datavorous/entrance-exam-dataset huggingface dataset:
Two plants can be conclusively said to belong to the same species if they

Options:
A Have more than 90 percent similar genes
B Look similar and possess identical secondary metabolites
C Have same number of chromosomes
D Can reproduce freely with each other and form seeds.

The correct answer is:
Can reproduce freely with each other and form seeds.

Solution:
When the members of a species are inter-fertile and produce fertile offsprings.

Topic: Biology, The Living World, NEET
Two poles are $10 \mathrm{~m}$ and $20 \mathrm{~m}$ high. The line joining their tips makes an angle of $15^{\circ}$ with the horizontal. What is the distance between the poles? $\quad$

Options:
A $10(\sqrt{3}-1) \mathrm{m}$
B $5(4+2 \sqrt{3}) \mathrm{m}$
C $20(\sqrt{3}+1) \mathrm{m}$
D $10(\sqrt{3}+1) \mathrm{m}$

The correct answer is:
$5(4+2 \sqrt{3}) \mathrm{m}$

Solution:
Let $\mathrm{AB}$ and $\mathrm{CD}$ be two poles of height $10 \mathrm{~m}$ and $20 \mathrm{~m}$ respectively. $\operatorname{In} \Delta \mathrm{AEC}$
$\frac{\mathrm{CE}}{\mathrm{AE}}=\tan 15^{\circ}$
$\frac{10}{\mathrm{AE}}=\tan \left(45^{\circ}-30^{\circ}\right)$
$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}$
$=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$\frac{10}{\mathrm{AE}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\mathrm{AE}=\frac{10(\sqrt{3}+1)}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=5(4+2 \sqrt{3})$

Topic: Mathematics, Heights and Distances, NDA, NDA 2012 (Phase 1)
Two poles are $10 \mathrm{~m}$ and $20 \mathrm{~m}$ high. The line joining their tops makes an angle of $15^{\circ}$ with the horizontal. The distance between the poles is approximately equal to

Options:
A $36.3 \mathrm{~m}$
B $37.3$ in
C $38.3 \mathrm{~m}$
D $39.3$ in

The correct answer is:
$37.3$ in

Solution:
$\tan 15^{\circ}=\mathrm{DE} / \mathrm{AE}$
$\mathrm{AE}=10 \cot 15^{\circ}$
$\cot \left(15^{\circ}\right)=\cot \left(45-30^{\circ}\right)$
$=\frac{\cot 45^{\circ} \cot 30^{\circ}+1}{\cot 30^{\circ}-\cot 45^{\circ}}$
$\cot 15^{\circ}=\frac{1 \cdot \sqrt{3}+1}{\sqrt{3}-1}=2+\sqrt{3}$
Putting $\cot 15^{\circ}$ in eq (1) $\begin{aligned} \mathrm{AE} &=10 \cot 15^{\circ} \\ &=10(2+\sqrt{3}) \\ &=10(3.73)=37.3 \mathrm{~m} \end{aligned}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2015 (Phase 2)
Two poles are 10 mand 20 mhigh. The line joining their tops makes an angle of $15^{\circ}$ with the horizontal. What is the approximate distance between the poles?

Options:
A $35.3 \mathrm{~m}$
B $37.3 \mathrm{~m}$
C $41 \mathrm{~m}$
D $44 \mathrm{~m}$

The correct answer is:
$37.3 \mathrm{~m}$

Solution:
Let $\mathrm{AC}$ and $\mathrm{ED}$ be two poles of height $20 \mathrm{~m}$ and $10 \mathrm{~m}$ respectively.
Let $\angle \mathrm{AEB}=15^{\circ}$
To find $: \mathrm{DC}$.
Now, $\mathrm{AB}=\mathrm{AC}-\mathrm{BC}=\mathrm{AC}-\mathrm{ED}=20-10=10 \mathrm{~m}$
Now in $\triangle \mathrm{ABE}, \quad(\because \mathrm{BC}=\mathrm{ED})$
$\tan 15^{\circ}=\frac{\mathrm{AB}}{\mathrm{BE}}$
$\Rightarrow \tan \left(45^{\circ}-30^{\circ}\right)=\frac{10}{\mathrm{BE}}$
$\Rightarrow \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{10}{\mathrm{BE}}$
$\left(\because \tan (\mathrm{A}-\mathrm{B})=\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A} \tan \mathrm{B}}\right)$
$\Rightarrow \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{10}{\mathrm{BE}}$
$\Rightarrow \quad \mathrm{BE}=10\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)=\frac{10 \times(\sqrt{3}+1)^{2}}{2}$
(Byrationalizing)
$5(3+1+2 \sqrt{3})=5(4+2 \times 1.73)=5(4+3.46)$
$\Rightarrow \mathrm{CD}=\mathrm{BE}=5 \times 7.46=37.3 \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2010 (Phase 2)
Two sets each of 20 observations, have the same standard deviation 5 . The first set has a mean 17 and the second a mean 22. Determine the standatd deviation of the set obtained by combining the given two sets.

Options:

None

Solution:
We have, $n_1=20, \sigma_1=5, \bar{x}_1=17$
$$
\begin{aligned}
n_2 &=20, \sigma_2=5, \bar{x}_2=22 \\
\bar{x}_{12} &=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2} \\
&=\frac{(20)(17)+(20)(22)}{20+20}=\frac{340+440}{40}=19.5 \\
d_1 &=\bar{x}_1-\bar{x}_{12}=17-19.5=-2.5
\end{aligned}
$$
$$
\begin{aligned}
d_2 &=\bar{x}_2-\bar{x}_{12}=22-19.5=2.5 \\
\sigma_{12} &=\sqrt{\frac{n_1\left(\sigma_1^2+d_1^2\right)+n_2\left(\sigma_2^2+d_2^2\right)}{n_1+n_2}} \\
&=\sqrt{\frac{20(25+6.25)+20(25+6.25)}{20+20}} \\
&=\sqrt{\frac{1250}{40}}=\sqrt{31.25}=5.59
\end{aligned}
$$

Topic: Mathematics, Statistics
Two similar boxes $B_{i}(\mathrm{i}=1,2)$ contain $(i+1)$ red and $(5-i-1)$ black balls. One box is chosen at random and two balls are drawn randomly. What is the probability that both the balls are of different colours?

Options:
A $\frac{1}{2}$
B $\frac{3}{10}$
C $\frac{2}{5}$
D $\frac{3}{5}$

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

Solution:
No Solution

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 1)
Two straight line paths are represented by the equation $2 \mathrm{x}-\mathrm{y}=2$ and $-4 \mathrm{x}+2 \mathrm{y}=6$. Then the paths will

Options:
A cross each other at one point
B not cross each other
C cross each other at two points
D cross each other at infinitely many points

The correct answer is:
not cross each other

Solution:
Given lines are $2 \mathrm{x}-\mathrm{y}=2$ and $2 \mathrm{x}-\mathrm{y}=-3$ since,
Both lines are parallel so they will never meet.

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 1)
Two straight lines passing through the point $\mathrm{A}(3,2)$ cut the line $2 \mathrm{y}=\mathrm{x}+3$ and $\mathrm{x}$ -axis perpendicularly at $\mathrm{P}$ and $\mathrm{Q}$ respectively. The equation of the line PQ is

Options:
A $7 x+y-21=0$
B $x+7 y+21=0$
C $2 x+y-8=0$
D $x+2 y+8=0$

The correct answer is:
$7 x+y-21=0$

Solution:
$\because$ Coordinates of $Q$ are $(3,0) \&$ it passes through $P Q$.
$\therefore$ Putting the values of $(x=3) \&(y=0)$ in options we get:
Equation of line $\mathrm{PQ}=7 \mathrm{x}+\mathrm{y}-21=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 2)
Two straight lines $x-3 y-2=0$ and $2 x-6 y-6=0$

Options:
A never intersect
B intersect at a single point
C intersect at infinite number of points
D intersect at more than one point (but finite number of points)

The correct answer is:
never intersect

Solution:
Given equation of straight lines are $x-3 y-2=0$ and $2 x$ $-6 y-6=0$
Here, $a_{1}=1, a_{2}=2, b_{1}=-3, b_{2}=-6$,
$c_{1}=-2, c_{2}=-6$
Now, $\frac{a_{1}}{a_{2}}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{1}{3}$
$\Rightarrow \quad \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\therefore \quad$ Both straight lines never intersect.

