datavorous/entrance-exam-dataset huggingface dataset:
Two cylinders $A$ and $B$ of equal capacity are connected to each other via stopcock. A contains gas at standard temperature and pressure, $B$ is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in $A$ and $B$ ?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its $P-V-T$ surface?

Options:

None

Solution:
(a) If the stopcock opens suddenly, volume of gas would become double at 1 atmospheric pressure, therefore, pressure would become half (i.e. $1 / 2$ atmosphere).
(b) As no work is done on the system by the gas, there will be no change in internal energy.
(c) Gas does not do any work in expanding, thereby, no change in temperature of gas would occur.
(d) No, the process of the expansion is rapid and cannot be controlled. The intermediate state would be a nonequilibrium state, which does not follow gas equation.
In due course, gas will come back to equilibrium state.

Topic: Physics, Thermodynamics
Two dice are thrown. What is the probability that the sumof the faces equals or exceeds $10 ?$

Options:
A $1 / 12$
B $1 / 4$
C $1 / 3$
D $1 / 6$

The correct answer is:
$1 / 6$

Solution:
Let $\mathrm{E}$ be the sum of the faces equals or exceeds
Then, $\mathrm{E}=\{(5,5),(4,6),(6,4),(5,6),(6,5),(6,6)\}$
$\therefore \quad n(E)=6$
Hence, $P(E)=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 1)
Two dice each numbered from 1 to 6 are thrown together. Let $A$ and $\mathrm{B}$ be two events given by
A: even number on the first die.
B: number on the second die is greater than 4
What is $\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ equal to ?

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{2}{3}$
D $\frac{1}{6}$

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

Solution:
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{6}{36}=\frac{1}{6}$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
Two dice each numbered from 1 to 6 are thrown together. Let $A$ and $\mathrm{B}$ be two events given by
A: even number on the first die.
B: number on the second die is greater than 4
What is $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$ equal to $?$

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{2}{3}$
D $\frac{1}{6}$

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

Solution:
In A there are 18 events possible.
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{18}{36}=\frac{1}{2}$
In B there are 12 events are possible
$\therefore \quad \mathrm{P}(\mathrm{B})=\frac{12}{36}=\frac{1}{3}$
In $A \cap B$ there are 6 events are possible. Now,
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{3+2-1}{6}=\frac{4}{6}=\frac{2}{3}$

Topic: Mathematics, Probability, NDA, NDA 2012 (Phase 2)
Two different species cannot live for long duration in the same niche or habitat. This law is:

Options:
A Allen's law
B Gause's law
C Competitive exclusion principal
D Weiseman's theory

The correct answer is:
Competitive exclusion principal

Solution:
The Principle of Competitive Exclusion states that two species or populations cannot inhabit the same niche: one will consistently outcompete the other.
Allen rule states that in a warmblooded animal species having distinct geographic populations, limbs, ears and other appendages of animals living in cold climate tend to be shorter than in animals of same species living in warm climates.
Weismann proposed the theory of the continuity of germ-plasm, a theory of heredity.

Topic: Biology, Organisms and Populations, NEET, NEET 2002
Two distributions $A$ and $B$ have the same mean. If their coefficients of variation are 6 and 2 respectively and $\sigma_{A^{\prime}} \sigma_B$ are their standard deviations, then

Options:
A $\sigma_A=3 \sigma_B$
B $3 \sigma_A=\sigma_B$
C $\sigma_A=2 \sigma_B$
D $2 \sigma_A=\sigma_B$

The correct answer is:
$\sigma_A=3 \sigma_B$

Solution:
Let $\bar{x}_A=\bar{x}_B=x$
$$
\begin{aligned}
& \frac{\sigma_A}{\bar{x}_A} \times 100=6, \frac{\sigma_B}{\bar{x}_B} \times 100=2 \\
& \frac{\sigma_A}{x} \times 100=6, \frac{\sigma_B}{x} \times 100=2 \\
& \frac{\sigma_A \times 100}{6}=x, \frac{\sigma_B \times 100}{2}=x \\
\Rightarrow \quad & \frac{\sigma_A \times 100}{6}=\frac{\sigma_B \times 100}{2} \Rightarrow \sigma_A=3 \sigma_B
\end{aligned}
$$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
Two events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{not} \mathrm{B})=0.8, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=$
$0.5$ and $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=0.4$. Then $\mathrm{P}(\mathrm{A})$ is equal to

