datavorous/entrance-exam-dataset huggingface dataset:
An amine compound on reaction with benzenesulphonyl chloride gives an organic compound (P) which is highly acidic in nature. The compound P is

Options:
A
B
C
D

The correct answer is:

Solution:
We know that the reaction of the benzene sulfonyl chloride with primary amines gives a sulfonamide product that is soluble in alkali while with secondary amines it gives a sulfonamide product that is not soluble in alkali. Secondary amine is acidic in nature because amine contains one hydrogen with Nitrogen but in the case of primary amine it contains two hydrogen atoms with the nitrogen molecule that's why it is more acidic in nature

Hence, we conclude that primary amine is more acidic than secondary amine so the compound P will be:

Topic: Chemistry, Amines, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 2)
An amine $(\mathrm{X})$ reacts with p-toluene sulphonyl chloride to give the product, $\mathrm{Y}$, which is insoluble in alkali. The product of $\mathrm{X}$ with benzoyl chloride is

Options:
A

B

C $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCOC}_6 \mathrm{H}_5$
D

The correct answer is:

Solution:
A secondary amine reacts with hinsberg reagent (benzene sulphonyl chloride) and forms a corresponding sulphonamide that is insoluble in aqueous $\mathrm{NaOH}$. So, ' $\mathrm{X}$ ' must be a secondary amine.

Topic: Chemistry, Amines, TS EAMCET, TS EAMCET 2023 (13 May Shift 2)
An analysis of monthly wages paid to the workers in two firms A and B belonging to the same industry the following result:
\begin{array}{|l|c|c|}
\hline & Firm A & Firm B \\
\hline Number of workers & 500 & 600 \\
\hline Average monthly wage & â‚¹ 1860 & â‚¹ 1750 \\
\hline Variance of distribution of wages & 81 & 100 \\
\hline
\end{array}
The average of monthly wages and variance of distribution of wages of all the workers in the firms A and B taken together are

Options:
A $â‚¹ 1860,100$
B $â‚¹ 1750,100$
C $â‚¹ 1800 \mathrm{~m}, 81$
D None of above

The correct answer is:
None of above

Solution:
Combined Average $=\frac{500 \times 1860+600 \times 1750}{1100}=1800$
Combined Variance
$$
\begin{array}{l}
=\frac{500(81+3600)+600(100+2500)}{1100} \\
=\frac{(5 \times 3681)+(6 \times 2600)}{1100} \approx 3092
\end{array}
$$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 2)
An analysis of monthly wages paid to workers in two firms $A$ and $B$, belonging to the same industry, gives the following results:
(i) Which firm $A$ or $B$ pays larger amount as monthly wages ?
(ii) Which firm $A$ or $B$ shows greater variability in individual wages?

Options:

None

Solution:
For Firm $A$ :
No. of wage earners $=586$
Mean of monthly wages $\bar{x}=₹ 5253$
Amount paid by firm $A=₹(586 \times 5253)$
$=$ ₹ 3078258
Variance of distribution of wages $=100$
Standard deviation $=\sigma=\sqrt{\text { Variance }}$ $=\sqrt{100}=10$
Coefficient of variation $=\frac{\sigma}{\bar{x}} \times 100$ $=\frac{10}{5253} \times 100=0.19$
For firm B
No. of wage earners $=648$
Mean of monthly wages, $\bar{x}=₹ 5253$
Amount paid by firm $B=648 \times 5253=₹ 3403944$
$\mathrm{S} . \mathrm{D}=\sigma=\sqrt{\text { Variance }}=\sqrt{121}=11$
$\therefore$ Coefficients of variation $=\frac{\sigma}{\bar{x}} \times 100$
$=\frac{11}{5253} \times 100=0.21$
Monthly wage paid by firm $A=₹ 3078258$
Monthly wage paid by firm $B=₹ 3403944$
Firm $B$ pays out larger amount as monthly wages
Coefficient of variation of wages, of firm $\mathrm{A}=0.19$
Coefficient of variation of wages of firm $\mathrm{B}=0.21$
Therefore, firm B shows greater variability in individual ways.

