontal velocity $v$ at position $A$ is just sufficient to make it reach the point $B$. The angle $\theta$ at which the speed of the bob is half of that at $A$, satisfies

Options:
A
$\theta=\frac{\pi}{4}$

B
$\frac{\pi}{4} < \theta < \frac{\pi}{2}$
C
$\frac{\pi}{2} < \theta < \frac{3 \pi}{4}$
D
$\frac{3 \pi}{4} < \theta < \pi$

The correct answer is:
$\frac{3 \pi}{4} < \theta < \pi$

Solution:
$v=\sqrt{5 g L}$
Solving Eqs. (i), (ii) and (iii) we get,
$$
\cos \theta=-\frac{7}{8} \quad \text { or } \theta=\cos ^{-1}\left(-\frac{7}{8}\right)=151^{\circ}
$$
$\therefore$ correct option is (d).

Topic: Physics, Motion In Two Dimensions, JEE Advanced, JEE Advanced 2008 (Paper 2)
A bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m, suspended by a string of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio l1l2 is

Options:

The correct answer is:
5

Solution:
As shown in the figure shown above, the minimum speed required by the bob of mass m attached to the string of length l1 in order to complete the full circle is 5gl1.
When the bob goes to the highest point, its speed becomes gl1. Here, it strikes with the identical mass, which is tied to the string of length l2. The speed delivered to the object is gl1. This should be equal to 5gl2 so that this bob could complete a full circle.
gl1=5gl2
⇒gl1=5gl2
⇒l1l2=5

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2013 (Paper 1)
A boy is pushing a ring of mass $2 \mathrm{~kg}$ and radius $0.5 \mathrm{~m}$ with a stick as shown in the figure. The stick applies a force of $2 \mathrm{~N}$ on the ring and rolls it without slipping with an acceleration of $0.3 \mathrm{~m} / \mathrm{s}^2$. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is $\frac{P}{10}$. The value of $P$ is

Options:

The correct answer is:
4

Solution:
There is no slipping between ring and ground. Hence, $f_2$ is not maximum. But there is slipping between ring and stick. Therefore, $f_1$ is maximum. Now, let us write the equations.
$$
\begin{aligned}
I & =m R^2=(2)(0.5)^2 \\
& =\frac{1}{2} \mathrm{kgm}^{-2} \\
N_1-F_2 & =m a \\
\text { or } N_1-F_2 & =(2)(0.3)=0.6 \mathrm{~N} \\
a & =R \alpha=\frac{R \tau}{I} \\
& =\frac{R\left(f_2-f_1\right) R}{I}=\frac{R^2\left(f_2-f_1\right)}{I} \\
\therefore \quad 0.3 & =\frac{(0.5)^2\left(f_2-f_1\right)}{(1 / 2)} \\
\text { or } \quad f_2-f_1 & =0.6 \mathrm{~N} \\
N_1^2+f_1^2 & =(2)^2=4
\end{aligned}
$$
Further $\quad F_1=\mu N_1=\left(\frac{P}{10}\right) N_1$
Solving above four equation, we get $P \simeq 3.6$
Therefore, the correct answer should be 4 .
Analysis of Question
(i) Question is moderately tough from concept point of view. But calculations are lengthy.
(ii) One has to think about the two components of the force applied by the stick.
(iii) Answer comes an integer when you consider only $N_1$.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2011 (Paper 1)
A circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed, the total amount of charge that flows from $Y$ to $X$ is

Options:
A
zero

B
$54 \mu \mathrm{C}$

C
$27 \mu \mathrm{C}$

D
$81 \mu \mathrm{C}$

The correct answer is:
$27 \mu \mathrm{C}$

Solution:
From $Y$ to $X$ charge flows to plates $a$ and $b$.
$$
\left(q_0+q_b\right)_i=0,\left(q_0+q_b\right)_f=27 \mu \mathrm{C}
$$
$\therefore 27 \mu \mathrm{C}$ charge flows from $Y$ to $X$.
$\therefore$ Correct option is (c)

Topic: Physics, Capacitance, JEE Main
A circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed, the total amount of charge that flows from $Y$ to $X$ is

Options:
A
zero

B
$54 \mu \mathrm{C}$

C
$27 \mu \mathrm{C}$

D
$81 \mu \mathrm{C}$

The correct answer is:
$27 \mu \mathrm{C}$

Solution:
From $Y$ to $X$ charge flows to plates $a$ and $b$.
$$
\left(q_0+q_b\right)_i=0,\left(q_0+q_b\right)_f=27 \mu \mathrm{C}
$$
$\therefore 27 \mu \mathrm{C}$ charge flows from $Y$ to $X$.
$\therefore$ Correct option is (c)

Topic: Physics, Capacitance, JEE Advanced, JEE Advanced 2007 (Paper 1)
A circular disc with a groove along its diameter is placed horizontally. A block of mass $1 \mathrm{~kg}$ is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is $\mu=2 / 5$. The disc has an acceleration of $25 \mathrm{~m} / \mathrm{s}^2$. Find the acceleration of the block with respect to disc.

