datavorous/entrance-exam-dataset huggingface dataset:
A straight line intersects $\mathrm{x}$ and $\mathrm{y}$ axes at $\mathrm{P}$ and Qrespectively. If $(3,5)$ is the middle point of $\mathrm{PQ}$, then what is the area of the triangle OPQ?

Options:
A 12 square units
B 15 square units
C 20 square units
D 30 square units

The correct answer is:
30 square units

Solution:
As we know that line $P Q$ intersects $x$ -axis and $y$ -axis at $P$ and $Q$
$\because \mathrm{M}$ is the mid point of $\mathrm{PQ}$
$\therefore \frac{x+0}{2}=3$ and $\frac{0+y}{2}=5$
$\Rightarrow x=6$ and $y=10$
Hence area of triangle $O P Q=\frac{1}{2} \times 6 \times 10=30$ sq. unit

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 1)
A straight line is passing through the points represented by
the complex numbers a $+\mathrm{ib}$ and $\frac{1}{-\mathrm{a}+\mathrm{ib}}$, where $(\mathrm{a}, \mathrm{b}) \neq$ $(0,0)$
Which one of the following is correct?

Options:
A It passes through the origin
B It is parallel to the $\mathrm{x}$ -axis
C It is parallel to the y-axis
D It passes through $(0, b)$

The correct answer is:
It passes through the origin

Solution:
Complex numbers are : $(\mathrm{a}+\mathrm{ib})$ and $\frac{1}{-\mathrm{a}+\mathrm{ib}}$
Second number is rationalized as,
$\frac{-a-i b}{(-a+i b)(-a-i b)}=\frac{-a-i b}{a^{2}+b^{2}}$
These two complex numbers represent two points,
$(\mathrm{a}, \mathrm{b})$ and $\left(\frac{-\mathrm{a}}{\mathrm{a}^{2}+\mathrm{b}^{2}}, \frac{-\mathrm{b}}{\mathrm{a}^{2}+\mathrm{b}^{2}}\right)$
Eq. of line passing through these point is
$(\mathrm{y}-\mathrm{b})=\frac{-\frac{\mathrm{b}}{\mathrm{a}^{2}+\mathrm{b}^{2}}-\mathrm{b}}{\frac{-\mathrm{a}}{\mathrm{a}^{2}+\mathrm{b}^{2}}-\mathrm{a}}(\mathrm{x}-\mathrm{a})$
$\Rightarrow \mathrm{y}-\mathrm{b}=\frac{\mathrm{b}}{\mathrm{a}}(\mathrm{x}-\mathrm{a}) \Rightarrow \mathrm{ay}-\mathrm{ab}=\mathrm{bx}-\mathrm{ab}$
$\Rightarrow \mathrm{y}=\frac{\mathrm{b}}{\mathrm{a}} \mathrm{x}$
So, line passes through the origin.

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 1)
A straight line passes through $(1,-2,3)$ and perpendicular to the plane $2 x+3 y-z=7$
What are the direction ratios of normal to plane?

Options:
A $ < 2,3,-1>$
B $ < 2,3,1>$
C $ < -1,2,3>$
D None of these

The correct answer is:
$ < 2,3,-1>$

Solution:
Direction ratios of normal to plane $2 \mathrm{x}+3 \mathrm{y}-\mathrm{z}=7$ is $ < 2,3,-1>$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 1)
A straight line passes through $(1,-2,3)$ and perpendicular to the plane $2 x+3 y-z=7$
What is the image of the point $(1,-2,3)$ in the plane?

Options:
A $(2,-1,5)$
B $(-1,2,-3)$
C $(5,4,1)$
D None of these

The correct answer is:
$(5,4,1)$

Solution:
Let $\mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is the image of $(1,-2,3)$ in the plane
$\frac{x+1}{2}=3 \Rightarrow x=5$
$\frac{y-2}{2}=1 \Rightarrow y=4$
$\frac{\mathrm{z}+3}{2}=2 \Rightarrow \mathrm{z}=1$
Image of $(1,-2,3)$ are $(5,4,1)$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 1)
A straight line passes through $(1,-2,3)$ and perpendicular to the plane $2 x+3 y-z=7$
Where does the line meet the plane?

