datavorous/entrance-exam-dataset huggingface dataset:
Consider the reaction equilibrium :
\[
2 \mathrm{SO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{SO}_{3(\mathrm{~g})} ; \Delta \mathrm{H}^{0}=-198 \mathrm{~kJ}
\]
On the basis of Le Chatelier's principle, the condition favourable for the forward reaction is

Options:
A Lowering of temperture as well as pressure.
B Increasing temperature as well as pressure.
C Lowering the temperature and increasing the pressure.
D Any value of temperature and pressure.

The correct answer is:
Lowering the temperature and increasing the pressure.

Solution:
Le Chatelier's principle states that when a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts in such a way as to counteract the change to re-establish the equilibrium.

The conversion of SO2 to SO3 is an exothermic reaction, hence, decrease in temperature will favour the forward reaction. