datavorous/entrance-exam-dataset huggingface dataset:
In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30° when light of wavelength 500 nm is incident on the slit. The first secondary maximum is observed at an angle of

Options:
A sin−1⁡12
B sin−1⁡34
C sin−1⁡14
D sin−1⁡23

The correct answer is:
sin−1⁡34

Solution:
Given that for slit of width a the first minimum form at θ=30°, the wavelength of light is 500 nm

Using Bragg's law of diffraction, nλ=asinθ, here n is the whole number and for the first minimum n=1

⇒1λ=asin30°⇒a=2λ

For the first secondary maximum, n+12λ=asinθ' where n=1

⇒32λ=asinθ'⇒32λ=2λsinθ'⇒θ'=sin-134

Topic: Physics, Wave Optics, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
In a diffraction pattern due to single slit of width 'a', the first minimum is observed at an angle of $30^{\circ}$ when the light of wavelength $5400 Å$ is incident on the slit. The first secondary maximum is observed at an angle of $\left(\sin 30^{\circ}=\frac{1}{2}\right)$

Options:
A $\sin ^{-1}\left(\frac{3}{4}\right)$
B $\sin ^{-1}\left(\frac{2}{3}\right)$
C $\sin ^{-1}\left(\frac{1}{2}\right)$
D $\sin ^{-1}\left(\frac{1}{4}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{3}{4}\right)$

Solution:
For $\mathrm{n}^{\text {th }}$ secondary minimum, path difference $=a \sin \theta_{\mathrm{n}}=\mathrm{n} \lambda$ For $\mathrm{n}^{\text {th }}$ secondary maximum, path difference $=a \sin \theta_n=(2 n+1) \frac{\lambda}{2}$
$\therefore \quad$ For $1^{\text {st }}$ minimum, a $\sin 30^{\circ}=\lambda$
For $2^{\text {nd }}$ maximum, a $\sin \theta_{\mathrm{n}}=(2+1) \frac{\lambda}{2}=\frac{3 \lambda}{2}$
$\therefore \quad$ Dividing equation (i) by equation (ii),
$\frac{\left(\frac{1}{2}\right)}{\sin \theta_n}=\frac{2}{3} \Rightarrow \theta_n=\sin ^{-1}\left(\frac{3}{4}\right)$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2023 (09 May Shift 1)
In a factory, there are 30 men and 20 women employees. If the average salary of men is Rs 4050 and the average salary of all the employees is Rs 3550 , then what is the averago salary of women?

Options:
A $\operatorname{Rs} 3800$
B $\mathrm{Rs} \underline{3300}$
C Rs 3000
D Rs 2800

The correct answer is:
Rs 2800

Solution:
Let average salary of women be $\mathrm{x}$. According to question, $\Rightarrow 50 \times 3550=30 \times 4050+20 \mathrm{x}$
$\Rightarrow 177500-121500=20 \mathrm{x}$
$\Rightarrow x=2800$
Hence, average salary of women $=\operatorname{Rs} 2800$

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 1)
In a fertilised embryo sac, the haploid, diploid and triploid structure are:

Options:
A
synergid, zygote and primary endosperm nucleus

B
synergid, antipodal and polar nuclei

C
antipodal, synergid and primary endosperm nucleus

D
synergid, polar nuclei and zygote

The correct answer is:
synergid, zygote and primary endosperm nucleus

Solution:
(i) Synergid - haploid
(ii) Polar nuclei-haploid
(iii) Antipodal-haploid
(iv) Zygote-diploid
(v) Primary Endosperm Nucleus (PEN).
Diploid secondary nucleus fertilises with a haploid male gamete to form a triploid PEN.

Topic: Biology, Sexual Reproduction in Flowering Plants
In a football championship 153 matches were played. Every team played one match with each other team. How many teams participated in the championship?