Options:
A $(2,3,-1)$
B $(1,2,3)$
C $(2,1,3)$
D $(3,1,2)$

The correct answer is:
$(3,1,2)$

Solution:
Equation of line, $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-1}$ Let $\mathrm{P}(2 \mathrm{r}+1,3 \mathrm{r}-2,-\mathrm{r}+3)$ of the line meets the plane. Then, $2(2 \mathrm{r}+1)+3(3 \mathrm{r}-2)-(-\mathrm{r}+3)=0$
$4 r+2+9 r-6+r-3=7$
$14 r=14$
$\mathbf{r}=1$
$\mathrm{P}(3,1,2)$ meets the plane.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2014 (Phase 1)
A straight line passes through the points $(5,0)$ and $(0,3)$. The length of the perpendicular from the point $(4,4)$ on the line is

Options:
A $\frac{\sqrt{17}}{2}$
B $\sqrt{\frac{17}{2}}$
C $\frac{15}{\sqrt{34}}$
D $\frac{17}{2}$

The correct answer is:
$\sqrt{\frac{17}{2}}$

Solution:
Suppose equation of line is
$\frac{x}{5}+\frac{y}{3}=1$
$\Rightarrow \quad 3 x+5 y-15=0$
Now, length of perpendicular from $(4,4)$ on $3 \mathrm{x}+5 \mathrm{y}-15$ $=0$ is
$P=\left|\frac{3.4+5.4-15}{\sqrt{34}}\right|=\left|\frac{17}{\sqrt{34}}\right|=\frac{\sqrt{17} \cdot \sqrt{17}}{\sqrt{17} \cdot \sqrt{2}}$
$P=\sqrt{\frac{17}{2}}$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 1)
A straight line with direction cosines $(0,1,0)$ is

Options:
A parallel to $\mathrm{x}$ -axis
B parallel to $y$ -axis
C parallel to z-axis
D equally inclined to all the axes

The correct answer is:
parallel to $y$ -axis

Solution:
$\mathrm{D} \cdot \mathrm{C}^{\prime} \mathrm{s}(0,1,0)$
Since $x$ and $z$ are zero, the straight line is parallel to $\mathrm{y}-$ axis

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 1)
A straight line $\mathrm{x}=\mathrm{y}+2$ touches the circle $4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{r}^{2}$. The value of r is $\quad[2015-I I]$

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

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

Solution:
$\because 4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{r}^{2}$
$\Rightarrow x^{2}+y^{2}=\left(\frac{r}{2}\right)^{2}$
Center $(0,0)$ and radius $\frac{\mathrm{r}}{2}$ Eq. of line is $; \mathrm{x}-\mathrm{y}-2=0$ $\because$ Line touches the circle.
$$
\therefore \frac{\mathrm{r}}{2}=\frac{|0-0-2|}{\sqrt{(1)^{2}+(-1)^{2}}}
$$
$\mathrm{r}=2 \sqrt{2}$

Topic: Mathematics, Circle, NDA, NDA 2015 (Phase 2)
A stretched uniform wire of length L under tension $\mathrm{T}$ is vibrating with frequency
'n'. A closed pipe of same length is also vibrating with same fundamental frequency
'n'. If $\mathrm{T}$ is increased by $16 \mathrm{~N}$, it is in resonance with $2^{\text {nd }}$ harmonic of same closed
pipe. The initial tension in the wire is

Options:
A $1 \mathrm{~N}$
B $2 \mathrm{~N}$
C $1 \cdot 5 \mathrm{~N}$
D $0 \cdot 5 \mathrm{~N}$

The correct answer is:
$2 \mathrm{~N}$

Solution:
$\mathrm{n}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{\mathrm{v}}{4 \ell}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}+16}{\mathrm{n}}}=\frac{3 \mathrm{v}}{4 \ell}$
$\sqrt{\frac{T}{T+16}}=\frac{1}{3}$
$\frac{T}{T+16}=\frac{1}{9}$
$9 \mathrm{T}=\mathrm{T}+16 \quad \therefore 8 \mathrm{T}=16 \quad \therefore \mathrm{T}=2 \mathrm{~N}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (20 Oct Shift 1)
A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is

Options:
A 320  m s-1,  120  Hz
B 320â m s-1,ââ80 âHz
C 180  m s-1,  80  Hz
D 180  m s-1,  120  Hz

The correct answer is:
320â m s-1,ââ80 âHz

Solution:
For a string fixed at both the ends,
f=n2Lv
For third harmonic, 240=32(2)v
⇒v=240×43
=320  m s-1
Fundamental frequency f0=12Lv
=14×320 = 80 Hz

Topic: Physics, Waves and Sound, NEET
A string has a length of 5 cm between fixed points and has a fundamental frequency of 20 Hz. What is the frequency of the second overtone?

