datavorous/entrance-exam-dataset huggingface dataset:
If the charge of electron \(e\), mass of electron \(m\), speed of light in vacuum \(c\) and Planck's constant \(h\) are taken as fundamental quantities, then the permeability of vacuum \(\mu_0\) can be expressed as

Options:
A \(\frac{h}{m c^2}\)
B \(\frac{h c}{m e^2}\)
C \(\frac{h}{c e^2}\)
D \(\frac{m c^2}{h e^2}\)

The correct answer is:
\(\frac{h}{c e^2}\)

Solution:
We can expressed the permeability of vacuum,
\(\mu_0 \propto e^a m^b c^c h^d\)
or \(\mu_0=k e^a m^b c^c h^d\) ...(i)
Where, \(k\) is a dimensional constant.
As we know that,
Dimension of
\(\begin{aligned}
\mu_0 & =\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{~A}^{-2}\right], \\
e & =[\mathrm{AT}], \\
m & =[\mathrm{M}], \\
c & =\left[\mathrm{L} \mathrm{T}^{-1}\right], \\
h & =\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right],
\end{aligned}\)
and
Putting the dimension of various physical quantities in Eq. (i), we get
\(\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{~A}^{-2}\right]=k[\mathrm{~A} \mathrm{~T}]^a\left[\mathrm{M}^b\left[\mathrm{~L} \mathrm{~T}^{-1}\right]^c\right]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]^d\)
\(\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{~A}^{-2}\right]=k[\mathrm{M}]^{b+d}[\mathrm{~L}]^{c+2 d}[\mathrm{~T}]^{a-c-d}[\mathrm{~A}]^a\)
Comparing the powers of \(\mathrm{M}, \mathrm{L}, \mathrm{T}\) and \(\mathrm{A}\) on the both sides, we get
\(\begin{aligned}
b+d & =1 \quad \ldots (ii) \\
c+2 d & =1 \quad \ldots (iii) \\
a-c-d & =-2 \quad \ldots (iv) \\
a & =-2 \quad \ldots (v)
\end{aligned}\)
After solving the Eqs. (ii), (iii), (iv) and (v), we get
\(\begin{aligned}
a & =-2, \\
b & =0, \\
c & =-1, \\
\text{and } d & =1, \\
\therefore \mu_0 & =\frac{h}{c e^2}
\end{aligned}\)
So, the permeability of vacuum \(\mu_0\) can be expressed as, \(\frac{h}{c e^2}\).

Topic: Physics, Units and Dimensions, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
If the circle $x^{2}+y^{2}+2 g x+2 f y+c=0(c>0)$ touches the $y$ axis, then which one of the following is correct?

Options:
A $\mathrm{g}=-\sqrt{\mathrm{c}}$ only
B $\mathrm{g}=\pm \sqrt{\mathrm{c}}$
C $\mathrm{f}=\sqrt{\mathrm{c}}$ only
D $\mathrm{f}=\pm \sqrt{\mathrm{c}}$

The correct answer is:
$\mathrm{f}=\pm \sqrt{\mathrm{c}}$

Solution:
As given, the circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ touches y-axis. then $\mathrm{r}=\pm \mathrm{g}$ and $\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}=\mathrm{g}^{2}$
$\Rightarrow \mathrm{f}^{2}-\mathrm{c}=0 \Rightarrow \mathrm{f}^{2}=\mathrm{c} \Rightarrow \mathrm{f}=\pm \sqrt{\mathrm{c}}$

Topic: Mathematics, Circle, NDA, NDA 2008 (Phase 1)
If the coefficient of mutual induction of the primary and secondary coils of an induction coil is $5 \mathrm{H}$ and a current of $10 \mathrm{~A}$ is cut off in $5 \times 10^{-4}$ second, the $e m f$ induced(in volt) in the secondary coil is

Options:
A $5 \times 10^{4}$
B $1 \times 10^{5}$
C $25 \times 10^{5}$
D $5 \times 10^{6}$

