datavorous/entrance-exam-dataset huggingface dataset:
What is the main use of dinitrogen?

Options:
A To manufacture ammonia and other industrial chemicals containing nitrogen.
B To provide an inert atmosphere.
C As a refrigerant.
D To preserve biological materials, food items, and in cryosurgery.

The correct answer is:
To preserve biological materials, food items, and in cryosurgery.

Solution:
The paragraph states that the main use of dinitrogen is to manufacture ammonia and other industrial chemicals containing nitrogen, to provide an inert atmosphere, as a refrigerant, and to preserve biological materials, food items, and in cryosurgery.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the main use of helium in gas-cooled nuclear reactors?

Options:
A As a fuel for nuclear reactions
B As a cooling agent
C As a moderator
D As a shielding material

The correct answer is:
As a cooling agent

Solution:
Helium is used in gas-cooled nuclear reactors as a cooling agent.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the main use of liquid helium (helium at its boiling point) in cryogenics?

Options:
A Filling balloons
B Powering MRI machines
C Carrying out experiments at low temperatures
D In gas-cooled nuclear reactors

The correct answer is:
Carrying out experiments at low temperatures

Solution:
Liquid helium is used as a cryogenic agent for carrying out various experiments at low temperatures.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the main use of neon in discharge tubes and fluorescent bulbs?

Options:
A In filling balloons
B For meteorological observations
C For advertisement display purposes
D As a cryogenic agent

The correct answer is:
For advertisement display purposes

Solution:
Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
What is the main way that CO₂ is transported in the body?

Options:
A As a dissolved gas in the blood
B As carbamino-haemoglobin
C As oxyhaemoglobin
D As bicarbonate (HCO₃)

The correct answer is:
As bicarbonate (HCO₃)

Solution:
Nearly 70 per cent of carbon dioxide is transported as bicarbonate (HCO₃) with the help of the enzyme carbonic anhydrase. 20-25 per cent of carbon dioxide is carried by hemoglobin as carbamino-hemoglobin.

Topic: Biology, Breathing and Exchange of Gases, NEET
What is the main way that O₂ is transported in the body?

Options:
A As a dissolved gas in the blood
B As carbamino-haemoglobin
C As oxyhaemoglobin
D As bicarbonate (HCO₃)

The correct answer is:
As oxyhaemoglobin

Solution:
Oxygen is transported mainly as oxyhaemoglobin. In the alveoli where pO₂ is higher, O₂ gets bound to hemoglobin which is easily dissociated at the tissues where pO₂ is low and pCO₂ and H⁺ concentration are high.

Topic: Biology, Breathing and Exchange of Gases, NEET
What is the major product expected from the following reaction?
Where D is an isotope of hydrogen.

Options:
A

B

C

D

The correct answer is:

Solution:
The hydrohalogenation of alkenes involves breaking a carbon to carbon double bond, followed by the electrophilic addition of a hydrogen atom and halogen. The halide will add to the more substituted carbon following Markovnikov's rule.

anti addition reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2015 (11 Apr Online)
What is the major product formed by HI on reaction with

Options:
A
B
C
D

The correct answer is:

Solution:
When a protic acid HC (X = Cl, Br, I) is added to an asymmetrically substituted alkene, addition of acidic hydrogen takes place at the less substituted carbon atom of the double bond, while X is added to the more alkyl substituted carbon atom.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2021 (24 Feb Shift 1)
What is the major product of the following reaction?

Options:
A

B

C

D

The correct answer is:

Solution:
This reaction is an example of aldol reaction. The product formation is shown below.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (25 Jul Shift 2)
What is the major product of the Hofmann rearrangement of a primary amide?

Options:
A Primary amine
B Secondary amine
C Nitrile
D Aldehyde

The correct answer is:
Primary amine

Solution:
Primary amine
Explanation: The Hofmann rearrangement involves the conversion of a primary amide to a primary amine with one fewer carbon atom. This reaction typically uses bromine $\left(\mathrm{Br}_2\right)$ and a strong base like $\mathrm{NaOH}$.

