datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
Let $A$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1 . Five of these entries are 1 and four of them are 0 .Question:
The number of matrices $A$ in $A$ for which the system of linear equations $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ is is inconsistent, is

Options:
A
0

B
more than 2

C
2

D
1

The correct answer is:
more than 2

Solution:
Given, $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
Case I $\left[\begin{array}{lll}1 & a & b \\ a & 1 & c \\ b & c & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
$a, b, c$ are selected from $1,0,0$.
$\Rightarrow x+a y+b z=1 \Rightarrow a x+y+c z=0$ $b x+c y+z=0$
(i) If $a=1, b=c=0$, then $x+y=1$ Inconsistent system of equation $x+y=0$
(ii) If $a=0=c, b=1$, then $x+z=1$, $y=0$
Inconsistent system of equation
$$
x+z=0
$$
(iii) If $c=1, a=b=0$, then $x=1, z=0$, $y=0$
Case II
(i) $\left[\begin{array}{lll}1 & a & b \\ a & 0 & c \\ b & c & 0\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
$a, b, c$ are selected from $1,1,0$.
$$
\begin{aligned}
\Rightarrow x+a y+b z & =1 \Rightarrow a x+c z=0 \\
b x+c y & =0
\end{aligned}
$$
Clearly, in all three cases, solutions are possible, so system is consistent.
$$
\begin{gathered}
\text { (ii) }\left[\begin{array}{lll}
0 & a & b \\
a & 1 & c \\
b & c & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
0 \\
0
\end{array}\right] \\
\Rightarrow a y+b z=1 \Rightarrow a x+y+c z=0 \\
b x+c y=0
\end{gathered}
$$
Clearly, $b=0, a=c=1$ gives
$$
y=1 ; x+y+z=0
$$
Inconsistent system $y=0$
More than 2 matrices are possible.
Hence, option (b) is correct.

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
Let $p$ be an odd prime number and $T_p$ be the following set of $2 \times 2$ matrices
$$
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
$$Question:
The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and det $(A)$ is divisible by $p$ is

Options:
A
$(p-1)^2$

B
$2(p-1)$

C
$(p-1)^2+1$

D
$2 p-1$

The correct answer is:
$2 p-1$

Solution:
Given, $A=\left[\begin{array}{ll}a & b \\ c & a\end{array}\right]$,
$a, b, c \in\{0,1,2, \ldots, p-1\}$
If $A$ is skew-symmetric matrix, then $a=0, b=-c$
$\therefore \quad|A|=-b^2$.
Thus, $P$ divides $|A|$ only when $b=0$...(i)
Again, if $A$ is symmetric matrix, then $b=c$ and $|A|=a^2-b^2$.
Thus, $p$ divides $|A|$ if either $p$ divides $(a-b)$ or $p$ divides $(a+b)$. $p$ divides $(a-b)$, only when $a=b$ ie, $a=b \in\{0,1,2, \ldots,(p-1)\}$
ie, pchoices
$p$ divides $(a+b)$.
$\Rightarrow p$ choices, including $a=b=0$ included in (i)
$\therefore$ Total number of choices are $(p+p-1)=2 p-1$.
Hence, (c) is the correct option.

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
Let $p$ be an odd prime number and $T_p$ be the following set of $2 \times 2$ matrices
$$
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
$$Question:
The number of $A$ in $T_p$ such that $\operatorname{det}(A)$ is not divisible by $p$, is

Options:
A
$2 p^2$

B
$p^3-5 p$

C
$p^3-3 p$

D
$p^3-p^2$

The correct answer is:
$p^3-p^2$

Solution:
The number of matrices for which $p$ does not divide $\operatorname{Tr}(A)=(p-1) p^2$ of these $(p-1)^2$ are such that $p$ divides $|A|$. The number of matrices for which $p$ divides $\operatorname{Tr}(A)$ and $p$ does not divides $|A|$ are $(p-1)^2$
$$
\begin{aligned}
& \therefore \text { Required number } \\
& =(p-1) p^2-(p-1)^2+(p-1)^2 \\
& =p^3-p^2
\end{aligned}
$$

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
Let $p$ be an odd prime number and $T_p$ be the following set of $2 \times 2$ matrices
$$
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
$$Question:
The number of $A$ in $T_p$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $p$ is
[Note : The trace of a matrix is the sum of its diagonal entries.]