Options:
A 21
B 18
C 17
D 15

The correct answer is:
18

Solution:
Let total no. of team participated in a championship ben. Since, every team played one match with each other team.
$\therefore{ }^{n} C_{2}=153 \Rightarrow \frac{n !}{2 !(n-2) !}=153$
$\Rightarrow \frac{n(n-1)(n-2) !}{2 !(n-2) !}=153 \Rightarrow \frac{n(n-1)}{2}=153$
$\Rightarrow n(n-1)=306$
$\Rightarrow n^{2}-n-306=0$
$\Rightarrow n^{2}-18 n+17 n-306=0$
$\Rightarrow n(n-18)+17(n-18)=0$
$\Rightarrow n=18,-17$
$n$ cannot be negative $\therefore n \neq-17$
$\Rightarrow n=18

Topic: Mathematics, Permutation Combination, NDA, NDA 2009 (Phase 2)
In a Fraunhofer diffraction at a single slit of width $d$ and incident light of wavelength $5500 \mathrm{~A}$, the firs minimum is observed at an angle $30^{\circ}$. The first secondary maxima are observed at an angle $\theta=$

Options:
A $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
B $\sin ^{-1}\left(\frac{1}{4}\right)$
C $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
D $\sin ^{-1}\left(\frac{3}{4}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{3}{4}\right)$

Solution:
Slit width $=d$
$\lambda=5500 \AA=5.5 \times 10^{-7} \mathrm{~m}, \theta_1=30^{\circ}$
For first secondary minima, $d \sin \theta_1=\lambda$
$d=\frac{\lambda}{\sin \theta_1}=\frac{5.5 \times 10^{-7}}{\sin 30^{\circ}}=11 \times 10^{-7}$
For first secondary maxima, $d \sin \theta_1=\frac{3 \lambda}{2}$
$\begin{aligned} & \Rightarrow \sin \theta_1=\frac{3 \lambda}{2 d}=\frac{5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}}=\frac{3}{4} \\ & \theta_1=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2022 (10 Aug Shift 2)
In a Fraunhofer diffraction at single slit of width $d$ with incident light of wavelength 5500 $Å$, the first minimum is observed, at angle $30^{\circ}$. The first secondary maximum is observed at an angle $\theta=$

Options:
A $\sin ^{-1} \frac{1}{\sqrt{2}}$
B $\sin ^{-1} \frac{1}{4}$
C $\sin ^{-1} \frac{3}{4}$
D $\sin ^{-1} \frac{\sqrt{3}}{2}$

The correct answer is:
$\sin ^{-1} \frac{3}{4}$

Solution:
Slit width $=d$
$\lambda=5500 Å=5.5 \times 10^{-7} \mathrm{~m}, \theta_n=30^{\circ}$
For first secondary minima, $d \sin \theta_n=\lambda$
$d=\frac{\lambda}{\sin \theta_n}=\frac{5.5 \times 10^{-7}}{\sin 30^{\circ}}=11 \times 10^{-7} \mathrm{~m}$
For first secondary maxima, $d \sin \theta_n=\frac{3 \lambda}{2}$
i. e. $\sin \theta_n=\frac{3 \lambda}{2 d}=\frac{3 \times 5.5 \times 10^{-7}}{2 \times 11 \times 10^{-7}}$
$\sin \theta_n=\frac{3}{4}$ or $\theta_n=\sin ^{-1}(3 / 4)$

Topic: Physics, Wave Optics, AIIMS, AIIMS 2016
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength $400 \mathrm{~nm}$, the first minimum is formed at an angle of $30^{\circ}$. The direction $\theta$ of the first secondary maximum is given by

Options:
A $\sin ^{-1}\left(\frac{2}{3}\right)$
B $\sin ^{-1}\left(\frac{3}{4}\right)$
C $\sin ^{-1}\left(\frac{1}{4}\right)$
D $\tan ^{-1}\left(\frac{2}{3}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{3}{4}\right)$

Solution:
For first diffraction minimum
$$
\Rightarrow \quad \begin{aligned}
a \sin \theta &=\lambda \\
a &=\frac{\lambda}{\sin \theta}
\end{aligned}
$$
For first secondary maximum
$$
a \sin \theta^{\prime}=\frac{3 \lambda}{2}
$$
or $\sin \theta^{\prime}=\frac{3 \lambda}{2} \times \frac{1}{a}=\frac{3 \lambda}{2} \times \frac{\sin \theta}{\lambda}=\frac{3}{2} \times \sin 30^{\circ}=\frac{3}{4}$
or
$$
\theta^{\prime}=\sin ^{-1}\left(\frac{3}{4}\right)
$$