Topic: Chemistry, Amines, JEE Main
What is the major product of the reaction between an amide and $\mathrm{LiAlH}_4$ followed by hydrolysis?

Options:
A Primary amine
B Secondary amine
C Tertiary amine
D Alcohol

The correct answer is:
Primary amine

Solution:
Primary amine
Explanation: $\mathrm{LiAlH}_4$ reduces the amide to a primary amine, and subsequent hydrolysis of the reduced product yields the primary amine.

Topic: Chemistry, Amines, JEE Main
What is the major product of the reaction between ethylene (ethene) and a peracid?

Options:
A Ethanol
B Ethylene glycol
C Ethylene oxide (oxirane)
D Acetaldehyde

The correct answer is:
Ethylene oxide (oxirane)

Solution:
Ethylene oxide (oxirane)
Explanation: The reaction of ethylene with a peracid produces ethylene oxide (oxirane), a simple epoxide.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
What is the major product "P" of the following reaction ?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (25 Jul Shift 2)
What is the major product when 1,2 -dibromoethane is treated with aqueous $\mathrm{KOH}$ ?

Options:
A Ethylene
B Ethylene glycol
C Acetaldehyde
D Ethylene oxide

The correct answer is:
Ethylene glycol

Solution:
Ethylene glycol
Explanation: When 1,2-dibromoethane is treated with aqueous $\mathrm{KOH}$, it undergoes hydrolysis to form ethylene glycol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution?
(Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)

Options:
A 28 g
B 3.5 g
C 7 g
D 14 g

The correct answer is:
7 g

Solution:
16.9 g AgNO3 is present in 100 mL solution.
∴8.45 g AgNo3 is present in 50 mL solution
5.8 g NaCl is present in 100 mL solution
∴2.9 g NaCl is present in 50 mL solution
Mass of AgCl
=0.049×143.5 g
=7 g AgCl

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
What is the mass ratio of ethylene glycol (C2H6O2, molar mass =62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?

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

The correct answer is:
2:1

Solution:
The molality of aqueous ethylene glycol solution = 0.25 m

Mass of ethylene glycol required for 1000 g water = 624 = 15.5 gm

Mass of solution = 1015.5 gm

Mass of ethylene glycol in 500 gm solution = 15.5 × 5001015.5 = 7.63 gm

Assuming density of solution as 1 gm/mL.

Mass of ethylene glycol in 250 mL =7.632 = 3.815 gm

Mass ratio of ethylene glycol for making 500 gm of 0.25 m solution and 250 mL of 0.25 m solution = 2 : 1

So, the correct option is C

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (25 Jan Shift 2)
What is the maximum acceleration of the train required for a box lying on its floor to remain stationary, considering a coefficient of static friction between the box and the train's floor of 0.15?

Options:
A 1.5 m/s²
B 0.15 m/s²
C 9.8 m/s²
D 0.98 m/s²

The correct answer is:
1.5 m/s²

Solution:
Correct Option is : (A)
1.5 m/s²

Topic: Chemistry, Laws of Motion, NEET
What is the maximum point on the curve $x=e^{x} y$ ?

Options:
A $(1, e)$
B $\left(1, e^{-1}\right)$
C $(e, 1)$
D $\left(e^{-1}, 1\right)$

The correct answer is:
$\left(1, e^{-1}\right)$

Solution:
Given curve is $x=e^{x} y$
Which can be rewritten as $y=x e^{-x}$ On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=-x e^{-x}+e^{-x}$
Put $\frac{d y}{d x}=0$ for maxima or minima
$\Rightarrow-x e^{-x}+e^{-x}=0$
$\Rightarrow e^{-x}(1-x)=0$
Since $e^{-x}$ can not be zero $\therefore \quad 1-x=0 \Rightarrow x=1$
Now, $\frac{d^{2} y}{d x^{2}}=-e^{-x}+x e^{-x}-e^{-x}=x e^{-x}-2 e^{-x}$
$=e^{-x}(x-2)$
$\Rightarrow\left(\frac{d^{2} y}{d x^{2}}\right) < 0$
$\therefore y$ is maximum at $x=1$. Thus, when $x=1$ then $\mathrm{y}=e^{-1}$
Hence, maximumpoint on the curve $x=e^{x} y$ is $\left(1, e^{-1}\right)$.