Options:
A
$(p-1)\left(p^2-p+1\right)$

B
$p^3-(p-1)^2$

C
$(p-1)^2$

D
$(p-1)\left(p^2-2\right)$

The correct answer is:
$(p-1)^2$

Solution:
Trace of $A=2 a$, will be divisible by $p$ iff $a=0$.
$|A|=a^2-b c$, for $\left(a^2-b c\right)$ to be divisible
by $p$. There are exactly $(p-1)$ ordered pairs $(b, c)$ for any value of $a$.
$\therefore$ Required number is $(p-1)^2$.
Hence, (c) is the correct option.

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2010 (Paper 1)
Paragraph:
Let $U_1$ and $U_2$ be two urns such that $U_1$ contains 3 white and 2 red balls and $U_2$ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from $U_1$ and put into $U_2$. However, if tail appears then 2 balls are drawn at random from $U_1$ and put into $U_2$. Now, 1 ball is drawn at random from $U_2$.Question:
Given that the drawn ball from $U_2$ is white, the probability that head appeared on the coin is

Options:
A
$\frac{17}{23}$

B
$\frac{11}{23}$

C
$\frac{15}{23}$

D
$\frac{12}{23}$

The correct answer is:
$\frac{12}{23}$

Solution:
$P$ (head appeared/white from $V_2$ )
$$
\begin{aligned}
& =P(H) \cdot \frac{\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\}}{\frac{23}{30}} \\
& =\frac{1}{2} \frac{\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\}}{\frac{23}{30}}=\frac{12}{23} \\
&
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Let $U_1$ and $U_2$ be two urns such that $U_1$ contains 3 white and 2 red balls and $U_2$ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from $U_1$ and put into $U_2$. However, if tail appears then 2 balls are drawn at random from $U_1$ and put into $U_2$. Now, 1 ball is drawn at random from $U_2$.Question:
The probability of the drawn ball from $U_2$ being white is

Options:
A
$\frac{13}{30}$

B
$\frac{23}{30}$

C
$\frac{19}{30}$

D
$\frac{11}{30}$

The correct answer is:
$\frac{23}{30}$

Solution:
Now, probability of the drawn ball from $\mathrm{V}_2$ being white is
$$
\Rightarrow P \text { (white/ } V_2 \text { ) }
$$
$$
\begin{aligned}
& \left.\Rightarrow P \text { (white/ } V_2\right) \\
& =P(H) \cdot\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\} \\
& +P(T)\left\{\frac{{ }^3 C_2}{{ }^5 C_2} \times \frac{{ }^3 C_2}{{ }^3 C_2}+\frac{{ }^2 C_2}{{ }^5 C_2} \times \frac{{ }^1 C_1}{{ }^3 C_2}\right. \\
& \left.+\frac{{ }^3 C_1 \cdot{ }^2 C_1}{{ }^5 C_2} \times \frac{{ }^2 C_1}{{ }^3 C_2}\right\} \\
& \text { Now, } P\left(\text { white } / V_2\right)=\frac{1}{2}\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\} \\
& +\frac{1}{2}\left\{\frac{3}{10} \times 1+\frac{1}{10} \times \frac{1}{3}+\frac{6}{10} \times \frac{2}{3}\right\}=\frac{23}{30} \\
&
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Let $V_r$ denotes the sum of the first $r$ terms of an arithmetic progression $(A P)$ whose first term is $r$ and the common difference is $(2 r-1)$. Let $T_r=V_{r+1}-V_r-2$ and $Q_r=T_{r+1}-T_r$ for $r=1,2, \ldots$Question:
$T_r$ is always

Options:
A
an odd number

B
an even number

C
a prime number

D
a composite number

The correct answer is:
a composite number

Solution:
$V_{r+1}-V_r=(r+1)^3-r^3-\frac{1}{2}\left[(r+1)^2-r^2\right]+\frac{1}{2}(1)$
$$
\begin{aligned}
& =3 r^2+2 r+1 \\
\therefore \quad T_r & =3 r^2+2 r-1=(r+1)(3 r-1)
\end{aligned}
$$
which is a composite number.

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Let $V_r$ denotes the sum of the first $r$ terms of an arithmetic progression $(A P)$ whose first term is $r$ and the common difference is $(2 r-1)$. Let $T_r=V_{r+1}-V_r-2$ and $Q_r=T_{r+1}-T_r$ for $r=1,2, \ldots$Question:
Which one of the following is a correct statement?