The correct answer is:
$1 \times 10^{5}$

Solution:
$\mathrm{M}=5 \mathrm{H}, \mathrm{I}=10 \mathrm{~A}, \mathrm{t}=5 \times 10^{-4} \mathrm{~s}$.
Now, emf induced in secondary is
$\begin{array}{l}
\mathrm{e}=-\mathrm{M} \frac{\mathrm{d}}{\mathrm{dt}}, \quad \text { (-ve sign shows direction) } \\
\Rightarrow \mathrm{e}=5 \times \frac{10}{5 \times 10^{-4}}=1 \times 10^{+5} \mathrm{~V}
\end{array}$

Topic: Physics, Current Electricity, VITEEE, VITEEE 2007
If the coefficient of variation and standard deviation are \( 60 \) and \( 21 \) respectively, the arithmetic
mean of distribution is

Options:
A \( 30 \)
B \( 21 \)
C \( 60 \)
D \( 35 \)

The correct answer is:
\( 35 \)

Solution:
We know that, coefficient of variation \(=\frac{\sigma}{\mu} \times 100 \%\)
\(\Rightarrow 60=\frac{21}{\mu} \times 100\), substitute the given values
\(\therefore \mu=\frac{2100}{60}=35\)

Topic: Mathematics, Linear Programming, KCET, KCET 2014
If the coefficients of $\mathrm{a}^{\mathrm{m}}$ and $\mathrm{a}^{\mathrm{n}}$ in the expansion of $(1+a)^{\mathrm{m}+n}$ are $\alpha$ and $\beta$, then which one of the following is correct?

Options:
A $\quad \alpha=2 \beta$
B $\alpha=\beta$
C $2 \alpha=\beta$
D $\alpha=(\mathrm{m}+\mathrm{n}) \beta$

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

Solution:
$(1+a)^{m+11}$
$\alpha=$ coefficient of $a^{\mathrm{m}}={ }^{m+n} C_{\mathrm{m}}$
$\beta=$ coefficient of $a^{\mathrm{n}}={ }^{m+n} C_{n}$
We know, ${ }^{n} C_{r}={ }^{n} C_{n-r}$
$\therefore \beta={ }^{m+n} C_{n}={ }^{m+n} C_{m+n-n}={ }^{m+n} C_{m}=\alpha$
$\therefore \alpha=\beta$

Topic: Mathematics, Binomial Theorem, NDA, NDA 2018 (Phase 1)
If the coefficients of variation of two distributions are 60 and 70 and their standard deviations are 21 and 16 respectively, then their arithmetic means are respectively

Options:
A $35,22.85$
B $32,25.85$
C $35,28.25$
D $35,25.25$

The correct answer is:
$35,22.85$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 2)
If the components of $\overrightarrow{\mathrm{b}}$ along and perpendicular to $\overrightarrow{\mathrm{a}}$ are
$\lambda \vec{a}$ and $\vec{b}-\lambda \vec{a}$ respectively, what is $\lambda$ equal to ?

Options:
A $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$
B $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
C $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}$
D $\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|^{2}}$

The correct answer is:
$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}$

Solution:
We know that the components of $\overrightarrow{\mathrm{b}}$ along $\overrightarrow{\mathrm{a}}$ is $\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|^{2}}\right\} \overrightarrow{\mathrm{a}}$ and perpendicular to $\overrightarrow{\mathrm{a}}$ is $\overrightarrow{\mathrm{b}}-\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|^{2}}\right\} \overrightarrow{\mathrm{a}}$
As given : $\left\{\begin{array}{l} \left.\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^{2}}\right\} \vec{a}=\lambda \vec{a}\end{array}\right.$
and $\overrightarrow{\mathrm{b}}-\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|^{2}}\right\} \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}-\lambda \overrightarrow{\mathrm{a}}$
$\Rightarrow \lambda=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|^{2}}\right\}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 2)
If the correlation coefficient between $\mathrm{x}$ and $\mathrm{y}$ is $0.6$, covariance is 27 and variance of $y$ is 25 , then what is the variance of x? $\quad$

