datavorous/entrance-exam-dataset huggingface dataset:
Two vibrating tuning forks produce progressive waves given by $y_1=4 \sin 500 \pi t$ and $y_2=2 \sin 506 \pi t$. Number of beats produced per minute is

Options:
A $360$
B $180$
C $3$
D $60$

The correct answer is:
$180$

Solution:
To reach the solution the given wave equations must be compared with standard equation of progressive wave.
So,
$\begin{aligned} & y_1=4 \sin 500 \pi t...(i) \\ & y_2=2 \sin 506 \pi t...(ii)\end{aligned}$
Comparing Eqs. (i) and (ii) with $y=a \sin \omega t...(iii)$
We have, $\omega_1=500 \pi$
$\Rightarrow f_1=\frac{500 \pi}{2 \pi}=250$ beats $/ \mathrm{s}$
and $\omega_2=506 \pi$
$\Rightarrow f_2=\frac{506 \pi}{2 \pi}=253$ beats $/ \mathrm{s}$
Thus, number of beats produced
$\begin{aligned} & =f_2-f_1=253-250 \\ & =3 \text { beats } / \mathrm{s} \\ & =3 \times 60 \text { beats } / \mathrm{min} \\ & =180 \text { beats } / \mathrm{min}\end{aligned}$

Topic: Physics, Waves and Sound, JIPMER, JIPMER 2007
Two wires $A$ and $B$ of lengths in the ratio $1: 2$ and masses in the ratio $2: 1$ are stretched by same tension. The ratio of the fundamental frequencies of wires $A$ and $B$ is

Options:
A $2 \sqrt{2}: 1$
B $1: \sqrt{2}$
C $1: 1$
D $\sqrt{2}: 1$

The correct answer is:
$1: 1$

Solution:
Fundamental frequency in a stretched wire is given as,
$f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
Ratio of fundamental frequencies of wire $A$ and $B$,
$\begin{aligned} \frac{f_A}{f_B} & =\frac{1}{2 L_A} \sqrt{\frac{T_A}{\mu_A}} \times \frac{2 L_B}{T} \sqrt{\frac{\mu_B}{T_B}} \\ & =\frac{L_B}{L_A} \cdot \sqrt{\frac{T_A}{T_B}} \sqrt{\frac{\mu_B}{\mu_A}}\end{aligned}$
Here $\frac{L_B}{L_A}=\frac{2}{1}$
$\begin{aligned} \frac{\mu_A}{\mu_B}=\frac{m_A / L_A}{m_B / L_B}=\frac{m_A}{m_B} \cdot \frac{L_B}{L_A}=\frac{2}{1} \times \frac{2}{1} & \\ & {\left[\because \frac{m_A}{m_B}=\frac{2}{1} \text { and } \frac{L_B}{L_A}=\frac{2}{1}\right] }\end{aligned}$
or $\quad \frac{\mu_A}{\mu_B}=\frac{2}{1} \times \frac{2}{1}=\frac{4}{1}$
$\Rightarrow \quad \frac{\mu_B}{\mu_A}=\frac{1}{4}$
So, $\quad \frac{f_A}{f_B}=\frac{2}{1} \cdot \sqrt{1} \cdot \sqrt{\frac{1}{4}}=1: 1$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 2)
Two wires, made up of same material, are of equal lengths but their radii are in the ratio 1:2. On stretching each of these two string by the same tension, the ratio between their fundamental frequencies is

Options:
A 1:2
B 2:1
C 1:4
D 4:1

The correct answer is:
2:1

Solution:
Here, Ï1=Ï2;r1r2=12,T1=T2
n1=12lr1T1ÏÏ1;n2=12lr2T2ÏÏ2
n1n2=r2r1=2

Topic: Physics, Waves and Sound, NEET
Two wooden blocks of mass $M_1$ and $M_2$ rest on a frictionless table. $A$ bullet of mass $m$ is fired at $M_1$ with speed $v$ which embedded in it and the two together finally collide with $M_2$. Find the velocity of $M_2$ after collision. (Ignore any energy loss and treat the problem to be one dimensional)