Options:
A $0.28$
B $0.32$
C $0.38$
D None of the above

The correct answer is:
$0.38$

Solution:
$\begin{aligned} & \because \mathrm{P}(\overline{\mathrm{B}})=0.8 \Rightarrow \mathrm{P}(\mathrm{B})=0.2 \\ & \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.5 \& \mathrm{P}(\mathrm{A} \mid \mathrm{B})=0.4 \\ & \because \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{B}) \times \mathrm{P}(\mathrm{A} \mid \mathrm{B})=0.2 \times 0.4=0.08 \\ & \& \mathrm{P}(\mathrm{A})=\mathrm{P}(\mathrm{A} \cup \mathrm{B})-\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\ & \mathrm{P}(\mathrm{A})=0.5-0.2+0.08=0.38 \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
Two fair dice are rolled. What is the probability of getting a sum of $7 ?$

Options:
A $\frac{1}{36}$
B $\frac{1}{6}$
C $\frac{7}{12}$
D $\frac{5}{12}$

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

Solution:
When two dice are rolled, the events where we get sum of 7 is $\mathrm{E}=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$
$\therefore \mathrm{n}(\mathrm{E})=6$
Total number of events, $\mathrm{n}(\mathrm{S})=36$.
$\therefore$ Probability $=\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 1)
Two forces are equal to $2 \overrightarrow{O A}$ and $3 \overrightarrow{B O}$, their resultant
being $\lambda \overrightarrow{O G}$, where $G$ is the point on $A B$ such that
$\frac{B G}{A G}=-\frac{2}{3} .$ What is the value of $\lambda ?$

Options:
A 1
B $-1$
C 2
D None of the above

The correct answer is:
$-1$

Solution:
$\overrightarrow{O G}=\frac{2 \overrightarrow{O A}-3 \overrightarrow{O B}}{2-3}$
$\overrightarrow{O G}=\frac{2 \overrightarrow{O A}-3 \overrightarrow{O B}}{-1}$
$-\overrightarrow{O G}=2 \overrightarrow{O A}-3 \overrightarrow{O B}$ ... (1)
$\lambda \overrightarrow{O G}=2 \overrightarrow{O A}+3 \overrightarrow{O B} \quad \cdots$ (2)
Adding (1) and (2)
$(\lambda-1) \overrightarrow{O G}=4 \overrightarrow{O A}$
$\Rightarrow \overrightarrow{O A}=\left(\frac{\lambda-1}{4}\right) \overrightarrow{O G}$ ... (3)
Subtracting (2) from (1)
$(-1-\lambda) \overrightarrow{O G}=-6 \overrightarrow{O B}$
$\overrightarrow{O B}=\frac{(1+\lambda)}{6} \overrightarrow{O G}$ ... (4)
From equ (2), (3) and (4)
$\lambda \overrightarrow{O G}=2\left(\frac{\lambda-1}{4}\right) \overrightarrow{O G}+3\left(\frac{\lambda+1}{6}\right) \overrightarrow{O G}$
$\Rightarrow \lambda=\frac{\lambda-1}{2}=\frac{\lambda+1}{3}$
$\therefore \lambda=\frac{\lambda-1}{2}$ or $\lambda=\frac{\lambda+1}{3}$
$\Rightarrow 2 \lambda-\lambda=-1$ or $3 \lambda-\lambda=1$
$\Rightarrow \lambda=-1$ or $\lambda=\frac{1}{2}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2008 (Phase 2)
Two half cell reactions are given below :
$\mathrm{CO}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Co}^{2+}, \mathrm{E}^{\circ} \mathrm{Co}^{2+} / \mathrm{Co}^{3+}=-1.81 \mathrm{~V}$
$2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al}(\mathrm{s}), \mathrm{E}^{\mathrm{o}} \mathrm{Al} / \mathrm{Al}^{3+}=+1.66 \mathrm{~V}$
The standard EMF of a cell with feasible redox reaction will be :

Options:
A $+7.09 \mathrm{~V}$
B $+0.15 \mathrm{~V}$
C $+3.47 \mathrm{~V}$
D $-3.47 \mathrm{~V}$

