datavorous/entrance-exam-dataset huggingface dataset:
A change in the relative abundance of an allele (the allelic frequency) within a population, over a succession of generations, is called

Options:
A Micro-evolution
B Macro-evolution
C Co-evolution
D Phylogenetic evolution

The correct answer is:
Micro-evolution

Solution:
Microevolution involves changes in allelic frequency within a gene pool for a given species in a relatively short period of geologic time by the alterations of successfully reproducing individuals within a population. The macroevolution involves large scale changes among groups of species. Co-evolution occurs when two or more species reciprocally affect each other's evolution through the process of natural selection. Phylogenetic evolution represents evolutionary relationships among organisms and reflects how species or other groups evolved from a series of common ancestors.

Topic: Biology, Evolution, NEET
A child took sugar cane and sucked its juice. Regarding this which of the following match is correct?

Options:
A \(\begin{array}{|c|c|c|c|} \hline \text{Substrate} & \text{Enzyme} & \text{Site of secretion of enzyme} & \text{Products formed} \\ \hline \text{Proteins} & \text{Pepsin} & \text{Duodenum} & \text{Polypeptides} \\ \hline \end{array}\)
B \(\begin{array}{|c|c|c|c|} \hline \text{Substrate} & \text{Enzyme} & \text{Site of secretion of enzyme} & \text{Products formed} \\ \hline \text{Starch} & \text{Amylase} & \text{Salivary glands} & \text{Glucose} \\ \hline \end{array}\)
C \(\begin{array}{|c|c|c|c|} \hline \text{Substrate} & \text{Enzyme} & \text{Site of secretion of enzyme} & \text{Products formed} \\ \hline \text{Lipids} & \text{Lipase} & \text{Pancreas} & \text{Fat globules} \\ \hline \end{array}\)
D \(\begin{array}{|c|c|c|c|} \hline \text{Substrate} & \text{Enzyme} & \text{Site of secretion of enzyme} & \text{Products formed} \\ \hline \text{Sucrose} & \text{Invertase} & \text{Duodenum} & \text{Glucose + Fructose} \\ \hline \end{array}\)

The correct answer is:
\(\begin{array}{|c|c|c|c|} \hline \text{Substrate} & \text{Enzyme} & \text{Site of secretion of enzyme} & \text{Products formed} \\ \hline \text{Sucrose} & \text{Invertase} & \text{Duodenum} & \text{Glucose + Fructose} \\ \hline \end{array}\)

Solution:
Sucrose (cane sugar; beet sugar; saccharose) is a sugar comprising one molecule of glucose linked to a fructose molecule. It occurs widely in plants and is particularly abundant in sugar cane and sugar beet \((15-20 \%)\), from which it is extracted and refined for table sugar. If heated to \(200^{\circ} \mathrm{C}\), sucrose becomes caramel. Sucrose is broken down into the monosaccharides glucose and fructose with the help of carbohydrate digesting enzyme, invertase, produced in the small intestine (duodenum).

Topic: Chemistry, Biomolecules, AIIMS, AIIMS 2007
A circle is drawn on the chord of a circle $x^{2}+y^{2}=a^{2}$ as diameter. The chord lies on the line $x+y=a$. What is the equation of the circle?

Options:
A $x^{2}+y^{2}-a x-a y+a^{2}=0$
B $x^{2}+y^{2}-a x-a y=0$
C $x^{2}+y^{2}+a x+a y=0$
D $x^{2}+y^{2}+a x+a y-2 a^{2}=0$

The correct answer is:
$x^{2}+y^{2}-a x-a y=0$

Solution:
Given, equation of circle $\Rightarrow x^{2}+y^{2}=a^{2}$ Equation of chord $\Rightarrow x+y=a$ (1) $\Rightarrow x^{2}+(a-x)^{2}=a^{2}$
$\Rightarrow x^{2}+a^{2}+x^{2}-2 a x=a^{2}$
$\Rightarrow 2 x^{2}=2 a x$
$\Rightarrow \mathrm{x}=0, \mathrm{a}$
When, $x=0, y=a$ and when $x=a, y=0$. $\therefore$ Points of intersection are $(0, \mathrm{a})$ and $(\mathrm{a}, 0)$ $\therefore$ Equation of circle with chord as diameter is $(x-0)(x-a)+(y-a)(y-0)=0$
$\Rightarrow x(x-a)+y(y-a)=0$
$\Rightarrow x^{2}-a x+y^{2}-a y=0$
$\Rightarrow x^{2}+y^{2}-a x-a y=0$

Topic: Mathematics, Circle, NDA, NDA 2019 (Phase 1)
A circle is drawn with the two foci of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
at the end of the diameter. What is the equation of the circle?

