datavorous/entrance-exam-dataset huggingface dataset:
The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \mathrm{~nm}$ to $0.5 \mathrm{~nm}$ is

Options:
A four times initial energy
B thrice the initial energy
C equal to the initial energy
D twice the initial energy

The correct answer is:
twice the initial energy

Solution:
$$
\lambda_1=1 \mathrm{~nm}
$$
Energy is given as, $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$
$$
\mathrm{E}_1=\frac{1240 \mathrm{nmev}}{1 \mathrm{~nm}}=1240 \mathrm{eV}
$$
for $0.5 \mathrm{~nm}$ wavelength,
$$
\mathrm{E}_2=\frac{1240 \mathrm{nmev}}{0.5 \mathrm{~nm}}=2(1240)=2 \mathrm{E}_1 \text {. }
$$
Thus it is twice the initial energy.

Topic: Physics, Dual Nature of Matter, JEE Main
The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \mathrm{~nm}$ to $0.5 \mathrm{~nm}$ is

Options:
A 2 times the initial kinetic energy
B 3 times the initial kinetic energy
C $0.5$ times the initial kinetic energy
D 4 times the initial kinetic energy

The correct answer is:
3 times the initial kinetic energy

Solution:
The de-Broglie wavelength,
$\lambda=\frac{h}{\sqrt{2 m E_{k}}} \quad \therefore \lambda \propto \frac{1}{\sqrt{E_{k}}}$
( $\because$ h and $m$ remains constant)
Here, in given condition
$\begin{array}{ll}
\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{k_{2}}}{E_{k_{1}}}}, \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{4 E}{E}} \\
\lambda_{1}=2 \lambda_{2}, & \lambda_{2}=\frac{\lambda_{1}}{2}
\end{array}$
So, if $E_{k}$ is increased by four times, then $\lambda$ becomes half.
$\therefore$ Additional kinetic energy that should be supplied to the electron
$=4 E_{k}-E_{k}=3 E_{k}$

Topic: Physics, Dual Nature of Matter, KCET, KCET 2013
The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$, to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2\left(R_2>R_1\right)$ is

Options:
A $\mathrm{GmM}\left(\frac{1}{\mathrm{R}_1^2}-\frac{1}{\mathrm{R}_2^2}\right)$
B $\mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
C $2 \mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
D $\frac{1}{2} \mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$

The correct answer is:
$\frac{1}{2} \mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$

Solution:
The kinetic energy changing the orbit of satellite
$$
\begin{aligned}
& \mathrm{KE}=\left(-\frac{\mathrm{GMm}}{2 \mathrm{R}_2}\right)-\left(-\frac{\mathrm{GMm}}{2 \mathrm{R}_1}\right) \\
& \mathrm{KE}=\frac{\mathrm{GMm}}{2}\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2010 (Mains)
The Adenosine deaminase deficiency results into:

Options:
A Dysfunction of Immune system
B Parkinson's disease
C Digestive disorder
D Addison's disease

The correct answer is:
Dysfunction of Immune system

Solution:
Adenosine deaminase enzyme deficiency causes SCID (Severe Combined Immunodeficiency). ADA enzyme is required for the survival of T-lymphocytes. Thus, deficiency of Adenosine deaminase enzyme causes dysfunction of immune system, due to loss of cell-mediated immunity.

The main symptoms of ADA deficiency are pneumonia, chronic diarrhea, and widespread skin rashes. Affected children also grow much more slowly than healthy children and some have developmental delay.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2021
The adjacent figure shows an infinite frame of two sides in a gravity-free space. What is the final constant kinetic energy expected of the ball of mass m projected as shown with an initial velocity v0? The coefficient of restitution for the collision between the ball and the frame is e=0.5.

Options:
A 0.5mv02
B 0.25mv02
C 0.125mv02
D 0.325mv02

The correct answer is:
0.25mv02

Solution:
The component v=v0sin45°=v02 perpendicular to the walls is finally lost after a long time.