Options:
A 40 Hz
B 50 Hz
C 60 Hz
D 30 Hz

The correct answer is:
60 Hz

Solution:
Given,
Length of string between fixed ends =5 m
Fundamental frequency f0=20 Hz
For second overtone, n=3
â´ f3=3f0
=3Ã20
=60 Hz

Topic: Physics, Waves and Sound, NEET
A string in a musical instrument is $50 \mathrm{~cm}$ long and its fundamental frequency is $800 \mathrm{~Hz}$. If a frequency of $1000 \mathrm{~Hz}$ is to be produced, then required length of string is

Options:
A $62.5 \mathrm{~cm}$
B $40 \mathrm{~cm}$
C $50 \mathrm{~cm}$
D $37.5 \mathrm{~cm}$.

The correct answer is:
$40 \mathrm{~cm}$

Solution:
Initial length of string $\left(l_1\right)=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Initial fundamental frequency $\left(n_1\right)=800 \mathrm{~Hz}$ and final frequency $\left(n_2\right)=1000 \mathrm{~Hz}$.
Frequency of vibration of the string, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}} \propto \frac{1}{l}$
$$
\therefore \quad \frac{n_1}{n_2}=\frac{l_2}{l_1} \quad \text { or } \quad l_2=\frac{n_1 \times l_1}{n_2}=\frac{800 \times 50}{1000}=40 \mathrm{~cm} \text {. }
$$
(where $l_2=$ final length of the string).

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2002
A string in a musical instrument is $50 \mathrm{~cm}$ long and its fundamental frequency is $800 \mathrm{~Hz}$. Keeping the tension applied to the string same, the change in the length to produce sound note of fundamental frequency $1000 \mathrm{~Hz}$ will be.

Options:
A $10 \mathrm{~cm}$
B $20 \mathrm{~cm}$
C $60 \mathrm{~cm}$
D $40 \mathrm{~cm}$

The correct answer is:
$40 \mathrm{~cm}$

Solution:
For stretched string $\mathrm{v} \propto \frac{1}{l}$
$\therefore \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{l_2}{l_1} \Rightarrow l_2=\frac{800}{1000} \times 50=40 \mathrm{~cm}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (07 Aug Shift 1)
A string is divided into three segments, so that the segment possesses fundamental frequencies in the ratio $1: 2: 3$. Then, the length of the segments are in the ratio

Options:
A $6: 3: 2$
B $4: 3: 2$
C $4: 2: 1$
D $3: 2: 1$

The correct answer is:
$6: 3: 2$

Solution:
Given, ratio of fundamental frequencies be $f_1: f_2: f_3=1: 2: 3$
Let ratio of length be $l_1: l_2: l_3$.
Since, $\quad f=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}$
where, $f$ is frequency,
$l$ is length of string,
$T$ is tension
and $\mu$ is mass per unit length.
$\therefore \quad l \propto \frac{1}{f}$
So, $l_1: l_2: l_3=\frac{1}{f_1}: \frac{1}{f_2}: \frac{1}{f_3}$
$=\frac{1}{1}: \frac{1}{2}: \frac{1}{3}$
$=\frac{6}{6}: \frac{3}{6}: \frac{2}{6}$
$=6: 3: 2$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
A string is vibrating in its fifth overtone between two rigid supports $2.4 \mathrm{~m}$ apart. The distance between successive node and antinode is

Options:
A $0.8 \mathrm{~m}$
B $0.2 \mathrm{~m}$
C $0.1 \mathrm{~m}$
D $0.6 \mathrm{~m}$