Options:
A $x^{2}+y^{2}=a^{2}+b^{2}$
B $x^{2}+y^{2}=a^{2}-b^{2}$ (20) 10-I]
C $x^{2}+y^{2}=2\left(a^{2}+b^{2}\right)$
D $x^{2}+y^{2}=2\left(a^{2}-b^{2}\right)$

The correct answer is:
$x^{2}+y^{2}=a^{2}-b^{2}$ (20) 10-I]

Solution:
Foci of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ are given as $(a e, 0)$ and $(-a, 0)$.
Since, two foci are at the end of the diameter $\therefore$ Equation of circle, is
$(x-a e)(x+a e)+(y-0)(y-0)=0$
$\Rightarrow x^{2}-a^{2} e^{2}+y^{2}=0$
$\Rightarrow x^{2}+y^{2}-a^{2}\left(1-\frac{b^{2}}{a^{2}}\right)=0\left(\because e=\sqrt{1-\frac{b^{2}}{a^{2}}}\right)$
$\Rightarrow x^{2}+y^{2}-a^{2}+b^{2}=0$
$\Rightarrow x^{2}+y^{2}=a^{2}-b^{2}$

Topic: Mathematics, Ellipse, NDA, NDA 2010 (Phase 1)
A class consists of 3 sections $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ with 35,35 and 30 students respectively. The arithmetic means of the marks secured by students of sections $\mathrm{A}$ and $\mathrm{B}$, who appeared fol a test of 100 marks are 74 and 70 respectively. The arithmetic mean of the marks secured by students of section $\mathrm{C}$, who appeared for a test in the same subject which carried marks is 51 . What is the averagc perocntage of marks socuros by all the 100 student of the three sections? $\quad$

Options:
A $\quad 70.0$
B $70 . \mathrm{E}$
C 65
D 67

The correct answer is:
$70 . \mathrm{E}$

Solution:
Total no. of students in section A $=74 \times 35=2590$
Total no. of students in section $\mathrm{B}$ $=70 \times 35=2450$
Now, total no. of students in section $\mathrm{C}$
$=\frac{51}{75} \times 100 \times 30=2040$
Thus, total students in all $=2590+2450+2040$ $\begin{aligned} &=7080 \\ \text { Thus, Required percentage } &=7080 \div 100 \\ &=70.8 \end{aligned}$

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 2)
A class has 15 students whose ages are $14,17,15,14,21$,
$17,19,20,16,18,20,17,16,19$ and 20 years. One student
is selected in such a manner that each has the same chance
of being chosen and the age $\mathrm{X}$ of the selected student is
recorded.What is the probability distribution of the random
variable $\mathrm{X}$ ? Find mean, variance and standard deviation
of X?

Options:

None

Solution:
There are 15 students in a class. Each has the same chance
of being choosen.
The probability of each student to be selected $=\frac{1}{15}$
First and third columns from the probability distribution Mean $=\mathrm{E}(\mathrm{X})=\sum \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=\frac{293}{15}=17.53333$
Varience $=E\left(X^2\right)-[E(X)]^2=\sum p_i x_i{ }^2-\left(\sum p_i x_i\right)^2$
$$
\begin{aligned}
&=\frac{4683}{15}-(17.53333)^2 \\
&=312.20-307.41766=4.78
\end{aligned}
$$
Standard deviation
$$
=\sqrt{\text { Variance }}=\sqrt{4 \cdot 78}=2 \cdot 19
$$
probability distribution is
Mean $=17 \cdot 53$, variance $=4 \cdot 78$, S.D. $(X)=2 \cdot 19$

Topic: Mathematics, Probability
A closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. The ratio of their lengths is