Options:
A $\frac{2 m v}{M_1+M_2+m}$
B $\frac{m v}{M_1+M_2+m}$
C $\frac{\left(M_1+M_2+m\right) v}{M_1+M_2+m}$
D $\frac{M_1+M_2}{M_1+M_2+m} v$

The correct answer is:
$\frac{2 m v}{M_1+M_2+m}$

Solution:
( ) For an elastic collision of 2 masses $m_1$ and $m_2$ with initial velocities $u_1$ and $u_2$, final velocities of masses after collision are
$$
v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\frac{2 m_2 u_2}{\left(m_1+m_2\right)}
$$
and
$$
v_2=\left(\frac{m_2-m_1}{m_1+m_2}\right) u_2+\frac{2 m_1 u_2}{m_1+m_2}
$$
These are obtained as standard results by applying conservation of energy and momentum equations. Now, in given case at first bullet is embedded in mass $M_1$. Let speed of bullet + Block system be $v_1$ after collision.
Then by conservation of momentum we have.
$$
\begin{aligned}
& m v=\left(m+M_1\right) v_1 \\
& \text { or } \quad v_1=\left(\frac{m v}{m+M_1}\right)
\end{aligned}
$$
Note Here we are not using standard result as first collision is perfectly inelastic.
Now, bullet $+M_1$ strikes $M_2$ and it is given in problem that this collision is elastic.
Now, by formula we have,
$$
\begin{aligned}
v_2 & =\left(\frac{\left(m+M_1\right)-M_2}{m+M_1+M_2}\right) v_1+0 \\
\text { and, } v_3 & =\left(\frac{M_2-\left(m+M_1\right)}{M_2+m+M_1} \times 0\right)+\left(\frac{2\left(m+M_1\right) v_1}{m+M_1+M_2}\right) \\
\Rightarrow \quad v_3 & =\frac{2\left(m+M_1\right)}{m+M_1+M_2} \cdot \frac{m v}{\left(m+M_1\right)} \\
& =\frac{2 m v}{m+M_1+M_2}
\end{aligned}
$$
Note You can remember these standard and useful results for finding final velocities in case of elastic collisions.

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
Tyloses an out growth from ray or axial parenchyma cell into the lumen of a vessel, which partially or completely blocks the cavity are present in

Options:
A periderm
B heart wood
C sap wood
D secondary cortex

The correct answer is:
heart wood

Solution:
Tyloses are found in heart wood, sometimes the xylem parenchyma develop balloon-like structure to which penetrates into the adjacent xylem. Vessels, which are called tyloses, block the luman of the xylem.

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2014
Tyndall effect is observed due to

Options:
A neutralization of charge on colloidal particles
B Zig-zag motion of colloidal particles
C precipitation of colloidal particles
D Scattering of light by colloidal particles

The correct answer is:
Scattering of light by colloidal particles

Solution:
No solution

Topic: Chemistry, Surface Chemistry, MHT CET, MHT CET 2020 (15 Oct Shift 1)
Under standard conditions, the density of a gas is \(\frac{1400}{1089} \mathrm{~kg} \mathrm{-} \mathrm{m}^{-3}\) and the speed of sound propagation in it is \(330 \mathrm{~ms}^{-1}\), then the number of degrees of freedom of the gas molecules is