Topic: Mathematics, Statistics
An Andalusian fowl, with the heterozygous condition of alleles for black plumage B and white b is blue coloured. A blue Andalusian fowl is bred to a black Andulasian fowl, the proportion of the offspring will be:

Options:
A 50 % white : 50 % blue
B 50 % black : 50 % blue
C 50 % black : 50 % white
D 25 % black : 25 % blue : 50 % white

The correct answer is:
50 % black : 50 % blue

Solution:
Incomplete dominance is a deviation of Mendels law of dominance where the heterozygous individual instead of expressing the dominant trait, it expresses a trait which is a mix of the parent traits. On crossing, black form fowl with white form fowl, the offspring is blue form.
When a black form fowl is crossed with blue form fowl we get-

B	B

B	BB
(Black)	BB
(Black)

b	Bb
(Blue)	Bb
(Blue)

Genotype of F1 - BB and Bb 1:1
Phenotype of F1 - Black and Blue 1:1 50% black and 50% blue.

Topic: Biology, Principles of Inheritance and Variation, NEET
An aqueous solution containing 1 M each of Au3+,Cu2+,Ag+,Li+ is being electrolysed by using inert electrodes. The value of standard potentials are
EAg+/Ago=0.80V,ECu+/Cuo=0.34V and EAu3+/Auo=1.50V,ELi+/Lio=-3.03V
with increasing voltage, the sequence of deposition of metals on the cathode will be

Options:
A Li, Cu, Ag, Au
B Cu, Ag, Au
C Au, Ag, Cu
D Au, Ag, Cu, Li

The correct answer is:
Au, Ag, Cu

Solution:
Only Au3+,Ag+ and Cu2+ will deposit at cathode. Li will not deposit at cathode because SRP of H2 is 0.00V, so after Cu2+ , H2 will evolve at cathode.

Topic: Chemistry, Electrochemistry, NEET
An aqueous solution containing 1 M each of Au3+ , Cu2+ , Ag+ , Li+ is being electrolyzed by using inert electrodes. The value of standard potentials are
EAg+/AgÂ°=0.80V,ECu+/CuÂ°=0.34V, EAu3+/AuÂ°=1.50V,ELi+/LiÂ°=-3.03V
With increasing voltage, the sequence of deposition of metals on the cathode will be

Options:
A Li,Cu,Ag,Au
B Cu,Ag,Au
C Au,Ag,Cu
D Au,Ag,Cu,Li

The correct answer is:
Au,Ag,Cu

Solution:
The most unreactive metal that the ion which has the most positive reduction potential will be released to first. Further Li+ has a very negative reduction potential hence it will not be released in aqueous medium.

Topic: Chemistry, Electrochemistry, NEET
An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge up to a radius R. The atom as a whole is neutral. For this model, what is the electric field at a distance r from the nucleus? (If r

ChemistryElectrostaticsNEET

Options:

A $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{2}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$
B $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$
C $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{R}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$
D $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{2r}{{{R}^{3}}} \right)$

Solution:

1124 Upvotes

Verified Answer

The correct answer is:
$\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$

Correct Option is : (B)
$\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$

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Options:
A $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{2}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$
B $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$
C $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{R}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$
D $\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{2r}{{{R}^{3}}} \right)$

The correct answer is:
$\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$

Solution:
Correct Option is : (B)
$\frac{Ze}{4\pi {{\epsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}-\frac{r}{{{R}^{3}}} \right)$

Topic: Chemistry, Electrostatics, NEET
An element \( X \) decays into element \( Z \) by two-step process.
\[
\begin{array}{l}
X \rightarrow Y+4 e \\
Y \rightarrow Z+2 e^{-} \text {then }:
\end{array}
\]

Options:
A \( X \& Z \) are isobars.
B \( X \& Y \) are isotopes
C \( X \& Z \) are isotones
D \( X \& Z \) are isotopes

The correct answer is:
\( X \& Z \) are isotopes

Solution:
The two processes are as follows:
\[
\mathrm{z}^{\mathrm{A}} X \rightarrow_{\mathrm{Z}-2}^{\mathrm{A}-{ }^{4} Y+{ }_{2}{ }^{4} \mathrm{He}}
\]
Since \( X \) and \( Z \) have same atomic number, that is, same number of protons but different atomic mass, that is, sum of
number of protons and neutrons.
We know isotopes are atoms with same number of protons but different number of neutrons. Therefore, \( X \) and \( Z \) are
isotopes.