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

The correct answer is:
1:2

Solution:
Fundamental frequency of close pipe,
v1=v4l1
Fundamental frequency of open pipe,
v2=v2l2
But v1=v2
∴l1l2=12

Topic: Physics, Waves and Sound, NEET
A closed organ pipe and an open organ pipe of same length produce 2 beats per second when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced per second will be

Options:
A 4
B 3
C 8
D 7

The correct answer is:
7

Solution:
Fundamental frequency of closed pipe, $\mathrm{n}_{\mathrm{c}}=\frac{\mathrm{V}}{4 \mathrm{~L}}$ Fundamental frequency of open pipe, $\mathrm{n}_0=\frac{\mathrm{v}}{2 \mathrm{~L}}$ They produce 2 beats per second
$\begin{aligned} & \therefore \mathrm{n}_{\mathrm{o}}-\mathrm{n}_{\mathrm{c}}=2, \\ & \frac{\mathrm{V}}{2 \mathrm{~L}}-\frac{\mathrm{V}}{4 \mathrm{~L}}=2 \\ & \frac{\mathrm{V}}{4 \mathrm{~L}}=2 \text { or } \frac{\mathrm{v}}{\mathrm{L}}=8\end{aligned}$
When length of open pipe is halved
$\mathrm{n}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{2\left(\frac{\mathrm{L}}{2}\right)}=\frac{\mathrm{v}}{\mathrm{L}}$
When length of closed pipe is doubled
$\mathrm{n}_{\mathrm{c}}^{\prime}=\frac{\mathrm{v}}{4 \times 2 \mathrm{~L}}=\frac{\mathrm{v}}{8 \mathrm{~L}}$
New beat frequency $=\mathrm{n}_{\mathrm{o}}^{\prime}-\mathrm{n}_{\mathrm{c}}^{\prime}=\frac{\mathrm{v}}{\mathrm{L}}-\frac{\mathrm{v}}{8 \mathrm{~L}}=\frac{7 \mathrm{v}}{8 \mathrm{~L}}=\frac{7}{8} \times 8=7$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (24 Sep Shift 2)
A closed organ pipe and an open organ pipe of same length produce 2 beats sec-1 when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced will be

Options:
A 7
B 4
C 8
D 2

The correct answer is:
7

Solution:
Fundamental frequency of closed organ pipe fc=v4l

Fundamental frequency of open organ pipe fo=v2l

Initial beats frequency

fo-fc=2

⇒v2l-v4l=2

⇒v4l=2

⇒vl=8

After changing of length

Fundamental frequency of closed organ pipe fc'=v42l=v8l

Fundamental frequency of open organ pipe fo'=v2l2=vl

New beats frequency

fo'-fc'

=vl-v8l

=7v8l

=7×88=7

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2020 (23 Sep Shift 1)
A closed organ pipe and an open organ pipe of same length produce 2 beats/second while vibrating in their fundamental modes. The length of the open organ pipe is halved and that of closed pipe is doubled. Then the number of beats produced per second while vibrating in the fundamental mode is

Options:
A 2
B 6
C 8
D 7

The correct answer is:
7

Solution:
Given, $\quad \mathrm{f}_{\mathrm{o}}-\mathrm{f}_{\mathrm{c}}=2 \quad \text{...(i)}$
Frequency of fundamental mode for a closed Frequency of fundar organ pipe, $f_{c}=\frac{v}{4 L_{c}}$
Similarly frequency of fundamental mode for an open organ pipe, $\mathrm{f}_{\mathrm{o}}=\frac{\mathrm{V}}{2 \mathrm{~L}_{\mathrm{o}}}$
Given $\quad \mathrm{L}_{\mathrm{c}}=\mathrm{L}_{\mathrm{o}}$
$\Rightarrow$ $f_{0}=2 f_{c} \quad \text{...(ii)}$
From Eqs. (i) and (ii), we get
$\mathrm{f}_{\mathrm{o}}=2 \mathrm{f}_{\mathrm{c}}$ $\mathrm{f}_{\mathrm{o}}=4 \mathrm{~Hz}$ and $\mathrm{f}_{\mathrm{c}}=2 \mathrm{~Hz}$
When the length of the open pipe is halved, frequency of fundamental mode is $$ \mathrm{f}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{2\left[\frac{\mathrm{L}_{\mathrm{o}}}{2}\right]}=2 \mathrm{f}_{\mathrm{o}}=2 \times 4 \mathrm{~Hz}=8 \mathrm{~Hz} $$
When the length of the closed pipe is doubled, its frequency of fundamental mode is
$$
\mathrm{f}_{\mathrm{o}}^{\prime}=\frac{\mathrm{v}}{4\left(2 \mathrm{~L}_{\mathrm{c}}\right)}=\frac{1}{2} \mathrm{f}_{\mathrm{c}}=\frac{1}{2} \times 2=1 \mathrm{~Hz}
$$
Hence, number of beats produced per second is
$$
\mathrm{f}_{\mathrm{o}}^{\prime}-\mathrm{f}_{\mathrm{c}}^{\prime}=8-1=7
$$