Topic: Mathematics, Application of Derivatives, JEE Main
What is the maximum slope of the curve $y=-x^{3}+3 x^{2}+2 x-27 ?$

Options:
A 1
B 2
C 5
D $-23$

The correct answer is:
5

Solution:
The equation of curve is given as:
$y=-x^{\frac{1}{3}}+3 x^{2}+2 x-27$
On differentiating w.r.t.x, we get
$\frac{d y}{d x}=-3 x^{2}+6 x+2$
This represents slope of the curve at any point. Let $\quad A=\frac{d y}{d x}=-3 x^{2}+6 x+2$
$\Rightarrow \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=-6 \mathrm{x}+6$
and $\frac{d^{2} A}{d x^{2}}=-6$
Put $\frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}}=0$ for maxima or minima. $-6 x+6=0$
$\Rightarrow \mathrm{x}=1$
Now, $\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=1} 1=-6 < 0$
$\therefore \quad$ A is maximumat $x=1$ $\therefore \quad$ Maximum slope of curve $=-3+6+2=5$

Topic: Mathematics, Application of Derivatives, JEE Main
What is the maximum value of the force F (in newton) such that the block shown in the arrangement, does not move?

Options:

The correct answer is:
20

Solution:
Free body diagram (FBD) of the block (shown by a dot) is shown in figure.
For vertical equilibrium of the block,
N=mg+Fsinâ¡60Â°=3g+3F2 â¦.i
For no motion, force of friction
fâ¥Fcosâ¡60Â°
or Â°Î¼ Nâ¥Fcosâ¡60Â°
or1233 g+3 F2â¥F2
or gâ¥F2 or Fâ¤2 g or 20 N
Therefore, maximum value of F is 20 N.

Topic: Physics, Laws of Motion, JEE Main
What is the maximum value of the force F such that the block shown in the arrangement, does not move?

Options:
A 15 N
B 5 N
C 10 N
D 20 N

The correct answer is:
20 N

Solution:
The forces acting on the block are shown. Since the block is not

moving forward for the maximum force F applied, therefore

Fcos60°=f=μN …(i) (Horizontal Direction)

and Fsin60°+mg=N …(ii)

From (i) and (ii)

Fcos60°=μ[ Fsin60°+mg ]

⇒    F=μmgcos60°−μsin60°=123×3×1012−123×32=514=20 N

Topic: Physics, Laws of Motion, JEE Main
What is the maximum value of the function $\log x-x ?$

Options:
A $-1$
B 0
C 1
D $\infty$

The correct answer is:
$-1$

Solution:
Let $y=\log x-x$ $\therefore \frac{d y}{d x}=\frac{1}{x}-1$ and $\frac{d^{2} y}{d x^{2}}=-\frac{1}{x^{2}}$
For maximum and minimum value of $y$
$\frac{d y}{d x}=\frac{1}{x}-1=0 \Rightarrow \frac{1}{x}=1 \Rightarrow x=1$
For $x=1, \frac{d^{2} y}{d x^{2}}=-v e$
Thus, the value of given function is maximum for $x=1$
So, the maximum value of the function $=\log (1)-1=-1$

Topic: Mathematics, Application of Derivatives, JEE Main
What is the maximum value of $\mathrm{x}$. y subject to the condition $\mathrm{x}+\mathrm{y}=8 ?$