Topic: Physics, Nuclear Physics, KCET, KCET 2016
An e.m.f. $E=4 \cos (1000 t)$ volt is applied to an LR circuit of inductance $3 \mathrm{mH}$ and resistance $4 \Omega$. The maximum current in the circuit is

Options:
A $\frac{4}{\sqrt{7}} \mathrm{~A}$
B $1.0 \mathrm{~A}$
C $\frac{4}{7} \mathrm{~A}$
D $0.8 \mathrm{~A}$

The correct answer is:
$0.8 \mathrm{~A}$

Solution:
For LR circuit,
$Z=\sqrt{R^2+(\omega L)^2}$
Comparing equation $E=4 \cos (1000 t)$ with standard equation, $\mathrm{E}=\mathrm{E}_0 \cos \omega \mathrm{t}$, we get $\omega=1000$ units and $E_0=4 \mathrm{~V}$
$\begin{aligned}
\therefore \quad Z & =\sqrt{16+\left(1000 \times 3^2 \times 10^{-6}\right)} \\
& Z=\sqrt{16+9}=5 \Omega \\
\therefore \quad & I=\frac{E_0}{Z}=\frac{4}{5}=0.8 \mathrm{~A}
\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2023 (09 May Shift 2)
An equilateral triangle has one vertex at $(0,0)$ and another at $(3, \sqrt{3})$. What are the coordinates of the third vertex?

Options:
A $(0,2 \sqrt{3})$ only
B $(3,-\sqrt{3})$ only
C $(0,2 \sqrt{3})$ or $(3,-\sqrt{3})$
D Neither $(0,2 \sqrt{3})$ nor $(3,-\sqrt{3})$

The correct answer is:
$(0,2 \sqrt{3})$ or $(3,-\sqrt{3})$

Solution:
Let $A B C$ is equilateral triangle with $A(0,0)$ and $B(3, \sqrt{3})$ and $C$ to be known.
$A B=\sqrt{(3-0)^{2}+(\sqrt{3}-0)^{2}}=\sqrt{9+3}=\sqrt{12}$
Take option (a) i.e. $C(0,2 \sqrt{3})$
$C A=\sqrt{0^{2}+(2 \sqrt{3})^{2}}=\sqrt{12}$
$C B=\sqrt{(3)^{2}+(\sqrt{3})^{2}}=\sqrt{12}$
Take option (b) i.e. $C(3,-\sqrt{3})$
$C A=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12}$
$C B=\sqrt{(0)^{2}+(2 \sqrt{3})^{2}}=\sqrt{12}$
Both option (a) and (b) are correct.

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)
An equilateral triangle is inscribed in the circle $x^{2}+y^{2}=a^{2}$ with one of the vertices at $(a, 0)$. What is the equation of the $\begin{array}{ll}\text { side opposite to this vertex ? }

Options:
A $2 x-a=0$
B $x+a=0$
C $2 \mathrm{x}+\mathrm{a}=0$
D $\quad 3 x-2 a=0$

The correct answer is:
$2 \mathrm{x}+\mathrm{a}=0$

Solution:
Since the equilateral triangle is inscribed in the circle with centre at the origin, centroid lies on the origin.
So, $\frac{A O}{O D}=\frac{2}{1}$
and $\mathrm{OD}=\frac{1}{2} \mathrm{AO}=\frac{\mathrm{a}}{2}$
So, other vertices of triangle have coordinates,
$\left(-\frac{\mathrm{a}}{2}, \frac{\sqrt{3 \mathrm{a}}}{2}\right)$ and $\left[-\frac{\mathrm{a}}{2},-\frac{\sqrt{3}}{2} \mathrm{a}\right]$
$\left(-\frac{a}{2} \frac{\sqrt{3 a}}{2}\right) y$
$$
\left(-\frac{a}{2}, \frac{\sqrt{3 a}}{2}\right)
$$
$\therefore \quad$ Equation of line $\mathrm{BC}$ is
$$
\begin{aligned}
x &=-\frac{a}{2} \\
\Rightarrow \quad 2 x+a &=0
\end{aligned}
$$

Topic: Mathematics, Circle, NDA, NDA 2006 (Phase 1)
An example of a seed with endosperm, perisperm and caruncle is

Options:
A lily.
B castor.
C cotton.
D coffee.

The correct answer is:
castor.

