datavorous/entrance-exam-dataset huggingface dataset:
Three sides of a trapezium are each equal to $6 \mathrm{~cm}$. Let $\alpha \in\left(0, \frac{\pi}{2}\right)$ be the angle between a pair of adjacent sides.
If the area of the trapezium is the maximum possible, then what is $\alpha$ equal to?

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{2 \pi}{5}$

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

Solution:
Length of three sides of trapezium $=6 \mathrm{~cm}$ Let $\mathrm{AE}$ be height of trapezium.
In $\Delta \mathrm{ADE}, \mathrm{AD}^{2}=\mathrm{AE}^{2}+\mathrm{DE}^{2}$
$\Rightarrow 6^{2}=\mathrm{AE}^{2}+\mathrm{x}^{2}$
$\Rightarrow \mathrm{AE}^{2}=36-\mathrm{x}^{2}$
$\Rightarrow \mathrm{AE}=\sqrt{36-\mathrm{x}^{2}}$
Area of trapezium $=\frac{1}{2} \mathrm{~h}(\mathrm{a}+\mathrm{b})$
$=\frac{1}{2} \sqrt{36-\mathrm{x}^{2}}(6+6+2 \mathrm{x})$
$A=(6+x) \sqrt{36-x^{2}}$
Given, Area of trapezium is maximum. $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[(6+\mathrm{x}) \cdot \sqrt{36-\mathrm{x}^{2}}\right]$
$=(6+x) \cdot\left(\frac{-2 x}{2 \sqrt{36-x^{2}}}\right)+\sqrt{36-x^{2}}$
$=\sqrt{36-x^{2}}-\frac{x(6+x)}{\sqrt{36-x^{2}}}$
$=\frac{36-x^{2}-6 x-x^{2}}{\sqrt{36-x^{2}}}=\frac{36-6 x-2 x^{2}}{\sqrt{36-2 x^{2}}}$
$\frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow 36-6 \mathrm{x}-2 \mathrm{x}^{2}=0$
$\Rightarrow 2 x^{2}+6 x-36=0$
$\Rightarrow x^{2}+3 x-18=0$
$\Rightarrow x^{2}+6 x-3 x-18=0$
$\Rightarrow x(x+6)-3(x+6)=0$
$(x+6)(x-3)=0$
$\Rightarrow x=-6$ or 3 . Since $x$ cannot be negative. So, $x=3$ $\therefore \operatorname{In} \triangle \mathrm{ADE}, \cos \alpha=\frac{\mathrm{x}}{6}=\frac{3}{6}=\frac{1}{2} \Rightarrow \alpha=\frac{\pi}{3}$.

Topic: Mathematics, Area Under Curves, NDA, NDA 2018 (Phase 1)
Three sides of a trapezium are each equal to $6 \mathrm{~cm}$. Let $\alpha \in\left(0, \frac{\pi}{2}\right)$ be the angle between a pair of adjacent sides.
What is the maximum area of the trapezium?

Options:
A $36 \sqrt{3} \mathrm{~cm}^{2}$
B $30 \sqrt{3} \mathrm{~cm}^{2}$
C $27 \sqrt{3} \mathrm{~cm}^{2}$
D $24 \sqrt{3} \mathrm{~cm}^{2}$

The correct answer is:
$27 \sqrt{3} \mathrm{~cm}^{2}$

Solution:
Area $=(6+x) \sqrt{36-x^{2}}$
$=(6+3) \sqrt{36-3^{2}}$
$=9 \sqrt{36-9}$
$=9 \sqrt{27}$
$=9 \sqrt{9 \times 3}$
$=27 \sqrt{3}$

Topic: Mathematics, Area Under Curves, NDA, NDA 2018 (Phase 1)
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to $B O D$ test. The samples were labelled $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$; but the laboratory attendant did not note which was which. The BOD values of the three samples $A, B$ and $C$ were recorded as $20 \mathrm{mg} / \mathrm{L}, 8 \mathrm{mg} / \mathrm{L}$ and $400 \mathrm{mg} / \mathrm{L}$, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?