Options:
A $\frac{9}{5}$
B $\frac{81}{25}$
C 9
D 81

The correct answer is:
81

Solution:
Given, $r=0.6$, covariance $=27, \sigma_{(y)}^{2}=25 \Rightarrow \sigma(y)=5$
We know, $\mathrm{r}=\frac{\text { covariance }(\mathrm{x}, \mathrm{y})}{\sigma(\mathrm{x}) \cdot \sigma(\mathrm{y})}$
$\Rightarrow \sigma(\mathrm{x})=\frac{\operatorname{covariance}(\mathrm{x}, \mathrm{y})}{\mathrm{r} . \sigma(\mathrm{y})}=\frac{27}{\left(\frac{6}{10}\right) \cdot 5}$
$=\frac{27 \times 2}{6}=9$
$\Rightarrow \sigma^{2}(\mathrm{x})=81 .$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
If the covariance between $x$ and $y$ is 30 , variance of $x$ is 25 and variance of $y$ is 144 , then what is the correlation coefficient? $\quad$

Options:
A $0.4$
B $0.5$
C $0.6$
D $0.7$

The correct answer is:
$0.5$

Solution:
$\operatorname{cov}(x, y)=30$
$$
\begin{array}{l}
\operatorname{var}(x)=25 ; \operatorname{var}(y)=144 \\
\Rightarrow r(x, y)=\frac{\operatorname{cov}(x, y)}{\sqrt{\operatorname{var}(x) \cdot \operatorname{var}(y)}} \\
\Rightarrow r(x, y)=\frac{30}{\sqrt{25 \times 144}}=\frac{30}{5 \times 12}=0.5
\end{array}
$$

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 2)
If the current 3 A flowing in the primary coil is made zero in 0.1 sec, the emf induced in the secondary coil is 1.5 volt. The mutual inductance between the coils is

Options:
A 0.05 H
B 1.05 H
C 0.1 H
D 0.2 H

The correct answer is:
0.05 H

Solution:
Here: $e=1.5$ volt
$\Delta i=0-3=-3 \mathrm{~A}, \Delta t=0.1 \mathrm{sec}$
Now the emf induced in secondary coil
$e=\frac{M \Delta i}{\Delta t} \quad$ (numerically)
or $\quad 1.5=\frac{M \times 3}{0.1}$
or $\quad M=\frac{1.5 \times 0.1}{3}=0.05 \mathrm{H}$

Topic: Physics, Electromagnetic Induction, JIPMER, JIPMER 2004
If the data are moderately non-symmetrical, then which onc of the following empirical relationships is correct?

Options:
A $2 \times$ Standard deviation $=5 \times$ Mean deviation
B $5 \times$ Standard deviation $=2 \times$ Mean deviation
C $4 \times$ Standard deviation $=5 \times$ Mean deviation
D $5 \times$ Standard deviation $=4 \times$ Mean deviation

The correct answer is:
$4 \times$ Standard deviation $=5 \times$ Mean deviation

Solution:
We know, $\frac{4}{5}$ standard deviation = Mean deviation $\Rightarrow 4 \times$ standard deviation $=5 \times$ mean deviation

Topic: Mathematics, Statistics, NDA, NDA 2017 (Phase 1)
If the derivative of the function
$f(x)=\left\{\begin{array}{cc}a x^{2}+b & x < -1 \\ b x^{2}+a x+4 & x \geq-1\end{array}\right.$
is every where continuous, then what are the values of a and $\mathrm{b}$ ?

