datavorous/entrance-exam-dataset huggingface dataset:
The eccentricity of the hyperbola $16 x^{2}-9 y^{2}=1$ is

Options:
A $\frac{3}{5}$
B $\frac{5}{3}$
C $\frac{4}{5}$
D $\frac{5}{4}$

The correct answer is:
$\frac{5}{3}$

Solution:
$\mathrm{r}+\mathrm{s}=\frac{13}{5}+\frac{37}{5}=\frac{50}{5}=10$ unit
$\because 16 \mathrm{x}^{2}-9 \mathrm{y}^{2}=1$
or $\frac{x^{2}}{\left(\frac{1}{4}\right)^{2}}-\frac{y^{2}}{\left(\frac{1}{3}\right)^{2}}=1$
Comparing with $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\therefore \mathrm{a}^{2}=\frac{1}{16}$
$\& b^{2}=\frac{1}{9}$
$\because \mathrm{e}=\sqrt{1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}$
$\therefore \mathrm{e}=\sqrt{1+\frac{1 / 9}{1 / 16}}$
$\Rightarrow \mathrm{e}=\frac{5}{3}$

Topic: Mathematics, Hyperbola, NDA, NDA 2015 (Phase 2)
The electric field part of an electromagnetic wave in a medium is represented by
\(\begin{aligned} & E_x=0 \\ & E_y=2.5 \mathrm{NC}^{-1} \cos \left[\left(2 \pi \times 10^6 \mathrm{rad} \mathrm{s}^{-1}\right) \mathrm{t}-\left(\pi \times 10^{-2} \mathrm{rad} \mathrm{m}^{-1}\right) \mathrm{x}\right] \\ & E_z=0\end{aligned}\)
The wave is

Options:
A moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.
B moving along $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $100 \mathrm{~m}$
C moving along $x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$
D moving along $-x$ direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$

The correct answer is:
moving along $y$ direction with frequency $2 \pi \times 10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.

Solution:
Comparing the given equation
$$
\begin{aligned}
& \mathrm{E}_{\mathrm{y}}=2.5 \frac{\mathrm{N}}{\mathrm{C}} \cos \left[\left(2 \pi \times 10^6 \frac{\mathrm{rad}}{\mathrm{m}}\right) \mathrm{t}\right. \\
&\left.-\left(\pi \times 10^{-2} \frac{\mathrm{rad}}{\mathrm{sec}}\right) \mathrm{x}\right]
\end{aligned}
$$
With the standard equation
$$
\mathrm{E}_{\mathrm{y}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kx})
$$
we get
$$
\begin{array}{ll}
& \omega=2 \pi \mathrm{f}=2 \pi \times 10^6 \\
\therefore & \mathrm{f}=10^6 \mathrm{~Hz}
\end{array}
$$
Moreover, we know that
$$
\Rightarrow \quad \begin{aligned}
\frac{2 \pi}{\lambda} & =\mathrm{k}=\pi \times 10^{-2} \mathrm{~m}^{-1} \\
\Rightarrow \quad \lambda & =200 \mathrm{~m}
\end{aligned}
$$
Hence, the wave is moving along positive $x$-direction with frequency $10^6 \mathrm{~Hz}$ and wavelength $200 \mathrm{~m}$.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2009 (Screening)
The electrode potential for
$$
\begin{aligned}
& M^{2+}(a q)+e^{-} \longrightarrow M^{+}(a q) \\
& M^{+}(a q)+e^{-} \longrightarrow M(s)
\end{aligned}
$$
are $+0.15 \mathrm{~V}$ and $+0.50 \mathrm{~V}$ respectively. The value of $E^{\circ}{ }_{M^{2+} / M}$ will be

Options:
A 0.150 V
B 0.300 V
C 0.325 V
D 0.650 V

The correct answer is:
0.325 V

Solution:
Given,
$$
\begin{aligned}
& M^{2+}(a q)+e^{-} \longrightarrow M^{+}(a q), E_1^{\circ}=+0.15 \mathrm{~V} \\
& M^{+}(a q)+e^{-} \longrightarrow M(s), E_2^{\circ}=0.5 \mathrm{~V}
\end{aligned}
$$
$\because$ Free energy, i.e. $\left[\Delta G^{\circ}\right]=-n F E^{\circ}$
where, $n=$ number of electron involved
$$
E^{\circ}=\text { standard electrode potential. }
$$
$F=$ Faraday's constant
To find $E_3^{\circ}$ for $E_{\left(M^{2+} / M\right)}^{\circ}$
Thus,
$$
\begin{array}{lr}
-\Delta G_1^{\circ}=n E_1^{\circ}=E_1^{\circ} & (\because n=1) \\
-\Delta G_2^{\circ}=n E_2^{\circ}=E_{{ }_2}{ }_2 & (\because n=1) \\
-\Delta G_3^{\circ}=n E_3=2 E_3^{\circ} & (\because n=2) \\
\text { [For reaction : } M^{2+}(a q)+2 e^{-} & \longrightarrow M(s) \text { ] }
\end{array}
$$
Therefore,
$$
\begin{aligned}
-\Delta G_3^{\circ} & =(-) \Delta G_1^{\circ}+(-) \Delta G_2^{\circ} \\
-2 E_3^{\circ} & =-\left(n E_1^{\circ}+n E_2^{\circ}\right) \\
2 E_3^{\circ} & =E_1^{\circ}+E_2^{\circ} \\
2 E_3^{\circ} & =0.15+0.5 \\
2 E_3^{\circ} & =0.65 \mathrm{~V} \\
E_3^{\circ} & =\frac{0.65}{2}=0.325 \mathrm{~V}
\end{aligned}
$$
Hence, option (c) is the correct answer.

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 1)
The electrostatic potential on the surface of a charged conducting sphere is $100 \mathrm{~V}$. Two statements are made in this regard $S_1$ at any point inside the sphere, electric intensity is zero. $\mathrm{S}_2$ at any point inside the sphere, the electrostatic potential is $100 \mathrm{~V}$. Which of the following is a correct statement?

Options:
A
$\mathrm{S}_1$ is true but $\mathrm{S}_2$ is false

B
Both $\mathrm{S}_1$ and $\mathrm{S}_2$ are false

C
$\mathrm{S}_1$ is true, $\mathrm{S}_2$ is also true and $\mathrm{S}_1$ is the cause of $\mathrm{S}_2$

D
$\mathrm{S}_1$ is true, $\mathrm{S}_2$ is also true but the statements are independant

The correct answer is:
$\mathrm{S}_1$ is true, $\mathrm{S}_2$ is also true and $\mathrm{S}_1$ is the cause of $\mathrm{S}_2$

Solution:
As we know that the relation between electric field intensity $E$ and electric potential $V$ is
$$
\mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dr}}
$$
Electric field intensity $E=0$ then $\frac{\mathrm{dV}}{\mathrm{dr}}=0$
This imply that $\mathrm{V}=$ constant
Thus, $\mathrm{E}=0$ inside the charged conducting sphere then the constant electrostatic potential $100 \mathrm{~V}$ at every where inside the sphere and it verifies the shielding effect also.

Topic: Physics, Electrostatic Potential and Capacitance
The element which is the most abundant in the earth crust is

Options:
A \(\mathrm{O}\)
B \(\mathrm{S}\)
C \(\mathrm{Al}\)
D \(\mathrm{H}\).

The correct answer is:
\(\mathrm{O}\)

Solution:
Oxygen is the most abundant element in the earth's crust. (above \(45.5 \mathrm{wt}~ \%\))

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), AIIMS, AIIMS 2007
The element with highest standard reduction potential (in volt) $\left[\mathrm{M}^{2+} \rightarrow M\right]$ among the first row of transition elements is

Options:
A Ti
B Ni
C Cr
D Cu

The correct answer is:
Cu

Solution:
In the electrochemical series, $\mathrm{Cu}$ is below hydrogen. Hence, it has positive value of standard reduction potential $\left(E_{M^{2+} / M}\right)$.
The standard reduction potentials are
$$
\begin{gathered}
\mathrm{Ni}(Z=28) \Rightarrow-0.257 \mathrm{~V} \\
\mathrm{Cu}(Z=29) \Rightarrow+0.337 \mathrm{~V} \\
\operatorname{Cr}(Z=24) \Rightarrow-0.90 \mathrm{~V} \\
\operatorname{Ti}(Z=21) \Rightarrow-1.63 \mathrm{~V}
\end{gathered}
$$
Highest reduction potential is of $\mathrm{Cu}$ as, it has positive value of reduction potential, while $\mathrm{Ni}, \mathrm{Cr}$ and Ti have negative reduction potential value.

Topic: Chemistry, Electrochemistry, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
The ellipse $\frac{x^{2}}{169}+\frac{y^{2}}{25}=1$ has the same eccentricity as the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. What is the ratio of $a$ to $b$ ?

Options:
A $5 / 13$
B $13 / 5$
C $7 / 8$
D 87

The correct answer is:
$13 / 5$

Solution:
Given ellipse is $\frac{x^{2}}{169}+\frac{y^{2}}{25}=1$
$e=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$
Also, standard equation of ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
nd eccentricity, $e=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$\frac{12}{13}=\sqrt{1-\frac{b^{2}}{a^{2}}} \Rightarrow \frac{b^{2}}{a^{2}}=1-\frac{144}{169}=\frac{25}{169}$
$\Rightarrow \frac{b}{a}=\frac{5}{13} \Rightarrow \frac{a}{b}=\frac{13}{5}$

Topic: Mathematics, Ellipse, NDA, NDA 2009 (Phase 2)
The embryo sac of an angiosperm is made up of

Options:
A
8 cells

B
7 cells and 8 nuclei

C
8 nuclei

D
7 cells and 7 nuclei

The correct answer is:
7 cells and 8 nuclei

Solution:
Embryo sac in angiosperm contains 2 synergids, 1 egg cell, 3 antipodal cells and one secondary nucleus.

Topic: Biology, Plant Kingdom
The emf induced in the secondary coil due to unit rate of change of current in the primary is called

Options:
A Impedance of two coils
B Reactance of two coils
C Mutual induction of two coils
D Self induction of a coil

The correct answer is:
Mutual induction of two coils

Solution:
No solution. Refer to answer key.

Topic: Physics, Electromagnetic Induction, AP EAMCET, AP EAMCET 2020 (07 Oct Special)
The emf of a galvanic cell constituted with the electrodes $\mathrm{Zn}^{2+} \mid \mathrm{Zn}(-0.76 \mathrm{~V})$ and $\mathrm{Fe}^{2+} \mid \mathrm{Fe}(-0.41 \mathrm{~V})$ is

Options:
A $-0.35 \mathrm{~V}$
B $+1.17 \mathrm{~V}$
C $+0.35 \mathrm{~V}$
D $-1.17 \mathrm{~V}$

The correct answer is:
$+0.35 \mathrm{~V}$

Solution:
The cell reaction may be split into two half reactions as
$\begin{aligned}
&\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \quad E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{~V} \\
&\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \quad E_{\mathrm{red}}^{\circ}=-0.44 \mathrm{~V}
\end{aligned}$
As $\mathrm{Zn}^{2+}$ ion has lower reduction potential than $\mathrm{Fe}^{2+}$ ion. Thus, oxidation takes place on zinc electrode standard EMF of the cell
$=\left[\begin{array}{l}\text { Standard reduction potential of the } \\ \text { the reduction half reaction }\end{array}\right]-$
$\left[\begin{array}{l}\text { Standard reduction potential of } \\ \text { the oxidation half reaction }\end{array}\right]$
$=-0.44-(-0.76)$
$=+0.35 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, KCET, KCET 2013
The end product in Hofmann's exhaustive of amines is

Options:
A Primary amine
B Secondary amine
C Tertiary amine
D Quaternary ammonium halide

The correct answer is:
Quaternary ammonium halide

Solution:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2+\underset{\text { Excess }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}} \underset{-\mathrm{HI}}{\longrightarrow}\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_4 \mathrm{~N}^{+} \mathrm{I}^{-}$

Topic: Chemistry, Amines, MHT CET, MHT CET 2022 (07 Aug Shift 1)
The energy available to the organism at the third trophic level is called

Options:
A gross primary productivity.
B net primary productivity.
C secondary productivity.
D primary productivity.

The correct answer is:
secondary productivity.

Solution:
Carnivores are organisms which obtain nutrient by killing and eating other animals. These form the secondary consumers present in the third trophic level and are preceded by the herbivore in the food chain. The rate of assimilation of food energy by consumers is called secondary productivity and this energy is available to carnivores as they are the consumers.
Total production of organic matter is called gross primary productivity and the energy left after utilization of producers is termed net primary productivity. Community productivity is the difference between GPP and the community respiration.

Topic: Biology, Ecosystem, NEET
The energy $(E)$, angular momentum $(L)$ and universal gravitational constant $(G)$ are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula of Planck's constant $(h)$ is

Options:
A Zero
B $-1$
C $\frac{5}{3}$
D $1$

The correct answer is:
Zero

Solution:
$\begin{aligned}
& h \propto G^x L^y E^z \\
& {\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right] \propto\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]^z} \\
& {\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]=k\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]^y}
\end{aligned}$
$\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]^z$
Comparing the powers, we get
$\begin{aligned}
1 & =-x+y+z \\
2 & =3 x+2 y+2 z \\
-1 & =-2 x-y-2 z
\end{aligned}$
On solving Eq. (i), (ii) and (iii), we get $x=0$

Topic: Physics, Units and Dimensions, AP EAMCET, AP EAMCET 2008
The energy $(E)$, angular momentum $(L)$ and universal gravitational constant $(G)$ are chosen as fundamental quantities. The dimensions of universal gravitational constant in the dimensional formula of Planck's constant $(h)$ is

Options:
A Zero
B $-1$
C $\frac{5}{3}$
D $1$

The correct answer is:
Zero

Solution:
$\begin{aligned}
& h \propto G^x L^y E^z \\
& {\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right] \propto\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]^z} \\
& {\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]=k\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-1}\right]^y}
\end{aligned}$
$\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]^z$
Comparing the powers, we get
$\begin{aligned}
1 & =-x+y+z \\
2 & =3 x+2 y+2 z \\
-1 & =-2 x-y-2 z
\end{aligned}$
On solving Eq. (i), (ii) and (iii), we get $x=0$

Topic: Physics, Units and Dimensions, TS EAMCET, TS EAMCET 2008
The enthalpies of all elements in their standard states are :
(i) unity
(ii) zero
(iii) \( < 0\)
(iv) different for each element

Options:

None

Solution:
(ii) zero

Topic: Chemistry, Thermodynamics (C)
The enthalpies of all elements in their standard states are

Options:
A zero
B $\gt0$
C $ \lt 0$
D unity

The correct answer is:
zero

Solution:
The enthalphy of elements are zero in their standard state.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JIPMER, JIPMER 2004
The enthalpies of formation of gaseous $\mathrm{N}_2 \mathrm{O}$ and $\mathrm{NO}$ at $298 \mathrm{~K}$ are 82.0 and $90.0 \mathrm{KJ} \mathrm{mol}^{-1}$ respectively. The enthalpy change of the reaction $\mathrm{N}_2 \mathrm{O}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow$ $2 \mathrm{NO}(\mathrm{g})$ is

Options:
A $-74 \mathrm{~kJ}$
B +98 kJ
C +89 kJ
D -47 kJ

The correct answer is:
+98 kJ

Solution:
$$
\Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[2 \Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{NO})\right]-\left[\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{N}_2 \mathrm{O}\right)+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{O}_2\right)\right]
$$
$\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left(\mathrm{O}_2\right)=0$ (standard state of an element)
$$
\begin{gathered}
\Rightarrow \Delta \mathrm{H}_{\mathrm{r}}{ }^{\circ}=[2 \times 90.0]-[82.0+0.0] \\
\quad=180.0-82.0=+98.0 \mathrm{~kJ} .
\end{gathered}
$$

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
The enthalpies of the elements in their standard states are assumed to be

Options:
A Zero at 298 K
B Unit at 298 K
C Zero at all temperature
D Zero at 273 K

The correct answer is:
Zero at 298 K

Solution:
The enthalpies of the elements in their standard states are assumed to be zero at 298 K

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy of reaction for the reaction : 2H2(g) + O2(g) → 2H2O(l) is ∆rH⊖= – 572 kJ mol–1. What will be standard enthalpy of formation of H2O(l)?