Options:

The correct answer is:
10

Solution:
Normal reaction in vertical direction $N_1=m g$
Normal reaction from side to the groove $N_2=m a \sin 37^{\circ}$
Therefore, acceleration of block with respect to discs
$$
a_r=\frac{m a \cos 37^{\circ}-\mu N_1-\mu N_2}{m}
$$
Substituting the values we get, $\quad a_r=10 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Laws of Motion, JEE Advanced, JEE Advanced 2006
A circular wire loop of radius $R$ is placed in the $x-y$ plane centered at the origin $O$. A square loop of side $a(a< < R)$ having two turns is placed with its centre at $z=\sqrt{3} R$ along the axis of the circular wire loop, as shown in figure.

The plane of the square loop makes an angle of $45^{\circ}$ with respect to the $z$-axis. If the mutual inductance between the loops is given by $\frac{\mu_{0} a^{2}}{2^{p / 2} R}$, then the value of $p$ is

Options:

The correct answer is:
7

Solution:
The magnetic field due to current carrying wire at the location of square loop is

$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+3 R^{2}\right)^{3 / 2}}=\frac{\mu_{0} i}{16 R}$

The mutual induction

$\begin{array}{l}

M=\frac{N \phi}{i}=\frac{2}{i}\left[\frac{\mu_{0} i}{16 R} \times a^{2} \cos 45^{\circ}\right] \\

\therefore \quad M=\frac{\mu_{0} a^{2}}{2^{\frac{7}{2}} R}

\end{array}$

Topic: Physics, Electromagnetic Induction, JEE Main
A circular wire loop of radius $R$ is placed in the $x-y$ plane centered at the origin $O$. A square loop of side $a(a< < R)$ having two turns is placed with its centre at $z=\sqrt{3} R$ along the axis of the circular wire loop, as shown in figure.

The plane of the square loop makes an angle of $45^{\circ}$ with respect to the $z$-axis. If the mutual inductance between the loops is given by $\frac{\mu_{0} a^{2}}{2^{p / 2} R}$, then the value of $p$ is

Options:

The correct answer is:
7

Solution:
The magnetic field due to current carrying wire at the location of square loop is

$B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+3 R^{2}\right)^{3 / 2}}=\frac{\mu_{0} i}{16 R}$

The mutual induction

$\begin{array}{l}

M=\frac{N \phi}{i}=\frac{2}{i}\left[\frac{\mu_{0} i}{16 R} \times a^{2} \cos 45^{\circ}\right] \\

\therefore \quad M=\frac{\mu_{0} a^{2}}{2^{\frac{7}{2}} R}

\end{array}$

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2012 (Paper 1)
A composite block is made of slabs $A, B, C, D$ and $E$ of different thermal conductivities (given in terms of a constant, $K$ ) and sizes (given in terms of length, $L$ ) as shown in the figure. All slabs are of same width. Heat $Q$ flows only from left to right through the blocks. Then, in steady state

Options:
A
heat flow through $A$ and $E$ slabs are same

B
heat flow through slab $E$ is maximum

C
temperature difference across slab $E$ is smallest

D
heat flow through $C=$ heat flow through $B$ + heat flow through $D$

The correct answers are:
heat flow through $A$ and $E$ slabs are same
,
temperature difference across slab $E$ is smallest
,
heat flow through $C=$ heat flow through $B$ + heat flow through $D$