Options:

None

Solution:
The sample with $\mathrm{BOD}$ value of $400 \mathrm{mg} / \mathrm{h}$ is most polluted and should be levelled as $\mathrm{B} \rightarrow$ untreated sewage river $\mathrm{H}_2 \mathrm{O}$ will be $\rightarrow$ less $\mathrm{BOD} \rightarrow 8 \mathrm{mg} / \mathrm{L}-$ sample $A$.
Untreated water $>2^{\circ}$ effluent $\mathrm{BOD}>$ River water $400>20>8$ and so should be labelled as sample C.

Topic: Biology, Microbes in Human Welfare
Three-digit numbers are formed from the digits 1,2 and 3 in such a way that the digits are not repeated. What is the sum of such three-digit numbers?

Options:
A 1233
B 1322
C 1323
D 1332

The correct answer is:
1332

Solution:
Sum of the numbers $=$ Sum of given numbers. $(\mathrm{n}-1) !\left[10^{0}+10^{1}+10^{2}+\ldots . .\right]$
Here, sum of th ree digit numbers $=\underline{\text { Sum }}$ of the numbers $(3-1) !\left[10^{\circ}+10^{1}+10^{2}\right]$
$=(1+2+3)(3-1) !\left[10^{0}+10^{1}+10^{2}\right]$
$=6 \times 2 \times 1[1+10+100]$
$=12 \times 111$
$=1332$

Topic: Mathematics, Permutation Combination, NDA, NDA 2017 (Phase 1)
Through which cell of the embryo sac, does the pollen tube enter the embryo sac?

Options:
A egg cell
B persistent synergid
C degenerated synergid
D central cell.

The correct answer is:
degenerated synergid

Solution:
Pollen tube grows through the style and reaches the micropylar end. At the micropylar end, there are two synergids around the egg. The synergids have filiform apparatus that guides the pollen tube towards the egg. Thus the pollen tube enters embryo sac through one of the synergids.
Related Theory
The fully developed embryo sac at consists of three antipodal cells, two synergids, one egg cell and one secondary nucleus. The pollen tube pierces through the nucellus and penetrates the wall of the embryo sac. The tip of the pollen tube disintegrates and the two male gametes are set free in the embryo sac. Of the two male gametes, one fuses with the egg cell and the other with the secondary nucleus on the polar nuclei. The fertilised egg becomes zygote and the secondary nucleus fuse to forms primary endosperm nucleus, which is the result of triple fusion.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2005
To enter or leave a cell, the substance must pass through __________ and its structure was explained by ___________ in the year __________.

Options:
A

P	Q	R

the cell wall	the cell theory	1978

B

P	Q	R

the plasma membrane,	the cell theory	1942

C

P	Q	R

the cell wall	the fluid mosaic model	1975

D

P	Q	R

the plasma membrane,	the fluid mosaic model	1972

The correct answer is:

P	Q	R

the plasma membrane,	the fluid mosaic model	1972

Solution:
The plasma membrane is the outermost boundary of the cell and it is selectively permeable. It regulates the movement of substance within a cell. The most accepted model regarding the structure of the plasma membrane is the fluid mosaic model, which was first proposed in 1972. According to this model, the plasma membrane components are arranged in a mosaic pattern and primarily made up of phospholipids, cholesterol, and proteins that move freely by showing fluidly.

Topic: Biology, Cell - The Unit of Life, NEET
To resist the coagulation of $100 \mathrm{cc}$ gold sol; 1 cC of $10 \% \mathrm{NaCl}$ is added to it in the presence of $10^{-4} \mathrm{~g}$ gelatin. The gold number of gelatin is

Options:
A 0.00001
B 0.0001
C 0.001
D 0.01

The correct answer is:
0.01

Solution:
Gold number is the amount (in milligrams) of aprotective colloid, which is just sufficient to prevent the coagulation of $10 \mathrm{~mL}$ of standard hydro gold sol, when $1 \mathrm{~mL}$ of $10 \% \mathrm{NaCl}$ solution is added to it.
$\because 10 \mathrm{cc}$ gold sol $=1 \mathrm{cc}$ of $10 \% \mathrm{NaCl}$
$\therefore 100 \mathrm{~mL}$ of gold $\mathrm{sol}=1 \mathrm{cc}$ of gelatin
Therefore, gold number of gelatin is
$\frac{1 \times 10}{100}=0.01$

Topic: Chemistry, Surface Chemistry, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about its axis passing through its centre. Which of the two will acquire a greater angular speed?