Options:
A 2
B 7
C 5
D 3

The correct answer is:
5

Solution:
Given, density of gas, \(\rho=\frac{1400}{1089} \mathrm{~kg} / \mathrm{m}^3\) speed of sound, \(v=330 \mathrm{~m} / \mathrm{s}\) and under standard condition, Pressure of gas, \(p=1 \times 10^5 \mathrm{~N} / \mathrm{m}^2\)
If \(\gamma\) be the ratio of \(C_p\) and \(C_V\) of a gas, then the speed of sound in gas is given by
\(\begin{aligned}
& \qquad v=\sqrt{\frac{\gamma P}{\rho}} \text { or } \frac{\gamma P}{\rho}=v^2 \\
& \quad \gamma=\frac{v^2 \rho}{p}=\frac{330 \times 330}{10^5} \times \frac{1400}{1089} \\
& \quad \gamma=1.4 \\
& \text { So, } \quad \gamma=\frac{C_p}{C_V}=1.4
\end{aligned} \quad\left[\begin{array}{l}
\because \gamma=1+\frac{2}{f} \\
\text { Since, for diatomic gas, the volume of }
\end{array}\right] \begin{aligned}
& \gamma \text { is 1.4. } \\
& \text { Hence, the degree of freedom for }
\end{aligned}\)
So, \(\gamma=\frac{C_p}{C_V}=1.4\)
Since, for diatomic gas, the volume of \(\left[\begin{array}{l}
\because \gamma=1+\frac{2}{f} \\
1.4=1+\frac{2}{f} \\
\therefore f=5
\end{array}\right]\)\(\gamma\) is 1.4.
Hence, the degree of freedom for
diatomic gas is equal to 5.

Topic: Physics, Kinetic Theory of Gases, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)
Under what condition are the two lines
$\mathrm{y}=\frac{\mathrm{m}}{\ell} \mathrm{x}+\alpha, \mathrm{z}=\frac{\mathrm{n}}{\ell} \mathrm{x}+\beta ;$ and $\mathrm{y}=\frac{\mathrm{m}^{\prime}}{\ell^{\prime}} \mathrm{x}+\alpha^{\prime}, \mathrm{z}=\frac{\mathrm{n}^{\prime}}{\ell^{\prime}} \mathrm{x}+\beta^{\prime}$
orthogonal ?

Options:
A $\alpha \alpha^{\prime}+\beta \beta^{\prime}+1=0$
B $\left(\alpha+\alpha^{\prime}\right)+\left(\beta+\beta^{\prime}\right)=0$
C $\ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=1$
D $\quad \ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=0$

The correct answer is:
$\quad \ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=0$

Solution:
Given two lines are: $\mathrm{y}=\frac{\mathrm{mx}}{\ell}+\alpha, \mathrm{z}=\frac{\mathrm{n}}{\ell} \mathrm{x}+\beta$ and
$\mathrm{y}=\frac{\mathrm{m}^{\prime}}{\ell^{\prime}} \mathrm{x}+\alpha^{\prime}, \mathrm{z}=\frac{\mathrm{n}^{\prime}}{\ell^{\prime}} \mathrm{x}+\beta^{\prime}$
These two lines can be represented as
$\frac{\mathrm{y}-\alpha}{\mathrm{m} / \ell}=\frac{\mathrm{x}-0}{1}=\frac{\mathrm{z}-\beta}{\mathrm{n} / \ell}$ and $\frac{\mathrm{y}-\alpha^{\prime}}{\mathrm{m}^{\prime} / \ell^{\prime}}=\frac{\mathrm{x}-0}{1}=\frac{\mathrm{z}-\beta^{\prime}}{\mathrm{n} 1 / \ell^{\prime}}$
They are orthogonal, if $\frac{\mathrm{m}}{\ell} \times \frac{\mathrm{m}^{\prime}}{\ell^{\prime}}+1 \times 1+\frac{\mathrm{n}}{\ell} \times \frac{\mathrm{n}^{\prime}}{\ell^{\prime}}=-1 \Rightarrow \ell \ell^{\prime}+\mathrm{mm}^{\prime}+\mathrm{nn}^{\prime}=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 1)
Under what condition do $\left\langle\frac{1}{\sqrt{2}}, \frac{1}{2}, \mathrm{~K}\right\rangle$ represent direction cosines of a line?