Options:
A ∆fH⊖(H2O(l)) = +286 kJ / mol
B ∆fH⊖(H2O(l)) = -214 kJ / mol
C ∆fH⊖(H2O(l)) = -365 kJ / mol
D ∆fH⊖(H2O(l)) = -286 kJ / mol

The correct answer is:
∆fH⊖(H2O(l)) = -286 kJ / mol

Solution:
Correct Option is : (D)
∆fH⊖(H2O(l)) = -286 kJ / mol

Topic: Chemistry, Thermodynamics (C), NEET
The equation $|1-x|+x^{2}=5$ has

Options:
A a rational root and an irrational root
B two rational roots
C two irrational roots
D no real roots

The correct answer is:
a rational root and an irrational root

Solution:
$\begin{aligned} &|1-x|+z^{2}=5 \\ & \Rightarrow|x-1|+x^{2}=5 \\ & \text { First case: If } x < 1,|x-1| \text { is negative. } \end{aligned}$
$\begin{aligned} & \therefore-(x-1)+x^{2}=5 \\ & \Rightarrow-x+1+x^{2}=5 \\ \Rightarrow & x^{2}-x+1=5 \\ \Rightarrow & x^{2}-x-4=0 \end{aligned}$
Roots are $\frac{-(-1) \pm \sqrt{(-1)^{2}-4(1)(-4)}}{2(1)}$
$=\frac{1 \pm \sqrt{1+16}}{2}=\frac{1 \pm \sqrt{17}}{2}$
Since, $x < 1$, root cannot be $\frac{1+\sqrt{17}}{2} .$ So, the root is
$\frac{1-\sqrt{17}}{2}$, which is irrational.
Second case: If $x>1,|x-1|=x-1$ $\mathrm{x}-1 \mid+2^{2}=5$
$\Rightarrow \quad x-1+x^{2}=5$
$\Rightarrow \quad x^{2}+x-6=0$
$\Rightarrow \quad x^{2}+3 x-2 x-6=0$
$\Rightarrow \quad x(x-3)-2(x+3)=0$
$\Rightarrow \quad x=2,-3$
Since $x>1$, root cannot be $-3 .$ So, root is 2 which is rational.
$\therefore$ Given expression has one irrational root and One rational root

Topic: Mathematics, Quadratic Equation, NDA, NDA 2018 (Phase 1)
The equation $2 \mathrm{x}^{2}-3 \mathrm{y}^{2}-6=0$ represents

Options:
A a circle
B a parabola
C an ellipse
D a hyperbola

The correct answer is:
a hyperbola

Solution:
Given equation, $2 x^{2}-3 y^{2}-6=0$ $\Rightarrow 2 x^{2}-3 y^{2}=6$
$\Rightarrow \frac{x^{2}}{3}-\frac{y^{2}}{2}=1$
This equation represents hyperbola.

Topic: Mathematics, Hyperbola, NDA, NDA 2019 (Phase 1)
The equation $12 x^{2}+7 x y+a y^{2}+13 x-y+3=0$
represents a pair of perpendicular lines. Then the value of ' $a$ ' is

Options:
A $\frac{7}{2}$
B $-19$
C $-12$
D 12

The correct answer is:
$-12$

Solution:
Comparing the given equation with standard equation, we get $a=12$ and $b=a$, for perpendicular lines coefficient of $x^{2}+$ coefficient of $y^{2}=0$ $\therefore \quad 12+a=0$
$\Rightarrow \quad a=-12$

Topic: Mathematics, Pair of Lines, MHT CET, MHT CET 2008
The equation $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$ has equal roots. Which one of the following is correct about $\mathrm{a}, \mathrm{b}$, and $\mathrm{c}$ ?