Options:

None

Solution:
$I_1, M . I$ of a hollow cylinder about its axis $=M R^2$
$I_2, M . I$ of a solid sphere about an axis passing through its centre $=\frac{2}{5} M R^2$
$\because \quad$ Applied torque on the cylinder and the sphere is
$$
\begin{array}{ll}
& \text { same } \\
\therefore \quad \tau=I_1 \alpha_1=I_2 \alpha_2 \\
\therefore & \frac{\alpha_2}{\alpha_1}=\frac{I_1}{I_2}=\frac{M R^2}{\frac{2}{5} M R^2}=\frac{5}{2} \\
\Rightarrow & \therefore \alpha_2>\alpha_1\left[\alpha_2=2.5 \alpha_1\right] \\
\because & \omega=\omega_0+\alpha t, \text { where } \omega_0 \text { and } t \text { are constants } \\
\therefore \quad & \omega_2>\omega_1 \\
\end{array}
$$
$\therefore \quad$ Angular speed of the solid sphere is greater than that of the cylinder.

Topic: Physics, System of Particles and Rotational Motion
Trachea is a straight tube extending upto the mid thoracic cavity, which divides into right and left bronchi at the level of

Options:
A 3rd cervical vertebra
B 5th cervical vertebra
C 5th thoracic vertebra
D 12th thoracic vertebra

The correct answer is:
5th thoracic vertebra

Solution:
Human respiratory system consists of external nostrils, nasal chamber, pharynx, trachea, bronchi, bronchioles and Lungs. Pharynx opens into trachea through larynx. Trachea is a straight tube in the thoracic cavity and at the level of fifth thoracic vertebra it divides into right and left bronchi which enters the respective Lung. The region at which the division takes place is known as Carina. Primary bronchi when enters the lung further divides into secondary and tertiary bronchi which then forms bronchioles which further extends as alveoli.

Topic: Biology, Breathing and Exchange of Gases, NEET
Transgenic animals are the animals in which a foreign gene is expressed. Such animals can be used to study the fundamental biological process, phenomenon as well as for producing products useful for mankind. Give one example for each type.

Options:

None

Solution:
The animals in which foreign genes are expressed are transgenic animals. The fundamental biological process/ phenomenon, like using model organisms like mouse and determining how genes are regulated (gene regulation), how they affect the normal functions of the body and its development, etc. are studied using such animals.
Transgenic animals are also used for producing products useful for mankind, e.g., transgenic cow (rosie) produced human protein enriched milk $(2.4 \mathrm{~g} / \mathrm{L})$. The milk contained the human alpha-lactalbumin protein and was nutritionally a more balanced product for human babies that cow-milk.

Topic: Biology, Biotechnology and Its Applications
Trunks of some of the aged tree species appear to be composed of several fused trunks. Is it a physiological or anatomical abnormality? Explain in detail.

Options:

None

Solution:
The appearance of several fused trunks is anatomical abnormality. It is an abnormal type of secondary growth where a regular vascular cambium or cork cambium is not formed in its normal position. Anomalous secondary growth produces cortical and medullary vascular bundles in case of old tree trunks.
Thus, the additional or accessory vascular bundles given appurtenance of several fused trunks.

Topic: Biology, Anatomy of Flowering Plants
Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube $A$ and $B$ is

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

The correct answer is:
$2: 1$

Solution:
Fundamental freq. for,
Tube $\mathrm{A} \Rightarrow \mathrm{f}_{\mathrm{A}}=\frac{\mathrm{v}}{2 \mathrm{~L}}$
Tube $\mathrm{B} \Rightarrow \mathrm{f}_{\mathrm{B}}=\frac{\mathrm{v}}{4 \mathrm{~L}}$
Now,
$$
\begin{array}{l}
\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}=\frac{\mathrm{v}}{2 \mathrm{~L}} \times \frac{4 \mathrm{~L}}{\mathrm{v}}=\frac{2}{1} \\
\mathrm{f}_{\mathrm{A}}: \mathrm{f}_{\mathrm{B}}=2: 1
\end{array}
$$

Topic: Physics, Waves and Sound, VITEEE, VITEEE 2019
Two angles of a triangle are $\tan ^{-1} \frac{1}{2}$ and $\tan ^{-1} \frac{1}{3}$. What is the third angle?

Options:
A $30^{\circ}$
B $45^{\circ}$
C $90^{\circ}$
D $135^{\circ}$

The correct answer is:
$135^{\circ}$

Solution:
In any $\Delta \mathrm{ABC}$ $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=\pi$
$\Rightarrow$ let $A=\tan ^{-1} \frac{1}{2}$ and $B=\tan ^{-1} \frac{1}{3}$
$\Rightarrow \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}+\angle \mathrm{C}=\pi$
$\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)+\angle \mathrm{C}=\pi$
$\left[\because \tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)\right]$
$\Rightarrow \tan ^{-1}\left(\frac{5 / 6}{5 / 6}\right)+\angle \mathrm{C}=\pi \Rightarrow \frac{\pi}{4}+\angle \mathrm{C}=\pi$
$\Rightarrow \angle \mathrm{C}=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}=135^{\circ}$

Topic: Mathematics, Inverse Trigonometric Functions, NDA, NDA 2007 (Phase 2)
Two bacteria found to be very useful in genetic engineering experiments are:

Options:
A Nitrobacter and Azotobacter
B Rhizobium and Diplococcus
C Nitrosomonas and Kliebsiella
D Escherichia and Agrobacterium

The correct answer is:
Escherichia and Agrobacterium

Solution:
E.coli contains many important standard cloning vectors widely used in gene cloning experiments like pBR322 contains origin of replication (ori). Other cloning vectors like PACYC177, pBR324, PRK 64.6 contain ampicillin resistance gene they are also found in E.coli. Among higher plants, Ti plasmid of Agrobacterium tumefaciens and Ri plasmid of A. rhizogenes is the best-known vector. TDNA from Ti or Ri plasmid of Agrobacterium is considered to be a very potential vector for cloning experiments with higher plants.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Two balls are selected from a box containing 2 blue and 7 red balls. What is the probability that at least one ball is blue?

Options:
A $\frac{2}{9}$
B $\frac{7}{9}$
C $\frac{5}{12}$
D $\frac{7}{12}$

The correct answer is:
$\frac{2}{9}$

Solution:
No. of blue balls $=2$ No. of red balls $=7$
Total no. of balls $=9$
Required probability $=P($ one ball is blue $)+P$ (both ball is blue)
$$
=\frac{2}{9} \times \frac{7}{8}+\frac{2}{9} \times \frac{1}{8}=\frac{14}{72}+\frac{2}{72}=\frac{16}{72}=\frac{2}{9}
$$

Topic: Mathematics, Probability, NDA, NDA 2010 (Phase 1)
Two cards are chosen at random from a deck of 52 playig cards. What is the probability that both of them have the same value?