Options:
A $\quad a=2, b=3$
B $a=3, b=2$
C $a=-2, b=-3$
D $a=-3, b=-2$

The correct answer is:
$\quad a=2, b=3$

Solution:
Derivative of $f(x)=\left\{\begin{array}{ll}a x^{2}+b & x < -1 \\ b x^{2}+a x+a & x \geq-1\end{array}\right.$ is
$f^{\prime}(x)=\left\{\begin{array}{ll}2 a x & x < -1 \\ 2 b x+a, & x \geq-1\end{array}\right.$
If $\mathrm{f}^{\prime}(\mathrm{x})$ is continuous everywhere then it is also continuous at $\mathrm{x}=-1$
$\left.\mathrm{f}^{\prime}(\mathrm{x})\right|_{\mathrm{x}=-1}=-2 \mathrm{a}=-2 \mathrm{~b}+\mathrm{a}$
or, $3 \mathrm{a}=2 \mathrm{~b}$
From the given choice $\mathrm{a}=2, \mathrm{~b}=3$ satisfied this equation.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2007 (Phase 1)
If the difference between the roots of $a x^{2}+b x+c=0$ is 1 , then which one of the following is correct?

Options:
A $\quad b^{2}=a(a+4 c)$
B $a^{2}=b(b+4 c)$
C $a^{2}=c(a+4 c)$
D $b^{2}=a(b+4 c)$

The correct answer is:
$\quad b^{2}=a(a+4 c)$

Solution:
Let $\alpha$ and $\beta$ be the roots of $a x^{2}+b x+c=0$ $\alpha+\beta=-\mathrm{b} / \mathrm{a}$
$\alpha \beta=\mathrm{c} / \mathrm{a}$
Given $\alpha-\beta=1$
Consider $(\alpha+\beta)^{2}-(\alpha-\beta)^{2}=4 \alpha \beta$
$\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}-1=\frac{4 \mathrm{c}}{\mathrm{a}}$
$b^{2}-a^{2}=4 a c$
$b^{2}=a(a+4 c)$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2012 (Phase 1)
If the differential equation for a simple harmonic motion is $\frac{d^{2} y}{d t^{2}}+2 y=0$, the time-period of the motion is

Options:
A $\pi \sqrt{2} \mathrm{sec}$
B $\frac{\sqrt{2} s}{\pi} \mathrm{sec}$
C $\frac{\pi}{\sqrt{2}} \mathrm{sec}$
D $2 \pi \mathrm{sec}$

The correct answer is:
$\pi \sqrt{2} \mathrm{sec}$

Solution:
The differential equation of simple harmonic motion is
$$
\frac{d^{2} y}{d t^{2}}+2 y=0 \text { or } \frac{d^{2} y}{d t^{2}}=-2 y......(i)
$$
Standard equation of simple harmonic motion is
$$
\frac{d^{2} y}{d t^{2}}=-\omega^{2} y......(ii)
$$
Comparing eq. (i) and (ii), $\omega^{2}=2$ or $\omega=\sqrt{2}$
As we know, $\omega=\frac{2 \pi}{T}$
$\therefore$ Time period, $T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{2}}=\pi \sqrt{2} \mathrm{sec}$

Topic: Physics, Oscillations, VITEEE, VITEEE 2019
If the diploid number of a flowering plant is 36 . What would be the chromosome number in its endosperm?

Options:
A
36

B
18

C
54

D
72

The correct answer is:
54

Solution:
Endosperm is a product of triple fusion. One male nuclei $(n=18)$ fuses with diploid secondary nucleus $(2 n=$ $36)$, so it becomes triploid $(3 n=54)$. Thus, ploidy of endosperm is $(3 n)$ and chromosomes will be 54.

Topic: Biology, Plant Kingdom
If the direction ratios of the normal to a plane are $ < 1, \mathrm{~m}, \mathrm{n}>$ and the length of the normal is $\mathrm{p}$, then what is the sum of intercepts cut-off by the plane from the coordinate axes?