Topic: Physics, Waves and Sound, KCET, KCET 2010
A closed organ pipe and an open pipe of same length produce 2 beats s-1 when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced will be

Options:
A 7
B 4
C 8
D 2

The correct answer is:
7

Solution:
nc=v4l, n0=v2l
As n0-nc=2
â´v2l-v4l=2 or vl=8
Now
Number of beats s-1=n0â²-ncâ²
=vl-v8l=7v8l=78Ã8=7

Topic: Physics, Waves and Sound, NEET
A closed organ pipe of length ' $L_c$ ' and an open organ pipe of length ' $L_0$ ' contain different gases of densities ' $\rho_1$ ' and ' $\rho_2$ ' respectively. The compressibility of the gases is the same in both the pipes. The gases are vibrating in their first overtone with the same frequency. What is the length of open organ pipe?

Options:
A $\frac{4 L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}$
B $\frac{3 L_c}{4} \sqrt{\frac{\rho_2}{\rho_1}}$
C $\frac{4 L_c}{3} \sqrt{\frac{\rho_2}{\rho_1}}$
D $\frac{2 L_c}{3} \sqrt{\frac{\rho_2}{\rho_1}}$

The correct answer is:
$\frac{4 L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}$

Solution:
For open organ pipe:
Fundamental frequency $\mathrm{n}=\frac{\mathrm{V}}{2 \mathrm{~L}_{\text {o }}}$
First overtone $\mathrm{n}_1=2 \mathrm{n}=\frac{\mathrm{V}}{\mathrm{L}_{\text {。 }}}$
For closed organ pipe
Fundamental frequency $n^{\prime}=\frac{V^{\prime}}{4 L_c}$
First overtone $n_1{ }^{\prime}=3 n^{\prime}=\frac{3 \mathrm{~V}^{\prime}}{4 \mathrm{~L}_{\mathrm{c}}}$
$$
\begin{aligned}
& \because \mathrm{n}_1=\mathrm{n}_1^{\prime} \\
& \therefore \frac{\mathrm{V}}{\mathrm{L}_{\mathrm{o}}}=\frac{3 \mathrm{~V}^{\prime}}{4 \mathrm{~L}_{\mathrm{c}}} \\
& \therefore \frac{\mathrm{L}_{\mathrm{o}}}{\mathrm{L}_{\mathrm{c}}}=\frac{4}{3} \frac{\mathrm{V}}{\mathrm{V}^{\prime}} \\
& \mathrm{V}=\sqrt{\frac{\mathrm{k}}{\rho_2}}, \mathrm{~V}^{\prime}=\sqrt{\frac{\mathrm{k}}{\rho_1}}
\end{aligned}
$$
where $\mathrm{k}$ is the adiabatic bulk modulus, which is reciprocal of compressibility.
$$
\frac{L_0}{L_c}=\frac{4}{3} \sqrt{\frac{\rho_1}{\rho_2}}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (21 Sep Shift 1)
A closed orgon pipe and an open orgon pipe are tuned to the same fundamental frequency. What is the ratio of their lengths?