Solution:
Thermal resistance
$$
\begin{aligned}
& R=\frac{l}{K A} \\
& \therefore \quad R_A=\frac{L}{(2 K)(4 L w)}=\frac{1}{8 K w} \\
& \text { (Here, } w=\text { width) } \\
& R_B=\frac{4 L}{3 K(L w)}=\frac{4}{3 K w} \\
& R_C=\frac{4 L}{(4 K)(2 L w)}=\frac{1}{2 K w} \\
& R_D=\frac{4 L}{(5 K)(L w)}=\frac{4}{5 K w} \\
& R_E=\frac{L}{(6 K)(L w)}=\frac{1}{6 K w} \\
& R_A: R_B: R_C: R_D: R_E \\
&=15: 160: 60: 96: 12
\end{aligned}
$$
So, let us write, $R_A=15 R, R_B=160 R$ etc and draw a simple electrical circuit as shown in figure.
$H=$ Heat current $=$ Rate of heat flow.
$$
H_A=H_E=H
$$
[let]
$\therefore$ Option (a) is correct.
In parallel, current distributes in inverse ratio of resistance.
$$
\begin{aligned}
& \therefore H_B: H_C: H_D=\frac{1}{R_B}: \frac{1}{R_C}: \frac{1}{R_D} \\
& =\frac{1}{160}: \frac{1}{60}: \frac{1}{96} \\
& =9: 24: 15 \\
& \therefore \quad H_B=\left(\frac{9}{9+24+15}\right) H=\frac{3}{16} H \\
& H_C=\left(\frac{24}{9+24+15}\right) H=\frac{1}{2} H \\
& \text { and } \quad H_D=\left(\frac{15}{9+24+15}\right) H=\frac{5}{16} H \\
&
\end{aligned}
$$
$$
H_C=H_B+H_D
$$
$\therefore$ Option (d) is correct.
Temperature difference (let us call it $T$ )
$=($ Heat current $) \times($ Thermal resistance $)$
$T_A=H_A R_A=(H)(15 R)=15 \mathrm{HR}$
$T_B=H_B R_B=\left(\frac{3}{16} H\right)(160 R)=30 H R$
$T_C=H_C R_C=\left(\frac{1}{2} H\right)(60 R)=30 H R$
$T_D=H_D R_D=\left(\frac{5}{16} H\right)(96 R)=30 H R$
$T_E=H_E R_E=(H)(12 R)=12 \mathrm{HR}$
Here, $T_E$ is minimum. Therefore, option
(c) is also correct.
$\therefore$ Correct options are (a), (c) and (d).
Analysis of Question
(i) From calculation point of view, question is difficult otherwise question is simple.
(ii) In heat transfer, questions are mainly asked from conduction and radiation topic.

Topic: Physics, Thermal Properties of Matter, JEE Advanced, JEE Advanced 2011 (Paper 1)
A compound $\mathrm{M}_{\mathrm{p}} \mathrm{X}_{\mathrm{q}}$ has cubic close packing (ccp) arrangement of $\mathrm{X}$. Its unit cell structure is shown below. The empirical formula of the compound is

Options:
A $\mathrm{MX}$
B $\mathrm{MX}_{2}$
C $\mathrm{M}_{2} \mathrm{X}$
D $\mathrm{M}_{5} \mathrm{X}_{14}$

The correct answer is:
$\mathrm{MX}_{2}$

Solution:
No. of $M$ atoms $=\frac{1}{4} \times 4+1=1+1=2$

No. of $X$ atoms $=\frac{1}{2} \times 6+\frac{1}{8} \times 8=3+1=4$

So, formula $=M_{2} X_{4}=M X_{2}$

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2012 (Paper 1)
A cubical region of side $a$ has its centre at the origin. It encloses three fixed point charges, $-q$ at $(0,-a / 4,0),+3 q$ at $(0,0,0)$ and $-q$ at $(0,+a / 4,0)$. Choose the correct options(s)

Options:
A The net electric flux crossing the plane $x=+a / 2$ is equal to the net electric flux crossing the plane $x=-$ $a / 2$
B The net electric flux crossing the plane $y=+a / 2$ is more than the net electric flux crossing the plane $y=-a / 2$.
C The net electric flux crossing the entire region is $\frac{q}{\varepsilon_{0}}$
D The net electric flux crossing the plane $z=+a / 2$ is equal to the net electric flux crossing the plane $x=+a / 2$.

The correct answers are:
The net electric flux crossing the plane $x=+a / 2$ is equal to the net electric flux crossing the plane $x=-$ $a / 2$, The net electric flux crossing the entire region is $\frac{q}{\varepsilon_{0}}$, The net electric flux crossing the plane $z=+a / 2$ is equal to the net electric flux crossing the plane $x=+a / 2$.

Solution:
Due to symmetry, the net electric flux passing through $x=+\frac{a}{2}$, $x=-\frac{a}{2}, z=+\frac{a}{2}$ is same

According to Gauss's theorem, net electric flux net electric flux crossing through any closed surface $\phi=\frac{q_{\text {in }}}{\varepsilon_{0}}=$ $\frac{-q+3 q-q}{\varepsilon_{0}}=\frac{q}{\varepsilon_{0}}$.

