datavorous/entrance-exam-dataset huggingface dataset:
The focus of the curve $y^{2}+4 x-6 y+13=0$ is

Options:
A $(2,3)$
B $(-2,3)$
C $(2,-3)$
D $(-2,-3)$

The correct answer is:
$(-2,3)$

Solution:
The given equation of curve is:
$$
y^{2}+4 x-6 y+13=0
$$
which can be written as: $\mathrm{y}^{2}-6 \mathrm{y}+9+4 \mathrm{x}$
$$
\begin{array}{l}
+4=0 \\
\Rightarrow\left(y^{2}-6 y+9\right)=-4(x+1) \\
\Rightarrow(y-3)^{2}=-4(x+1)
\end{array}
$$
Put $Y=y-3$ and $X=x+1$
On comparing $Y^{2}=4 a X$
Length of focus from vertex, $a=-1$
At focus $X=a$ and $Y=0 \Rightarrow x+1=-1$
$$
\begin{array}{l}
\Rightarrow \mathrm{x}=-2 \\
\therefore \mathrm{y}-3=0 \Rightarrow \mathrm{y}=3 \\
\therefore \quad \text { Focus is }(-2,3)
\end{array}
$$

Topic: Mathematics, Parabola, JEE Main
The focus of the parabola $x^2=-16 y$ is

Options:
A $(4,0)$
B $(0,4)$
C $(-4,0)$
D $(0,-4)$

The correct answer is:
$(0,-4)$

Solution:
$a=4$, vertex $=(0,0)$, focus $=(0,-4)$.

Topic: Mathematics, Parabola, JEE Main
The following are the Vander Waal's parameters for four gases W, X, Y and Z

Gas	a(atmL2mol-2)	b(Lmol-1)

W	4.0	0.027

X	8.0	0.030

Y	6.0	0.032

Z	12.0	0.027

Based on given values identify, which one of these gases has the highest critical temperature

Options:
A W
B X
C Y
D Z

The correct answer is:
Z

Solution:
Critical temperature is the temperature at and above which gases cannot liquefied, no matter how much pressure is applied. For Z gas of given gases, critical temperature is highest
T c  =  8a 27Rb ⇒ T c ∝ a b ⇒It is highest for Z

Topic: Chemistry, States of Matter, JEE Main
The following arrangement performs the logic function of

Options:
A AND
B OR
C NAND
D NOR

The correct answer is:
NOR

Solution:
Y=A¯⋅B¯¯¯=A¯⋅B¯=A+B¯

Topic: Physics, Semiconductors, JEE Main
The following bodies,
1 a ring
2 a disc
3 a solid cylinder
4 a solid sphere,

of same mass m and radius R are allowed to roll down without slipping simultaneously from the top of the inclined plane. The body which will reach first at the bottom of the inclined plane is

[Mark the body as per their respective numbering given in the question]

Options:

The correct answer is:
4

Solution:
MgsinθR=mk2+mR2α

α=Rgsinθk2+R2⇒a=gsinθ1+k2R2

t=2 sa=2 s gsinθ1+k2R2

for least time, k should be least and we know k, is least for the solid sphere.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane: (i) a ring of radius R, (ii) a solid cylinder of radius R2 and (iii) a solid sphere of radius R4. If, in each case, the speed of the center of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is:

Options:
A 2:3:4
B 20:15:14
C 4:3:2
D 14:15:20

The correct answer is:
20:15:14

Solution:
From conservation of mechanical energy.
12mRv2+12IRωr2=mRghR⇒V22+V22=ghR
12mdv2+12Idωd2=mdghd⇒V22+V24=ghd
12msv2+12Isωs2=msghs⇒V22+V25=ghs
∴ hR:hd:hs=1:34:710=20:15:14

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The following chain-growth polymer is made up of how many difluoroethylene monomer units ?

Options:

The correct answer is:
4.00

Solution:
4 momomer units

Topic: Chemistry, Polymers, JEE Main
The following compound

can readily be prepared by Williamson ether synthesis by reaction between

Options:
A
B
C
D .