The correct answer is:
$+3.47 \mathrm{~V}$

Solution:
The cell reaction is written as :
$\begin{aligned}
& \mathrm{E}^{\circ} \mathrm{Co}^{2+} / \mathrm{Co}^{3+}=-1.81 \mathrm{~V}(\mathrm{SOP}) \\
& \therefore \quad \mathrm{E}^{\circ} \mathrm{Co}^{3+} / \mathrm{Co}^{2+}=+1.81 \mathrm{~V} \text { (SRP) } \\
& \mathrm{E}_{\mathrm{A} / / \mathrm{Al}^{3+}}^{\circ}=+1.66 \mathrm{~V}(\mathrm{SOP}) \\
& \therefore \mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\circ}=-1.66 \mathrm{~V}(\mathrm{SRP}) \\
&
\end{aligned}$
Here $\mathrm{E}^{\circ}$ is given in standard Reduction Potential
$\therefore \mathrm{Co}^{3+} / \mathrm{Co}$ will undergo reduction acting as cathode and $\mathrm{Al}^{3+} / \mathrm{Al}$ will undergo oxidation acting as anode.
$\begin{aligned}
\therefore \quad \mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\text {cathode (Red.) }}^{\circ}-\mathrm{E}_{\text {Anode (oxide) }}^{\circ} \\
\mathrm{E}_{\text {cell }}^{\circ} & =1.81-(-1.66) \\
\mathrm{E}_{\text {cell }}^{\circ} & =+3.47 \mathrm{~V}
\end{aligned}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2022 (Phase 2)
Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

Options:
A $0.01$
B $0.02$
C $0.03$
D $0.04$

The correct answer is:
$0.02$

Solution:
When both the wires vibrate simultaneously, beats per second,
$$
\begin{array}{c}
n_{1} \pm n_{2}=6 \\
\text { or } \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600=6 \Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=606...(i)
\end{array}
$$
Given that fundamental frequency
$$
\frac{1}{2 l} \sqrt{\frac{T}{m}}=600...(ii)
$$
Dividing Eq. (i) by(ii),
$$
\begin{array}{l}
\frac{\frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}=\frac{606}{600} \\
\Rightarrow \sqrt{\frac{T^{\prime}}{T}}=(1.01) \Rightarrow \frac{T^{\prime}}{T}=(1.02) \% \\
\Rightarrow T^{\prime}=T(1.02)
\end{array}
$$
Increase in tension
$$
\begin{array}{l}
\Delta T^{\prime}=T \times 1.02-T=(0.02 T) \\
\therefore \Delta T=0.02
\end{array}
$$

Topic: Physics, Waves and Sound, VITEEE, VITEEE 2009
Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

Options:
A 0.01
B 0.02
C 0.03
D 0.04

The correct answer is:
0.01

Solution:
Beats per second when both the wires vibrate simultaneously.
$n_1 \pm n_2=6$
$\begin{aligned} \text { or } & \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600 & =6\end{aligned}$
Given that fundamental frequency
Dividing Eq. (i) by Eq. (ii), we get
$\frac{\frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}=\frac{606}{600}$
$\begin{array}{ll}\text { or } & \sqrt{\frac{T^{\prime}}{T}}=(1.01) \\ \text { or } & \frac{T^{\prime}}{T}=(1.02) \% \\ \text { or } & T^{\prime}=T(1.02)\end{array}$
Increase in tension
$\begin{aligned}
\Delta T^{\prime} & =T \times 1.02-T \\
& =(0.02 T) \\
\text{Hence,} \quad \Delta T^{\prime}=0.02
\end{aligned}$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2009
Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

Options:
A 0.01
B 0.02
C 0.03
D 0.04

The correct answer is:
0.01

Solution:
Beats per second when both the wires vibrate simultaneously.
$n_1 \pm n_2=6$
$\begin{aligned} \text { or } & \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}} & =6 \\ \text { or } & \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600 & =6\end{aligned}$
Given that fundamental frequency
Dividing Eq. (i) by Eq. (ii), we get
$\frac{\frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}=\frac{606}{600}$
$\begin{array}{ll}\text { or } & \sqrt{\frac{T^{\prime}}{T}}=(1.01) \\ \text { or } & \frac{T^{\prime}}{T}=(1.02) \% \\ \text { or } & T^{\prime}=T(1.02)\end{array}$
Increase in tension
$\begin{aligned}
\Delta T^{\prime} & =T \times 1.02-T \\
& =(0.02 T) \\
\text{Hence,} \quad \Delta T^{\prime}=0.02
\end{aligned}$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2009
Two identical strings of length ' $\ell$ ' and '2 $\ell$ ' vibrate with fundamental frequencies 'N' hertz and '1:5 N' hertz, respectively. The ratio of tensions for smaller length to larger length is