Options:
A 8
B 16
C 24
D 32

The correct answer is:
16

Solution:
The given condition is:
$x+y=8 \Rightarrow y=8-x$
and let, the product of $\mathrm{x}$ and $\mathrm{y}$ be, $\mathrm{P}=\mathrm{xy}$
$\Rightarrow P=x(8-x)=8 x-x^{2}$
Differentiating w.r.t. $\mathrm{x}$
$\Rightarrow \frac{\mathrm{dP}}{\mathrm{dx}}=8-2 \mathrm{x}$
and $\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}=-2 < 0$
Put $\frac{\mathrm{dP}}{\mathrm{dx}}=0$, for maxima or minima
$8-2 x=0$
$\Rightarrow x=\frac{8}{2}=4$ and $y=4$
and $\left(\frac{\mathrm{d}^{2} \mathrm{P}}{\mathrm{dx}^{2}}\right) < 0$ when $\mathrm{x}=\mathrm{y}$
$\therefore$ P is maximumat $\mathrm{x}=4$ Maximum value of $\mathrm{P}=4 \cdot(8-4)=4.4=16$

Topic: Mathematics, Application of Derivatives, JEE Main
What is the meaning of stability in context to a biological community?

Options:
A A stable community should not show too much variation in productivity from year to year.
B A stable community must be either resistant or resilience to occasional disturbances (natural or man-made).
C A stable community must be resistant to invasion by alien species.
D All of the above.

The correct answer is:
All of the above.

Solution:
Various groups of interacting organisms in a particular region are included in community. A community is said to be stable when its populations size of all species remains same over a period of time. In other words, the population is not showing much change or variation and is able to withstand any hardships or changes in the environment. A stable community is not affected by invasion of other organisms including alien species. It resists loss of biodiversity.

Topic: Biology, Biodiversity and its Conservation, NEET
What is the measure of the angle $114^{\circ} 35^{\prime} 30^{\prime}$ in radian?

Options:
A $1 \mathrm{rad}$
B $2 \mathrm{rad}$
C $3 \mathrm{rad}$
D $4 \mathrm{rad}$

The correct answer is:
$2 \mathrm{rad}$

Solution:
From relation between minute 6 seconds measure $60^{13}=\mathrm{I}^{\circ}$
$\Rightarrow 30^{\prime \prime}=\frac{1^{\prime}}{2}$
$\Rightarrow 35^{\prime} 30^{\prime \prime}=\left(35+\frac{1}{2}\right)^{\prime}=\left(\frac{71}{2}\right)^{\prime}$
Also, $60^{\circ}=1^{\circ}$
$\therefore 1^{\prime}=\left(\frac{1}{60}\right)^{\circ}$
$\Rightarrow\left(\frac{71}{2}\right)^{\prime}=\left(\frac{71}{2} \times \frac{1}{60}\right)^{\circ}=\left(\frac{71}{120}\right)^{\circ}$
$\therefore \quad 114^{\circ} 35^{\circ} 30^{\prime}=\left(114+\frac{71}{120}\right)^{\circ}=\left(\frac{13751}{120}\right)^{\circ}$
We know that, $2 \pi \mathrm{rad}=360^{\circ}$
$\Rightarrow 1^{\circ}=\frac{2 \pi}{360} \mathrm{rad}$
$\Rightarrow\left(\frac{13751}{120}\right)^{\circ}=\frac{2 \pi}{360^{\circ}} \times \frac{13751}{120} \mathrm{rad}$
$=\frac{2 \times 22 \times 13751}{7 \times 360 \times 120} \mathrm{rad}=2.0008069 \mathrm{rad}$
$\Rightarrow 114^{\circ} 35^{\prime} 30^{\prime \prime}=2 \mathrm{rad}($ approx. $)$

Topic: Mathematics, Trigonometric Equations, JEE Main
What is the middle lamella in plants and fungi?

Options:
A A layer of cellulose and hemicellulose
B A layer of calcium pectate that holds neighbouring cells together
C A layer of proteins that helps in cell-to-cell interaction
D A layer of minerals like calcium carbonate that gives shape to the cell

The correct answer is:
A layer of calcium pectate that holds neighbouring cells together

Solution:
The middle lamella is a layer mainly of calcium pectate that holds or glues the different neighbouring cells together. It serves as a cementing layer between the cell walls of adjacent cells.

Topic: Biology, Cell - The Unit of Life, NEET
What is the middle term in the expansion of $\left(1-\frac{x}{2}\right)^{8}$ ?