Options:
A
$Q_1, Q_2, Q_3, \ldots$ are in AP with common difference 5

B
$Q_1, Q_2, Q_3, \ldots$ are in AP with common difference 6

C
$Q_1, Q_2, Q_3, \ldots$ are in AP with common difference 11

D
$Q_1=Q_2=Q_3=\ldots$

The correct answer is:
$Q_1, Q_2, Q_3, \ldots$ are in AP with common difference 6

Solution:
Since, $T_r=3 r^2+2 r-1$
$$
\begin{array}{rlrl}
& \therefore & T_{r+1} & =3(r+1)^2+2(r+1)-1 \\
& \therefore & Q_r & =T_{r+1}-T_r=3[2 r+1]+2[1] \\
\Rightarrow & Q_r & =6 r+5 \\
\Rightarrow & Q_{r+1} & =6(r+1)+5
\end{array}
$$
Common difference $=Q_{r+1}-Q_r=6$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Let $V_r$ denotes the sum of the first $r$ terms of an arithmetic progression $(A P)$ whose first term is $r$ and the common difference is $(2 r-1)$. Let $T_r=V_{r+1}-V_r-2$ and $Q_r=T_{r+1}-T_r$ for $r=1,2, \ldots$Question:
The sum $V_1+V_2+\ldots+V_n$ is

Options:
A
$\frac{1}{12} n(n+1)\left(3 n^2-n+1\right)$

B
$\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$

C
$\frac{1}{2} n\left(2 n^2-n+1\right)$

D
$\frac{1}{3}\left(2 n^3-2 n+3\right)$

The correct answer is:
$\frac{1}{12} n(n+1)\left(3 n^2+n+2\right)$

Solution:
$$
\begin{aligned}
& V_r=\frac{r}{2}[2 r+(r-1)(2 r-1)]=\frac{1}{2}\left(2 r^3-r^2+r\right) \\
& \therefore \quad \Sigma V_r=\frac{1}{2}\left[2 \Sigma r^3-\Sigma r^2+\Sigma r\right] \\
& =\frac{1}{2}\left[2\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right] \\
& =\frac{1}{12} n(n+1)\left(3 n^2+n+2\right) \text {. } \\
&
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Modern trains are based on Maglev technology in which trains are magnetically leviated, which runs its EDS Maglev system.
There are coils on both sides of wheels. Due to motion of train, current induces in the coil of track which levitate it. This is in accordance with Lenz's law. If trains lower down then due to Lenz's law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force disc to gravity. The advantage of Maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds.
Disadvantage of Maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it leviated and as it moves forward according to Lenz law, there is an electromagnetic drag force.Question:
What is the disadvantage of this system?

Options:
A
Train experiences upward force according to Lenz's law

B
Friction force create a drag on the train

C
Retardation

D
By Lenz's law train experience a drag

The correct answer is:
By Lenz's law train experience a drag

Solution:
No solution is required.

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2006
Paragraph:
Modern trains are based on Maglev technology in which trains are magnetically leviated, which runs its EDS Maglev system.
There are coils on both sides of wheels. Due to motion of train, current induces in the coil of track which levitate it. This is in accordance with Lenz's law. If trains lower down then due to Lenz's law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force disc to gravity. The advantage of Maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds.
Disadvantage of Maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it leviated and as it moves forward according to Lenz law, there is an electromagnetic drag force.Question:
What is the advantage of this system?

Options:
A
No friction hence no power consumption

B
No electric power is used

C
Gravitation force is zero

D
Electrostatic force draws the train

The correct answer is:
No friction hence no power consumption

Solution:
No solution is required.

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2006
Paragraph:
Modern trains are based on Maglev technology in which trains are magnetically leviated, which runs its EDS Maglev system.
There are coils on both sides of wheels. Due to motion of train, current induces in the coil of track which levitate it. This is in accordance with Lenz's law. If trains lower down then due to Lenz's law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force disc to gravity. The advantage of Maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds.
Disadvantage of Maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it leviated and as it moves forward according to Lenz law, there is an electromagnetic drag force.Question:
Which force causes the train to elevate up?

Options:
A
Electrostatic force

B
Time varying electric field

C
Magnetic force

D
Induced electric field

The correct answer is:
Magnetic force

Solution:
No solution is required.