Topic: Physics, Wave Optics, KCET, KCET 2008
In a Fraunhofer diffraction experiment at a single slit using a light of wavelength $400 \mathrm{~nm}$, the first minimum is formed at an angle of $30^{\circ}$. The direction $\theta$ of the first secondary maximum is given by

Options:
A $\sin ^{-1}(2 / 3)$
B $\sin ^{-1}(3 / 4)$
C $\sin ^{-1}(1 / 4)$
D $\tan ^{-1}(2 / 3)$

The correct answer is:
$\sin ^{-1}(1 / 4)$

Solution:
Given, wavelength of used light,
$$
\lambda=400 \mathrm{~nm}
$$
For minima in Fraunhofer diffraction,
$\therefore \quad d \sin 30^{\circ}=1 \times 400$
$\Rightarrow \quad d=800 \mathrm{~nm}$
For maxima in diffraction,
$$
d \sin \theta=(2 n+1) \frac{\lambda}{2}
$$
For first maxima $n=0$
$$
\begin{array}{ll}
\therefore & d \sin \theta=(0+1) \frac{\lambda}{2} \\
\Rightarrow & d \sin \theta=\frac{\lambda}{2} \\
\Rightarrow & 800 \sin \theta=\frac{400}{2} \\
\Rightarrow & \sin \theta=\frac{1}{4} \\
\Rightarrow & \theta=\sin ^{-1}\left(\frac{1}{4}\right)
\end{array}
$$

Topic: Physics, Wave Optics, COMEDK, COMEDK 2018
In a Fraunhoffer diffraction experiment at a single slit using light of wavelength 400 nm , the first minimum is formed at an angle of 30Â° . Then the direction Î¸ of the first secondary maximum is

Options:
A tan-1⁡43
B 60o
C sin-1⁡34
D tan-1⁡34

The correct answer is:
sin-1⁡34

Solution:
For first minimum,
sin⁡30o=λa
12=λa
a=2λ
Now, For 1st secondary maxima
sin⁡θ=3λ2a
sin⁡θ=3λ2×2λ
sin⁡θ=34
θ=sin-1⁡34

Topic: Physics, Wave Optics, NEET
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is
CH3OHl+32O2gâCO2g+2H2Ol
At 298 K standard Gibbâs energies of formation for CH3OHl, H2Ol and CO2g are -166.2, -237.2 and -394.4 kJ mol-1 respectively. If standard enthalphy of combustion of methanol is -726 kJmol â1 , efficiency of the fuel cell will be

Options:
A 87%
B 90%
C 97%
D 80%

The correct answer is:
97%

Solution:
CH3OHl+32O2g→CO2g+2H2Ol
ΔH=-726kJmol-1
Also ΔGfoCH3OHI=-166.2kJmol-1
ΔGfoH2OI=-166.2kJmol-1
ΔGfoCO2I=-394.4kJmol-1
ΔGf=∑ΔGfoproducts-∑ΔGforeactants
=-394.4-2237.2+166.2=702.6kJmol-1
now efficiency of fuel cell
=ΔGΔH×100=702.6726×100=97%

Topic: Chemistry, Electrochemistry, NEET
In a geometric progression with first term $a$ and common ratio $r$, what is the arithmetic mean of first five terms?

Options:
A $a+2 r$
B $a r^{2}$
C $a\left(r^{5}-1\right) /(r-1)$
D $a\left(r^{5}-1\right) /[5(r-1)]$

The correct answer is:
$a\left(r^{5}-1\right) /[5(r-1)]$

Solution:
Let the Geometric progression be a, ar, ar $^{2}, \mathrm{ar}^{3}, \mathrm{ar}^{4}, \ldots$ First five terms of a geometric progression are $\mathrm{a}, \mathrm{ar}, \mathrm{ar}^{2}$, $\mathrm{ar}^{3}, \mathrm{ar}^{4}$
$\begin{array}{ll}\therefore \quad & \text { Mean }=\frac{a+a r+a r^{2}+a r^{3}+a r^{4}}{5} \\ & =\frac{a\left(r^{5}-1\right)}{5(r-1)} \quad\left(\because \text { Sum of G.P }=\frac{a\left(r^{n}-1\right)}{r-1}\right)\end{array}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2009 (Phase 1)
In a GP of positive terms, any term is equal to one-third of the sum of next two terms. What is the common ratio of the GP?