Options:
A $\frac{35 x^{4}}{8}$
B $\frac{17 x^{5}}{8}$
C $\frac{35 x^{5}}{8}$
D None of these

The correct answer is:
$\frac{35 x^{4}}{8}$

Solution:
Since $n=8$ is even number therefore middle term
$=\left(\frac{n}{2}+1\right)^{\mathrm{th}}$ term $=(4+1)=5^{\mathrm{th}}$ term
Hence, $T_{5}={ }^{8} C_{4}(1)^{4}\left(-\frac{x}{2}\right)^{4}$
$=\frac{8 !}{4 ! 4 !} \times \frac{x^{4}}{16}=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot \frac{x^{4}}{16}=\frac{70 x^{4}}{16}=\frac{35 x^{4}}{8}$

Topic: Mathematics, Binomial Theorem, JEE Main
What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?

Options:
A GmM 2 R
B GmM 3 R
C 5 GmM 6 R
D 2 GmM 3 R

The correct answer is:
5 GmM 6 R

Solution:
As we know,
Gravitational potential cnergy =-GMmr
and orbital velocity, v0=GM/R+h
Ef=12mv02-GMm3R=12 mGM3R-GMm3R
=GMm3R12-1=-GMm6R
Ei=-GMmR+K
Ei=Ef
Therefore minimum required energy,
K=5GMm6R

Topic: Physics, Gravitation, JEE Main, JEE Main 2013 (07 Apr)
What is the minimum length of antenna required to transmit a radio signal of frequency 20 MHz ?

Options:
A 5.0 m
B 7.5 m
C 2.0 m
D 3.75 m

The correct answer is:
2.0 m

Solution:
Given
\(v=20 \mathrm{mH}_2=20 \times 10^6 \mathrm{~Hz}\)
The length of transmiting antenna solved be approximately equal to wavelength of wave
\(\begin{aligned}
& \text { wavelength }=\frac{\text { velocity }}{\text { frequency }} \mathrm{m} \\
& \lambda=\frac{3 \times 10^8}{20 \times 10^6} \mathrm{~m} \\
& \lambda=15 \mathrm{~m}
\end{aligned}\)
minimum length of antenna
\(\begin{aligned}
& \frac{\lambda}{4}=\frac{15}{4} \\
& =3.75 \mathrm{~m}
\end{aligned}\)

Topic: Physics, Communication System, JEE Main
What is the minimum pH needed to prevent the precipitation of ZnS in a solution i.e. 0.01 M  ZnCl 2 and saturated with 0.1 M  H 2 S ?
Given:
K sp of ZnS =  10 −21
Ka of H 2 S =  10 −20

Options:

The correct answer is:
1

Solution:
Ksp(ZnS)=[Zn2+][S2-]
S2-=KspZn2+=10-210.01=10-19
KH2S=H+2S-2H2S
10-20=H+210-190.1
H+=0.1
pH=-log.⁡H+
=1

Topic: Chemistry, Ionic Equilibrium, JEE Main
What is the minimum quantity (in grams) of methyl iodide required for preparing one mole of ethane by Wurtz reaction?
(Atomic weight of iodine $=127$ )

Options:
A $142$
B $568$
C $326$
D $284$

The correct answer is:
$284$

Solution:
Wurtz reaction for the preparation of ethane is
$$
\underset{1}{\mathrm{CH}_3 \mathrm{I}+2 \mathrm{Na}}+\underset{\mathrm{ICH}}{\mathrm{ICH}_3} \stackrel{\text { dry ether }}{\longrightarrow} \underset{1 \mathrm{~mole}}{\mathrm{C}_2 \mathrm{H}_6}+2 \mathrm{NaI}
$$
$1 \mathrm{~mole} \quad 1 \mathrm{~mole} \quad 1 \mathrm{~mole}$
Number of moles of $\mathrm{CH}_3 \mathrm{I}$ required for the manufacture of 1 mole $_2 \mathrm{H}_6=2$ moles
Weight of 2 moles of $\mathrm{CH}_3 \mathrm{I}$
$$
\begin{aligned}
& =2 \times(12+3+127) \\
& =2 \times(142) \\
& =284 \mathrm{~g}
\end{aligned}
$$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
What is the minimum value of $2 x^{2}-3 x+5 ? \quad$