Topic: Physics, Electrostatics, JEE Advanced, JEE Advanced 2012 (Paper 1)
A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are)

Options:
A The emf induced in the loop is zero if the current is constant.
B The emf induced in the loop is finite if the current is constant.
C The emf induced in the loop is zero if the current decreases at a steady rate.
D The emf induced in the loop is infinite if the current decreases at a steady rate.

The correct answers are:
The emf induced in the loop is zero if the current is constant., The emf induced in the loop is zero if the current decreases at a steady rate.

Solution:
If the current is constant, the emf induced in the loop zero. Emf will be induced in the circular wire loop when flux through it changes with time.

$\mathrm{e}=-\frac{\Delta \phi}{\Delta \mathrm{t}}$

when the current is constant, the flux changing through it will be zero.

Also, if the current decreases at steady rate, the emf. induced in the loop is zero.

When the current is decreasing at a steady rate then the change in the flux (decreasing inwards) on the right half of the wire is equal to the change in flux (decreasing outwards) on the left half of the wire such that $\Delta \phi$ through the circular loop is zero.

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2012 (Paper 2)
A cylindrical cavity of diameter a exists inside a cylinder of diameter $2 a$ as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density $J$ flows along the length. If the magnitude of the magnetic field at the point $P$ is given by $\frac{N}{12} \mu_{0} a J$, then the value of $N$ is

Options:

The correct answer is:
5

Solution:
Current density $J=\frac{\text { current }}{\text { area }}=\frac{I}{A} \Rightarrow I=\mathrm{JA}$ Magnetic field $\mathrm{B}_{\mathrm{R}}$ after removing cavity $(\mathrm{C})$ $\mathrm{B}_{\mathrm{R}}=\mathrm{B}_{\text {total }}-\mathrm{B}_{\text {cavity }}$

$\begin{array}{l}

\frac{\mu_{0} I_{t}}{2 \pi a}-\frac{\mu_{0} I_{c}}{2 \pi\left(\frac{3}{2} a\right)} \\

\quad=\frac{\mu_{0}}{\pi a}\left[\frac{I_{t}}{2}-\frac{I_{c}}{3}\right]\left(\text { here } I_{t}=J\left(\pi a^{2}\right) I_{c}=J\left(\frac{\pi a^{2}}{4}\right)\right) \\

=\frac{\mu_{0}}{\pi a}\left[\frac{\pi a^{2} J}{2}-\frac{\pi a^{2} J}{12}\right] \\

\text { or, } \mathrm{B}_{\mathrm{R}}=\frac{5 \mu_{0} a J}{12}

\end{array}$

Comparing it with $\frac{N}{12} \mu_{0} a J$ We get $N=5$

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2012 (Paper 1)
A cylindrical furnace has height H and diameter D both 1 m. It is maintained at temperature 360 K. The air gets heated inside the furnace at constant pressure Pa and its temperature becomes T=360 K. The hot air with density ρ rises up a vertical chimney of diameter d=0.1 m and height h=9 m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density ρa=1.2 kg m-3, pressure Pa and temperature Ta=300 K enters the furnace. Assume air as an ideal gas, neglect the variations in ρ and T inside the chimney and the furnace. Also ignore the viscous effects.

[Given: The acceleration due to gravity g=10 m s-2 and π=3.14]

When the chimney is closed using a cap at the top, a pressure difference ∆P develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of ∆P is _____ N m−2.

Options:

The correct answer is:
7850

Solution:
Applying Bernoulli’s theorem between top and bottom points of furnace :

Pa+12ρaV2=Pa+ρgH+12ρV2 ... 1

Also, since PM=ρRT

⇒ ρa×300=ρ×360

⇒ ρ=1 kg m-3 ... 2

From equation(1), we can write

⇒ V220.2=1×10×1

⇒ V=10 m s-1

Now, let V' be the speed of air in chimney.

⇒ By continuity equation,

πD24·V=πd24·V'

⇒ V'=100V=1000 m s-1

Therefore, pressure difference ∆p

=ρV'2·area

=1×10002×πd24=1×10002×3.14×0.124=7850 N m2

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2023 (Paper 2)
A cylindrical furnace has height H and diameter D both 1 m. It is maintained at temperature 360 K. The air gets heated inside the furnace at constant pressure Pa and its temperature becomes T=360 K. The hot air with density ρ rises up a vertical chimney of diameter d=0.1 m and height h=9 m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density ρa=1.2 kg m-3, pressure Pa and temperature Ta=300 K enters the furnace. Assume air as an ideal gas, neglect the variations in ρ and T inside the chimney and the furnace. Also ignore the viscous effects.