Hence, final remaining K.E.=12mv022

Topic: Physics, Center of Mass Momentum and Collision, NEET
The adjacent graph shows the extension $(\Delta l)$ of a wire of length $1 m$ suspended from the top of a roof at one end with a load $W$ connected to the other end. If the cross sectional area of the wire is ${ }^{10^{-6}} \mathrm{~m}^2$, calculate the young's modulus of the material of the wire

Options:
A $2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
B $2 \times 10^{-11} \mathrm{~N} / \mathrm{m}^2$
C $3 \times 10^{-12} \mathrm{~N} / \mathrm{m}^2$
D $2 \times 10^{-13} \mathrm{~N} / \mathrm{m}^2$

The correct answer is:
$2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Solution:
From the graph $l=10^{-4} \mathrm{~m}, F=20 \mathrm{~N}$
$A=10^{-6} \mathrm{~m}^2, L=1 \mathrm{~m}$
$\therefore Y=\frac{F L}{A l}=\frac{20 \times 1}{10^{-6} \times 10^{-4}}=20 \times 10^{10}=2 \times 10^{11} \mathrm{~N}/\mathrm{~m}^2$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The adjoining diagram shows three soap bubbles A, B and C prepared by blowing the capillary tube fitted with stop cocks, S1,S2 and S3. The bubble B is the largest and the bubble C is the smallest. With stop cock S closed, what will happen when we open the stop cocks S1,S2 and S3 simultaneously?

Options:
A B will start collapsing with volumes of A and C increasing
B C will start collapsing with volumes of A and B increasing
C C and A both will start collapsing with the volume of B increasing
D Volumes of A, B and C will become equal at equilibrium

The correct answer is:
B will start collapsing with volumes of A and C increasing

Solution:
Excess pressure inside soap bubble is inversely proportional to the radius of bubble i.e. ÎPâ1r
This means that bubbles A and C posses greater pressure inside it than B. So the air will move from A and C towards B.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The adjoining figure shows the graph of $y=a x^2+b x+c$.
Then

Options:
A $a \lt 0$
B $b^2 \lt 4 a c$
C $c\gt0$
D $a$ and $b$ are of opposite signs

The correct answers are:
$a \lt 0$, $a$ and $b$ are of opposite signs

Solution:
It is evident from the figure that the function $y=a x^2+b x+c$ has a maximum between $x_1$ and $x_2$.
$\therefore \frac{d^2 y}{d x^2} \lt 0 \Rightarrow a \lt 0$
Obviously, $x_1, x_2\gt0 \Rightarrow x_1+x_2\gt0$
$\Rightarrow$ Sum of the roots $\gt0$
$\Rightarrow \frac{-b}{a}\gt0 \Rightarrow \frac{b}{a} \lt 0 \Rightarrow a$ and $b$ are of opposite signs.

Topic: Mathematics, Quadratic Equation, JEE Main
The adjoint matrix of $\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$ is

Options:
A $\left[\begin{array}{lll}4 & 8 & 3 \\ 2 & 1 & 6 \\ 0 & 2 & 1\end{array}\right]$
B $\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$
C $\left[\begin{array}{ccc}11 & 9 & 3 \\ 1 & 2 & 8 \\ 6 & 9 & 1\end{array}\right]$
D $\left[\begin{array}{ccc}1 & -2 & 1 \\ -1 & 3 & 3 \\ -2 & 3 & -3\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$

Solution:
Let $A=\left[\begin{array}{lll}3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1\end{array}\right]$
Then, $A_{11}=1, A_{12}=-2, A_{13}=-2$
$A_{21}=-1, A_{22}=3, A_{23}=3$
$A_{31}=0, A_{32}=-4, A_{33}=-3$
$\operatorname{adj}(A)=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 0 \\ -2 & 3 & -4 \\ -2 & 3 & -3\end{array}\right]$.

Topic: Mathematics, Matrices, JEE Main
The adrenal cortical hormone that reduces inflammation and suppresses the immune response is

Options:
A Cortisol
B Deoxycorticosterone
C Corticostone
D Aldosterone

The correct answer is:
Cortisol

Solution:
Glucocorticoids stimulate gluconeogenesis, lipolysis and proteolysis; and inhibit cellular uptake and utilisation of amino acids. Cortisol is also involved in maintaining the cardio-vascular system as well as the kidney functions. Glucocorticoids, particularly cortisol, produces antiinflammatory reactions and suppresses the immune response. Cortisol stimulates the RBC production.