Options:
A 0
B $3 / 4$
C $31 / 4$
D $31 / 8$

The correct answer is:
$31 / 8$

Solution:
Let $y=2 x^{2}-3 x+5$
$\Rightarrow \frac{d y}{d x}=4 x-3$
and $\frac{d^{2} y}{d x^{2}}=4$
For maximum and minimum value of $y$
$\frac{d y}{d x}=0$
$\Rightarrow 4 x-3=0$
$\Rightarrow \quad x=\frac{3}{4}$
Since, for every value of $x, \frac{d^{2} y}{d x^{2}}=+v e$ So, minimum value of $2 x^{2}-3 x+5$ is minimum
$x=\frac{3}{4}$
$\therefore \quad$ The minimum value $=2\left(\frac{3}{4}\right)^{2}-3\left(\frac{3}{4}\right)+5$
$=\frac{9}{8}-\frac{9}{4}+5=\frac{9-18+40}{8}=\frac{31}{8}$

Topic: Mathematics, Application of Derivatives, JEE Main
What is the minimum value of $\cos \theta+\cos 2 \theta$ ?

Options:
A $-2$
B $-\frac{9}{8}$
C 0
D $-\frac{9}{16}$

The correct answer is:
$-\frac{9}{8}$

Solution:
Let $\mathrm{A}=\cos \theta+\cos 2 \theta$
$\therefore$ On differentiating w.r.t. to $\theta$, we get
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{d} \theta}=-\sin \theta-2 \sin 2 \theta$
Put $\frac{\mathrm{dA}}{\mathrm{d} \theta}=0$ for maximaor minima. $\sin \theta+2 \sin 2 \theta=0 \Rightarrow \sin \theta+4 \sin \theta \cos \theta$
$\Rightarrow \sin \theta(1+4 \cos \theta)=0$
$\Rightarrow \sin \theta=0$, or $4 \cos \theta+1=0$
$\Rightarrow \cos \theta=1$ or $\cos \theta=-\frac{1}{4}$
Now, $\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}=-\cos \theta-4 \cos 2 \theta$
$=-\cos \theta-4\left(2 \cos ^{2} \theta-1\right)$
For $\cos \bar{\theta}=1$
$\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}=-\cos \theta-4\left(2 \cos ^{2} \theta-1\right)$
$=-1-4(2(1)-1)=-1-4=-5 < 0$
So, A is maximum at $\cos \theta=1$
$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{~d} \theta^{2}}\right)_{\cos \theta=\frac{-1}{4}}=\frac{1}{4}-4\left(2 \cdot \frac{1}{16}-1\right)>0$
$\quad\left[\right.$ Since $\left.\cos 2 \theta=2 \cos ^{2} \theta-1\right]$
$\therefore \mathrm{A}$ is minimum at $\theta=\cos ^{-1}\left(\frac{-1}{4}\right)$
Now minimum value of $\cos \theta+\cos 2 \theta$
or of $\cos \theta+2 \cos ^{2} \theta-1$
$=\left(\frac{-1}{4}\right)+2\left(\frac{1}{16}\right)-1$
$=\frac{-1}{4}+\frac{1}{8}-1=\frac{-2+1-8}{8}=\frac{-9}{8}$

Topic: Mathematics, Trigonometric Equations, JEE Main
What is the mode of transport of the remaining 3% of O₂ in the blood?

Options:
A Dissolved state through plasma
B Transport by RBCs
C Transport as bicarbonate
D None of the above

The correct answer is:
Dissolved state through plasma

Solution:
The remaining 3% of O₂ is carried in a dissolved state through the plasmA)

Topic: Biology, Breathing and Exchange of Gases, NEET
What is the modulation index of an FM signal having a carrier swing of 100 kHz when the modulating signal has a frequency of 8 kHz?