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

The correct answer is:
1:9

Solution:
$\mathrm{N}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~m}}}$
$1.5 \mathrm{~N}=\frac{1}{4 \ell} \sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~m}}}$
$\therefore \sqrt{\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}}=\frac{\mathrm{N} \times 2 \ell}{1.5 \mathrm{~N} \times 4 \ell}=\frac{1}{3}$
$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{1}{9}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (16 Oct Shift 1)
Two identical wires have a fundamental frequency $f_0$ when kept under the same tension $T$. If the tension of one wire is increased by $\Delta T$, then the $N$ beats occur when both wires oscillate simultaneously.

Options:
A $\frac{\Delta T}{T}=\left(\frac{f_0+N}{f_0}\right)^2-1$
B $\frac{\Delta T}{T}=\left(\frac{f_0}{f_0-N}\right)^2-1$
C $\frac{\Delta T}{T}=\left(\frac{f_0-N}{f_0}\right)^2+1$
D $\frac{\Delta T}{T}=\left(\frac{f_0+N}{f_0}\right)^2$

The correct answer is:
$\frac{\Delta T}{T}=\left(\frac{f_0+N}{f_0}\right)^2-1$

Solution:
Fundamental frequency of the wire,
$$
f_0=\frac{v}{2 L}, \text { where } v=\sqrt{\frac{T}{\mu}}
$$
$\begin{array}{ll}\Rightarrow & f_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \\ \Rightarrow & f_0 \propto \sqrt{T}\end{array}$
Dividing Eq. (ii) by Eq. (i), we get
$$
\begin{aligned}
\frac{\left(f_0+N\right)^2}{f_0^2} & =\frac{T+\Delta T}{T} \\
\Rightarrow \quad \frac{\Delta T}{T} & =\frac{\left(f_0+N\right)^2}{f_0^2}-1=\left(\frac{f_0+N}{f_0}\right)^2-1
\end{aligned}
$$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 2)
Two identical wires have the same fundamental frequency of 400 Hz when kept under the same tension. If the tension in one wire is increased by 2%, the number of beats produced will be

Options:
A 4
B 8
C 2
D 1

The correct answer is:
4

Solution:
n∝T⇒Δnn=12ΔTT
Beat frequency =Δn=12ΔTTn=12×2100×400=4

Topic: Physics, Waves and Sound, NEET
Two independent events $\mathrm{A}$ and $\mathrm{B}$ are such that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=$
$\frac{2}{3}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}$. If $\mathrm{P}(\mathrm{B}) < \mathrm{P}(\mathrm{A})$, then what is $\mathrm{P}(\mathrm{B})$
equal to ?

Options:
A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
$P(A \cup B)=\frac{2}{3}$
$P(A \cap B)=\frac{1}{6}$
Since, $A$ and $B$ are independent events.
$P(A \cap B)=P(A) \cdot P(B)=\frac{1}{6}$ ... (1)
$P(A \cup B)=\frac{2}{3} \Rightarrow P(A)+P(B)-P(A \cap B)=\frac{2}{3}$
$\Rightarrow P(A)+P(B)-\frac{1}{6}=\frac{2}{3}$
$\Rightarrow P(A)+P(B)=\frac{2}{3}+\frac{1}{6}=\frac{5}{6}$ .... (2)
from $(1),(2), P(B)=\frac{1}{3}$ or $\frac{1}{2}$
$\because P(B) < P(A), P(B)=\frac{1}{3}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
Two independent events $\mathrm{A}$ and $\mathrm{B}$ have $\mathrm{P}(\mathrm{A})=\frac{1}{3}$
and $\mathrm{P}(\mathrm{B})=\frac{3}{4}$. What is the probability that exactly one of the two events A or B occurs?