The correct answer is:
$0.2 \mathrm{~m}$

Solution:
The correct option is (B).
If the string is clamped between two supports separated by L.
For the fundamental overtone separation between successive nodes is $\mathrm{L}$.
For the first overtone separation between successive nodes is $\frac{L}{2}$.
For the second overtone the separation between successive nodes will be $\frac{L}{3}$.
So, for nth overtone the separation between successive nodes will be $\frac{\mathrm{L}}{(\mathrm{n}+1)}$.
Therefore, for the fifth over separation between successive nodes will be $\frac{\mathrm{L}}{6}$.
In the separation between successive node and antinode is half of the separation between two successive nodes.
So, $\frac{\mathrm{L}}{12}=\frac{2.4 \mathrm{~m}}{12}=0.2 \mathrm{~m}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (05 Aug Shift 1)
A string of length $100 \mathrm{~cm}$ has three resonant frequencies, $120 \mathrm{~Hz}, 200 \mathrm{~Hz}$ and $280 \mathrm{~Hz}$. If a node is formed at the end of the string, the speed of the transverse wave on this string is :

Options:
A $60 \mathrm{~m} / \mathrm{s}$
B $80 \mathrm{~m} / \mathrm{s}$
C $100 \mathrm{~m} / \mathrm{s}$
D $120 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$80 \mathrm{~m} / \mathrm{s}$

Solution:
Given, length of string, $l=100 \mathrm{~cm}=1 \mathrm{~m}$
Three resonant frequencies are given as $f_1=120 \mathrm{~Hz}, f_2=200 \mathrm{~Hz}, f_3=280 \mathrm{~Hz}$
$\because$ Fundamental frequency $\left(f_0\right)$ is equal to the greatest common factor of $f_1, f_2$ and $f_3$.
$\therefore f_0=$ greatest common factor of $120 \mathrm{~Hz}, 200 \mathrm{~Hz}$ and $280 \mathrm{~Hz}$ i.e., $f_0=40 \mathrm{~Hz}$
Now, we know that, speed of transverse wave is given by $v=2 l f_0=2 \times 1 \times 40=80 \mathrm{~m} / \mathrm{s}$.

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2019 (04 May Shift 2)
A structure that forms when the secondary coiled polypeptides are folded to produce a hollow, wollen-like structure is called___ structure of proteins.

Options:
A Primary
B Secondary
C Tertiary
D Quaternary

The correct answer is:
Tertiary

Solution:
A structure that forms when the secondary coiled polypeptides are folded to produce a hollow, wollen-like structure is called tertiary structure of proteins.

Topic: Biology, Biomolecules (B), NEET
A student appears for tests I, II and III. The student is considered successful if he passes in tests I, II or III or all the three. The probabilities of the student passing in tests I, II and III are $\mathrm{m}, \mathrm{n}$ and $1 / 2$ respectively. If the probability of the student to be successful is $1 / 2$, then which one of the following is correct?

Options:
A $\quad \mathrm{m}(1+\mathrm{n})=1$
B $\mathrm{n}(1+\mathrm{m})=1$
C $\mathrm{m}=1$
D $\mathrm{mn}=1$

The correct answer is:
$\quad \mathrm{m}(1+\mathrm{n})=1$

Solution:
$\frac{1}{2}=\mathrm{m} \times \mathrm{n} \times \frac{1}{2}+\mathrm{m} \times \frac{1}{2} \times(1-\mathrm{n})+\mathrm{m} \times \mathrm{n} \times 1-\frac{1}{2}$
$\Rightarrow 1=\mathrm{m}(\mathrm{n}+1)$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
A survey of 850 students in a University yields that 680 students like music and 215 like dance. What is the least number of students who like both music and dance?

Options:
A 40
B 45
C 50
D 55

The correct answer is:
45

Solution:
Number of students who like music, $\mathrm{n}(\mathrm{m})=680$ Number of students who like dance, $n(d)=215$ Total number of students, $n(m \cup \mathrm{d})=850$
$n(m \cup d)=n(m)+n(d)-n(m \cap d)$
$\Rightarrow 850=680+215-\mathrm{n}(\mathrm{m} \cap \mathrm{d})$
$\Rightarrow n(m \cap d)=895-850=45$
$0 < a < 1$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
A survey was conducted among 300 students. If was found that 125 students like to play cricket, 145 students like to play football and 90 students like to play tennis, 32 students like to play exactly two games out of the three games.
How many students like to play exactly only one game?