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

The correct answer is:
$1: 2$

Solution:
Avoiding end correction, the length of the closed organ pipe is
$l_1=\frac{\lambda_1}{4}$ or $\lambda_1=4 l_1$
The length of open organ pipe is
$l_2=\frac{\lambda_2}{2}$ or $\lambda_2=2 l_2$
Here, $n_1=n_2$
So, $\frac{v}{\lambda_1}=\frac{v}{\lambda_2}$ or $\frac{v}{4 l_1}=\frac{v}{2 l_2}$
Therefore, $l_1: l_2=1: 2$

Topic: Physics, Waves and Sound, JIPMER, JIPMER 2005
A closed pipe is suddenly opened and changed to an open pipe of same length. The fundamental frequency of the resulting open pipe is less than that of 3rd harmonic of the earlier closed pipe by $55 \mathrm{~Hz}$. Then, the value of fundamental frequency of the closed pipe is

Options:
A $165 \mathrm{~Hz}$
B $110 \mathrm{~Hz}$
C $55 \mathrm{~Hz}$
D $220 \mathrm{~Hz}$

The correct answer is:
$55 \mathrm{~Hz}$

Solution:
Frequency of first mode of vibration
$$
n_1=\frac{3 v}{4 /}
$$
(closed pipe) and frequency of first mode of vibration
$$
n_2=\frac{v}{2 l}
$$
(open pipe)
According to the question,
$$
n_1-n_2=\frac{3 v}{4 l}-\frac{v}{2 l}=55
$$
So $\quad \frac{3 v-2 v}{4 !}=55$
$$
\Rightarrow \quad \frac{v}{4 l}=55 \mathrm{~Hz}
$$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2014
A closed pipe is suddenly opened and changed to an open pipe of same length. The fundamental frequency of the resulting open pipe is less than that of 3rd harmonic of the earlier closed pipe by $55 \mathrm{~Hz}$. Then, the value of fundamental frequency of the closed pipe is

Options:
A $165 \mathrm{~Hz}$
B $110 \mathrm{~Hz}$
C $55 \mathrm{~Hz}$
D $220 \mathrm{~Hz}$

The correct answer is:
$55 \mathrm{~Hz}$

Solution:
Frequency of first mode of vibration
$$
n_1=\frac{3 v}{4 /}
$$
(closed pipe) and frequency of first mode of vibration
$$
n_2=\frac{v}{2 l}
$$
(open pipe)
According to the question,
$$
n_1-n_2=\frac{3 v}{4 l}-\frac{v}{2 l}=55
$$
So $\quad \frac{3 v-2 v}{4 !}=55$
$$
\Rightarrow \quad \frac{v}{4 l}=55 \mathrm{~Hz}
$$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2014
A coin is tossed 5 times. The probability that tail appears an odd number of times, is

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

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

Solution:
Let $\mathrm{p}$ denote the probability of getting tail in a single of a coin.
$\therefore \mathrm{p}=\frac{1}{2} \Rightarrow \mathrm{q}=\frac{1}{2} \& \mathrm{n}=5$
Let $X$ denot no. of tails in 5 tosses of coin. $\begin{aligned} \therefore \text { Required probability } &=\mathrm{P}(\mathrm{x}=1)+\mathrm{P}(\mathrm{x}=3)+\mathrm{P}(\mathrm{x}=5) \\ &=\frac{1}{2^{5}}\left[{ }^{5} \mathrm{C}_{1}+{ }^{5} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{5}\right] \\ &=\frac{1}{2^{5}}[5+10+1]=\frac{16}{32}=\frac{1}{2} \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
A coin is tossed 10 times. The number of heads minus the number of tails in 10 tosses is considered as the outcome of the experiment. What is the number of points in the sample space

Options:
A 10
B 11
C 21
D 99

The correct answer is:
11

Solution:
\begin{array}{c|c}
Head & Tail \\
10 & 0 \\
9 & 1 \\
8 & 2 \\
7 & 3 \\
6 & 4 \\
5 & 5 \\
4 & 6 \\
3 & 7 \\
2 & 8 \\
1 & 9 \\
0 & 10
\end{array}
Hence, total number of points in the sample space is
$11 .$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 2)
A coin is tossed five times. What is the probability that heads are observed more than three times?