Options:
A

6.25

B

6.23

C

6.55

D

6.33

The correct answer is:

6.25

Solution:
Given: Carrier Swing =100 kHz
fa=8 kHz
From the defining equation,
mf=âffa
First determining âf ,
âf=Carrier Swing2
=100Ã1032
=50 kHz
Now substituting into the equation for mf
mf=50Ã1038Ã103
=6.25

Topic: Physics, Communication System, JEE Main
What is the molecular formula of borazole?

Options:
A $\mathrm{B}_2 \mathrm{H}_6$
B $\mathrm{B}_6 \mathrm{~N}_6 \mathrm{H}_6$
C $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_6$
D $\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_3$

The correct answer is:
$\mathrm{B}_3 \mathrm{~N}_3 \mathrm{H}_3$

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
What is the moment of inertia of a solid sphere of density $\rho$ and radius $\mathrm{R}$ about its diameter?

Options:
A $\frac{105}{176} \mathrm{R}^{5} \rho$
B $\frac{105}{176} \mathrm{R}^{2} \rho$
C $\frac{176}{105} \mathrm{R}^{5} \rho$
D $\frac{176}{105} \mathrm{R}^{2} \rho$

The correct answer is:
$\frac{176}{105} \mathrm{R}^{5} \rho$

Solution:
For solid sphere
$\begin{array}{l}
I=\frac{2}{5} M R^{2}=\frac{2}{5}\left(\frac{4}{3} \pi R^{3} \rho\right) R^{2} \\
\rho=\frac{176}{105} R^{5} \rho
\end{array}$

Topic: Physics, Rotational Motion, JEE Main
What is the name of the inactive stage of the cell cycle where cells remain metabolically active but do not proliferate?

Options:
A G₁ phase
B S phase
C G₂ phase
D Gₒ phase

The correct answer is:
Gₒ phase

Solution:
The cells that do not divide further exit G₁ phase to enter an inactive stage called quiescent stage (Gₒ) of the cell cycle.

Topic: Biology, Cell Cycle and Cell Division, NEET
What is the name of the last region of the small intestine?

Options:
A Duodenum
B Jejunum
C Ileum
D Pyloric

The correct answer is:
Ileum

Solution:
The small intestine is divided into three regions, the last of which is the highly coiled ileum. The ileum is responsible for absorbing the remaining nutrients from the food before it reaches the large intestine.

Topic: Biology, Digestion and Absorption, NEET
What is the nature of Gaussian surface involved in Gauss's law of electrostatic?

Options:
A Scalar
B Electrical
C Magnetic
D Vector

The correct answer is:
Vector

Solution:
The Gauss's law in electrostatic given a relation between electric flux through any closed hypothetical surface (called a Gaussian surface) and the charge enclosed by the surface.
So, the nature is area vector.

Topic: Physics, Electrostatics, JEE Main
What is the negation of \( q \vee \sim(p \wedge r) \) ?

Options:
A \( \sim q \wedge \sim(p \wedge r) \)
B \( \sim q \wedge(p \wedge r) \)
C \( \sim q \vee(p \wedge r) \)
D None of these

The correct answer is:
\( \sim q \wedge(p \wedge r) \)

Solution:
As we know from De Morgan's law, that

~p∨q≡~p∧~q
Therefore, ~q∨~p∧r≡~q∧~~p∧r
Hence, ~q∨~p∧r≡~q∧p∧r

Topic: Mathematics, Mathematical Reasoning, JEE Main
What is the normality of a $1 \mathrm{M}$ solution of $\mathrm{H}_3 \mathrm{PO}_4$

Options:
A $0.5 \mathrm{~N}$
B $1.0 \mathrm{~N}$
C $2.0 \mathrm{~N}$
D $3.0 \mathrm{~N}$

The correct answer is:
$3.0 \mathrm{~N}$

Solution:
$\mathrm{H}_3 \mathrm{PO}_4$ is tribasic so $\mathrm{N}=3 \mathrm{M}=3 \times 1=3$.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main