Options:
A $\frac{1}{4}$
B $\frac{5}{6}$
C $\frac{5}{12}$
D $\frac{7}{12}$

The correct answer is:
$\frac{7}{12}$

Solution:
$A$ and $B$ are independent.
$P(A)=\frac{1}{3}$
$P(B)=\frac{3}{4}$
We want to find probability that exactly one of the two events $A$ or $B$ occurs i.e. when $A$ occurs $B$ does not and vice-versa. Lets take desired prob. is $P$. $\therefore P=P(A)(1-P(B))+P(B)(1-P(A))$
$$
\begin{array}{l}
=\frac{1}{3}\left(1-\frac{3}{4}\right)+\frac{3}{4}\left(1-\frac{1}{3}\right) \\
=\frac{1}{3} \times \frac{1}{4}+\frac{3 \times 2}{12} \\
P=\frac{7}{12}
\end{array}
$$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 1)
Two integers $\mathrm{x}$ and $\mathrm{y}$ are chosen with replacement from the
set $(0,1,2, \ldots \ldots, 10)$. The probability that $|\mathrm{x}-\mathrm{y}|>5$ is

Options:
A $\frac{6}{11}$
B $\frac{35}{121}$
C $\frac{30}{121}$
D $\frac{25}{121}$

The correct answer is:
$\frac{30}{121}$

Solution:
Probability $=\frac{2 \times \frac{1}{2} \times 5 \times 6}{11 \times 11}=\frac{30}{121}$

Topic: Mathematics, Probability, NDA, NDA 2018 (Phase 2)
Two letters are drawn at random from the word 'HOME'. What is the probability that both the letters are vowels?

Options:
A $1 / 6$
B $5 / 6$
C $1 / 2$
D $1 / 3$

The correct answer is:
$1 / 2$

Solution:
Total number of letters $=4$ Total number of vowels $=2(\mathrm{O}$ and $\mathrm{E})$
Required Probability $=\frac{2}{4}=\frac{1}{2}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 2)
Two men hit at a target with probabilities $\frac{1}{2}$ and $\frac{1}{3}$ respectively. What is the probability that exactly one of them hits the target?

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

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

Solution:
$\begin{aligned} & \mathrm{P}=\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\overline{\mathrm{E}}_{2}\right)+\mathrm{P}\left(\overline{\mathrm{E}}_{1}\right) \mathrm{P}\left(\mathrm{E}_{2}\right) \\ &=\frac{1}{2}\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{2}\right) \frac{1}{3} \\ &=\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3} \\ &=\frac{1}{3}+\frac{1}{6}=\frac{1}{2} \\ \therefore & \text { Option (a) is correct. } \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 1)
Two moles of ideal helium gas is in a rubber balloon at 30 Â°C.The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35 Â°C. The amount of heat required in raising the temperature is nearly ( take R = 8.31 J mol-1 K-1 )

Options:
A 62 J
B 140 J
C 124 J
D 208 J

The correct answer is:
208 J

Solution:
Δ Q = n C p Δ T Isobaric process
= 2 × 5 2 R × 3 5 - 3 0
= 2 0 8 J

Topic: Physics, Thermodynamics, NEET
Two numbers are successively drawn from the set $\{1,2,3,4$, 5. 6. 7\} without replacement and the outcomes recorded in
that order. What is the number of elementary events in the randomexperiment?

Options:
A 49
B 42
C 21
D 14

The correct answer is:
42

Solution:
Given set is $\{1,2,3,4,5,6,7\}$ Total numbers in set $=7$ Since, two numbers are drawn without replacement $\therefore$ Total number of elementary events $={ }^{7} C_{1} \times{ }^{6} C_{1}$ $=7 \times 6$
$=42$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 2)
Two numbers are successively drawn from the set $\mathrm{U}=\{1,2,3,4,5,6,7,8\}$, the second being drawn without replacing the first. The number of elementary events in the sample is:

Options:
A 64
B 56
C 32
D 14

The correct answer is:
56

Solution:
Required no. of elementary events $={ }^{8} \mathrm{C}_{2} \times 2 !=56$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 1)
Two numbers $X$ and $Y$ are simultaneously drawn from the set $\{1,2,3,4,5,6,7,8,9,10\}$. What is the conditional probability of exactly one of the two numbers $X$ and $Y$ bems even, given $(X+Y)=15$ ?