The correct answer is:

Solution:
Williamson ether synthesis is the reaction in which an alkyl halide is allowed to react with alkoxide to give ether. The reaction involves SN2 attack of an alkoxide ion on primary alkyl halide. In option a and b there are alkoxide and alkyl halide, but option b is preferred because I is better leaving group than Cl-

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The following compounds have been arranged in the order of increasing thermal stabilities. Identify the correct order.
K 2 CO 3 I ,  MgCO 3 II ,  CaCO 3 III ,  BeCO 3 IV

Options:
A I < II < III < IV
B IV < II < III < I
C IV < II < I < III
D II < IV < III < I

The correct answer is:
IV < II < III < I

Solution:
Carbonates of group 2 decompose to give oxides. Thermal stabilities of carbonates increase from Mg to Ba. Beryllium carbonate in unstable and is kept only in the atmosphere of CO2 .
Group 1 carbonates (and also hydroxides) except lithium do not decompose on heating. In this problem, K2CO3 remains unchanged on heating.

Topic: Chemistry, s Block Elements, NEET
The following concentrations were observed at 500 K for the formation of NH3 from N2 and H2. At equilibrium : N2=2×10-2M,H2=3×10-2M and NH3=1.5×10-2M. Equilibrium constant for the reaction is ______.

Options:

The correct answer is:
417

Solution:
The equilibrium constant of concentration (denoted by Kc) of a chemical reaction at equilibrium can be defined as the ratio of the concentration of products to the concentration of the reactants, each raised to their respective stoichiometric coefficients.

The equilibrium reaction is

N2g+3H2g⇋2NH3g

Kc=NH32H23N2

=32×10-223×10-23×2×10-2

=32×10-422×33×10-6×2×10-2

=14×2×3×104=1000024=416.6≈417

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The following constituent radiations of electromagnetic spectrum and their use is mentioned below
(i) Produce intense heating effect.
(ii) Is absorbed by the ozone layer in the atmosphere-ultraviolet radiation.
(iii) Is used for studying crystal structure X-rays.
How many of them are correct?

Options:
A one
B two
C three
D none

The correct answer is:
three

Solution:
(i) Infra-red waves: These are used in green houses to keep the plants warm.

(ii) Ultraviolet radiations: These are used in food preservation.

(iii) X-rays: These are used in medical diagnosis.

Topic: Physics, Electromagnetic Waves, JEE Main
The following conversion can be obtained by using
NH 3 (excess) → ? N 2 H 4

Options:
A OCl-
B HSO3-
C HCO3-
D PO4-3

The correct answer is:
OCl-

Solution:
2  NH 3 excess  + NaOCl→ N 2 H 4  + NaCl +  H 2 O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The following conversion is known as

Options:
A Etard reaction
B Rosenmund reaction
C Stephen reaction
D Gattermann-Koch reaction

The correct answer is:
Rosenmund reaction

Solution:
Hydrogenation of Acyl chloride in the presence of catalyst palladium on barium sulphate to form aldehyde is known as Rosenmund reaction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume.

SO2Cl2gâSO2g+Cl2g

Experiment	Time / s-1	Total pressure/atm

1	0	0.5

2	100	0.6

What is the rate of reaction when total pressure is 0.65 atm?

Options:
A 0.35 atm s-1
B 2.235×10-3 atm s-1
C 7.8×10-4 atm s-1
D 1.55×10-4 atm s-1

The correct answer is:
7.8×10-4 atm s-1

Solution:
SO2Cl2g→SO2g+Cl2g
k=2.303tlog⁡p02p0-Pt
=2.303100log⁡0.52×0.5-0.6=2.303100log⁡1.25
=2.2318×10-3s-1
Pressure of SO2Cl2 at time t pSO2Cl2
=2p0-Pt=2×0.50-0.65 atm
=0.35 atm
Rate of that time =k×pSO2Cl2
=2.2318×10-3×0.35=7.8×10-4 atm s-1

Topic: Chemistry, Chemical Kinetics, JEE Main
The following diagrams represent the types of aestivation in corolla. Identify the correct combination of labelling.