Options:
A $\frac{\sqrt{13}+1}{2}$
B $\frac{\sqrt{13}-1}{2}$
C $\frac{\sqrt{13}+1}{3}$
D $\sqrt{13}$

The correct answer is:
$\frac{\sqrt{13}-1}{2}$

Solution:
Let a, ar and ar $^{2}$ be three positive terms of G.P.
According to question,
$\mathrm{a}=\frac{1}{3}\left(\mathrm{ar}+\mathrm{ar}^{2}\right)$
$\Rightarrow 3=\mathrm{r}+\mathrm{r}^{2}$
$\Rightarrow \mathrm{r}^{2}+\mathrm{r}-3=0$
$\Rightarrow \mathrm{r}=\frac{-1 \pm \sqrt{1+4 \times 3}}{2}$
$\Rightarrow \mathrm{r}=\frac{-1 \pm \sqrt{13}}{2}=\frac{\sqrt{13}-1}{2},-\left(\frac{1+\sqrt{13}}{2}\right)$
Since, $r$ can not be negative.
$\therefore \quad r=\frac{\sqrt{13}-1}{2}$

Topic: Mathematics, Sequences and Series, NDA, NDA 2011 (Phase 1)
In a group of 50 people, two tests were conducted, one for diabetes and one for blood pressure. 30 people were diagnosed with diabetes and 40 people were diagnosed with high blood pressure. what is the minimum number of people who were having diabetes and high blood pressure

Options:
A 0
B 10
C 20
D 30

The correct answer is:
20

Solution:
$\mathrm{n}(\mathrm{T})=50$
$\mathrm{n}(\mathrm{D})=30$
$\mathrm{n}(\mathrm{H})=40$
$\mathrm{n}(\mathrm{T})=\mathrm{n}(\mathrm{D})+\mathrm{n}(\mathrm{H})-\mathrm{n}(\mathrm{DnH})$
$50=30+40-\mathrm{n}(\mathrm{D} \cap \mathrm{H})$
$\mathrm{n}(\mathrm{D} \cap \mathrm{H})=70-50=20$
Number of people having diabetes and high blood pressure $=20$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
In a group of 500 students, there are 475 students who can speak Hindi and 200 can speak Bengali. What is the number of students who can speak Hindi only ?

Options:
A 275
B 300
C 325
D 350

The correct answer is:
300

Solution:
Total number of students $=500$ Let $\mathrm{H}$ be the set showing number of students who can speak Hindi $=475$ and $\mathrm{B}$ be the set showing number of students who can speak Ben gali $=200$ So, $\mathrm{n}(\mathrm{H})=475$ and $\mathrm{n}(\mathrm{B})=200$ and given that $\mathrm{n}(\mathrm{B} \cup$
$\mathrm{H})=500$
we have $n(B \cup H)=n(B)+n(H)-n(B \cap H)$
$\Rightarrow 500=200+475-n(B \cap H)$
$0, \mathrm{n}(\mathrm{B} \cap \mathrm{H})=175$
Hence, persons who speak Hindi only $=\mathrm{n}(\mathrm{H})-\mathrm{n}(\mathrm{B} \cap \mathrm{H})$
$=475-175=300$

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
In a hyperbola, if the length of transverse axis is twice that of the conjugate axis, then the distance between its directrices is..... units.