Solution:
In the central sac of an embryo, a primary endosperm nucleus is present from which endosperm is developed. The primary endosperm nucleus is formed when one male gamete fuses with a secondary nucleus and this fusion is called triple fusion. They provide an essential nutrient to the embryo. The residual nucellus that persists in the seed is called perisperm. The caruncle is formed when cells present at the tip of the outer integument become enlarged and form a fleshy outgrowth at the micropylar side of the seed. It helps in seeds dispersal. The endosperm, perisperm is found in all four options mentioned above but the caruncle is present only in castor.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
An experiment consists of flipping a coin and then flipping it a second time if head occurs. If a tail occurs on the first flip, then a six-faced die is tossed once. Assuming that the outcomes are equally likely, what is the probability of getting one head and one tail?

Options:
A $1 / 4$
B $1 / 36$
C $1 / 6$
D $1 / 8$

The correct answer is:
$1 / 8$

Solution:
Events when a coin is flipped and head occurs are $\{\mathrm{HT}, \mathrm{HH}\}$
Events when a coin is flipped and tail occurs are $\{\mathrm{T} 1$, $\mathrm{T} 2, \mathrm{~T} 3, \mathrm{~T} 4, \mathrm{~T} 5, \mathrm{~T} 6\}$
$(\because$ dice are rolled after tail appears) So, Total number of events $=8$ Favourable event $=\{\mathrm{H} \mathrm{T}\}=1$
$\therefore$ Required probability $=\frac{1}{8}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 1)
An ideal transformer converts 220 V AC to 3.3 kV AC. It transmits the power of 4.4 kW. If the primary coil has 600 turns, then alternating current in the secondary coil is

Options:
A 13 A
B 43 A
C 53 A
D 73 A

The correct answer is:
43 A

Solution:
V in =220 V, V out =3.3Ã 10 3 V
Power=4.4 kW
N p =600
P= V in Ã I in .
I in = 4.4Ã1000 220 = 44Ã10 22 =20âA
e s e p = I p I s
I s = I p = e p e s Ã 20Ã220 3.3Ã1000 = 44 33 = 4 3 A

Topic: Physics, Alternating Current, MHT CET, MHT CET 2018
An ideal transformer converts 220 V AC to 3.3 kV AC. It transmits the power of 4.4 kW. If the primary coil has 600 turns, then alternating current in the secondary coil is

Options:
A 13 A
B 43 A
C 53 A
D 73 A

The correct answer is:
43 A

Solution:
V in =220 V, V out =3.3Ã 10 3 V
Power=4.4 kW
N p =600
P= V in Ã I in .
I in = 4.4Ã1000 220 = 44Ã10 22 =20âA
e s e p = I p I s
I s = I p = e p e s Ã 20Ã220 3.3Ã1000 = 44 33 = 4 3 A

Topic: Physics, Alternating Current, NEET
An ideal transformer is used to set up an alternating e.m.f. of $220 \mathrm{~V}$ to $4.4 \mathrm{kV}$ to transmit $6.6 \mathrm{~kW}$ of power. The primary coil has 100 turns. What is current rating of secondary?

Options:
A $1.0 \mathrm{~A}$
B 0.75 A
C $2.5 \mathrm{~A}$
D $1.5 \mathrm{~A}$

The correct answer is:
$1.5 \mathrm{~A}$

Solution:
$\begin{aligned} & P_0=V_s i_s \\ & 6.6=4.4 \times i_s \Rightarrow i_s=1.5 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2022 (06 Aug Shift 1)
An important evidence in favour of organic evolution is the occurrence of:

Options:
A homologous and analogous organs
B homologous and vestigial organs
C analogous and vestigial organs
D homologous organs only

The correct answer is:
homologous and vestigial organs

Solution:
An important evidence in favour of organic evolution is the occurrence of homologous and vestigial organs. Homologous organs have a common origin and are built on the same fundamental pattern, but they perform different functions and have different appearances. Vestigial organs in animals have no function now, but they had important functions in ancestors. Analogous organs are quite different in fundamental structure and embryonic origin but perform the same function. The study of analogous organs illustrates the occurrence of convergent evolution.

Topic: Biology, Evolution, NEET, NEET 2006
An individual and a population has certain characteristics. Name these attributes with definitions.