Topic: Biology, Chemical Coordination and Integration, NEET
The adsorbent used in adsorption chromatography is/are -
A. silica gel
B. alumina
C. quick lime
D. magnesia
Choose the most appropriate answer from the options given below :

Options:
A A only
B B only
C C and D only
D A and B only

The correct answer is:
A and B only

Solution:
The most common polar and acidic support used is adsorption chromatography is silica. The surface silanol groups on their supported to adsorb polar compound and work particularly well for basic substances. Alumina is the example of polar and basic adsorbent that is used in adsorption chromatography.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The adsorption isotherm for a gas is given by the relation $x=a p /(1+b p)$ where $x$ is moles of gas adsorbed per gram of the adsorbent, $p$ is the pressure of the gas, and a and $b$ are constants. Then $x$

Options:
A increases with $\mathrm{p}$
B remains unchanged with $\mathrm{p}$
C decreases with $\mathrm{p}$
D increases with $\mathrm{p}$ at low pressures and then remains the same at high pressure

The correct answer is:
increases with $\mathrm{p}$ at low pressures and then remains the same at high pressure

Solution:
According to Lagmuir curve
$\mathrm{x}=\frac{\mathrm{ap}}{1+\mathrm{bp}}$
$\begin{array}{ll}\mathrm{p} \rightarrow \infty & \mathrm{x}=\frac{\mathrm{a}}{\mathrm{b}} \\ \mathrm{p} \rightarrow 0 & \mathrm{x} \propto \mathrm{p}\end{array}$

Topic: Chemistry, Surface Chemistry, KVPY, KVPY 2011 (SB/SX)
The adsorption of a gas at the boiling point of the gas follows the isotherm shown in the figure. Identify the correct thermodynamic properties at point $C$.

Options:
A $\Delta H=T \Delta S, \Delta G=0$
B $\Delta H>T \Delta S, \Delta G=+$ ve
C $\Delta H < T \Delta S, \Delta G=-v e$
D $\Delta H=T \Delta S=\Delta G=0$

The correct answer is:
$\Delta H=T \Delta S, \Delta G=0$

Solution:
Amount of gas adsorbed remains constant at point $C$ irrespective of increase in pressure.
So, the system is in equilibrium
$$
\therefore \quad \Delta G=0
$$
As we know,
$$
\Delta G=\Delta H-T \Delta S
$$
$$
\therefore \quad \Delta H=T \Delta S .
$$

Topic: Chemistry, Surface Chemistry, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
The adsorption of a gas on a solid surface varies with pressure of the gas in which of the following manner

Options:
A Fast $\rightarrow$ slow $\rightarrow$ independent of the pressure
B Slow $\rightarrow$ fast $\rightarrow$ independent of the pressure
C Independent of the pressure $\rightarrow$ fast $\rightarrow$ slow
D Independent of the pressure $\rightarrow$ slow $\rightarrow$ fast

The correct answer is:
Fast $\rightarrow$ slow $\rightarrow$ independent of the pressure

Solution:
The extent of adsorption is measured as \(\mathrm{x} / \mathrm{m}\) where m is the mass of adsorbent and \(x\) that of adsorbate.
At low pressure: \(\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{p}^1 \Rightarrow \frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kp}\) \(\frac{\mathrm{x}}{\mathrm{m}}\) varies linearly with p with faster rate
At intermediate pressure: \(\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{p} \frac{1}{\mathrm{n}}(\mathrm{n}>1)\) or, \(\frac{\mathrm{x}}{\mathrm{m}}=\operatorname{kp} \frac{1}{\mathrm{n}}\)
\(\frac{\mathrm{x}}{\mathrm{m}}\) changes slowly with pressure.
At high pressures: \(\frac{\mathrm{x}}{\mathrm{m}} \propto \mathrm{p}^{\circ}\)
\(\frac{\mathrm{x}}{\mathrm{m}}\) is independent of p
The adsorption varies with pressure in fast, slow and independent way.

Topic: Chemistry, Surface Chemistry, JEE Main
The adsorption of water by hydrophilic compounds like cellulose and pectin in root hair cell wall is called _______.