Options:
A $\frac{8 b}{\sqrt{5}}$
B $\frac{8 a}{\sqrt{5}}$
C $\frac{2 a}{\sqrt{5}}$
D $\frac{2 b}{\sqrt{5}}$

The correct answer is:
$\frac{8 b}{\sqrt{5}}$

Solution:
Given, in a hyperbola, the length of transverse axis is twice that of the conjugate axis. The equation of standard hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\because$ Length of transverse axis $=2 \times$ length of conjugate axis
$2 a=2 \times 2 b$
$\begin{array}{lll}\Rightarrow & a & =2 b \\ \because & b^2 & =a^2\left(e^2-1\right)\end{array}$
$\begin{array}{ll}\Rightarrow & \frac{a^2}{4}=a^2\left(e^2-1\right) \\ \Rightarrow & e^2=1+\frac{1}{4}=\frac{5}{4}\end{array}$
$\Rightarrow \quad e=\frac{\sqrt{5}}{2}$
$\therefore$ Distance between its directrix $=\frac{2 a}{e}=\frac{2(2 b)}{\sqrt{5} / 2}$
$=\frac{8 b}{\sqrt{5}}$

Topic: Mathematics, Hyperbola, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
In a lottery, 16 tickets are sold and 4 prizes are awarded. If a person buys 4 tickets,what is the probability of his winning a prize?

Options:
A $\frac{4}{16^{4}}$
B $\frac{175}{256}$
C $\frac{1}{4}$
D $\frac{81}{256}$

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

Solution:
$16$ tickets are sold and 4 prizes are awarded. A person
buys 4 tickets, then required probability $=\frac{4}{16}=\frac{1}{4}$

Topic: Mathematics, Probability, NDA, NDA 2007 (Phase 1)
In a manufacture of ready-made garments, which average is used to find the most frequent size ?

Options:
A Arithmetic mean
B Geometric mean
C Mode
D Harmonic mean

The correct answer is:
Mode

Solution:
Mode is most suitable for this.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 1)
In a mature 7-celled or 8-nucleate embryo sac, the ploidy level of the secondary nucleus is the same as that of the:

Options:
A Synergids
B Antipodals
C Nucellus
D Polars

The correct answer is:
Nucellus

Solution:
The embryo sac is the mature female gametophyte of the angiosperm which is a 7 celled 8 nuclei structure. In it, at the chalazal end there are cells called antipodal cells which act as nutritional centre. Female gamete is surrounded by two synergid cells which help in the fertilization process and present at the micropylar end of the sac. In the centre of the sac, polar nuclei is found which is a cell with two nuclei. The ploidy level of synergids, antipodal cells and female gamete 'n' but the polar nuclei is '2n'. It is similar to the nucellus (2n) tissue cells which have not participated in the megasporogenesis process.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In a mature dicot stem which has undergone secondary growth, the youngest layer of secondary xylem is situated ________________.

Options:
A in between pith and primary xylem
B just outside the vascular cambium
C just inner to the vascular cambium
D just inner to the phellogen

The correct answer is:
just inner to the vascular cambium

Solution:
Dicot plants are sub group of the division Angiosperms which have two cotyledons or embryonic leaves in their seeds. Meristematic tissue occurring between the mature regions of root and shoots of angiosperms are called secondary meristem. These generally form by dedifferentiation of the permanent tissues. The vascular cambium tissues found within the vascular tissues like xylem and phloem are secondary meristem which help in lateral growth of the plant helping in increasing the girth of the stem. The cambium cells between the xylem and phloem is called as intra fascicular cambium. The new cambium cells that form towards the pith and mature into secondary xylem that is inner to the vascular cambium and the cells forming towards the periphery on the outer side is called secondary phloem. The younger cambium cells that will develop into xylem grow towards the central or inner side of the stem.