Options:
A 196
B 228
C 254
D 268

The correct answer is:
254

Solution:
Exactly one $=|\mathrm{A}|+|\mathrm{B}|+|\mathrm{C}|-$
$2[[\mathrm{~A} \cap \mathrm{B}|+| \mathrm{B} \cap \mathrm{C}|+| \mathrm{A} \cap \mathrm{C} \mid]+3|\mathrm{~A} \cap \mathrm{B} \cap \mathrm{C}|$
$=125+145+90-2[32+3 \times 14]+3 \times 14$
$=360-106=254$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 2)
A survey was conducted among 300 students. If was found that 125 students like to play cricket, 145 students like to play football and 90 students like to play tennis, 32 students like to play exactly two games out of the three games. How many students like to play all the three games?

Options:
A 14
B 21
C 28
D 35

The correct answer is:
14

Solution:
$\begin{aligned} 300=125+145+90-\\ &(|A \cap B|+|B \cap C|+|A \cap C|)+|A \cap B \cap C| \\ &|A \cap B|+|B \cap C|+|A \cap C| \\ &=60+|A \cap B \cap C|...(i) \end{aligned}$
Again, $|A \cap B|+|B \cap C|+|A \cap C|-3|A \cap B \cap C|=32$
$\begin{aligned} \Rightarrow|A \cap B|+\mid B & \cap C|+| A \cap C \mid \\ &=32+3|A \cap B \cap C|...(ii) \end{aligned}$
From (i) and (ii) $|A \cap B \cap C|=14$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 2)
A taxon, which is facing an extremely high risk of extinction in the wild in immediate future is known as

Options:
A Exotic
B Vulnerable
C Endangered
D Critically endangered

The correct answer is:
Critically endangered

Solution:
A taxon is critically endangered when it is facing an extremely high risk of extinction in the wild in the near future.

Topic: Biology, Biodiversity and its Conservation, NEET
A tea party is arranged for 16 people along two sides of a long table with eight chairs on each side. Four particular men wish to sit on one particular side and two particular men on the other side. The number of ways they can be seated is

Options:
A $24 \times 8 ! \times 8 !$
B $(81)^{3}$
C $210 \times 8 ! \times 8 !$
D $16 !$

The correct answer is:
$210 \times 8 ! \times 8 !$

Solution:
Number of ways $=\frac{8 ! 8 !}{4 ! 6 !} \times 10 !$
$=(8 !)^{2} \times \frac{10 !}{4 ! 6 !}$
$=(8 !)^{2} \times \frac{10 \times 9 \times 8 \times 7}{4 !}$
$=(8 !)^{2} \times \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$
$=(210) \times(8 !)^{2}$

Topic: Mathematics, Permutation Combination, NDA, NDA 2017 (Phase 2)
A team of 8 players is to be chosen from a group of 12 players. Out of the eight players one is to be elected as captain and another vice-captain. In how many ways can this be done?

Options:
A 27720
B 13860
C 6930
D 495

The correct answer is:
27720

Solution:
Total no. of players $=12$ No. of chosen players $=8$ Number of ways to choose 8 players from 12 players
$={ }^{12} C_{8}=\frac{12 !}{8 ! 4 !}=\frac{12 \times 11 \times 10 \times 9 \times 8 !}{8 ! 4 !}=495$
Since, out of the 8 players 1 is to be elected as captain and another vice-captain Therefore number of ways to choose a captain and a vice-captain
$={ }^{8} C_{1} \times{ }^{7} C_{1}=8 \times 7=56 \quad\left(\because{ }^{n} C_{1}=n\right)$
Hence, required number of ways $=495 \times 56=27720$

Topic: Mathematics, Permutation Combination, NDA, NDA 2010 (Phase 1)
A tertiary butyl carbocation is more stable than a secondary butyl carbocation because of which of the following?