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
Two strings $A$ and $B$ of lengths, $L_A=80 \mathrm{~cm}$ and $L_B=x \mathrm{~cm}$ respectively are used separately in a sonometer. The ratio of their densities $\left(d_A / d_B\right)$ is 0.81 . The diameter of $B$ is one-half that of $A$. If the strings have the same tension and fundamental frequency the value of $x$ is :

Options:
A 33
B 102
C 144
D 130

The correct answer is:
144

Solution:
Given, $\frac{T_A}{T_B}=1, \quad \frac{L_A}{L_B}=\frac{80}{x}$
$\frac{D_A}{D_B}=\frac{2}{1}, \quad \frac{d_A}{d_B}=\frac{0.81}{1}$
Let $\mu_1$ and $\mu_2$ be the linear densities.
$\therefore \quad \frac{\mu_A}{\mu_B}=\left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B}=\left(\frac{2}{1}\right)^2 \times 0.81$
$=4 \times 0.81=3.24$
$\therefore \quad \frac{f_1}{f_2}=\frac{L_B}{L_A} \times \sqrt{\frac{T_A}{T_B} \times \frac{\mu_B}{\mu_{\mathrm{A}}}}$
$1=\frac{x}{80} \times \sqrt{1 \times \frac{1}{3.24}}$
$\therefore \quad x=80 \times \sqrt{3.24}$
or $\quad x=144$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2006
Two strings $A$ and $B$ of lengths, $L_A=80 \mathrm{~cm}$ and $L_B=x \mathrm{~cm}$ respectively are used separately in a sonometer. The ratio of their densities $\left(d_A / d_B\right)$ is 0.81 . The diameter of $B$ is one-half that of $A$. If the strings have the same tension and fundamental frequency the value of $x$ is :

Options:
A 33
B 102
C 144
D 130

The correct answer is:
144

Solution:
Given, $\frac{T_A}{T_B}=1, \quad \frac{L_A}{L_B}=\frac{80}{x}$
$\frac{D_A}{D_B}=\frac{2}{1}, \quad \frac{d_A}{d_B}=\frac{0.81}{1}$
Let $\mu_1$ and $\mu_2$ be the linear densities.
$\therefore \quad \frac{\mu_A}{\mu_B}=\left(\frac{D_A}{D_B}\right)^2 \times \frac{d_A}{d_B}=\left(\frac{2}{1}\right)^2 \times 0.81$
$=4 \times 0.81=3.24$
$\therefore \quad \frac{f_1}{f_2}=\frac{L_B}{L_A} \times \sqrt{\frac{T_A}{T_B} \times \frac{\mu_B}{\mu_{\mathrm{A}}}}$
$1=\frac{x}{80} \times \sqrt{1 \times \frac{1}{3.24}}$
$\therefore \quad x=80 \times \sqrt{3.24}$
or $\quad x=144$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2006
Two strings A and B of lengths, LA=80 cm and LB=x cm respectively are used separately in a sonometer. The ratio of their densities is 0.81. the diameter of B is one-half that of A. if the strings have the same tension and fundamental frequency the value of x is

Options:
A 33
B 102
C 144
D 130

The correct answer is:
144

Solution:
Given,
TATB=1, LALB=80x
DADB=21,dAdB=0.811
Let μ1and μ2 be the linear densities.
∴μAμB=DADB2×dAdB
=212×0.81
=4×0.81=3.24
∴ v1v2=LBLA×TATB×μBμA
1=x80×1×13.21
or x=144

Topic: Physics, Waves and Sound, NEET
Two strings of same material having lengths as 'L', '2L', and radii '2r', 'r' respectively, are vibrating in the fundamental mode. Tension applied to both the
strings is same. The ratio of their respective fundamental frequencies is

Options:
A $4: 3$
B $1: 2$
C $1: 1$
D $3: 4$

The correct answer is:
$1: 1$

Solution:
No Solution

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (12 Oct Shift 2)
Two students $X$ and $Y$ appeared in an examination. The probability that $\mathrm{X}$ will qualify the examination is $0.05$ and $\mathrm{Y}$ will qualify the examination is $0.10$. The probability that both will qualify the examination is $0.02$. What is the probability that only one of them will qualify the examination?

Options:
A $\quad 0.15$
B $0.14$
C $0.12$
D $0.11$

The correct answer is:
$0.11$

Solution:
Let $\mathrm{A}$ and $\mathrm{B}$ be the events $=\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})+\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{A}})$
$=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.05+0.1-2(0.02)$
$=0.15-0.04=0.11$
Hence the pobability that only one of them will qualify the examination is $0.11$.