Options:
A $\frac{1}{17}$
B $\frac{3}{17}$
C $\frac{5}{17}$
D $\frac{7}{17}$

The correct answer is:
$\frac{1}{17}$

Solution:
Number of ways of selecting 2 cards from deck of cards $={ }^{52} \mathrm{C}_{2}$
favourable cases $={ }^{13} \mathrm{C}_{2}$
Probability $=\frac{13}{52} C_{2}=\frac{\frac{13 \times 12}{2}}{\frac{52 \times 51}{2}}=\frac{13 \times 12}{52 \times 51}=\frac{1}{17}$

Topic: Mathematics, Probability, NDA, NDA 2019 (Phase 1)
Two cards are drawn successively without replacement from a wellshuffled pack of 52 cards. The probability of drawing two aces is

Options:
A $\frac{1}{26}$
B $\frac{1}{221}$
C $\frac{4}{223}$
D $\frac{1}{13}$

The correct answer is:
$\frac{1}{221}$

Solution:
For the nexttwo (02) items that follow
$\begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{{ }^{4} \mathrm{C}_{1}}{{ }^{52} \mathrm{C}_{1}}=\frac{4}{52}=\frac{1}{13} \\ & \mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{{ }^{3} \mathrm{C}_{1}}{{ }^{51} \mathrm{C}_{1}}=\frac{3}{51}=\frac{1}{17} \\ \therefore \quad & \text { Required probability }=\frac{1}{13} \times \frac{1}{17}=\frac{1}{221} \\ \therefore \quad & \text { Option (b) is correct. } \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2015 (Phase 1)
Two cells $A$ and. $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing length are respectively $400 \mathrm{~cm}$ and $440 \mathrm{~cm}$. The emf of the cell $A$ is 1.08 volt. The emf of the second cell $B$ in volts is

Options:
A $1.08$
B $1.188$
C $11.88$
D $12.8$

The correct answer is:
$1.188$

Solution:
$\begin{array}{lc}\frac{E_1}{E_2}=\frac{l_1}{l_2} & \\ \Rightarrow & \frac{1.08}{E_2}=\frac{400}{440} \\ \Rightarrow & E_2=\frac{440 \times 1.08}{400}=1.188 \mathrm{~V}\end{array}$

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2004
Two cells $A$ and. $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing length are respectively $400 \mathrm{~cm}$ and $440 \mathrm{~cm}$. The emf of the cell $A$ is 1.08 volt. The emf of the second cell $B$ in volts is

Options:
A $1.08$
B $1.188$
C $11.88$
D $12.8$

The correct answer is:
$1.188$

Solution:
$\begin{array}{lc}\frac{E_1}{E_2}=\frac{l_1}{l_2} & \\ \Rightarrow & \frac{1.08}{E_2}=\frac{400}{440} \\ \Rightarrow & E_2=\frac{440 \times 1.08}{400}=1.188 \mathrm{~V}\end{array}$

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2004
Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively $360 \mathrm{~cm}$ and $420 \mathrm{~cm}$. If emf of $A$ is $2.4 \mathrm{~V}$, the emf of the second cell $B$ is

Options:
A $2.8 \mathrm{~V}$
B $3.2 \mathrm{~V}$
C $3.0 \mathrm{~V}$
D $2.6 \mathrm{~V}$

The correct answer is:
$2.8 \mathrm{~V}$

Solution:
If $E_1$ and $E_2$ are emf's of cells with $l_1$ and $l_2$ balance lengths in a potentiometer experiment, then, $\frac{E_1}{E_2}=\frac{l_1}{l_2}$
or $\quad E_2=E_1 \cdot \frac{l_2}{l_1}$
Here, $E_1=24 \mathrm{~V}, l_1=360 \mathrm{~cm}$ and $l_2=420 \mathrm{~cm}$ Substituting values in Eq. (i), we get
$$
E_2=24 \times \frac{420}{360}=28 \mathrm{~V}
$$
So, emf of cell $B$ is $2.8 \mathrm{~V}$.