Options:
A $\mathrm{p}\left(\frac{1}{\ell}+\frac{1}{\mathrm{~m}}+\frac{1}{\mathrm{n}}\right)$
B $\mathrm{p} \sqrt{\left(\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)}$
C $\mathrm{p} \sqrt{\left(\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)}\left(\frac{1}{\ell}+\frac{1}{\mathrm{~m}}+\frac{1}{\mathrm{n}}\right)$
D $\frac{\mathrm{p}}{\sqrt{\left(\ell^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)}}\left(\frac{1}{\ell}+\frac{1}{\mathrm{~m}}+\frac{1}{\mathrm{n}}\right)$

The correct answer is:
$\mathrm{p}\left(\frac{1}{\ell}+\frac{1}{\mathrm{~m}}+\frac{1}{\mathrm{n}}\right)$

Solution:
Since, the direction ratio's of normal to a plane are $ < l, \mathrm{~m}, \mathrm{n}>$ and the length of normal is $\mathrm{p}$, then intercept on $\mathrm{x}$ -axis is $\mathrm{p} / /$ and that on $\mathrm{y}$ -axis is $\mathrm{p} / \mathrm{m}$ and on $\mathrm{z}$ -axis it is $\mathrm{p} / \mathrm{n}$, hence sum of intercepts cut off by the plane from the coordinate axes
$=p\left(\frac{1}{\ell}+\frac{1}{m}+\frac{1}{n}\right)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2006 (Phase 2)
If the distance between the points $(7,1,-3)$ and $(4,5, \lambda)$ is 13 units, then what is one of the values of $\lambda$ ? $\quad$

Options:
A 20
B 10
C 9
D 8

The correct answer is:
9

Solution:
We have,
$13=\sqrt{(4-7)^{2}+(5-1)^{2}+(\lambda+3)^{2}}$
$169=9+16+\lambda^{2}+9+6 \lambda$
$\Rightarrow \quad \lambda^{2}+6 \lambda-135=0$
$\Rightarrow \quad \lambda^{2}+15 \lambda-9 \lambda-135=0$
$\Rightarrow \quad \lambda(\lambda+15)-9(\lambda+15)=0$
$\Rightarrow(\lambda+15)(\lambda-9)=0$
$\Rightarrow \quad \lambda=-15$ or $\lambda=9$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
If the DNA strand undergoing transcription has a following base pair sequence. What would be the sequence of mRNA formed from this strand?

3' GAT CTT AGC GGG AAT TCG 5'

Options:
A 5' GAU CUU AGC GGG AAU UCG 3'
B 5' CAU GAA UGC CCC UUA AGC 3'
C 5' CUA GAA UCG CCC UUA AGC 3'
D 3' CUA GAA UGG CCC UUA AGC 5'

The correct answer is:
5' CUA GAA UCG CCC UUA AGC 3'

Solution:
Protein synthesis is the process of manufacturing protein from DNA involving two steps - transcription and translation. Transcription is the process of making mRNA from the DNA and this mRNA is translated into a long chain polypeptide with help of ribosome and tRNA.
The strand of NDA from which the new mRNA is synthesized is called as template strand or non-coding or anti-sense strand. It is in the 3' → 5' direction. The mRNA is synthesized on it in the 5' → 3' with the help of RNA polymerase enzyme which leads in the same direction.
The new mRNA sequence transcribed from the DNA sequence - 3' GAT CTT AGC GGG AAT TCG 5' will be 5' CUA GAA UCG CCC UUA AGC 3'

Topic: Biology, Molecular Basis of Inheritance, NEET
If the eccentricity and length of latus rectum of a hyperbola
are $\frac{\sqrt{13}}{3}$ and $\frac{10}{3}$ units respectively, then what is the length
of the transverse axis?