Topic: Biology, Anatomy of Flowering Plants, NEET
In a meter bridge experiment, with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap, the balancing length obtained is l. If the temperature of the water is increased, the new balancing length

Options:
A <l
B >l
C =0
D =l

The correct answer is:
>l

Solution:
According to the question,
RunknownRstandard=l1-l
If temperature increases than resistance increases. Hence the new balancing is greater than l

Topic: Physics, Current Electricity, NEET
In a meter bridge the balancing length from the left end is found to be $25 \mathrm{~cm}$. The value of the unknown resistance is (assume, standard resistance of $1 \Omega$ is in the right gap)

Options:
A $0.25 \Omega$
B $0.33 \Omega$
C $0.20 \Omega$
D $0.50 \Omega$

The correct answer is:
$0.33 \Omega$

Solution:
According to given situation, circuit diagram of meter bridge is given as
At balance condition of meter bridge,
$\frac{S}{R}=\frac{25}{(100-25)} \Rightarrow \frac{S}{1}=\frac{25}{75} \Rightarrow S=\frac{1}{3} \Omega=0.33 \Omega$

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)
In a metre-bridge, the balancing length from the left end when a standard resistance of 1 Î© is in the right gap is found to be 20 cm. The value of the unknown resistance is

Options:
A 0.4 Ω
B 0.5 Ω
C 4 Ω
D 0.25 Î©

The correct answer is:
0.25 Î©

Solution:
Let the unknown resistance be X. Then the condition of Wheatstone's bridge gives XR=20r80r, where r is the resistance of wire per cm.

â´X=2080ÃR=14Ã1=0.25 Î©

Topic: Physics, Current Electricity, NEET
In a multiple-choice test, an examinee either knows the correct answer with probability $\mathrm{p}$, or guesses with probability $1-\mathrm{p}$. The probability of answering a question
correctly is $\frac{1}{\mathrm{~m}}$, if he or she merely guesses. If the examinee
answers a question correctly, the probability that he or she really knows the answer is

Options:
A $\frac{\mathrm{mp}}{1+\mathrm{mp}}$
B $\frac{m p}{1+(m-1) p}$
C $\frac{(\mathrm{m}-1) \mathrm{p}}{1+(\mathrm{m}-1) \mathrm{p}}$
D $\frac{(\mathrm{m}-1) \mathrm{p}}{1+\mathrm{mp}}$

The correct answer is:
$\frac{m p}{1+(m-1) p}$

Solution:
Probability of knowing correct answer $=\mathrm{p}$ Probability to guess correct answer $=(1-\mathrm{p})\left(\frac{1}{\mathrm{~m}}\right)$
Probability to answer correctly $=\mathrm{p}+\frac{1-\mathrm{p}}{\mathrm{m}}$
So, required probability $=\frac{p}{p+\frac{1-p}{m}}=\frac{m p}{m p+1-p}$.
$=\frac{\mathrm{mp}}{1+\mathrm{p}(\mathrm{m}-1)}$

Topic: Mathematics, Probability, NDA, NDA 2017 (Phase 2)
In a non-leap year, the probability of getting 53 Sunday or 53 Tuesday or 53 Thursday is

Options:
A $\frac {1}{7}$
B $\frac {2}{7}$
C $\frac {3}{7}$
D $\frac {4}{7}$

The correct answer is:
$\frac {3}{7}$

Solution:
In non-leap year, there are 365 days i.e. 52 complete weeks and $1$ extra day.
That day can be any of 7 days.
$\therefore \quad P(53 \text { Sunday })=P(\text { Last } 1 \text { day Sunday })=\frac{1}{7}$
Similarly, $P(53$ Tuesday $)=\frac{1}{7}$
$\begin{aligned} & P(53 \text { Thursday })=\frac{1}{7} \\ & \therefore \quad \text { Probability }=\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{3}{7} .\end{aligned}$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2023 (12 May Shift 1)
In a pea flower, five petals are arranged in a specialized manner with one posterior, two lateral and two anterior. These are named as ___________, ___________ and ___________ respectively.

Options:
A Keel, Standard and Carina
B Standard, Wings and Keel
C Keel, Wings and Standard
D Vexillum, Keel and Standard

The correct answer is:
Standard, Wings and Keel

Solution:
In pea flowers, vexillary aestivation is found. In this aestivation, the largest petal, i.e., standard overlaps the two smaller lateral petals that are wings which in turn overlap. The two smallest anterior petals are keel.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2023 (Manipur)
In a periodic table the average atomic mass of magnesium is given as $24.312 \mathrm{u}$. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are ${ }_{12}^{24} \mathrm{Mg}(\mathbf{2 3 . 9 8 5 0 4 \mathrm { u } ) \text { , }}$ ${ }_{12}^{25} \mathrm{Mg}(\mathbf{2 4 . 9 8 5 8 4 u})$ and ${ }_{12}^{26} \mathrm{Mg}(\mathbf{2 5 . 9 8 2 5 9} \mathrm{u})$. The natural abundance of ${ }_{12}^{24} \mathrm{Mg}$ is $78.99 \%$ by mass. Calculate the abundances of other two isotopes.