Options:
A They are in AP
B Theyare in GP
C They are in $\mathrm{HP}$
D They are neither in AP, nor in GP, nor in HP

The correct answer is:
Theyare in GP

Solution:
The given equation $\left(a^{2}+b^{2}\right) x^{2}-2 b(a+c) x+\left(b^{2}+c^{2}\right)=0$
has equal roots, so, discriminant $=0$ Hence, $\{2 \mathrm{~b}(\mathrm{a}+\mathrm{c})\}^{2}-4\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)\left(\mathrm{b}^{2}+\mathrm{c}^{2}\right)=0$
$\Rightarrow 4 b^{2}\left(a^{2}+c^{2}+2 c a\right)-4\left(a^{2} b^{2}+a^{2} c^{2}+b^{4}+b^{2} c^{2}\right)$
$=0$
$\Rightarrow b^{2} a^{2}+b^{2} c^{2}+2 b^{2} c a-a^{2} b^{2}-a^{2} c^{2}-b^{4}-b^{2} c^{2}=0$
$\Rightarrow 2 b^{2} c a=b^{4}+a^{2} c^{2}$
$\Rightarrow b^{4}-2 b^{2} c a+a^{2} c^{2}=0$
$\Rightarrow\left(b^{2}\right)^{2}-2\left(b^{2}\right)(a c)+(a c)^{2}=0$
$\Rightarrow\left(b^{2}-a c\right)^{2}=0$
$\Rightarrow \mathrm{b}^{2}=\mathrm{ac}$
$\Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in GP.

Topic: Mathematics, Sequences and Series, NDA, NDA 2006 (Phase 2)
The equation $b y+c z+d=0$ represents a plane parallel to which one of the following?

Options:
A $x$ -axis
B $y$ -axis
C $z$ -axis
D None of these

The correct answer is:
$x$ -axis

Solution:
Direction cosines of the normal to the given plane $b y+c z+d=0$ are $0, b, c$
Direction cosines of the $\mathrm{x}$ -axis are $1,0,0$
Since, $0 \times 1+b \times 0+c \times 0=0$
Hence $x$ -axis is perpendicular to normal to the given plane. Therefore $x$ -axis is parallel to the given plane

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2008 (Phase 2)
The equation formed by multiplying each root of $a x^{2}+b x+c=0$ by 2 is $x^{2}+36 x+24=0$
What is b:c equal to?

Options:
A 3 :1
B $1: 2$
C $1: 3$
D $3: 2$

The correct answer is:
3 :1

Solution:
Let $\alpha, \beta$ be roots of equation $a x^{2}+b x+c$ $\alpha+\beta=-\mathrm{b} / \mathrm{a}, \alpha \beta=\mathrm{c} / \mathrm{a}$
are roots of equation $x^{2}+36 x+24$ Now, $2 \alpha+2 \beta=-36$
$\Rightarrow \alpha+\beta=-18 \Rightarrow-\frac{\mathrm{b}}{\mathrm{a}}=-18 \Rightarrow \mathrm{b}=18 \mathrm{a}$
and $(2 \alpha)(2 \beta)=24$
$\Rightarrow \alpha \beta=6 \Rightarrow \frac{\mathrm{c}}{\mathrm{a}}=6 \Rightarrow \mathrm{c}=6 \mathrm{a}$
$\mathrm{c}=3: 1$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 1)
The equation formed by multiplying each root of $a x^{2}+b x+c=0$ by 2 is $x^{2}+36 x+24=0$
Which one of the following is correct?

Options:
A $b c=a^{2}$
B $b c=36 a^{2}$
C $b c=72 a^{2}$
D $b c=108 a^{2}$

The correct answer is:
$b c=108 a^{2}$

Solution:
$\therefore \mathrm{bc}=(18 \mathrm{a})(6 \mathrm{a})=108 \mathrm{a}^{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 1)
The equation of a circle whose end points of a diameter are $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is

Options:
A $\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=x^{2}+y^{2}$
B $\left(\mathrm{x}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}-\mathrm{y}_{1}\right)^{2}=\mathrm{x}_{2} \mathrm{y}_{2}$
C $x^{2}+y^{2}+2 x_{1} x_{2}+2 y_{1} y_{2}=0$
D $\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$

The correct answer is:
$\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$

Solution:
Equation of circle is
$\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$

Topic: Mathematics, Circle, NDA, NDA 2018 (Phase 2)
The equation of a progressive wave can be given by $y=15 \sin (660 \pi t-0.02 \pi x) \mathrm{cm}$. The frequency of the wave is

Options:
A $330 \mathrm{~Hz}$
B $342 \mathrm{~Hz}$
C $365 \mathrm{~Hz}$
D $660 \mathrm{~Hz}$