Options:
A $\mathrm{k}=\frac{1}{2}$
B $\mathrm{k}=-\frac{1}{2}$
C $\mathrm{k}=\pm \frac{1}{2}$
D $\mathrm{k}$ can take any value

The correct answer is:
$\mathrm{k}=\pm \frac{1}{2}$

Solution:
For $\left(\frac{1}{\sqrt{2}}, \frac{1}{2}, \mathrm{k}\right)$ to represent direction cosines
$\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\mathrm{k}^{2}=1$
or, $\frac{1}{2}+\frac{1}{4}+\mathrm{k}^{2}=1$
$\frac{3}{4}+\mathrm{k}^{2}=1 \Rightarrow \mathrm{k}=\pm \frac{1}{2}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 1)
Under what condition do the planes

Options:
A $a+b+c=\mathbf{0}$
B $a=b=c$
C $a l+b m+c n=0$
D $l+m+n=0$

The correct answer is:
$a l+b m+c n=0$

Solution:
The planes $b x-a y=n, c y-b z=l$ and $a z-c x=m$ intersect in a line, if $a l+b m+c n=0$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 1)
Under what condition does $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$ hold, where $\mathrm{A}$, $\mathrm{B}, \mathrm{C}$ are three matrices?

Options:
A $\mathrm{AB}$ and $\mathrm{BC}$ both must exist
B Only Ab must exist
C Only BC must exist
D Always true

The correct answer is:
$\mathrm{AB}$ and $\mathrm{BC}$ both must exist

Solution:
$\mathrm{AB}$ and $\mathrm{BC}$ both must exit, to hold the condition $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$

Topic: Mathematics, Matrices, NDA, NDA 2008 (Phase 1)
Under what condition does the equation $x^{2}+y^{2}+z^{2}+2 u x+2 u y+2 w z+d=0$ represent a real sphere?

Options:
A $u^{2}+v^{2}+w^{2}=d^{2}$
B $u^{2}+v^{2}+w^{2}>d$
C $u^{2}+v^{2}+w^{2} < d$
D $u^{2}+v^{2}+w^{2} < d^{2}$

The correct answer is:
$u^{2}+v^{2}+w^{2}>d$

Solution:
Equation $x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 w z+d=0$
represent a real sphere if $u^{2}+v^{2}+w^{2}>d$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 1)
Under what condition is ECell= 0 or ∆rG = 0?

Options:
A ECell= 0 when the cell is operating at standard conditions (25°C and 1 atm pressure).
B ECell= 0 when the cell potential is greater than 1 V.
C ECell= 0 when the concentration of reactants is equal to the concentration of products.
D When the cell reaction reaches equilibrium

The correct answer is:
When the cell reaction reaches equilibrium

Solution:
Correct Option is : (D)
When the cell reaction reaches equilibrium

Topic: Chemistry, Electrochemistry, NEET
Under what condition will the angle between two regression lines become zero?

Options:
A $\mathrm{r}=0$
B Only when $\mathrm{r}=+1$
C Only when $r=-1 \quad$
D $\quad r=\pm 1 \quad$

The correct answer is:
$\quad r=\pm 1 \quad$

Solution:
The angle between two regression lines becomes zero if $\mathrm{r}=\pm 1$

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 1)
Under what conditions does sulphur dioxide liquefy at room temperature?

Options:
A Under standard atmospheric pressure
B Under a pressure of two atmospheres
C Under reduced pressure
D Under increased pressure

The correct answer is:
Under a pressure of two atmospheres

Solution:
Under a pressure of two atmospheres

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Under what conditions is the tangent to a given curve at a point perpendicular to $\mathrm{x}$ -axis ?

Options:
A $\frac{d y}{d x}=0$
B $\frac{d y}{d x}=1$
C $\frac{d x}{d y}=0$
D $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=1 \quad$

The correct answer is:
$\frac{d x}{d y}=0$

Solution:
Tangent to a given curve at a point is perpendicular to
$\mathrm{x}$ -axis gives $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \frac{\pi}{2}$
$\Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{dy}}=\cot \frac{\pi}{2}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
Under which of the following conditions does a general second degree equation $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$ $(a \neq 0)$ represents a circle?