Topic: Mathematics, Probability, NDA, NDA 2014 (Phase 2)
Two symmetric dice flipped with each dice having two sides painted red, two painted black, one painted yellow and the other painted white. What is the probability that both land on the same colour?

Options:
A $\frac{3}{18}$
B $\frac{2}{9}$
C $\frac{5}{18}$
D $\frac{1}{3}$

The correct answer is:
$\frac{5}{18}$

Solution:
Sides of dice $=\mathrm{R}, \mathrm{R}, \mathrm{B}, \mathrm{B}, \mathrm{Y}, \mathrm{W}$
Total events with dice $=6 \times 6=36$ Favourable events $={ }^{2} \mathrm{C}_{1} \cdot{ }^{2} \mathrm{C}_{1}+{ }^{2} \mathrm{C}_{1} \cdot{ }^{2} \mathrm{C}_{1}+{ }^{1} \mathrm{C}_{1} \cdot{ }^{1} \mathrm{C}_{1}+$
${ }^{1} \mathrm{C}_{1}{ }^{1} \mathrm{C}_{1}$
$=4+4+2+2=10$
$\therefore$ Probability $=\frac{10}{36}=\frac{5}{18}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
Two teams $A$ and $B$ have the same mean and their coefficients of variation are 4,2 , respectively. If $\sigma_A, \sigma_B$ are the standard deviations of teams $A, B$ respectively, then the relation between them is

Options:
A $\sigma_A=\sigma_B$
B $\sigma_B=2 \sigma_A$
C $\sigma_A=2 \sigma_B$
D $\sigma_B=4 \sigma_A$

The correct answer is:
$\sigma_A=2 \sigma_B$

Solution:
We have,
$$
\bar{X}_A=\bar{X}_B \text { and } \mathrm{CV}_A=4, \mathrm{CV}_B=2
$$
Let $\sigma_A$ and $\sigma_B$ be the standard deviations of teams $A$ and $B$ respectively.
Then,
$$
\mathrm{CV}=\frac{\sigma}{\bar{X}} \times 100 \text { or } \sigma=\frac{\mathrm{CV} \times \bar{X}}{100}
$$
$$
\therefore \quad \sigma_A=\frac{4 \bar{X}_A}{100} \text { and } \sigma_B=\frac{2 \bar{X}_B}{100}
$$
$$
\Rightarrow \quad \bar{X}_A=25 \sigma_A \text { and } \bar{X}_B=50 \sigma_B
$$
But $\quad \bar{X}_A=\bar{X}_B$
$$
\begin{array}{rlrl}
& \therefore & 25 \sigma_A & =50 \sigma_B \\
\Rightarrow & \sigma_A & =2 \sigma_B
\end{array}
$$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2015
Two transparent media A and B are separated by a plane boundary. The speed of light in those media are 1.5×108 m s-1 and 2.0×108 m s-1, respectively. The critical angle for a ray of light for these two media is:

Options:
A sin-10.750
B tan-10.500
C tan-10.750
D sin-10.500

The correct answer is:
sin-10.750

Solution:
Refractive index μ in terms of speed of light c in vacuum and wave speed v in medium is given by, μ=cv.

So, for medium A, μA=cvA=3×1081.5×108=2 and for medium B, μB=cvB=3×1082×108=1.5

For critical angle, angle of refraction r=90°.

Using Snell's law,

μ1sini=μ2sinr

⇒2sini=1.5sin90°⇒sini=1.52=0.750⇒i=sin-10.750

Topic: Physics, Ray Optics, NEET, NEET 2022 (Phase 1)
Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is :