Options:
A A-Valvate, B-Vexillary, C-Twisted, D-Imbricate
B A-Vexillary, B-Imbricate C-Twisted, D-Valvate,
C A-Valvate, B-Imbricate C-Twisted, D-Vexillary
D A-Valvate, B-Twisted, C-Imbricate D-Vexillary

The correct answer is:
A-Valvate, B-Twisted, C-Imbricate D-Vexillary

Solution:
The mode of arrangement of sepals or petals in floral bud with respect to the other members of the same whorl is known as aestivation. The main types of aestivation are valvate, twisted, imbricate and vexillary.
In valvate, sepals or petals just touch one another at the margin, without overlapping, e.g., Calotropis.
In twisted, one margin of sepal or petal overlaps that of the next one and so on, e.g., China rose, lady’s finger, cotton, etc.
In imbricate, The margins of sepal or petals overlap one another but not in any particular direction e.g., Cassia, Goldmohur.
In vexillary, the largest posterior petal (vexillum or standard) overlaps two lateral petals (alae or wings) which in turn overlaps the two smallest, anterior but united petals (keel or carina), e.g., pea, bean etc.

Topic: Biology, Morphology of Flowering Plants, NEET
The following equilibrium constants are given:

N 2 + 3 H 2 ⇌ 2 NH 3 ; K⁡ 1

N 2 + O 2 ⇌ 2 NO ; K⁡ 2

H 2 + 1 / 2 O 2 ⇌ H 2 O ; K⁡ 3

The equilibrium constant for the oxidation of 2 mole NH 3 by oxygen to give NO is

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

The correct answer is:
K 2 K 3 3 K 1

Solution:
N 2 +3 H 2 ⇌ 2NH 3
K 1 = NH 3 2 N 2    H 2 3                 … i
N 2 +O 2 ⇌2NO  ;   K 2 = NO 2 N 2    O 2                 … ii
H 2 +1/2   O 2 ⇌ H 2 O  ;   K 3 = H 2 O H 2    O 2 1/2                 … iii
For the reaction, 2 NH 3 + 5 2 O 2 ⇌ 2 NO + 3 H 2 O
K⁡ = NO 2 H 2 O 3 NH 3 2 O 2 5 / 2
= NO 2 N 2 O 2 × H 2 O 3 H 2 3 O 2 3 / 2 × N 2 H 2 3 NH 3 2
= K 2 × K 3 3 K 1

Topic: Chemistry, Chemical Equilibrium, JEE Main
The following equilibrium constants are given
$\begin{array}{l}
\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; K_{1} \\
\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} ; K_{2}
\end{array}$
$\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O} ; K_{3}$
The equilibrium constant for the oxidation of 2 mole of $\mathrm{NH}_{3}$ to give NO is

Options:
A $K_{1} \cdot \frac{K_{2}}{K_{3}}$
B $K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$
C $K_{2}, \frac{K_{2}^{2}}{K_{1}}$
D $K_{2}^{2} \frac{K_{3}}{K_{1}}$

The correct answer is:
$K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

Solution:
$\begin{array}{l}
\text {Given, } \\
\begin{aligned}
\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; K_{1} &=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}} \\
\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons & 2 \mathrm{NO} ; K_{2}=\frac{[\mathrm{NO}]^{2}}{\left[\mathrm{N}_{2}\right]\left[\mathrm{O}_{2}\right]} \\
\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons & \mathrm{H}_{2} \mathrm{O} ; K_{3}=\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]}{\left[\mathrm{H}_{2}\right]\left[\mathrm{O}_{2}\right]^{1 / 2}}
\end{aligned}
\end{array}$
The chemical equation for the oxidation of 2 mol of
$\mathrm{NH}_{3}$ to give NO is
$2 \mathrm{NH}_{3}+\frac{5}{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$
(ii) and To get the equation (iv) from equation (i),
(iii) following steps are followed:
Reversing equation (i), we get
$2 \mathrm{NH}_{3} \rightleftharpoons \mathrm{N}_{2}+3 \mathrm{H}_{2} \text { so } K^{\prime}=\frac{1}{K_{1}}$
Multiplying cquation (iii) by 3, we get
$3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2} \rightleftharpoons 3 \mathrm{H}_{2} \mathrm{O} \mathrm{so} \mathrm{K}^{\prime}=K_{3}^{3}$
Adding equation (ii) and (vi)
$\mathrm{N}_{2}+3 \mathrm{H}_{2}+\frac{5}{2} \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}+3 \mathrm{H}_{2} \mathrm{O}$
so, $K^{\prime}=K_{2} \cdot K_{3}^{3}$
On combining (v) and (vii), we get the required equation having equilibrium constant $K^{\prime}=K_{2} \cdot \frac{K_{3}^{3}}{K_{1}}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The following equilibrium exists in a closed system at \( 25^{\circ} \mathrm{C} \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \). When a sample of pure \( \mathrm{NH}_{4} \mathrm{HS} \) (s) is placed in an evacuated vessel and allowed to reach equilibrium at \( 25^{\circ} \mathrm{C} \), the total pressure is \( 0.66 \) atm. Find the value of \( \mathrm{K}_{\mathrm{p}} \). (Answer by rounding off up to \( 2 \) decimal places.)