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

The correct answer is:
1

Solution:
Given, $X+Y=15$ The total number of ordered pairs which satisfies $X+Y=15$ is
$\therefore(5,10),(6,9),(7,8),(8,7),(9,6),(10,5)$
$\therefore n(S)=6$ where $S$ denotes the sample space. In each above pairs exactly one of the two numbers is even number. Therefore $E=\{(5,10),(6,9),(7,8),(8,7),(9,6),(10,5)\}$
where $E$ is an event.
$\therefore$
$n(E)=6$
$\therefore$ Required probability $=\frac{n(E)}{n(S)}=\frac{6}{6}=1$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 2)
Two open organ pipes of fundamental frequencies $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ are joined in series. The fundamental frequency of the new pipe is

Options:
A $\mathrm{n}_{1}-\mathrm{n}_{2}$
B $\frac{n_{1} n_{2}}{\left(n_{1}+n_{2}\right)}$
C $\frac{1}{n_{1} n_{2}}$
D $\frac{n_{1}+n_{2}}{n_{1} n_{2}}$

The correct answer is:
$\frac{n_{1} n_{2}}{\left(n_{1}+n_{2}\right)}$

Solution:
$\mathfrak{n}=\frac{\mathrm{V}}{2 \ell_{1}} \quad \therefore \quad \ell_{1}=\frac{\mathrm{V}}{2 \mathrm{n}_{1}}$
$\mathfrak{n}_{2}=\frac{\mathrm{V}}{2 \ell_{2}} \quad \therefore \quad \ell_{2}=\frac{\mathrm{V}}{2 \mathrm{n}_{2}}$
$\mathfrak{n}=\frac{\mathrm{V}}{2 \ell} \quad \therefore \quad \ell=\frac{\mathrm{V}}{2 \mathrm{n}}$
$\ell=\ell_{1}+\ell_{2}$
$\frac{\mathrm{V}}{2 \mathrm{n}}=\frac{\mathrm{V}}{2 \mathrm{n}_{1}}+\frac{\mathrm{V}}{2 \mathrm{n}_{2}}$
$\therefore \frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_{1}}+\frac{1}{\mathrm{n}_{2}}$
$\therefore \mathrm{n}=\frac{\mathrm{n}_{1} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2020 (14 Oct Shift 1)
Two open organ pipes of lengths $50 \mathrm{~cm}$ and $51 \mathrm{~cm}$ are totally immersed in a medium. They are found to give 40 beats in $10 \mathrm{~s}$ when each is sounding at its fundamental note. The speed of sound in this medium is

Options:
A $275 \mathrm{~ms}^{-1}$
B $310 \mathrm{~ms}^{-1}$
C $258 \mathrm{~ms}^{-1}$
D $204 \mathrm{~ms}^{-1}$

The correct answer is:
$204 \mathrm{~ms}^{-1}$

Solution:
We know that, fundamental frequency
$$
\begin{aligned}
f_0 & =\frac{v}{2 l} \\
\text { Beat } & =f_1-f_2 \\
\Rightarrow \frac{40}{10} & =\frac{v}{2 l_1}-\frac{v}{2 l_2} \Rightarrow 4=\frac{v}{2}\left(\frac{1}{l_1}-\frac{1}{l_2}\right) \\
\Rightarrow \quad 4 & =\frac{v}{2}\left(\frac{1}{50}-\frac{1}{51}\right) \times \frac{1}{10^{-2}} \\
\Rightarrow 8 \times 10^{-2} & =v\left(\frac{51-50}{50 \times 51}\right) \\
\Rightarrow 8 \times 10^{-2} & =\frac{v}{2550} \\
\Rightarrow \quad v & =2550 \times 8 \times 10^{-2} \\
& =204 \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 2)
Two open pipes of different lengths and of same diameter in which the air column vibrates with fundamental frequencies ‘ n1 ’, and ‘ n2 ’ respectively. When both pipes are joined to form a single pipe, its fundamental frequency will be

Options:
A n1+n2n1n2
B n1n22n2+n1
C 2n2+n1n1n2
D n1n2n1+n2

The correct answer is:
n1n2n1+n2

Solution:
n1=v2l1            ;      n2= v2l2
∴ n = v2r+l2 =   v2r+2l2
n = vvn1+vn2   = n1n2n1+n2

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2019 (02 May Shift 1)
Two plants can be conclusively said to belong to the same species if they:

Options:
A have more than 90 per cent 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:
Two organisms are said to belong to same species if they can interbreed among themselves to produce fertile offspring. In the case of plants, seeds represent the embryonic structure of the plant which later develops into offspring. So, the correct option is 'Can reproduce freely with each other and form seeds'.

Topic: Biology, The Living World, NEET, NEET 2007