Options:
A $h=g, a=b$
B $h=g=f, a=b$
C $h=0, a=\mathrm{b}$
D $h=0, g^{2}+f^{2}-c=a+b$

The correct answer is:
$h=0, a=\mathrm{b}$

Solution:
The given equation represents a circle, if $a=b, h=0$.

Topic: Mathematics, Circle, NDA, NDA 2010 (Phase 2)
Under which of the following condition(s), will the matrix
$\mathrm{A}=\left[\begin{array}{lll}0 & 0 & \mathrm{q} \\ 2 & 5 & 1 \\ 8 & \mathrm{p} & \mathrm{p}\end{array}\right]$ be singular?
$1. \quad \mathrm{q}=0$
$2. \quad \mathrm{p}=0$
$3. \quad\mathrm{p}=20$
Select the correct answer using the code given below:

Options:
A 1 and 2
B 3 only
C 1 and 3
D 1 or 3

The correct answer is:
1 or 3

Solution:
$\mathrm{A}=\left[\begin{array}{lll}0 & 0 & \mathrm{q} \\ 2 & 5 & 1 \\ 8 & \mathrm{p} & \mathrm{p}\end{array}\right]$
Matrix A will be singular, when $|\mathrm{A}|=0,|\mathrm{~A}|$ will be zero when either one row or one column is zero.
(1) For $\mathrm{q}=0$ $\mathrm{A}=\left[\begin{array}{lll}0 & 0 & 0 \\ 2 & 5 & 1 \\ 8 & \mathrm{p} & \mathrm{p}\end{array}\right]$
$\Rightarrow|\mathrm{A}|=0$
$\therefore$ A is singular.
(2) For $\mathrm{p}=0$ $\mathrm{A}=\left[\begin{array}{lll}0 & 0 & \mathrm{q} \\ 2 & 5 & 5 \\ 8 & 0 & 0\end{array}\right]$
$\Rightarrow|\mathrm{A}|=\mathrm{q}\left|\begin{array}{ll}2 & 5 \\ 8 & 0\end{array}\right|=-40 \mathrm{q}$
$\therefore \quad$ A is not singular.
(3) For $\mathrm{p}=20$
$|\mathrm{A}|=\mathrm{q}\left|\begin{array}{cc}2 & 5 \\ 8 & 20\end{array}\right|=40-40=0$
$\therefore \quad$ A is singular. Thus codes (1) or (3) are correct.

Topic: Mathematics, Matrices, NDA, NDA 2007 (Phase 1)
Under which one of the following condition does the system of equations
$\begin{array}{l}
k x+y+z=k-1 \\
x+k y+z=k-1 \\
x+y+k z=k-1
\end{array}$
have no solution?

Options:
A $k=1$
B $k \neq-2$
C $k=1$ or $k=-2$
D $k=-2$

The correct answer is:
$k=1$ or $k=-2$

Solution:
The given system of equations are $\mathrm{kx}+\mathrm{y}+\mathrm{z}=\mathrm{k}-1$
$x+k y+z=k-1$
$\mathrm{x}+\mathrm{y}+\mathrm{kz}=\mathrm{k}-1$
$A=\left[\begin{array}{lll}k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k\end{array}\right], B=\left[\begin{array}{l}k-1 \\ k-1 \\ k-1\end{array}\right]$ and $\mathrm{x}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$
Now, $|A|=\left|\begin{array}{lll}k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k\end{array}\right|$
$=k\left(k^{2}-1\right)-1(k-1)+1(1-k)$
$=k^{3}-k-k+1+1-k$
$=k^{3}-3 k+2$
The given system of equations has no solution, if
$|A|=0$
$\Rightarrow k^{3}-3 k+2=0$
$\Rightarrow(k-1)^{2}(k+2)=0$
$\Rightarrow k=1$ or $k=-2$

Topic: Mathematics, Determinants, NDA, NDA 2009 (Phase 1)
Under which one of the following conditions are the lines $\mathrm{x}$ $=a y+b ; z=c y+d$ and $x=e y+f ; z=g y+h$ perpendicular?