Options:

The correct answer is:
0.11

Solution:
At equilibrium same moles of NH3 and H2S is formed.

NH4HS (s)⇌NH3 (g)+H2S (g)Initially: x 0 0At equilibrium - P P

Given: Total pressure =0.66 atm

P+P=0.66 atm

P=0.33 atm

KP=PNH3×PH2S

For pure solid and pure liquid, active mass is taken as unity.

KP=0.33×0.33=0.11

Topic: Chemistry, Chemical Equilibrium, JEE Main
The following figure gives the movement of an object. Select the correct statement from the given choices.

Options:
A The total distance travelled by the object is 975 m
B The maximum acceleration of the object is 2 m s-2
C The maximum deceleration happened between 25th and 35th seconds
D The object was at rest between 10th and 15th seconds

The correct answer is:
The total distance travelled by the object is 975 m

Solution:
The total distance travelled by the object will be area of given graph in the figure.
Total distance s=75+225+125+100+400+50
=975 m

Topic: Physics, Motion In One Dimension, JEE Main
The following figure represents two biconvex lenses $L_1$ and $L_2$ having focal length $10 \mathrm{~cm}$ and $15 \mathrm{~cm}$ respectively. The distance between $L_1 \& L_2$ is :

Options:
A $10 \mathrm{~cm}$
B $35 \mathrm{~cm}$
C $25 \mathrm{~cm}$
D $15 \mathrm{~cm}$

The correct answer is:
$25 \mathrm{~cm}$

Solution:
$D=f_1+f_2=25 \mathrm{~cm}$
Paraxial parallel rays pass through focus and ray from focus of convex lens will become parallel

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The following figure shows a logic gate circuit with two inputs $\mathrm{A}$ and $\mathrm{B}$ and the output $\mathrm{C}$. The voltage waveforms of $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$ are as shown below:
The logic circuit gate is:

Options:
A OR gate
B AND gate
C NAND gate
D NOR gate

The correct answer is:
AND gate

Solution:
Truth table
The circuit is AND- gate

Topic: Physics, Semiconductors, JEE Main
The following figure shows a logic gate circuit with two inputs $A$ and $B$ and the output $Y$. The voltage waveforms of $A, B$ and $Y$ are as given
The logic gate is

Options:
A NOR gate
B OR gate
C AND gate
D NAND gate

The correct answer is:
NAND gate

Solution:
Table of logic gate
The given truth table is for NAND gate.

Topic: Physics, Semiconductors, JEE Main
The following figure shows a logic gate circuit with two inputs $A$ and $B$ and the output $Y$. The voltage waveforms of $A, B$ and $Y$ are as given
The logic gate is

Options:
A NOR gate
B OR gate
C AND gate
D NAND gate

The correct answer is:
NAND gate

Solution:
Table of logic gate
The given truth table is for NAND gate.

Topic: Physics, Semiconductors, NEET, NEET 2010 (Mains)
The following figure shows the graph of a continuous function $\mathrm{y}=\mathrm{f}(\mathrm{x})$ on the interval $[1,3]$. The points $\mathrm{A}$, $\mathrm{B}, \mathrm{C}$ have coordinates $(1,1),(3,2),(2,3)$ respectively, and the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ are parallel, with $\mathrm{L}_{1}$ being tangent to the curve at $\mathrm{C}$. If the area under the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ from $\mathrm{x}=1$ to $\mathrm{x}=3$ is 4 square units, then the area of the shaded region is-