Options:
A $\frac{3}{16}$
B $\frac{5}{16}$
C $\frac{1}{2}$
D $\frac{3}{32}$

The correct answer is:
$\frac{3}{16}$

Solution:
Let P denote the probability of getting head in a single toss of a coin.
$\mathrm{p}=\frac{1}{2} \Rightarrow \mathrm{q}=\frac{1}{2}$
Let $X$ denote the no. of heads in 5 tosses of a coin. then, $\mathrm{X}$ is a binomial variate with parameters; $\mathrm{n}=5 \&$
$\mathrm{p}=\frac{1}{2}$.
$\therefore$ Req. probability $=\mathrm{P}(\mathrm{x}>3)$
$=1-\mathrm{P}(\mathrm{x} \leq 3)$
$=1-[\mathrm{P}(\mathrm{x}=0)+\mathrm{P}(\mathrm{x}=1)+\mathrm{P}(\mathrm{x}=2)+\mathrm{P}(\mathrm{x}=3)]$
$=1-\left[{ }^{5} \mathrm{C}_{0}+{ }^{5} \mathrm{C}_{1}+{ }^{5} \mathrm{C}_{2}+{ }^{5} \mathrm{C}_{3}\right] \frac{1}{2^{5}}$
$=1-[1+5+10+10] \frac{1}{32}$
$=\frac{32}{32}-\frac{26}{32}=\frac{6}{32}=\frac{3}{16}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
A coin is tossed three times. Consider the following events:
A: No head appears
B: Exactly one head appears
C: At least two heads appear Which one of the following is correct?

Options:
A $(\mathrm{A} \cup \mathrm{B}) \cap(\mathrm{A} \cup \mathrm{C})=\mathrm{B} \cup \mathrm{C}$
B $\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right) \cup\left(\mathrm{A} \cap \mathrm{C}^{\prime}\right)=\mathrm{B}^{\prime} \cup \mathrm{C}$
C $\mathrm{A} \cap\left(\mathrm{B}^{\prime} \cup \mathrm{C}^{\prime}\right)=\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}$
D $A \cap\left(B^{\prime} \cup C^{\prime}\right)=B^{\prime} \cap C$

The correct answer is:
$A \cap\left(B^{\prime} \cup C^{\prime}\right)=B^{\prime} \cap C$

Solution:
$\quad U=\{(H H H)(H H T)(H T H)(H T T)(T H H)(T H T)(T T H)(T T T)\}$
$A=\{(T T T)\}$
$B=\{(H T T)(T H T)(T T H)\}$
$C=\{(H H H)(H H T)(H T H)(T H H)\}$
By checking the options
(d) $A \cap\left(B^{\prime} \cup C^{\prime}\right)=B^{\prime} \cap C^{\prime}$ is correct.

Topic: Mathematics, Sets and Relations, NDA, NDA 2016 (Phase 2)
A coin is tossed three times. What is the probability of getting head and tail alternately?

Options:
A $\frac{1}{8}$
B $\frac{1}{4}$
C $\frac{1}{2}$
D $\frac{3}{4}$

The correct answer is:
$\frac{1}{4}$

Solution:
Coin is tossed three times i.e. total outcomes $=2^{3}=8$ $[(\mathrm{H}, \mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{H}, \mathrm{T}),(\mathrm{H}, \mathrm{T}, \mathrm{H}),(\mathrm{H}, \mathrm{T}, \mathrm{T}),(\mathrm{T}, \mathrm{H}, \mathrm{H}),$,
$(\mathrm{T}, \mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{T}, \mathrm{T})]$
Alternate head and tail are coming two times only.
Thus prob. of getting head and tail alternately $=\frac{2}{8}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)
A coin is tossed three times. What is the probability of getting head and tail (HTH) or tail and head (THT) alternatively?

Options:
A $1 / 4$
B $1 / 5$
C $1 / 6$
D $1 / 8$

The correct answer is:
$1 / 4$

Solution:
Total possible outcomes, $\mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{THT}$.
TTH. THH, TTT, HTT $\}$ and desired outcomes E $=\{\mathrm{HTH}, \mathrm{THT}\}$
$\Rightarrow \mathrm{n}(\mathrm{E})=2$ and $\mathrm{n}(\mathrm{S})=8$
Hence, required probability $=\mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{2}{8}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2006 (Phase 2)
A colloidal solution is kept in dark and is illuminated by a beam of light then, brightness appears at the right angle of direction of light. This effect is called

Options:
A Tyndall effect
B Brownian effect
C Hardy-Schulze effect
D none of the above.