Options:
A $\mathrm{ae}+\mathrm{cg}-1=0$
B $a e+b f-1=0$
C $\mathrm{ae}+\mathrm{cg}+1=0$
D $a g+c e+1=0$

The correct answer is:
$\mathrm{ae}+\mathrm{cg}+1=0$

Solution:
Given, lines $\mathrm{x}=\mathrm{ay}+\mathrm{b}$ and $\mathrm{z}=\mathrm{cy}+\mathrm{d}$ are perpendicular.
$\Rightarrow \frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}$
Also, $\mathrm{x}=\mathrm{e} \mathrm{y}+\mathrm{fand} \mathrm{z}=\mathrm{g} \mathrm{y}+\mathrm{h}$ are perpendicular.
$\Rightarrow \frac{x-f}{e}=\frac{y}{1}=\frac{z-h}{g}$
We know, for $\perp^{r}$ lines $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
$\Rightarrow \mathrm{ae}+1+\mathrm{cg}=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 1)
Under which one of the following conditions does the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ meet the $x$ -axis in two points on opposite sides of the origin?

Options:
A $\quad c>0$
B $\mathrm{c} < 0$
C $\mathrm{c}=0$
D $\mathrm{c} \leq 0$

The correct answer is:
$\mathrm{c} < 0$

Solution:
For a circle to meet $x$ -axis in two points on the opposite side of the origin its radius $\mathrm{r}$, should be more than the distance of its centre from the origin. Co-ordinate of centre of the circle $x^{2}+y^{2}+2 g x+2 f y$ $+\mathrm{c}=0$ is $(-\mathrm{g},-\mathrm{f})$
In the figure shown, $\mathrm{OQ}=\mathrm{OP}=\mathrm{r}$, and distance of centre $\mathrm{C}$, from origin, $\mathrm{O}$ is CO
$\mathrm{r}>\sqrt{\mathrm{OC}}$ i.e. $\mathrm{r}>\sqrt{(-\mathrm{g})^{2}+(-\mathrm{f})^{2}}$
or, $\sqrt{(-\mathrm{g})^{2}+(-\mathrm{f})^{2}-\mathrm{c}}>\sqrt{(-\mathrm{g})^{2}+(-\mathrm{f})^{2}}$
or, $\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}>\mathrm{g}^{2}+\mathrm{f}^{2}$
or, $-\mathrm{c}>0$
or, $\mathrm{c} < 0$

Topic: Mathematics, Circle, NDA, NDA 2007 (Phase 1)
Under which one of the following conditions does the
solution of $\frac{d y}{d x}=\frac{a x+b}{c y+d}$ represent a parabola?

Options:
A $\mathrm{a}=0, \mathrm{c}=0$
B $\mathrm{a}=1, \mathrm{~b}=2, \mathrm{c} \neq 0$
C $\mathrm{a}=0, \mathrm{c} \neq 0, \mathrm{~b} \neq 0$
D $\mathrm{a}=1, \mathrm{c}=1 \quad$