Options:

None

Solution:
The attributes of an individual and a population include pattern of distribution, dispersal biotic potential and gene pool. Inter population dispersion or internal distribution patterns or dispersion is phenomenon of distrubuion of individual within geographical boundaries of the population.
An individual is dispersed at one or another time during their life in a population which is revealed by immigration or emigration. This is known as dispersal.
(i) Immigrations refers to the number of individuals of the same species that have come into the habitat from elsewhere during a specified time period.
(ii) Emigration refers to the number of individuals of the population that exit or leave the habitat and go elsewhere during a specified time period.
Biotic Potential. The natural capacity of a population to increase its size under ideal environmental conditions is called biotic potential.
Gene pool. All the genotypes of all individuals in a breeding population is referred to as gene pool.

Topic: Biology, Organisms and Populations
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident involving a scooter driver, car driver and a truck driver are $0.01,0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. The probability that the person is a scooter driver is

Options:
A $\frac{1}{52}$
B $\frac{3}{52}$
C $\frac{15}{52}$
D $\frac{19}{52}$

The correct answer is:
$\frac{1}{52}$

Solution:
Let $\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3} \& \mathrm{~A}$ be events defined as follows. $\mathrm{E}_{1}=$ person chosen is a scooter driver $\mathrm{E}_{2}=$ person chosen is a car driver. $\mathrm{E}_{3}=$ person chosen is a truck driver $\&$ $\mathrm{A}=$ person meets with an accident $\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{2000}{12000}=\frac{1}{6} ; \mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{1}{3} \& \mathrm{P}\left(\mathrm{E}_{3}\right)=\frac{1}{2}$
$\because$ Probability that a person meets with an accident
given that he is a scooter driver $=\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)=0.01$
$\mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)=0.03 \& \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)=0.15$
- the person meets with an accident. $\therefore$ the probability that he was a scooter driver
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \mathrm{P}\left(\mathrm{A} / \mathrm{E}_{3}\right)}$
$=\mathrm{P}\left(\mathrm{E}_{1} / \mathrm{A}\right)=\frac{1 / 6 \times 0.01}{\left(\frac{1}{6} \times 0.01\right)+\left(\frac{1}{3} \times 0.03\right)+\left(\frac{1}{2} \times 0.15\right)}=\frac{1}{52}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 2)
An object is projected with a velocity of $20 \mathrm{~m} / \mathrm{s}$ making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory is $h=A x-B x^2$ where $h$ is height, $x$ is horizontal distance $A$ and $B$ are constants. The ratio $A: B$ is $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A $1: 5$
B $5: 1$
C $1: 40$
D $40: 1$

The correct answer is:
$40: 1$

Solution:
$u=20 \mathrm{~m} / \mathrm{s}, \theta=45^{\circ}$
Equation of trajectory $h=A x-B x^2$ Standard equation
$h=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$
On comparing the two equations
$\begin{aligned} A & =\tan \theta, B=\frac{g}{2 u^2 \cos ^2 \theta} \\ \frac{A}{B} & =\frac{\tan \theta}{\frac{g}{2} u^2 \cos ^2 \theta}=\frac{2 u^2 \cos ^2 \theta \tan \theta}{g} \\ & =\frac{2 \times(20)^2 \cos ^2 45^{\circ} \tan 45^{\circ}}{10} \\ & =\frac{2 \times 400 \times\left(\frac{1}{\sqrt{2}}\right)^2 \times 1}{10} \\ & =2 \times 40 \times \frac{1}{2}=40\end{aligned}$
$A: B=40: 1$

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2001
An object is projected with a velocity of $20 \mathrm{~m} / \mathrm{s}$ making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory is $h=A x-B x^2$ where $h$ is height, $x$ is horizontal distance $A$ and $B$ are constants. The ratio $A: B$ is $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A $1: 5$
B $5: 1$
C $1: 40$
D $40: 1$

The correct answer is:
$40: 1$

Solution:
$u=20 \mathrm{~m} / \mathrm{s}, \theta=45^{\circ}$
Equation of trajectory $h=A x-B x^2$ Standard equation
$h=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$
On comparing the two equations
$\begin{aligned} A & =\tan \theta, B=\frac{g}{2 u^2 \cos ^2 \theta} \\ \frac{A}{B} & =\frac{\tan \theta}{\frac{g}{2} u^2 \cos ^2 \theta}=\frac{2 u^2 \cos ^2 \theta \tan \theta}{g} \\ & =\frac{2 \times(20)^2 \cos ^2 45^{\circ} \tan 45^{\circ}}{10} \\ & =\frac{2 \times 400 \times\left(\frac{1}{\sqrt{2}}\right)^2 \times 1}{10} \\ & =2 \times 40 \times \frac{1}{2}=40\end{aligned}$
$A: B=40: 1$