Topic: Physics, Electromagnetic Induction, JEE Advanced, JEE Advanced 2006
Paragraph:
$P_{14-16}$ : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius $R$ and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface as shown in the figure. The atmospheric pressure is $p_0$.
Question:
While the piston is at a distance $2 L$ from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

Options:
A
$\left(\frac{2 p_0 \pi R^2}{\pi R^2 p_0+M g}\right)(2 L)$

B
$\left(\frac{p_0 \pi R^2-M g}{\pi R^2 p_0}\right)(2 L)$

C
$\left(\frac{p_0 \pi R^2+M g}{\pi R^2 p_0}\right)(2 L)$

D
$\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)$

The correct answer is:
$\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)$

Solution:
Let $p$ be the pressure in equilibrium
Then,
$$
\begin{aligned}
& \therefore \quad p=p_0-\frac{M g}{A}=p_0-\frac{M g}{\pi R^2} \\
& \text { Applying } \quad p_1 V_1=p_2 V_2 \\
& p_0(2 A L)=(p)\left(A L^{\prime}\right) \\
& L^{\prime}=\frac{2 p_0 L}{p}=\left(\frac{p_0}{p_0-\frac{M g}{\pi R^2}}\right) \text { (2L) } \\
& =\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L) \\
&
\end{aligned}
$$
$\therefore$ Option (d) is correct.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
$P_{14-16}$ : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius $R$ and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface as shown in the figure. The atmospheric pressure is $p_0$.
Question:
The piston is now pulled out slowly and held at a distance $2 L$ from the top. The pressure in the cylinder between its top and the piston will then be

Options:
A
$p_0$

B
$\frac{p_0}{2}$

C
$\frac{p_0}{2}+\frac{M g}{\pi R^2}$

D
$\frac{p_0}{2}-\frac{M g}{\pi R^2}$

The correct answer is:
$p_0$

Solution:
Since it is open from top, pressure will be $p_0$. $\therefore$ Option (a) is correct.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
$P_{14-16}$ : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius $R$ and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface as shown in the figure. The atmospheric pressure is $p_0$.
Question:
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium, the height $H$ of the water column in the cylinder satisfies

Options:
A
$\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)+L_0 p_0=0$

B
$\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)-L_0 p_0=0$

C
$\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0$

D
$\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)+L_0 p_0=0$

The correct answer is:
$\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0$

Solution:
$$
\begin{gathered}
p_1=p_2 \\
p_0+\rho g\left(L_0-H\right)=p
\end{gathered}
$$
Now, applying $p_1 V_1=p_2 V_2$
for the air inside the cylinder, we have
$$
p_0\left(L_0\right)=p\left(L_0-H\right)
$$
$\therefore \quad p=\frac{p_0 L_0}{L_0-H}$
Substituting in Eq. (i), we have
$$
p_0+\rho g\left(L_0-H\right)=\frac{p_0 L_0}{L_0-H}
$$
or $\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-p_0 L_0=0$
$\therefore$ Option (c) is correct.

Topic: Physics, Thermodynamics, JEE Main
Paragraph:
$P_{14-16}$ : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius $R$ and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface as shown in the figure. The atmospheric pressure is $p_0$.
Question:
While the piston is at a distance $2 L$ from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

Options:
A
$\left(\frac{2 p_0 \pi R^2}{\pi R^2 p_0+M g}\right)(2 L)$

B
$\left(\frac{p_0 \pi R^2-M g}{\pi R^2 p_0}\right)(2 L)$

C
$\left(\frac{p_0 \pi R^2+M g}{\pi R^2 p_0}\right)(2 L)$

D
$\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)$

The correct answer is:
$\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L)$

Solution:
Let $p$ be the pressure in equilibrium
Then,
$$
\begin{aligned}
& \therefore \quad p=p_0-\frac{M g}{A}=p_0-\frac{M g}{\pi R^2} \\
& \text { Applying } \quad p_1 V_1=p_2 V_2 \\
& p_0(2 A L)=(p)\left(A L^{\prime}\right) \\
& L^{\prime}=\frac{2 p_0 L}{p}=\left(\frac{p_0}{p_0-\frac{M g}{\pi R^2}}\right) \text { (2L) } \\
& =\left(\frac{p_0 \pi R^2}{\pi R^2 p_0-M g}\right)(2 L) \\
&
\end{aligned}
$$
$\therefore$ Option (d) is correct.