Options:
A $1: 2$
B $1: 3$
C $1: 4$
D $1: 5$

The correct answer is:
$1: 3$

Solution:
Fundamental frequency of string,
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
where, $m=$ mass per unit length of the string
$=\pi r^2 \rho$
$\therefore \quad n=\frac{1}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}}$
For string $A$ first overtone,
$n_A=\frac{1}{l_1} \sqrt{\frac{T_1}{m_1}}=\frac{1}{l_1} \sqrt{\frac{T_1}{\pi r_1^2 \rho}}$
For string $B$ second overtone,
$n_B=\frac{3}{2 l_2} \sqrt{\frac{T_2}{\pi r_2^2 \rho_2}}$
According to question,
$n_A=n_B$
$\frac{1}{l_1} \sqrt{\frac{T_1}{\pi r_1^2 l_1}}=\frac{3}{2 l_2} \sqrt{\frac{T_2}{\pi r_2^2 l_2}}$
but $\quad T_1=T_2$
$\therefore \quad \frac{2 l_2}{3 l_1}=\sqrt{\frac{\pi r_1^2 \rho}{\pi r_2^2 \rho}}$
$\frac{4 l_2^2}{9 l_1^2}=\frac{r_1^2}{r_2^2}$
$\left(\frac{l_1}{l_2}\right)^2=\frac{4 r_2^2}{9 r_1^2}$
$\frac{2 l_2}{3 l_1}=\frac{r_1}{r_2}$
$\frac{l_2}{l_1}=\frac{3 r_1}{2 r_2}$
$\frac{l_1}{l_2}=\frac{2 r_2}{3 r_1}$
but $\quad r_1=2 r_2$
$\therefore \quad \frac{l_1}{l_2}=\frac{2 r_2}{3\left(2 r_2\right)}$
$l_1: l_2=1: 3$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2003
Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is :

Options:
A $1: 2$
B $1: 3$
C $1: 4$
D $1: 5$

The correct answer is:
$1: 3$

Solution:
Fundamental frequency of string,
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
where, $m=$ mass per unit length of the string
$=\pi r^2 \rho$
$\therefore \quad n=\frac{1}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}}$
For string $A$ first overtone,
$n_A=\frac{1}{l_1} \sqrt{\frac{T_1}{m_1}}=\frac{1}{l_1} \sqrt{\frac{T_1}{\pi r_1^2 \rho}}$
For string $B$ second overtone,
$n_B=\frac{3}{2 l_2} \sqrt{\frac{T_2}{\pi r_2^2 \rho_2}}$
According to question,
$n_A=n_B$
$\frac{1}{l_1} \sqrt{\frac{T_1}{\pi r_1^2 l_1}}=\frac{3}{2 l_2} \sqrt{\frac{T_2}{\pi r_2^2 l_2}}$
but $\quad T_1=T_2$
$\therefore \quad \frac{2 l_2}{3 l_1}=\sqrt{\frac{\pi r_1^2 \rho}{\pi r_2^2 \rho}}$
$\frac{4 l_2^2}{9 l_1^2}=\frac{r_1^2}{r_2^2}$
$\left(\frac{l_1}{l_2}\right)^2=\frac{4 r_2^2}{9 r_1^2}$
$\frac{2 l_2}{3 l_1}=\frac{r_1}{r_2}$
$\frac{l_2}{l_1}=\frac{3 r_1}{2 r_2}$
$\frac{l_1}{l_2}=\frac{2 r_2}{3 r_1}$
but $\quad r_1=2 r_2$
$\therefore \quad \frac{l_1}{l_2}=\frac{2 r_2}{3\left(2 r_2\right)}$
$l_1: l_2=1: 3$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2003
Two uniform wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the $2^{\text {nd }}$ overtone of $2^{\text {nd }}$ wire and radius of the first wire is twice the radius of the $2^{\text {nd }}$ wire then the ratio of length of first wire to $2^{\text {nd }}$ wire is

Options:
A $1: 3$
B $3: 1$
C $1: 9$
D $9: 1$

The correct answer is:
$1: 3$

Solution:
Fundamental frequency of the first wire is
$\mathrm{n}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}_1^2 \rho}}=\frac{1}{2 l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
The first overtone $\mathrm{n}_1=2 \mathrm{n}=\frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
Similarly, the second overtone of the second wire will be,
$\mathrm{n}_2=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}}$
Given that $\mathrm{n}_1=\mathrm{n}_2$
$\begin{aligned}
& \therefore \quad \frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi \rho}}=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi \rho}} \\
& \therefore \quad 3 l_1 \mathrm{r}_1=2 l_2 \mathrm{r}_2 \\
& \\
& \quad \frac{l_1}{l_2}=\frac{2 \mathrm{r}_2}{3 \mathrm{r}_1}
\end{aligned}$
$=\frac{2 r_2}{3\left(2 r_2\right)} \quad \ldots .\left(\because r_1=2 r_2\right)$
$=\frac{1}{3}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (10 May Shift 2)
Two uniform wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and radius of the first wire is twice the radius of the second wire then the ratio of the lengths of the first wire to second wire is