Options:
A 2
B 3
C 4
D 5

The correct answer is:
2

Solution:
Equation of $\ell_{2} \rightarrow \mathrm{y}-1=\frac{2-1}{3-1}(\mathrm{x}-1)$
$2 y-2=x-1$ $2 y-x=1$
Equation of $\ell_{1}=y-3=\frac{1}{2}(x-2)$
$$
\begin{array}{l}
2 y-6=x-2 \\
2 y-x=4 \\
D \rightarrow\left(1, \frac{5}{2}\right), E \rightarrow\left(3, \frac{7}{2}\right)
\end{array}
$$
area under $f(x)=4$
$\begin{aligned} \text { shaded area } &=\text { area of trapezium DEFG }-\text { area under } \mathrm{f}(\mathrm{x}) \\ &=\frac{1}{2}\left(\frac{5}{2}+\frac{7}{2}\right) \times 2-4 \\ &=6-4=2 \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The following graph represents enzyme activity with changing

Options:
A pH
B temperature
C either a or b
D neither A nor B

The correct answer is:
either a or b

Solution:
The major factors on which enzyme activity depends are temperature and pH. Every enzyme has certain optimum temperature range at which it is active. At temperatures lower or higher the activity will not be seen. At lower temperatures, the enzyme remains totally inactive and at higher temperatures denaturation of proteins is seen and thus activity is hampered. Similar is the case with pH value. Every enzyme is active at a certain pH. At lower or higher pH this active is not seen. Hence when the graph of enzyme activity is plotted against temperature or pH a bell-shaped curve is obtained.

Topic: Chemistry, Biomolecules, NEET
The following graph represents the variation of photocurrent with anode potential for a metal surface. Here $I_{1}, I_{2}$ and $I_{3}$ represents intensities and $\gamma_{1}, \gamma_{2}, \gamma_{3}$ represent frequency for curves 1,2 and 3 respectively, then

Options:
A $\gamma_{1}=\gamma_{2}$ and $I_{1} \neq I_{2}$
B $\gamma_{1}=\gamma_{3}$ and $I_{1} \neq I_{3}$
C $\gamma_{1}=\gamma_{2}$ and $I_{1}=I_{2}$
D $\gamma_{2}=\gamma_{3}$ and $I_{1}=I_{3}$

The correct answer is:
$\gamma_{1}=\gamma_{2}$ and $I_{1} \neq I_{2}$

Solution:
The given figure can be shown as
From figure, we can see that the stopping potential $\left(V_{0}\right)$ is same for curve 1 and 2. But for curve 3 , it is greater than that for land 2 .
As, we know that, $e V_{0}=E_{\max }=h \gamma-\phi_{0}$
where, $\phi_{0}$ is the work function.
Since, $V_{0}$ is same for 1 and 2, so from above equation, we have $r_{1}=r_{2}$
Thus, for 1,2 and 3
$\gamma_{1}=\gamma_{2} \neq \gamma_{3}$
Also, the saturation current of curve 2 is greater than of curve 1 . But for curve 2 and 3 it is equal, so,
$I_{1} \neq I_{2}=I_{3}$

Topic: Physics, Dual Nature of Matter, JEE Main
The following graph shows how $\mathrm{T}_{1 / 2}$ (half-life) of a reactant $\mathrm{R}$ changes with the initial reactant concentration $a_0$.
The order of the reaction will be :

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$2$

Solution:
$t_{1 / 2} \propto \frac{1}{a^{n-1}}$
Hence (C)

Topic: Chemistry, Chemical Kinetics, JEE Main
The following is the illustration of the sequence of ovarian events “a” to “I” in a human female :

Identify developing follicle, mature follicles and corpus luteum.

Options:
A (a) to (d)- mature follicle, (e) to (f) developing follicle,(g)-(I) corpus luteum
B (a) to (d)- developing follicle, (e) to (f) mature follicle, (g)-(I) corpus luteum
C (a) to (d)- corpus luteum (e) to (f) mature follicle, (g)-(I) developing follicle
D None of these

The correct answer is:
(a) to (d)- developing follicle, (e) to (f) mature follicle, (g)-(I) corpus luteum

Solution:
a to d is developing follicle, e to f are mature follicles, g is developing corpus luteum. g to i is corpus luteum pituitary hormone-progesterone hormone. Corpus luteum secretes a large amount of progesterone hormone that is essential for the maintenance of the endometrium required for implantation of blastocyst leading to pregnancy. (i) In developing corpus luteum, the level of progesterone hormone is high but the level of estrogen hormone is lower. In regressing corpus luteum, the level of progesterone on the hormone is low but the level of estrogen hormone is high.