The correct answer is:
$\mathrm{a}=0, \mathrm{c} \neq 0, \mathrm{~b} \neq 0$

Solution:
Given: $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cy}+\mathrm{d}}$
or, $\quad(\mathrm{cy}+\mathrm{d}) \mathrm{d} \mathrm{y}=(\mathrm{ax}+\mathrm{b}) \mathrm{d} \mathrm{x}$
Integrating both the sides.
c. $\int \mathrm{ydy}+\mathrm{d} \int \mathrm{dy}=\mathrm{a} \int \mathrm{x} \mathrm{d} \mathrm{x}+\mathrm{b} \int \mathrm{dx}+\mathrm{K}$
$[\mathrm{K}$ is constant
integration]
or, c. $\frac{y^{2}}{2}+d \cdot y=a \frac{x^{2}}{2}+b \cdot x+K$
or, $\mathrm{cy}^{2}+2 \mathrm{~d} . \mathrm{y}=\mathrm{ax}^{2}+2 \mathrm{~b} \cdot \mathrm{x}+2 \mathrm{~K}$
This equation will represent a parabola when either, the coefficient of $\mathrm{x}^{2}$ or the coefficient of $\mathrm{y}^{2}$ is zero, but not both. Thus either $\mathrm{c}=0$ or $\mathrm{a}=0$ but not both. From the choice given, $\mathrm{a}=0, \mathrm{c} \neq 0$ and $\mathrm{b} \neq 0$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 1)
Under which one of the following conditions will the two planes $x+y+z=7$ and $\alpha x+\beta y+\gamma z=3$, be parallel (but not coincident)?

Options:
A $\alpha=\beta=\gamma=1$ only
B $\alpha=\beta=\gamma=\frac{3}{7}$ only
C $\alpha=\beta=\gamma$
D None of the above

The correct answer is:
$\alpha=\beta=\gamma$

Solution:
Given equation of planes are $x+y+z=7$
and $\alpha x+\beta y+\gamma z=3$
For these planes to be parallel, coefficients of $\mathrm{x}, \mathrm{y}$ and $z$ should be same i.e. $\Rightarrow \alpha=\beta=\gamma$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 1)
Unification of weak and electromagnetic interactions was done by

Options:
A Einstein
B Raman
C Salam
D Hubble

The correct answer is:
Salam

Solution:
As we know that, electromagnetic and weak nuclear forces are fundamental forces, which can be understood as a single interaction, this conclusion was given by three physicists Glashow, Weinberg and Salam.

Topic: Physics, Units and Dimensions, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
Universal set, $\mathrm{U}=\left\{\mathrm{x} \mid \mathrm{x}^{5}-6 \mathrm{x}^{4}+11 \mathrm{x}^{3}-6 \mathrm{x}^{2}=0\right\}$
$A=\left\{x \mid x^{2}-5 x+6=0\right\}$
$B=\left\{x \mid x^{2}-3 x+2=0\right\}$
What is $(A \cap B)$ equal to ?

Options:
A $\{1,3\}$
B $\{1,2,3\}$
C $\{0,1,3\}$
D $\{0,1,2,3\}$

The correct answer is:
$\{0,1,3\}$

Solution:
$U=\left\{x: x^{5}-6 x^{4}+11 x^{3}-6 x^{2}=0\right\}$
Solving for values of $\mathrm{x}$, we get $=\{0,1,2,3\}$
$\mathrm{A}=\left\{\mathrm{x}: \mathrm{x}^{2}-5 \mathrm{x}+6=0\right\}$
Solving for values of $\mathrm{x}$, we get $=\{2,3\}$
and $B=\left\{x: x^{2}-3 x+2=0\right\}$
Solving for values of $\mathrm{x}$, we get $=\{2,1\}$
$\mathrm{A} \cap \mathrm{B}=\{2\}$
$(A \cap B)^{\prime}=U-(A \cap B)$
$=\{0,1,2,3\}-\{2\}=\{0,1,3\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:
\(\begin{array}{lll}\text { Isotope } & \text { Isotopic molar mass } & \text { Abundance } \\ { }^{36} \mathrm{Ar} & 35.96755 \mathrm{~g} \mathrm{~mol}^{-1} & 0.337 \% \\ { }^{38} \mathrm{Ar} & 37.96272 \mathrm{~g} \mathrm{~mol}^{-1} & 0.063 \% \\ { }^{40} \mathrm{Ar} & 39.9624 \mathrm{~g} \mathrm{~mol}^{-1} & 99.600 \%\end{array}\)

Options:

None

Solution:
Molar mass of Ar
\(\begin{aligned}
&=35.96755 \times 0.00337+37.96272 \times 0.00063+39.96924 \\
&\times 0.99600 \\
&=39.948 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}\)