Options:
A $\frac{7}{2}$ unit
B 12 unit
C $\frac{15}{2}$ unit
D $\frac{15}{4}$ unit

The correct answer is:
$\frac{15}{2}$ unit

Solution:
Length of latus rectum of a hyperbola is $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}$ where
a is the half of the distance bet ween two vertex of the hyperbola.
Latus rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=\frac{10}{3}$
or, $\mathrm{b}^{2}=\frac{5 \mathrm{a}}{3}$...(1)
In case of hyperbola, $b^{2}=a^{2}\left(e^{2}-1\right)$...(2)
Putting value of $b^{2}$ from equation (1) and e $=\frac{\sqrt{13}}{3}$ in
equation (2),
$\frac{5 a}{3}=a^{2}\left(\frac{13}{9}-1\right)$
or, $\frac{5 a}{3}=\frac{4 a^{2}}{9}$
$\Rightarrow 4 a^{2}-15 a=0$ or $a(4-15 a)=0$
$a \neq 0$, hence, $a=\frac{15}{4}$
Length of transverse axis $=2 a=2 \times \frac{15}{4}=\frac{15}{2}$

Topic: Mathematics, Hyperbola, NDA, NDA 2007 (Phase 1)
If the ellipse $9 \mathrm{x}^{2}+16 \mathrm{y}^{2}=144$ intercepts the line $3 \mathrm{x}+4 \mathrm{y}=12$. then what is the length of the chord so formed?

Options:
A 5 units
B 6 units
C 8 units
D 10 units

The correct answer is:
5 units

Solution:
Here,
$9 x^{2}+16 y^{2}=144$ and $3 x+4 y=12$
$\Rightarrow x=\frac{12-4 y}{3}$
$9\left(\frac{12-4 y}{3}\right)^{2}+16 y^{2}=144$
On solving we get, $y=0,3$ For $y=0 ; x=4$
For $y=3 ; x=0$
$\Rightarrow$ Length of chord $=\sqrt{(0-3)^{2}+(4-0)^{2}}=\sqrt{9+16}$
$=\sqrt{25}=5$ units

Topic: Mathematics, Ellipse, NDA, NDA 2016 (Phase 2)
If the equation $3 x^2-k x y-3 y^2=0$ represents the bisectors of angles between the lines $x^2-3 x y-4 y^2=0$, then value of $k$ is

Options:
A -6
B -10
C 6
D 10

The correct answer is:
-10

Solution:
We have $x^2-3 x y-4 y^2=0$ and comparing it with standard equation, we write
$$
\mathrm{A}=1, \mathrm{H}=\frac{-3}{2}, \mathrm{~B}=-4
$$
Equation of bisector of angle of this line is
$$
\begin{aligned}
& \frac{x^2-y^2}{A-B}=\frac{x y}{H} \Rightarrow \frac{x^2-y^2}{1+4}=\frac{x y}{\left(\frac{-3}{2}\right)} \\
& \therefore-3 x^2+3 y^2=10 x y \Rightarrow 3 x^2+10 x y-3 y^2=0
\end{aligned}
$$
Comparing with given equation, we get $\mathrm{k}=-10$

Topic: Mathematics, Pair of Lines, MHT CET, MHT CET 2021 (20 Sep Shift 2)
If the equation $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$ represents a circle passing through the origin, then

Options:
A $a=b, c=0$
B $|a|=|b|, h=0=c$
C $a=b, h=c=0$
D $a=b, h=0$

The correct answer is:
$a=b, h=c=0$

Solution:
The standard equation of circle which passes through origin is $x^2+y^2+2 g x+2 f y=0$
If the equation $a x^2+b y^2+2 h x y+2 f y+c=0$ represent a circle passing through origin, then $a=b, h=0, c=0$.

Topic: Mathematics, Circle, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 2)
If the equation $x^{2}-b x+1=0$ does not possess real roots, then which one of the following is correct?