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
Two charge $q_1$ and $q_2$ are placed $30 \mathrm{~cm}$ apart, as shown in the figure. A third charge $q_3$ is moved along the are of a circle of radius $40 \mathrm{~cm}$ from $\mathrm{C}$ to $\mathrm{D}$. The change in the potential energy of the system is \(\frac{q_3 k}{4 \pi \in_0}\), where $k$ is:

Options:
A $8 q_1$
B $6 q_1$
C $8 q_2$
D $6 q_2$

The correct answer is:
$8 q_2$

Solution:
$$
\begin{aligned}
& V_i=\frac{1}{4 \pi \epsilon_0}\left[\frac{q_1 q_3}{(0.4)}+\frac{q_1 q_2}{(0.3)}+\frac{q_2 q_3}{(0.5)}\right] \\
& v_f=\frac{1}{4 \pi \epsilon_0}\left[\frac{q_1 q_3}{(0.4)}+\frac{q_1 q_2}{(0.3)}+\frac{q_2 q_3}{(0.1)}\right]
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \Delta V=V f-V_1 \\
& =\frac{1}{4 \pi \epsilon_0} q_2 q_3\left(\frac{1}{0.1}-\frac{1}{0.5}\right) \\
& =\frac{q_2 q_3}{\pi \epsilon_0}\left(10^{-2}\right)=\frac{q_3}{4 \pi \epsilon_0}\left(8 q_2\right) \\
& \Rightarrow k=8 q_2
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2005
Two closed organ pipes of length \(100 \mathrm{~cm}\) and \(101 \mathrm{~cm}\) produces 16 beats in \(20 \mathrm{sec}\). When each pipe is sounded in its fundamental mode calculate the velocity of sound.

Options:
A \(303 \mathrm{~ms}^{-1}\)
B \(332 \mathrm{~ms}^{-1}\)
C \(323.2 \mathrm{~ms}^{-1}\)
D \(300 \mathrm{~ms}^{-1}\).

The correct answer is:
\(323.2 \mathrm{~ms}^{-1}\)

Solution:
Frequency of fundamental mode of vibrations of $1^{\text {st }}$ closed organ pipe,
From question, $v_1-v_2=\frac{16}{20}$
$\begin{aligned} & \Rightarrow \frac{v}{4 \times 100}-\frac{v}{4 \times 101}=\frac{16}{20} \\ & \Rightarrow \frac{v}{4} \cdot\left[\frac{1}{100}-\frac{1}{101}\right]=\frac{16}{20} \\ & \Rightarrow \frac{v}{4}\left[\frac{101-100}{101 \times 100}\right]=\frac{16}{20} \Rightarrow \frac{v}{4} \times \frac{1}{101 \times 100}=\frac{16}{20} \\ & \Rightarrow v=\frac{16 \times 101 \times 4 \times 100}{20}=32320 \mathrm{~cm} / \mathrm{s} . \\ & =323.2 \mathrm{~m} / \mathrm{s} .\end{aligned}$

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2008
Two closed pipes have the same fundamental frequency. One is filled with oxygen and the other with hydrogen at the same temperature. Ratio of their lengths respectively is

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

The correct answer is:
$1:4$

Solution:
The fundamental frequency,
$\begin{aligned}
& f=\frac{v}{4 l}=\frac{1}{4 e} \sqrt{\frac{\gamma R T}{M_0}} \\
& \ell \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{\ell_1}{\ell_2}=\sqrt{\frac{M_1}{M_2}}=\sqrt{\frac{2}{32}}=\frac{1}{4}
\end{aligned}$
So, $\ell_1: \ell_2=1: 4$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2015
Two coaxial coils A and B of radii 'R ' and 'Rz' are placed in the same plane. (R $_{2}>$
$\mathrm{R}_{1}$ ). If a current is passed through coil $\mathrm{B}$, the coefficient of mutual inductance
between the coils is proportional to

Options:
A $\frac{1}{\mathrm{R}_{1} \mathrm{R}_{2}}$
B $\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}$
C $R_{1} R_{2}$
D $\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$

The correct answer is:
$\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}$

Solution:
Magnetic field at the centre of primary coil
$$
B=\mu_{0} i_{1} / 2 R_{1} .
$$
Considering it to be uniform, magnetic flux passing through secondary coil is
$$
\phi_{2}=B A=\frac{\mu_{0} i_{1}}{2 R_{1}}\left(\pi R_{2}^{2}\right)
$$
Now, $M=\frac{\phi_{2}}{i_{1}}=\frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}$
$$
\therefore \quad M \propto \frac{R_{2}^{2}}{R_{1}} .
$$