Options:
A Diffusion
B Imbibition
C Guttation
D Osmosis

The correct answer is:
Imbibition

Solution:
It is a physical process which involves adsorption and/or absorption of water by hydrophilic substances without forming a solution e.g., wooden doors absorb water and swell up in rainy season and seeds of pea and gram when placed in water swell up due to imbibition
The imbibition process involves diffusion and capillary action.
In imbibition, Liquid substance (water) is called imbibate, and solid colloidal substance is called imbibant. e.g. the gum, dry wood, cell protoplasm, seed coat, etc. Due to imbibition, water molecules remain held (absorbed) over the surface of hydrophilic particles and also remain held between the space of particles of imbibant.

Topic: Biology, Transport in Plants, NEET
The aerosol is a kind of colloid in which

Options:
A solid is dispersed in gas.
B liquid is dispersed in water.
C gas is dispersed in solid.
D gas is dispersed in liquid.

The correct answer is:
solid is dispersed in gas.

Solution:
An aerosol is a colloidal solution of solid or liquid particles in the air or other gas, for example smoke, dust etc. are considered as an aerosol.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The agriculture sector of India employs

Options:
A
$50 \%$ of the population

B
$70 \%$ of the population

C
$30 \%$ of the population

D
$60 \%$ of the population

The correct answer is:
$60 \%$ of the population

Solution:
India is mainly an agricultural country. Agriculture employs nearly $60 \%$ of the population.

Topic: Biology, Strategies for Enhancement in Food Production
The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of the sound in air is 330 m s-1. End corrections may be neglected. Let P0 denote the mean pressure at any point in the pipe and âP0 the maximum amplitude of pressure variation then the amplitude of pressure variation in the middle of the column is

Options:
A ∆p22
B ∆p03
C ∆p02
D ∆p03

The correct answer is:
∆p22

Solution:
P=∆P0sin⁡2πxλ
For the second overtone L=5λ4
At x=L2=5λ8
Pressure amplitude at middle
p=∆P0sin⁡2πλ5λ8
∆P0sin⁡5π4=∆P02

Topic: Physics, Waves and Sound, JEE Main
The alcohol obtained during the hydrolysis of ground nut oil is

Options:
A glycol
B butan-1-ol
C pentan-2-ol
D propane $1,2,3$-triol

The correct answer is:
propane $1,2,3$-triol

Solution:
The alcohol obtained during the hydrolysis of ground nut oil is propane $1,2,3$ - triol.
It's structure is

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The alcohol that produces turbidity immediately with $\mathrm{ZnCl}_2{ }^{+}$conc. $\mathrm{HCl}$ at room temperature

Options:
A 1-hydroxybutane
B 2-hydroxybutane
C 2-hydroxy-2-methylpropane
D 1-hydroxy-2-methylpropane

The correct answer is:
2-hydroxy-2-methylpropane

Solution:
1-hydroxy butane- $1^0$ alcohol
2-hydroxy butane- $2^{\circ}$ alcohol
1-hydoxy-2-methyl propane- $1^{\circ}$ alcohol

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The alcoholic solution of KOH is used in which type of the organic reactions?

Options:
A Decarboxylation
B Dehydrogenation
C Dehydrohalogenation
D Dehydration

The correct answer is:
Dehydrohalogenation

Solution:
CH3CH2Br+alcKOH→dehydrohalogenationCH2=CH2+KBr+H2O
In alcoholic KOH alkoxide ions RO- are present which is a strong base. They abstract proton from beta carbon of alkyl halide and favours elimination reaction
ROH+KOH→ROK+H2O
ROK→RO-+K+
RO-+H-CH2β-CH2α-Br→ROH+CH2=CH2+Br-

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The alcohols which have two hydroxyl \( (-\mathrm{OH}) \) groups are called

Options:
A glycols
B gemdiols
C dianes
D dimers

The correct answer is:
glycols

Solution:
The two hydroxyl groups containing alcohols can be classified into gemdiols and vicdiols. gemdiols not exist in nature. The vicinal diols are the alcohols containing two hydroxyl groups adjacent to each other. These are also called glycols.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The aleurone layer in maize grain is specially rich in:

Options:
A proteins
B starch
C lipids
D auxins

The correct answer is:
proteins

Solution:
The aleurone layer of maize grain is especially rich in proteins in mature seeds and tubers. It forms the outermost layer of the endosperm.
Related Theory
Aleurone lies between the pericarp and the hyaline layer of the endosperm. Unlike the cells of the starchy endosperm, aleurone cells remain alive at maturity. The ploidy of the aleurone is (3n) as a result of double fertilization.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2003
The algebraic sum of two co-initial vectors is 16 units. Their vector sum is 8 units and the resultant of the vectors are perpendicular to the smaller vector. Then magnitudes of the two vectors are -

Options:
A 2 unit & 14 unit
B 4 unit & 12 unit
C 6 unit & 10 unit
D 8 unit & 8 unit

The correct answer is:
6 unit & 10 unit

Solution:
P+Q=16 (i)
P2+Q2+2PQcos Î¸ =64 (ii)
tan 90 Î¿ = Q sin Î¸ P + Q cos Î¸
â P + Q cos Î¸ = 0
cos Î¸ = - P Q
From Eq. (ii)
P 2 + Q 2 + 2 PQ cos Î¸ = 64
P 2 + Q 2 + 2.P.Q - P Q = 64
- P 2 + Q 2 = 64
Q + P Q - P = 64
as P + Q = 16
Q - P = 4
Solving we get
2 Q = 20
Q = 10 units
P = 6 units

Topic: Physics, Mathematics in Physics, JEE Main
The alkali metals from salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders?

Options:
A $\mathrm{LiH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH}$
B $\mathrm{CsH}>\mathrm{RbH}>\mathrm{KH}>\mathrm{NaH}>\mathrm{LiH}$
C $\mathrm{KH}>\mathrm{NaH}>\mathrm{LiH}>\mathrm{CsH}>\mathrm{RbH}$
D $\mathrm{NaH}>\mathrm{LiH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH}$

The correct answer is:
$\mathrm{LiH}>\mathrm{NaH}>\mathrm{KH}>\mathrm{RbH}>\mathrm{CsH}$

Solution:
No Solution Available

Topic: Chemistry, s Block Elements, NEET, NEET 2008 (Mains)
The alkaline earth metal nitrate that does not crystallise with molecules is:

Options:
A BaNO32
B SrNO32
C MgNO32
D CaNO32

The correct answer is:
BaNO32

Solution:
Ba NO32 does not have any water of crystallization. This happens because the large size of Ba+2 leads to poor polarizing power and hence, unable to attract water molecules effectively.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The alkaline earth metal that imparts apple green colour to the bunsen flame when introduced in it in the form of its chloride is

Options:
A Barium
B Strontium
C Calcium
D Magnesium

The correct answer is:
Barium

Solution:
$Ba^{2+}$ imparts green colour to the flame.

Topic: Chemistry, Practical Chemistry, JEE Main
The alkaline earth metals $B a, S r, C a$ and $M g$ may be arranged in the order of their decreasing first ionisation potential as

Options:
A $M g, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$,
B Ca, $\mathrm{Sr}, \mathrm{Ba}, \mathrm{Mg}$
C Sr, Ba, Mg, Ca
D $\mathrm{Ba}, \mathrm{Mg}, \mathrm{Ca}, \mathrm{Sr}$,

The correct answer is:
$M g, \mathrm{Ca}, \mathrm{Sr}, \mathrm{Ba}$,

Solution:
$\begin{array}{lllll}\text {Element }- & \mathrm{Mg} & \mathrm{Ca} & \mathrm{Sr} & \mathrm{Ba} \\ \text {I.E - } & 737 & 590 & 549 & 503\end{array}$

Topic: Chemistry, s Block Elements, JEE Main
The alkene formed as a major product in the above elimination reaction is

Options:
A

B
$\mathrm{CH}_2=\mathrm{CH}_2$

C

D

The correct answer is:
$\mathrm{CH}_2=\mathrm{CH}_2$

Solution:
No Solution Available

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2006
The alkene that exhibits geometrical isomerism is:

Options:
A
propene

B
2-methyl propene

C
2-butene

D
2- methyl -2- butene

The correct answer is:
2-butene

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2009
The alkene which on ozonolysis yields acetone is

Options:
A $\mathrm{CH}_2=\mathrm{CH}_2$
B $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$
C $\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$
D $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$

The correct answer is:
$\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_3\right)_2$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The alloy used in the construction of aircrafts is:

Options:
A Mg-Al
B Mg-Mn
C Mg-Sn
D Mg-Zn

The correct answer is:
Mg-Al

Solution:
Aircraft manufacturers use high-strength alloys. Aircraft has Aluminium as main constituent.
Mg-Al alloy is used for the construction of aircraft.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The alpha and beta cells are found in:

Options:
A Endocrine part of pancreas
B Exocrine part of pancreas
C Accessory pancreatic duct
D Pancreatic acini

The correct answer is:
Endocrine part of pancreas

Solution:
The endocrine pancreas consists of âIslets of Langerhansâ. There are about 1 to 2 million Islets of Langerhans in a normal human pancreas representing only 1 to 2 percent of the pancreatic tissue. The two main types of cells in the Islet of Langerhans are called alpha-cells, beta-cells, delta cells and PP cells. The alpha-cells secrete a hormone called glucagon, while the beta-cells secrete insulin. Delta cells producing somatostatin (3â10%), while PP cells (gamma cells) producing pancreatic polypeptide

Topic: Biology, Chemical Coordination and Integration, NEET
The alternating current is given by, i=42sin2πTt+10 A. The R.M.S. value of this current is __________ A.

Options:

The correct answer is:
11

Solution:
The given function of current is, i=42sin2πTt+10 A

It means, i2=42sin22πtT+100+2042sin2πtT

⇒i2=422+100

RMS value of the current is, irms=i2=121=11 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The alternating e.m.f. induced in the secondary coil of a transformer is mainly due to

Options:
A varying electric field
B varying magnetic field
C the iron core
D heat produced in the coil

The correct answer is:
varying magnetic field

Solution:
The secondary coil experiences an emf as a result of the changing magnetic field. Therefore, a changing magnetic field is primarily responsible for the transformer voltage that is induced in a transformer's secondary coil.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2023 (14 May Shift 1)
The amine 'A', when treated with nitrous acid, gives a yellow oily substance. The amine A is

Options:
A Triethylamine
B Trimethylamine
C Aniline
D Methylphenylamine

The correct answer is:
Methylphenylamine

Solution:
2o/ Secondary amines âNitrous acid Yellow oily substance.
Out of these given four options, only methyl phenylamine is a secondary amine.

Topic: Chemistry, Amines, JEE Main
The amine that reacts will Hinsberg's reagent to give an alkali insoluble product is

Options:
A

B

C

D

The correct answer is:

Solution:
Secondary amines on reaction with Hinsberg's reagent (benzene sulphonyl chloride) forms $\mathrm{N}, \mathrm{N}$-dialkyl benzene sulphonamide. The product formed is not acidic as no $\mathrm{H}$-atom is attached to $\mathrm{N}$ thus it is insoluble in alkali.
The reaction of $2^{\circ}$ amine given in option (1) takes place as follows:
On the other hand, $1^{\circ}$ amines react with Hinsberg's reagent to yield $\mathrm{N}$-alkyl benzene sulphonamide which is soluble in alkali and $3^{\circ}$ amines do not react with $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$. The remaining options contain $1^{\circ}$ and $3^{\circ}$ amines.
Thus, option (1) is correct.

Topic: Chemistry, Amines, NEET, NEET 2019 (Odisha)
The amine which can react with $\mathrm{C}_6 \mathrm{H}_5-\mathrm{SO}_2-\mathrm{Cl}$ to form a product insoluble in alkali shall be

Options:
A Primary amine
B Secondary amine
C Tertiary amine
D Both primary and secondary amines

The correct answer is:
Secondary amine

Solution:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$ is called Hinsberg's reagent; they react with secondary amine to form a product that is soluble in alkalies. This reaction used to separate $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ amine from their mixture.

Topic: Chemistry, Amines, JEE Main