The correct answer is:
Tyndall effect

Solution:
No Solution Available

Topic: Chemistry, Surface Chemistry, AIIMS, AIIMS 2015
A colloidal solution is kept in dark and is illuminated by a beam of light then brightness appears at the right angle of direction of light. This effect is called

Options:
A Tyndall effect
B Brownian effect
C Hardy-Schulze effect
D None of these

The correct answer is:
Tyndall effect

Solution:
No Solution Available

Topic: Chemistry, Surface Chemistry, AIIMS, AIIMS 2012
A committee of three has to be chosen from a group of 4 men and 5 women. If the selection is made at random, what is the probability that exactly two members are men?

Options:
A $\frac{5}{14}$
B $\frac{1}{21}$
C $\frac{3}{14}$
D $\frac{8}{21}$

The correct answer is:
$\frac{5}{14}$

Solution:
$\mathrm{n}(\mathrm{E})={ }^{4} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{1}=6 \times 5=30$.
$\mathrm{n}(\mathrm{S})={ }^{9} \mathrm{C}_{3}=84$
$\therefore$ Probability, $\mathrm{P}(\mathrm{E})=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{30}{84}=\frac{5}{14}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)
A committee of two persons is contituted from two men and two women. What is the probability that the committee will have only women?

Options:
A $\frac{1}{6}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{2}{3}$

The correct answer is:
$\frac{1}{6}$

Solution:
$2 \mathrm{Men}+2$ Women Required probability $=\frac{{ }^{2} \mathrm{C}_{0} \times{ }^{2} \mathrm{C}_{2}}{{ }^{4} \mathrm{C}_{2}}=\frac{1 \times 1}{6}=\frac{1}{6}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 1)
A committee of two persons is selected from two men and two women. The probability that the committee will have exactly one woman is

Options:
A $\frac{1}{6}$
B $\frac{2}{3}$
C $\frac{1}{3}$
D $\frac{1}{2}$

The correct answer is:
$\frac{2}{3}$

Solution:
We know, $\mathrm{P}(\mathrm{E})=\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}$
Number of possible outcomes $={ }^{4} \mathrm{C}_{2}=\frac{4 \times 3}{2 \times 1}=6$
(selecting 2 people from 4 people $)$ Number of favourable outcomes
$={ }^{2} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{1}=2 \times 2=4$
(selecting 1 from 2 men, 1 from 2 women)
$\mathrm{P}(\mathrm{E})=\frac{4}{6}=\frac{2}{3}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)
A common structural feature of vessel elements and sieve tube elements is:

Options:
A enucleate condition
B thick secondary walls
C pores on lateral walls
D presence of P-protein

The correct answer is:
enucleate condition

Solution:
Enucleate condition means cells without nuclei. This is found in vessel elements and sieve tube elements. Thick secondary wall is the inner most layer of cell wall, it is found in the cell wall of vessel elements. The thickened areas present on the lateral walls of the vessel elements are called pores. P- proteins are the small tubular fibrillar, granular or crystalline sub microscopic structures appear in the cytoplasm of sieve tubes.
Caution
Xylem vessels become dead and lose their protoplasm due to deposition of lignified secondary wall mature sieve tube elements are living cells without nucleus.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2006
A common structural feature of vessel elements and sieve tube elements is ________.

Options:
A enucleate condition
B thick secondary walls
C pores on lateral walls
D presence of protoplasm

The correct answer is:
enucleate condition

Solution:
The food conducting tissue of vascular plants is phloem. The components of phloem are sieve elements, companion cells, phloem parenchyma, phloem fibers, and sclereids. Whereas xylem is the water-conducting tissue of vascular plants. The components of xylem are Tracheids, vessel, xylem parenchyma, and xylem fiber. There is one characteristic that is common in vessel element and sieve tube element is that they both are non-living at maturity or nucleus is absent in both of them.

Topic: Biology, Anatomy of Flowering Plants, NEET