Topic: Chemistry, Some Basic Concepts of Chemistry
Using dimensional analysis the resistivity in terms of fundamental constants $\mathrm{h}, \mathrm{m}_{\mathrm{e}}, \mathrm{c}, \mathrm{e}, \varepsilon_{0}$ can be expressed as

Options:
A $\frac{\mathrm{h}}{\varepsilon_{0} \mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$
B $\frac{\varepsilon_{0} \mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}{\mathrm{~h}}$
C $\frac{\mathrm{h}^{2}}{\mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$
D $\frac{\mathrm{m}_{\mathrm{e}} \varepsilon_{0}}{\mathrm{ce}^{2}}$

The correct answer is:
$\frac{\mathrm{h}^{2}}{\mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$

Solution:
$\rho=\frac{E}{J}$
$E=\left[M L T^{-3} I^{-1}\right]$
$\mathrm{J}=\left[\mathrm{I} \mathrm{L}^{-2}\right]$
$\mathrm{h}=\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-1}$
$\mathrm{~m}_{\mathrm{e}}=\mathrm{M}$
$\mathrm{c}=\mathrm{LT}^{-1}$
$\varepsilon_{0}=\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{I}^{2}$
$\therefore \rho=\frac{\mathrm{h}^{2}}{\mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$

Topic: Physics, Units and Dimensions, KVPY, KVPY 2017 (5 Nov SB/SX)
Using mass $(M)$, length $(L)$, time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permeability is

Options:
A $\mathrm{M}^{-1} \mathrm{LT}^{-2} \mathrm{~A}$
B $\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}$
C $\mathrm{MLT}^{-2} \mathrm{~A}^{-2}$
D $\mathrm{MLT}^{-1} \mathrm{~A}^{-1}$.

The correct answer is:
$\mathrm{MLT}^{-2} \mathrm{~A}^{-2}$

Solution:
We know that the force per unit length of a wire carrying current due to another parallel wire carrying current is given by
$$
\begin{gathered}
\frac{d F}{d l}=\frac{\mu_0 i_1 i_2}{2 \pi d} \Rightarrow \mu_0=\frac{2 \pi d}{i_1 i_2} \frac{d F}{d l} \\
\therefore \quad\left[\mu_0\right]=\frac{[\mathrm{L}]}{\left[\mathrm{A}^2\right]} \cdot \frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]} \\
\text { or }\left[\mu_0\right]=\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{A}^{-2}\right]=\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right] .
\end{gathered}
$$

Topic: Physics, Units and Dimensions, AIIMS, AIIMS 2003
Using mass $(M)$, length $(L)$, time $(T)$ and current $(A)$ as fundamental quantities, the dimension of permittivity is

Options:
A $\left[\mathrm{ML}^{-2} \mathrm{~T}^2 \mathrm{~A}\right]$
B $\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
C $\left[\mathrm{MLT}^{-2} \mathrm{~A}\right]$
D $\left[\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{~A}^2\right]$.

The correct answer is:
$\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$

Solution:
Force of attraction between two charges is given by
$$
\begin{aligned}
F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \\
\Rightarrow \quad \varepsilon_0 & =\frac{1}{4 \pi} \frac{1}{F} \frac{q_1 q_2}{r^2}=\frac{\mathrm{A}^2 \mathrm{~T}^2}{\mathrm{MLT}^{-2} \mathrm{~L}^2} \\
& =\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right] .
\end{aligned}
$$
(where $4 \pi$ is dimensionless)
$=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$

Topic: Physics, Units and Dimensions, AIIMS, AIIMS 2004
Using \(s, p, d\) notations, describe the orbital with the following quantum numbers:
\(\begin{array}{ll}\text { (a) } n=1, l=0 & \text { (b) } n=3, l=1\end{array}\)

Options:

None

Solution:
(a) \(1 s\) (b) \(3 p\) (c) \(4 d\) (d) \(4 f\).

Topic: Chemistry, Structure of Atom