Options:
A 13
B 14
C 15
D 16

The correct answer is:
13

Solution:
Fundamental frequency of the first wire
f=12L1 Tm =12L1TÏr1 2Ï =12L1r1TÏÏ
The first overtone f1=2f=22L1r1TÏÏ =1L1r1TÏÏ
The second overtone of the second wire
f2=32L2r2TÏÏ
f1=f2
1L1r1TÏÏ =32L2 r2 TÏÏ
â´ 3L1r1=2L2r2
L1L2=23.r2r1
=23.r22r2 [â´ r1=2r2]
=13

Topic: Physics, Waves and Sound, NEET
Two uniform wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and radius of the first wire is twice the radius of the second wire then the ratio of the lengths of the first wire to second wire is

Options:
A 13
B 14
C 15
D 16

The correct answer is:
13

Solution:
Fundamental frequency of the first wire
f=12L1 Tm =12L1TÏr1 2Ï =12L1r1TÏÏ
The first overtone f1=2f=22L1r1TÏÏ =1L1r1TÏÏ
The second overtone of the second wire
f2=32L2r2TÏÏ
f1=f2
1L1r1TÏÏ =32L2 r2 TÏÏ
â´ 3L1r1=2L2r2
L1L2=23.r2r1
=23.r22r2 [â´ r1=2r2]
=13

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2017
Two variates, $\mathrm{x}$ and $\mathrm{y}$, are uncorrelated and have standard
deviations $\sigma_{x}$ and $\sigma_{y}$ respectively. What is the correlation coefficient between $x+y$ and $x-y ?$

Options:
A $\frac{\sigma_{x} \sigma_{y}}{\sigma_{x}^{2}+\sigma_{y}^{2}}$
B $\frac{\sigma_{x}+\sigma_{y}}{2 \sigma_{x} \sigma_{y}}$
C $\frac{\sigma_{x}^{2}-\sigma_{y}^{2}}{\sigma_{x}^{2}+\sigma_{y}^{2}}$
D $\frac{\sigma_{y}-\sigma_{x}}{\sigma_{x} \sigma_{y}}$

The correct answer is:
$\frac{\sigma_{x}^{2}-\sigma_{y}^{2}}{\sigma_{x}^{2}+\sigma_{y}^{2}}$

Solution:
Let $u=(x+y) ; v=(x-y)$
$\therefore \bar{u}=(\bar{x}+\bar{y}) ; \bar{v}=(\bar{x}-\bar{y})$
$\operatorname{cov}(u, v)=E\{(u-\bar{u})(v-\bar{v})\}$
$=E\{(x-\bar{x})+(y-\bar{y})\} \cdot\{(x-\bar{x})-(y-\bar{y})\}$
$=E\left\{(x-\bar{x})^{2}-(y-\bar{y})^{2}\right\}=\sigma_{x}^{2}-\sigma_{y}^{2}$
$\operatorname{var}(u)=E(u-\bar{u})^{2}=E\{(x-\bar{x})+(y-\bar{y})\}^{2}=\sigma_{x}^{2}+\sigma_{y}^{2}$
Therefore $x$ and $y$ are uncorrelated. $E(x-\bar{x})(y-\bar{y})=0$
Similarly, $\operatorname{var}(v)=\sigma_{x}^{2}+\sigma_{y}^{2}$
Thus, $r(u, v)=\frac{\operatorname{cov}(u, v)}{\sqrt{\operatorname{var}(u) \cdot \operatorname{var}(v)}}=\frac{\sigma_{x}^{2}-\sigma_{y}^{2}}{\sigma_{x}^{2}+\sigma_{y}^{2}}$

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 2)
Two vector $2 \hat{\mathbf{i}}+m \hat{\mathbf{j}}-3 n \hat{\mathbf{k}}$ and $5 \hat{\mathbf{i}}+3 m \hat{\mathbf{j}}+n \hat{\mathbf{k}}$ are such
that their magnitudes are respectively $\sqrt{14}$ and $\sqrt{35}$, where
$\mathrm{m}, \mathrm{n}$ are integers. Which one of the following is correct?