[Given: The acceleration due to gravity g=10 m s-2 and π=3.14]

Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _____ gm s−1.

Options:

The correct answer is:
7850

Solution:
Applying Bernoulli’s theorem between top and bottom points of furnace :

Pa+12ρaV2=Pa+ρgH+12ρV2 ... 1

Also, since PM=ρRT

⇒ ρa×300=ρ×360

⇒ ρ=1 kg m-3 ... 2

From equation(1), we can write

⇒ V220.2=1×10×1

⇒ V=10 m s-1

Required mass flow rate =V×πD24×ρ

=10×3.144×12×1 kg s-1

=7.85 kg s-1

=7850 gm s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2023 (Paper 2)
A cylindrical vessel of height $500 \mathrm{~mm}$ has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it upto height $H$. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 $\mathrm{mm}$. Find the fall in height (in $\mathrm{mm}$ ) of water level due to opening of the orifice.
[Take atmospheric pressure $=1.0 \times 10^5 \mathrm{Nm}^{-2}$, density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g=10$ $\mathrm{ms}^{-2}$. Neglect any effect of surface tension.]

Options:

The correct answer is:
6

Solution:
In this question we will have to assume that temperature of enclosed air about water is constant (or $p V=$ constant)
$$
\begin{gathered}
p=p_0-\rho g h \\
p_0[A(500-H)]=p[A(200)]
\end{gathered}
$$
Solving these two equations, we get
$$
\begin{aligned}
H & =206 \mathrm{~mm} \\
\therefore \quad \text { Level fall } & =(206-200) \mathrm{mm} \\
& =6 \mathrm{~mm}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2009 (Paper 2)
A decapeptide (molecular weight 796) on complete hydrolysis gives glycine (molecular weight 75), alanine and phenylalanine. Glycine contributes $47.0 \%$ to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is

Options:

The correct answer is:
6

Solution:
$$
\text { A decapeptide has nine peptide (amide) linkage as }
$$
Therefore, no hydrolysis, it will absorb nine water molecules.
Hence, total mass of hydrolysis product $=796+18 \times 9=958$
$\Rightarrow$ mass of glycine in hydrolysis product $=\frac{958 \times 47}{100}=450$
$\Rightarrow$ number of glycine molecule in one molecule of decapeptide $=\frac{450}{75}=6$

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2011 (Paper 1)
A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_i$ $T_i$ (in kelvin) and the final temperature is $a T_i$, the value of $a$ is

Options:

The correct answer is:
4

Solution:
In adiabatic process,
$$
\begin{aligned}
& T V^{\gamma-1}=\text { constant } \\
& \therefore T_i V_i^{0.4}=T_f V_f^{0.4} \\
& \text { (as } \gamma=1.4 \text { for diatomic gas) } \\
& \text { or } \quad T_i V_i^{0.4}=\left(\alpha T_i\right)\left(\frac{V_i}{32}\right)^{0.4} \\
& \text { or } \quad \alpha(32)^{0.4}=4 \\
&
\end{aligned}
$$
$\therefore$ The correct answer is 4 .

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2010 (Paper 2)
A disk of radius $\frac{a}{4}$ having a uniformly distributed charge $6 \mathrm{C}$ is placed in the $x-y$ plane with its centre at $\left(\frac{-a}{2}, 0,0\right)$. A rod of length a carrying a uniformly distributed charge $8 \mathrm{C}$ is placed on the $x$-axis from $x=\frac{a}{4}$ to $x=\frac{5 a}{4}$. Two point charges $-7 \mathrm{C}$ and $3 \mathrm{C}$ are placed at $\left(\frac{a}{4}, \frac{-a}{4}, 0\right)$ and $\left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right)$, respectively. Consider a cubical surface formed by six surfaces $x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}$.
The electric flux through this cubical surface is

Options:
A
$\frac{-2 C}{\varepsilon_0}$

B
$\frac{2 C}{\varepsilon_0}$

C
$\frac{10 C}{\varepsilon_0}$

D
$\frac{12 C}{\varepsilon_0}$

The correct answer is:
$\frac{-2 C}{\varepsilon_0}$

Solution:
Total enclosed charge as already shown is
$$
q_{\text {net }}=\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C
$$
From Gauss-theorem, net flux,
$$
\varphi_{\text {net }}=\frac{q_{\text {net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}
$$

Topic: Physics, Electrostatics, JEE Main