Options:
A +R effect of –CH3 groups
B –R effect of –CH3 groups
C Hyperconjugation
D –I effect of –CH3 groups

The correct answer is:
Hyperconjugation

Solution:
The stability of given carbocations can be explained based on the hyperconjugation effect. The more the number of alpha hydrogen, the more is the stability of carbocations.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2020 (Phase 1)
A thin wire of length of $99 \mathrm{~cm}$ is fixed at both ends as shown in the figure. The wire is kept under a tension and is divided into three segments of lengths $l_1, l_2$ and $l_3$ as shown in figure. When the wire is made to vibrate, the segments vibrate respectively with their fundamental frequencies in the ratio $1: 2: 3$. Then, the lengths $l_1, l_2$ and $l_3$ of the segments respectively are (in cm)

Options:
A $27,54,18$
B $18,27,54$
C $54,27,18$
D $27,9,14$

The correct answer is:
$54,27,18$

Solution:
Ratio of fundamental frequency
$$
\left(n_1: n_2: n_3\right)=1: 2: 3
$$
$$
\begin{aligned}
l_1: l_2: l_3 & =\frac{1}{n_1}: \frac{1}{n_2}: \frac{1}{n_3} \\
& =\frac{1}{1}: \frac{1}{2}: \frac{1}{3}=6: 3: 2 \\
l_1 & =\frac{6 \times 99}{11}=54 \\
l_2 & =\frac{3 \times 99}{11}=27 \\
l_3 & =\frac{2 \times 99}{11}=18
\end{aligned}
$$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2015
A tower of height $15 \mathrm{~m}$ stands vertically on the ground. From a point on the ground the angle of elevation of the top of the tower is found to be $30^{\circ}$. What is the distance of the point from the foot of the tower? $\quad$

Options:
A $15 \sqrt{3} \mathrm{~m}$
B $10 \sqrt{3} \mathrm{~m}$
C $5 \sqrt{3} \mathrm{~m}$
D $30 \mathrm{~m}$

The correct answer is:
$15 \sqrt{3} \mathrm{~m}$

Solution:
Let $\mathrm{AB}$ be a tower of height $15 \mathrm{~m}$. Let $\mathrm{B}$ be the point on the ground. Let $\angle \mathrm{ACB}=30^{\circ}$ be the angle of elevation. To find $\mathrm{BC}$.
Let $\mathrm{BC}=x m$
$\operatorname{In} \Delta \mathrm{ABC}, \tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{15}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{15}{x} \Rightarrow x=15 \sqrt{3} \mathrm{~m}$

Topic: Mathematics, Heights and Distances, NDA, NDA 2011 (Phase 2)
A train covers the first $5 \mathrm{~km}$ of its journey at a speed of 30 $\mathrm{km} / \mathrm{hr}$ and the next $15 \mathrm{~km}$ at a speed of $45 \mathrm{~km} / \mathrm{hr}$. What is the average speed of the train?

Options:
A $35 \mathrm{~km} / \mathrm{hr}$
B $37.5 \mathrm{~km} / \mathrm{hr}$
C $39.5 \mathrm{~km} / \mathrm{hr}$
D $40 \mathrm{~km} / \mathrm{hr}$

The correct answer is:
$40 \mathrm{~km} / \mathrm{hr}$

Solution:
Time taken by train to cover first $5 \mathrm{~km}=\frac{\text { Distance }}{\text { Speed }}$
$=\frac{5}{30}=\frac{1}{6} \mathrm{hr}$
Time taken by train to cover next $15 \mathrm{~km}=\frac{15}{45}=\frac{1}{3} \mathrm{hr}$.
Average speed $=\frac{5+15}{\frac{1}{6}+\frac{1}{3}}$
$=\frac{20}{\frac{1+2}{6}}=\frac{120}{3}=40 \mathrm{~km} / \mathrm{h}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2018 (Phase 1)
A train whistling at constant frequency $n$ is moving towards a station at a constant speed $v$. The train goes past a stationary observer on the station. The frequency $n$ of the sound as heard by the observer is plotted as a function of time $t$. Identify the correct curve.

Options:
A

B

C

D

The correct answer is:

Solution:
Since, the observer is stationary, so when train is approaching towards the observer or moving towards the observer, apparent frequency is given as,
$n^{\prime}=n\left(\frac{v}{v-v_{s}}\right)$
where, $v_{s}$ is the speed of sound.
It is clear from the above expression that $n^{\prime}>n$. However, when the train is going away from the observer, then apparent frequency,
$\begin{array}{ll}
& n^{\prime}=n\left(\frac{v}{v+v_{s}}\right) \\
\therefore \quad & n^{\prime} < n
\end{array}$
Hence, graph represented in option (d) is correct.