Topic: Biology, Human Reproduction, NEET
The following ligand is:

Options:
A tridentate
B hex dentate
C bidentate
D tetradentate

The correct answer is:
tetradentate

Solution:
The number of lone pairs that can be used by ligands is a measure of its denticity. In the given compound, both oxygen atoms and both nitrogen atoms are used as ligands and have lone pairs. Hence denticity is tetradentate.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The following logic gate is equivalent to :

Options:
A NOR Gate
B OR Gate
C AND Gate
D NAND Gate

The correct answer is:
NOR Gate

Solution:
Truth table for the given logic gate :

A	B	C

0	0	1

0	1	1

1	0	1

1	1	0

The truth table is similar to that of a NOR gate.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The following mechanism has been proposed for the reaction of $\mathrm{NO}$ with $\mathrm{Br}_2$ to form $\mathrm{NOBr}$ : $\mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NOBr}_2(\mathrm{~g})$
$$
\mathrm{NOBr}_2(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g})
$$
If the second step is the rate determining step, the order of the reaction with respect to $N O(g)$ is

Options:
A
1

B
0

C
3

D
2

The correct answer is:
2

Solution:
$$
\mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NOBr}_2(\mathrm{~g})
$$
$$
\mathrm{NOBr}_2(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \longrightarrow 2 \mathrm{NOBr}(\mathrm{g})
$$
$$
\begin{aligned}
\mathrm{R} & =\mathrm{K}\left[\mathrm{NOBr}_2\right][\mathrm{NO}] \\
& =\mathrm{K} \cdot \mathrm{K}_{\mathrm{c}}[\mathrm{NO}]\left[\mathrm{Br}_2\right][\mathrm{NO}], \text { where } \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{NOBr}_2\right]}{\left[\mathrm{NO}_2\right]\left[\mathrm{Br}_2\right]} \\
& =\mathrm{K}^{\prime}[\mathrm{NO}]^2\left[\mathrm{Br}_2\right]
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2006
The following metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.

Options:
A Copper
B Calcium
C Potassium
D Iron

The correct answer is:
Potassium

Solution:
Potassium metal ion activates many enzymes, participates in the oxidation of glucose to produce ATP and with Na, is responsible for the transmission of nerve signals.

Potassium is the main intracellular ion for all types of cells, while having a major role in maintenance of fluid and electrolyte balance. Animals, in particular, employ sodium and potassium differentially to generate electrical potentials in animal cells, especially in nervous tissue.

Topic: Chemistry, Chemistry in Everyday Life, NEET, NEET 2020 (Phase 1)
The following molecule acts as an :

Options:
A Antiseptic
B Anti–depressant
C Anti–bacterial
D Anti–histamine

The correct answer is:
Anti–histamine

Solution:
Synthetic drugs, brompheniramine (Dimetapp) act as antihistamines.

Antihistamines are drugs which treat hay fever and other allergies. Typically, people take antihistamines as an inexpensive, not patented, drug that can be bought without a prescription and relieves from nasal congestion, sneezing, or hives caused by pollen, dust mites, or animal allergy with few side effects.

We know that drugs help us diagnose and cure numerous illnesses. But any drug consumed by humans can actually have two kinds of effects - therapeutic effects and side effects. The goal is to prescribe the ideal drug, the one that only has a therapeutic effect.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D=1.25×10-2 m
rise of water, h=1.45×10-2 m
Using g=9.80 m s-2 and the simplified relation T=rhg2×103 N m-1 the possible error in surface tension is closest to:

Options:
A 10%
B 0.15%
C 1.5%
D 2.4%

The correct answer is:
1.5%

Solution:
Percentage error in surface tension
∆TT=∆rr+∆hh
∆rr=∆DD=0.011.25
∆hh=0.011.45
∴% error in T=∆TT×100=0.011.25+0.011.45×100=1.489≈1.5

Topic: Physics, Mathematics in Physics, JEE Main
The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D=1.25×10-2 m
rise of water, h=1.45×10-2 m
Using g=9.80 m s-2 and the simplified relation T=rhg2×103 N m-1 the possible error in surface tension is closest to:

Options:
A 10%
B 0.15%
C 1.5%
D 2.4%

The correct answer is:
1.5%

Solution:
Percentage error in surface tension
∆TT=∆rr+∆hh
∆rr=∆DD=0.011.25
∆hh=0.011.45
∴% error in T=∆TT×100=0.011.25+0.011.45×100=1.489≈1.5

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2017 (02 Apr)