Topic: Physics, Thermodynamics, JEE Main
Paragraph:
$P_{14-16}$ : Paragraph for Questions Nos. 14 to 16 A fixed thermally conducting cylinder has a radius $R$ and height $L_0$. The cylinder is open at its bottom and has a small hole at its top. A piston of mass $M$ is held at a distance $L$ from the top surface as shown in the figure. The atmospheric pressure is $p_0$.
Question:
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium, the height $H$ of the water column in the cylinder satisfies

Options:
A
$\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)+L_0 p_0=0$

B
$\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)-L_0 p_0=0$

C
$\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0$

D
$\rho g\left(L_0-H\right)^2-p_0\left(L_0-H\right)+L_0 p_0=0$

The correct answer is:
$\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-L_0 p_0=0$

Solution:
$$
\begin{gathered}
p_1=p_2 \\
p_0+\rho g\left(L_0-H\right)=p
\end{gathered}
$$
Now, applying $p_1 V_1=p_2 V_2$
for the air inside the cylinder, we have
$$
p_0\left(L_0\right)=p\left(L_0-H\right)
$$
$\therefore \quad p=\frac{p_0 L_0}{L_0-H}$
Substituting in Eq. (i), we have
$$
p_0+\rho g\left(L_0-H\right)=\frac{p_0 L_0}{L_0-H}
$$
or $\rho g\left(L_0-H\right)^2+p_0\left(L_0-H\right)-p_0 L_0=0$
$\therefore$ Option (c) is correct.

Topic: Physics, Thermodynamics, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
$\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question:
The loss of kinetic energy during the above process is

Options:
A
$\frac{I \omega^2}{2}$

B
$\frac{i \omega^2}{3}$

C
$\frac{I \omega^2}{4}$

D
$\frac{I \omega^2}{6}$

The correct answer is:
$\frac{i \omega^2}{3}$

Solution:
Loss of kinetic energy $=K_i-K_f$
$$
\begin{aligned}
& =\left\{\frac{1}{2} I(2 \omega)^2+\frac{1}{2}(2 I)(\omega)^2\right\}-\frac{1}{2}(3 I)\left(\frac{4}{3} \omega\right)^2 \\
& =\frac{1}{3} I \omega^2
\end{aligned}
$$
$\therefore$ Option (b) is correct.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
$\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question:
The ratio $\frac{x_1}{x_2}$ is

Options:
A
2

B
$\frac{1}{2}$

C
$\sqrt{2}$

D
$\frac{1}{\sqrt{2}}$

The correct answer is:
$\sqrt{2}$

Solution:
$$
\begin{aligned}
\frac{1}{2} I(2 \omega)^2 & =\frac{1}{2} k x_1^2 \\
\frac{1}{2}(2 I)(\omega)^2 & =\frac{1}{2} k x_2^2
\end{aligned}
$$
From Eqs. (i) and (ii), we have
$$
\frac{x_1}{x_2}=\sqrt{2}
$$
$\therefore$ Option (c) is correct.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
$\mathbf{P}_{17 \text { - } 19}$ : Paragraph for Questions Nos. 17 to 19 Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2 \omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.Question:
When disc $B$ is brought in contact with disc $A$, they acquire a common angular velocity in time $t$. The average frictional torque on one disc by the other during this period is

Options:
A
$\frac{2 I \omega}{3 t}$

B
$\frac{9 I \omega}{2 t}$

C
$\frac{91 \omega}{4 t}$

D
$\frac{3 I \omega}{2 t}$

The correct answer is:
$\frac{2 I \omega}{3 t}$

Solution:
Let $\omega^{\prime}$ be the common velocity. Then from conservation of angular momentum, we have
$$
\begin{aligned}
(i+2 I) \omega^{\prime} & =I(2 \omega)+2 I(\omega) \\
\omega^{\prime} & =\frac{4}{3} \omega
\end{aligned}
$$
From the equation,
Angular impulse = change in angular momentum for any of the disc, we have
$$
\begin{aligned}
\tau t & =I(2 \omega)-I\left(\frac{4}{3} \omega\right)=\frac{2 I \omega}{3} \\
\tau & =\frac{2 I \omega}{3 t}
\end{aligned}
$$
$\therefore$ Option (a) is correct.

Topic: Physics, Rotational Motion, JEE Advanced, JEE Advanced 2007 (Paper 1)