Topic: Physics, Waves and Sound, KCET, KCET 2020
A transformer has 20 turns in the primary and 100 turns in the secondary coil. An ac voltage of $\mathrm{V}_{\text {in }}=600 \sin 314 \mathrm{t}$ is applied to primary terminal of transformer. Then maximum value of secondary output voltage obtained in volt is

Options:
A 600
B 300
C 3000
D 6000

The correct answer is:
3000

Solution:
We know,
$$
\frac{V_s}{V_p}=\frac{N_s}{N_p}
$$
$\therefore \quad$ The maximum value of secondary output voltage is:
$$
\begin{aligned}
V_s & =\frac{N_s}{N_p} \times V_p=\frac{100}{20} \times 600 \\
V_s & =3000 V
\end{aligned}
$$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2023 (13 May Shift 1)
A transformer has 100 turns in the primary coil and carries 8 A current. If the input power is 1 kW, the number of turns in secondary coil to have 500 V of output will be

Options:
A 100
B 200
C 400
D 300

The correct answer is:
400

Solution:
Given that,
Np=100, Ip=8 A
Vs=500 V, Pi=1 kW=100 W
Input power, Pi=VPIP
1000=VpÃ8
VP=125 V
Again, we know that
NsNP=VsVp
â Ns100=500125
Ns=500Ã100125
=400

Topic: Physics, Alternating Current, NEET
A transformer having an efficiency of 90% is working on 200 V, 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively, are

Options:
A 300 V, 15 A
B 450 V, 15 A
C 450 V, 13.5 A
D 600 V, 15 A

The correct answer is:
450 V, 15 A

Solution:
Power output =3 kW×90100=2.7 kW
Is=6 A
∴Vs=2.7 kW6 A=450 V
⇒Ip=3 kW200 V=15 A

Topic: Physics, Alternating Current, NEET
A transformer having an efficiency of 90% is working on 200 V, 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil, respectively, are

Options:
A 300 V, 15 A
B 450 V, 15 A
C 450 V, 13.5 A
D 600 V, 15 A

The correct answer is:
450 V, 15 A

Solution:
Power output =3 kW×90100=2.7 kW
Is=6 A
∴Vs=2.7 kW6 A=450 V
⇒Ip=3 kW200 V=15 A

Topic: Physics, Alternating Current, NEET
A transformer having efficiency of $75 \%$ is working on 220 V and 4.4 kW power supply. If the current in the secondary coil is 5 A . What will be the voltage across secondary coil and the current in primary coil?

Options:
A $V_S=220 \mathrm{~V}, i_P=20 \mathrm{~A}$
B $V_S=660 \mathrm{~V}, i_P=15 \mathrm{~A}$
C $V_S=660 \mathrm{~V}, i_P=20 \mathrm{~A}$
D $V_S=220 \mathrm{~V}, i_P=15 \mathrm{~A}$

The correct answer is:
$V_S=660 \mathrm{~V}, i_P=20 \mathrm{~A}$

Solution:
Given, $P_{\text {in }}=4.4 \mathrm{~kW}, V_p=220 \mathrm{~V}, i_s=5 \mathrm{~A}, V_s=$ ?
We know that, $\quad i_P=\frac{P_{\text {in }}}{V_P}=\frac{4.4 \times 10^3}{220}=20 \mathrm{~A}$
$\therefore$ Efficiency, $\eta=\frac{\text { Output power }}{\text { Input power }}$
Percentage efficiency $\eta=\frac{P_{\text {out }}}{P_{\text {in }}} \times 100$
$\begin{aligned} 75 \% & =100 \times \frac{P_{\text {out }}}{4.4 \times 10^3} \\ \therefore \quad P_{\text {out }} & =\frac{75 \times 4.4 \times 10^3}{100} \\ & =3300 \mathrm{~W} \\ \therefore \quad P_{\text {out }} & =V_S \cdot i_S \\ \Rightarrow \quad V_S & =\frac{3300}{5}=660 \mathrm{~V}\end{aligned}$

Topic: Physics, Alternating Current, JIPMER, JIPMER 2016