Options:
A $-3 < b < 3$
B $-2 < b < 2$
C $\bar{b} \geq 2$
D $b < -2$

The correct answer is:
$-2 < b < 2$

Solution:
Given quadratic equation is $x^{2}-b x+1=0$ It has no real roots. It means, equation has imaginary roots. Which is possible when $B^{2}-4 A C < 0$ Here, $B=-b, A=1, C=1$
$\Rightarrow b^{2}-4 < 0 \Rightarrow b^{2} < 4 \Rightarrow-2 < b < 2$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 1)
If the equation $x^{2}+k^{2}=2(k+1) x$ has equal roots, then what is the value of $\mathrm{k} ?$

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

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

Solution:
$x^{2}+k^{2}=2(k+1) x$
$\Rightarrow x^{2}-2(k+1) x+k^{2}=0$
For roots to beequal discriminant $=0$ So, $\{-2(k+1)\}^{2}-4 k^{2}=0$
or $4(k+1)^{2}-4 k^{2}=0$
or $(\mathrm{k}+1)^{2}-\mathrm{k}^{2}=0$
or, $2 \mathrm{k}+1=0$
$k=-\frac{1}{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2007 (Phase 1)
If the equation $x^{2}+k x+1=0$ has the roots $\alpha$ and $\beta$, then what is the value of $(\alpha+\beta) \times\left(\alpha^{-1}+\beta^{-1}\right) ?$

Options:
A $\mathrm{k}^{2}$
B $\frac{1}{\mathrm{k}^{2}}$
C $2 \mathrm{k}^{2}$
D $\frac{1}{\left(2 \mathrm{k}^{2}\right)}$

The correct answer is:
$\mathrm{k}^{2}$

Solution:
As given: Roots of the equations $x^{2}+k x+1=0$ are $\alpha$ and $\beta .$ Given expression $(\alpha+\beta)\left(\alpha^{-1}+\beta^{-1}\right)=(\alpha+\beta)\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$
$=(\alpha+\beta)\left(\frac{\alpha+\beta}{\alpha \beta}\right)=\frac{(\alpha+\beta)^{2}}{\alpha \beta}=\frac{(-k)^{2}}{1}=\mathrm{k}^{2}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2008 (Phase 1)
If the equations $x^{2}+k x+64=0$ and $x^{2}-8 x+k=0$ have real roots, then what is the value of $k ?$

Options:
A 4
B 12
C 8
D 16

The correct answer is:
16

Solution:
Givenequations are $x^{2}+k x+64=0$ and $x^{2}-8 x+k=0$...(i)
Since both the eqns have real roots, discriminant $\geq 0$...(ii)
$=b^{2} \geq 4 a c$
from eq $^{\mathrm{n}}(\mathrm{i})$, we have
$k^{2} \geq 4(64) \Rightarrow k^{2} \geq 256$
$\Rightarrow k \geq 16$...(A)
and from $\mathrm{eq}^{\mathrm{n}}$ (ii), we have
$\Rightarrow \quad 64 \geq 4 k \Rightarrow 4 k \leq 64$
$\Rightarrow k \leq 16$$\ldots(\mathrm{B})$
Hence, from eq $^{\mathrm{n}}(\mathrm{A})$ and $(\mathrm{B})$, we have $\mathrm{k}=16$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2010 (Phase 2)
If the equations $x^{2}-p x+q=0$ and $x^{2}-a x+b=0$ have a common root and the roots of the second equation are equal, then which one of the following is correct?

Options:
A $a q=2(b+p)$
B $a q=(b+p)$
C $a p=2(b+q)$
D $a p=b+q$

The correct answer is:
$a p=2(b+q)$

Solution:
Given equations are $x^{2}-p x+q=0$...(1)
and $x^{2}-a x+b=0$...(2)
Root of second equation is
$x=\frac{a \pm \sqrt{a^{2}-4 b}}{2}$
Since, Roots of second equation are equal discriminant
$=0$
$\sqrt{a^{2}-4 b}=0 \Rightarrow a^{2}=4 b$...(3)
$x=\frac{a}{2}$
Since, Equations (1) and (2) have common roots.
$x=\frac{a}{2}$ is the root of equation (1) also.
Thus, $x=\frac{a}{2}$ satisfies (1)
$\Rightarrow\left(\frac{a}{2}\right)^{2}-p\left(\frac{a}{2}\right)+q=0$
$\Rightarrow \frac{a^{2}}{4}=\frac{p a}{2}-q \Rightarrow 2 b=p a-2 q \quad($ from $(3)$
$\Rightarrow a p=2(b+q)$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 2)
If the equilibrium constant (K) is measured directly in the laboratory, what can be calculated?