Options:
A $\mathrm{m}$ takes 1 value, $\mathrm{n}$ takes 1 value
B $\mathrm{m}$ takes 1 value, $\mathrm{n}$ takes 2 values
C $\mathrm{m}$ takes 2 value, $\mathrm{n}$ takes 1 value
D $\mathrm{m}$ takes 2 value, $\mathrm{n}$ takes 2 values

The correct answer is:
$\mathrm{m}$ takes 2 value, $\mathrm{n}$ takes 2 values

Solution:
Vectors are $2 \mathrm{i}+\mathrm{mj}-3 \mathrm{nk}$ and $5 \mathrm{i}+3 \mathrm{mj}+\mathrm{nk}$
$|2 \mathrm{i}+\mathrm{mj}-3 \mathrm{nk}|=\sqrt{14}$
and $|5 \mathrm{i}+3 \mathrm{mj}+\mathrm{nk}|=\sqrt{35}$
$\sqrt{2^{2}+m^{2}+(-3 n)^{2}}=\sqrt{14}$
or, $4+\mathrm{m}^{2}+9 \mathrm{n}^{2}=14$
or, $\mathrm{m}^{2}+9 \mathrm{n}^{2}=10$
From (ii)
$\sqrt{5^{2}+(3 m)^{2}+n^{2}}=\sqrt{35}$
or, $25+9 \mathrm{~m}^{2}+\mathrm{n}^{2}=35$
or, $9 \mathrm{~m}^{2}+\mathrm{n}^{2}=10$
From (iii) and (iv) $m^{2}+9 n^{2}=9 m^{2}+n^{2}$
or, $8 \mathrm{n}^{2}=8 \mathrm{~m}^{2}$
or, $\mathrm{n}^{2}=\mathrm{m}^{2}$
$\mathrm{n}=\pm \mathrm{m}$
n takes 2 values and $\mathrm{m}$ takes 2 values.

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
Two vectors $\vec{a}$ and $\overrightarrow{\mathrm{b}}$ are non-zero and non-collinear. What
is the value of $\mathrm{x}$ for which the vectors $\overrightarrow{\mathrm{p}}=(\mathrm{x}-2) \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ and
$\overrightarrow{\mathrm{q}}=(\mathrm{x}+1) \overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$ are collinear?

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

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

Solution:
Since, $\overrightarrow{\mathrm{p}}$ and $\overrightarrow{\mathrm{q}}$ are collinear, then
$\overrightarrow{\mathrm{p}}=\mathrm{k} \overrightarrow{\mathrm{q}}$
[where $\mathrm{k}$ is a scalar $]$
$\Rightarrow(\mathrm{x}-2) \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\mathrm{k}(\mathrm{x}+1) \overrightarrow{\mathrm{a}}-\mathrm{k} \overrightarrow{\mathrm{b}}$
On equating the coefficients $x-2=k(x+1)$ and $-k=1$
putting value of $\mathrm{k}$ we get, $\mathrm{x}-2=-(\mathrm{x}+1)$
$\Rightarrow 2 \mathrm{x}=1$
$\Rightarrow \mathrm{x}=\frac{1}{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2007 (Phase 1)
Two vibrating strings of the same material but length L and 2L have radii 2r and r respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length L with frequency Ï1 and the other with frequency Ï2. The ratio Ï1Ï2 is given by

Options:
A 1
B 12
C 32
D 2

The correct answer is:
1

Solution:
Fundamental frequency f=12l⁡Tμ
∴ f1f2=l⁡1l⁡2μ2μ1 (since tension is same)

=2LLπr2ρπ·4r2ρ, (since the wires are of the same material)

Hence the desired ratio is 1

Topic: Physics, Waves and Sound, NEET
Two vibrating strings of the same material but lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $v_1$ and the other with frequency $v_2$. The ratio $v_1 / v_2$ is

Options:
A 2
B 4
C 8
D 1

The correct answer is:
1

Solution:
Fundamental frequency,
$\begin{array}{ll}
& v=\frac{1}{2 l} \sqrt{\frac{T}{m}}=\frac{1}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}} \\
\text { or } & v \propto \frac{1}{l r} \\
\therefore & \frac{v_1}{v_2}=\frac{l_2 r_2}{l_1 r_1}=\frac{2 L \times r}{L \times 2 r}=1
\end{array}$

Topic: Physics, Waves and Sound, JIPMER, JIPMER 2010