The correct answer is:
$330 \mathrm{~Hz}$

Solution:
Given, equation of progressive wave,
$$
y=15 \sin (660 \pi t-0.02 \pi x)
$$
Comparing this equation with standard equation of progressive wave as $y=A \sin \left(\frac{2 \pi}{T}, t-\frac{2 \pi}{\lambda}, x\right)$
We get,
$\begin{aligned} \frac{2 \pi}{T} &=660 \pi \\ \Rightarrow \quad T &=\frac{2 \pi}{660 \pi} \\ \Rightarrow \quad T &=\frac{1}{330} \\ \therefore \text { Frequency of the wave, } \\ v &=\frac{1}{T} \\ &=\frac{1}{1 / 330}=330 \mathrm{~Hz} \end{aligned}$

Topic: Physics, Waves and Sound, COMEDK, COMEDK 2020
The equation of a progressive wave is given by $y=5 \sin (100 \pi t-0.4 \pi x)$ where $y$ and $x$ are in $\mathrm{m}$ and $t$ is in $\mathrm{s}$.
(1) The amplitude of the wave is $5 \mathrm{~m}$.
(2) The wavelength of the wave is $5 \mathrm{~m}$.
(3) The frequency of the wave is $50 \mathrm{~Hz}$.
(4) The velocity of the wave is $250 \mathrm{~m} \mathrm{~s}^{-1}$.
Which of the following statements are correct?

Options:
A (1), (2) and (3)
B (2) and (3)
C (1) and (4)
D All are correct

The correct answer is:
All are correct

Solution:
The equation of a given progressive wave is
$$
y=5 \sin (100 \pi t-0.4 \pi x)
$$
The standard equation of a progressive wave is
$$
y=a \sin (\omega t-k x)
$$
Comparing (i) and (ii), we get
$$
a=5 \mathrm{~m}, \omega=100 \pi \mathrm{rad} \mathrm{s}^{-1}, k=0.4 \pi \mathrm{m}^{-1}
$$
(1) Amplitude of the wave, $a=5 \mathrm{~m}$
(2) Wavelength of the wave,
$$
\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.4 \pi}=5 \mathrm{~m}
$$
(3) Frequency of the wave,
$$
v=\frac{\omega}{2 \pi}=\frac{100 \pi}{2 \pi}=50 \mathrm{~Hz}
$$
(4) Velocity of the wave,
$$
v=v \lambda=\left(50 \mathrm{~s}^{-1}\right)(5 \mathrm{~m})=250 \mathrm{~m} \mathrm{~s}^{-1}
$$

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2015
The equation of a simple harmonic wave is given by $y=6 \sin 2 \pi(2 t-0 . l x)$, where $x$ and $y$ are in $\mathrm{mm}$ and $t$ is in seconds. The phase difference between two particles $2 \mathrm{~mm}$ apart at any instant is

Options:
A $18^{\circ}$
B $36^{\circ}$
C $54^{\circ}$
D $72^{\circ}$

The correct answer is:
$72^{\circ}$

Solution:
Equalion of simple harmonic wave is given as
$$
y=6 \sin 2 \pi(2 t-0.1 x)
$$
Path difference, $\Delta x=2 \mathrm{~mm}$
Comparing with standard equation of wave,
$$
\begin{aligned}
y &=A \sin (\omega t-k x) \\
&=A \sin 2 \pi\left(v t-\frac{k}{2 \pi} \cdot x\right) \\
\frac{k}{2 \pi} &=0.1 \\
k &=0.2 \pi
\end{aligned}
$$
Wavelength, $\lambda=\frac{2 \pi}{k}=\frac{2 \pi}{0.2 \pi}=10 \mathrm{~mm}$
$\therefore$ We know that, phase difference,
$$
\begin{aligned}
\phi &=\frac{2 \pi}{\lambda} \times \text { path difference }(\Delta x) \\
&=\frac{2 \pi}{10} \times 2=\frac{2 \pi}{5}=\frac{2 \times 180}{5}=72^{\circ}
\end{aligned}
$$

Topic: Physics, Waves and Sound, COMEDK, COMEDK 2018