Options:
A The value of ∆H°
B The value of ∆S°
C The value of ∆G° at any other temperature
D The value of ∆G° at standard conditions

The correct answer is:
The value of ∆G° at any other temperature

Solution:
If the equilibrium constant (K) is measured directly in the laboratory, the value of ∆G° at any other temperature can be calculated using the relationship between ∆rG° and K.

Topic: Chemistry, Thermodynamics (C), NEET
If the \(E_{\text {cell }}^{\circ}\) of an equilibrium reaction \(A(s)+2 B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+2 B(s)\)
at \(298 \mathrm{~K}\) is \(0.59 \mathrm{~V}\), the equilibrium constant \(K_c\) is

Options:
A \(1.0 \times 10^{10}\)
B \(1.0 \times 10^2\)
C \(1.0 \times 10^{-20}\)
D \(1.0 \times 10^{20}\)

The correct answer is:
\(1.0 \times 10^{20}\)

Solution:
For the reaction,
\(\begin{gathered}
A+2 B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+2 B(s) \\
\Delta G^{\circ}=-R T \ln K_c=-n F E_{\text {cell }}^{\circ} \quad \ldots (i)
\end{gathered}\)
where,
\(\begin{aligned}
\Delta G^{\circ} & =\text { Gibbs free energy } \\
n & =\text { number of electrons involved }(=2) \\
F & =\text { Faraday's constant }=96500 \mathrm{C} \text { charge. } \\
E^{\circ} & =\text { standard electrode cell potential }(=0.59 \mathrm{~V}) \\
R & =\text { gas constant } \\
T & =\text { temperature }(=298 \mathrm{~K})
\end{aligned}\)
\(\therefore\) From Eq. (i), we have
\(\begin{aligned}
\text {or } \log K_c & =\frac{n F E^{\circ}}{2.303 R T} \\
& =\frac{2 \times 96500 \times 0.59}{2303 \times 8.314 \times 298}=-19.96 \approx 20 \\
\therefore \quad K_c & =1.0 \times 10^{20}
\end{aligned}\)
Hence, option (c) is the correct answer.
Alternatively
At, \(T=298 \mathrm{~K}\)
\(\ln \quad K_c=\frac{n E_{\text {ccll }}^{\circ}}{0.059}=\frac{2 \times 0.59}{0.059}=20\)
\(\therefore \quad K_c=1 \times 10^{20}\)

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)
If the extremities of a diameter of a circle are $(0,0)$ and
$\left(\mathrm{a}^{3}, 1 / \mathrm{a}^{3}\right)$, then the circle passes through which one of the
following points?

Options:
A $\left(a^{2}, 1 / a^{2}\right)$
B $(\mathrm{a}, \mathrm{l} / \mathrm{a})$
C $(\mathrm{a},-\mathrm{a})$
D $(1 / a, a)$

The correct answer is:
$(1 / a, a)$

Solution:
Given the extremities of a diameter of a circle as
$(0,0)$ and $\left(\mathrm{a}^{3}, \frac{1}{\mathrm{a}}\right)$, equation of circle is
$(x-0)\left(x-a^{3}\right)+(y-0)\left(y-\frac{1}{a^{3}}\right)=0$
$\Rightarrow \quad x^{2}-x a^{3}+y^{2}-\frac{y}{a^{3}}=0$
$\Rightarrow \quad x^{2}+y^{2}-x a^{3}-\frac{y}{a^{3}}=0$
Putting $\mathrm{x}=\frac{1}{\mathrm{a}}$ and $\mathrm{y}=\mathrm{a}$, the equation is satisfied.
Thus, the circle passes through the point $\left(\frac{1}{\mathrm{a}}, \mathrm{a}\right)$.

Topic: Mathematics, Circle, NDA, NDA 2006 (Phase 2)