Topic: Physics, Electromagnetic Induction, MHT CET, MHT CET 2020 (12 Oct Shift 1)
Two common characters found in centipede, cockroach and crab are:

Options:
A book lungs and antennae
B compound eyes and anal cerci
C jointed legs and chitinous exoskeleton
D green gland and tracheae

The correct answer is:
jointed legs and chitinous exoskeleton

Solution:
Crab, centipede and cockroach belong to Phylum Arthropoda. These have jointed appendages and a chitinous exoskeleton.

Topic: Biology, Animal Kingdom, NEET, NEET 2006
Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1>R_2$, the mutual inductance $M$ between them will be directly proportional to

Options:
A $\frac{R_1}{R_2}$
B $\frac{R_2}{R_1}$
C $\frac{R_1^2}{R_2}$
D $\frac{R_2^2}{R_1}$

The correct answer is:
$\frac{R_2^2}{R_1}$

Solution:
Magnetic field at the centre of primary coil $B=\mu_0 i_1 / 2 R_1$. Considering it to be uniform, magnetic flux passing through secondary coil is
$\phi_2=B A=\frac{\mu_0 i_1}{2 R_1}\left(\pi R_2^2\right)$
Now, $M=\frac{\phi_2}{i_1}=\frac{\mu_0 \pi R_2^2}{2 R_1} \quad \therefore \quad M \propto \frac{R_2^2}{R_1}$.

Topic: Physics, Electromagnetic Induction, AIIMS, AIIMS 2014
Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1>R2, the mutual inductance M between them will be directly proportional to

Options:
A R1R2
B R2R1
C R12R2
D R22R1

The correct answer is:
R22R1

Solution:
Let the magnetic field is B at the centre of the primary coil. Considering it to be uniform, the magnetic flux passing through the secondary coil is
ϕ=BA=μ0i12R1πR22
Now, M=ϕ2i1=μ0πR222R1
∴ M∝R22R1

Topic: Physics, Electromagnetic Induction, NEET
Two consecutive harmonics of an air column in pipe closed at one end are of frequencies $150 \mathrm{~Hz}$ and $250 \mathrm{~Hz}$. The fundament frequency of an air column is

Options:
A $25 \mathrm{~Hz}$
B $75 \mathrm{~Hz}$
C $100 \mathrm{~Hz}$
D $50 \mathrm{~Hz}$

The correct answer is:
$50 \mathrm{~Hz}$

Solution:
In a pipe closed at one end only odd harmonics of the fundamental are produced, i.e., $50 \mathrm{~Hz}, 150 \mathrm{~Hz}, 250 \mathrm{~Hz}$. Hence fundamental frequency is $50 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (20 Sep Shift 1)
Two cross-sections of stem and root appear simple, when viewed by naked eye. But under microscope, they can be differentiated by

Options:
A Exarch condition of root and stem
B Endarch condition of stem and root
C Endarch condition of root and exarch condition of stem
D Endarch condition of stem and exarch condition of root

The correct answer is:
Endarch condition of stem and exarch condition of root

Solution:
Characteristic feature of stem is endarch condition of xylem tissue. In endarch, protoxylem is present towards the centre of stem, while metaxylem towards the pericycle. Characteristic feature of root is exarch condition of xylem tissue. In exarch condition, protoxylem is present towards the pericycle and metaxylem towards the center of the root.

Topic: Biology, Anatomy of Flowering Plants, NEET
Two cylinders A and B of equal capacity are connected to each other via a stop cock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stop cock is suddenly opened. The process is:

Options:
A adiabatic
B isochoric
C isobaric
D isothermal

The correct answer is:
adiabatic

Solution:
Free expansion i.e. expansion against vacuum is adiabatic in nature for all type of gases. It should be noted that temperature final temperature is equal to initial temperature for adiabatic process.

Topic: Physics, Thermodynamics, NEET, NEET 2020 (Phase 1)