Options:

None

Solution:
Let the abundance of ${ }_{12}^{25} \mathrm{Mg}$ be $\mathrm{x} \%$
$$
\begin{aligned}
\therefore \quad & \text { Abundance of }{ }_{12}^{26} \mathrm{Mg} \\
&=\left(100-\text { abundance of }{ }_{12}^{24} \mathrm{Mg}-\mathrm{x}\right) \\
&=100-8.99-\mathrm{x}=(21.01-\mathrm{x}) \%
\end{aligned}
$$
Average atomic mass $=24.312 \mathrm{u}$
$$
\begin{aligned}
=& \frac{23.98504 \times 78.99+24.98584 \times x+25.98259(21.01-x)}{100} \\
\Rightarrow & 24.312=\frac{1894.578+24.98584 x+545.89-25.98259 x}{100} \\
\Rightarrow & 2431.2=2440.47-0.9967 x \\
\Rightarrow & x=\frac{2440.47-2431.2}{0.9967}=\frac{9.270}{0.9967}=9.3 \% \\
\therefore \quad & \% \text { of }{ }_{12}^{25} \mathrm{Mg}=9.3 \% \\
& \% \text { of }{ }_{12}^{26} \mathrm{Mg}=21.01-9.3=11.71 \%
\end{aligned}
$$

Topic: Physics, Nuclear Physics
In a pipe closed at one end, air column is vibrating in the fourth overtone, so the vibrating air column has

Options:
A 4 nodes and 5 antinodes.
B 5 nodes and 4 antinodes.
C 5 nodes and 5 antinodes.
D 4 nodes and 4 antinodes.

The correct answer is:
5 nodes and 5 antinodes.

Solution:
For a pipe with one closed end:
Fundamental mode has one node and one antinode.
First overtone has two nodes and two antinodes.
Second overtone has three nodes and three antinodes.
So, the fifth overtone has five nodes and five antinodes.

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (06 Aug Shift 1)
In a plant organ which is covered by periderm and in which the stomata are absent, some gaseous exchange still takes place through

Options:
A aerenchyma
B trichomes
C pneumatophores
D lenticels.

The correct answer is:
lenticels.

Solution:
Due to continued formation of secondary tissues in the older stem and roots however the epidermis gets stressed and ultimately tends to rupture and followed by death of epidermal cells and outer tissues, and a new protective layer is developed called periderm. Structurally it consists of three parts Phellogen or Cork cambium, phellem or cork and phelloderm.
Usually in the periderm of most plants, certain areas with loosely arranged cells have been found, which possess more or less raised and corky spots where the underneath tissues break through the epidermis. Such areas are universally found in the stems of woody plants. These broken areas are called lenticels, which perform the function of exchange of gases in the absence of stomata.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2004
In a poisson distribution, If P (X) = 2 is twice
P (X), = 1 then the standard deviation of the
distribution is

Options:
A 1
B 2
C 4
D 8

The correct answer is:
2

Solution:
$\begin{aligned} & \text { } P(x=2)=2 P(x=1) \\ & P(X=x)=\frac{e^{-\lambda} \cdot \lambda^x}{x !} \Rightarrow \frac{e^{-x} \cdot \lambda^2}{2 !}=2 \times \frac{e^{-x} \cdot \lambda^1}{1 !} \\ & \Rightarrow \quad \frac{\lambda}{2}=2 \Rightarrow \lambda=4 \\ & \because \quad \mu=n p=\lambda \Rightarrow \mu=4 \\ & \text { So, } \quad \sigma=\sqrt{\mu}=\sqrt{4}=2 \\ & \end{aligned}$

Topic: Mathematics, Probability, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 2)