Topic: Physics, Motion In Two Dimensions, TS EAMCET, TS EAMCET 2001
An object of density $2000 \mathrm{~kg}-\mathrm{m}^{-3}$ is hung from a thin light wire. The fundamental frequency of the transverse waves in the wire is $200 \mathrm{~Hz}$. If the object is immersed in water such that half of its volume is submerged, then the fundamental frequency of the transverse waves in the wire is

Options:
A 200 Hz
B 173.2 Hz
C 100 Hz
D 141.4 Hz

The correct answer is:
173.2 Hz

Solution:
Fundamental frequency, initially
$$
n_1=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}=\frac{1}{2 l} \sqrt{\frac{V g(2000)}{\mu}}
$$
Fundamental frequency, finally
$$
n_2=\frac{1}{2 l} \sqrt{\frac{V g\left(2000-\frac{1000}{2}\right)}{\mu}}
$$
(here we applied loss of weight due to upthrust)
$$
\begin{aligned}
& =\frac{1}{2 l} \sqrt{\frac{V g \times 1500}{\mu}} \\
\Rightarrow \quad \frac{n_1}{n_2} & =\sqrt{\frac{2000}{1500}}=\sqrt{\frac{4}{3}} \\
\Rightarrow \quad n_2 & =\frac{n_1 \times \sqrt{3}}{2}=100 \sqrt{3} \mathrm{~Hz}=173.2 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 2)
An object of specific gravity Ï is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency (in Hz) is:

Options:
A 3002ρ-12ρ12
B 3002ρ2ρ-112
C 3002ρ2ρ-1
D 3002ρ-12ρ

The correct answer is:
3002ρ-12ρ12

Solution:
The diagrammatic representation of the given problem is shown in the figure. The expression of the fundamental frequency is v=12lTμ
T=mg=vρ′g ; f=12lvρ′gμ
T′=mg-V2ρwg
f′=12lvρ′-v2ρwgμ
f′f=2ρ′-ρw2ρ′ρ′ρw=specific gravity=ρ
f′=3002ρ-12ρ Hz

Topic: Physics, Waves and Sound, NEET
An observed event $B$ can occur after one of the three events $A_{1}, A_{2}, A_{3} .$ If
$P\left(A_{1}\right)=P\left(A_{2}\right)=0.4, P\left(A_{3}\right)=0.2$ and $P\left(B / A_{1}\right)=0.25, P\left(B / A_{2}\right)$
$=0.4, P\left(B / A_{3}\right)=0.125$, what is the probability of $A_{1}$ after observing $B$ ?

Options:
A $\frac{1}{3}$
B $\frac{6}{19}$
C $\frac{20}{57}$
D $\frac{2}{5}$

The correct answer is:
$\frac{20}{57}$

Solution:
By Baye's theorem Required probability $=P\left(A_{1} / B\right)$ $=\frac{P\left(A_{1}\right) P\left(B / A_{1}\right)}{P\left(A_{1}\right) P\left(B / A_{1}\right)+P\left(A_{2}\right) P\left(B / A_{2}\right)+P\left(A_{3}\right) P\left(B / A_{3}\right)}$
$=\frac{0.4 \times 0.25}{0.4 \times 0.25+0.4 \times 0.4+0.2 \times 0.125}$
$=\frac{0.1}{0.1+0.16+0.025}=\frac{0.1}{0.285}=\frac{20}{57}$

Topic: Mathematics, Probability, NDA, NDA 2010 (Phase 1)
An old piece of dicot stem and a dicot root is given to a candidate during practical examination. Which of the following anatomical structures can be used to distinguish between the two?

Options:
A Secondary xylem
B Secondary phloem
C Protoxylem
D Cortical cells

The correct answer is:
Protoxylem

Solution:
The first formed primary xylem elements are called protoxylem and the later formed primary xylem is called metaxylem. In stems, the protoxylem lies towards the centre (pith) and the metaxylem lies towards the periphery of the organ. This type of primary xylem is called endarch. In roots, the protoxylem lies towards periphery and metaxylem lies towards the centre. Such an arrangement of primary xylem is called an exarch.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2014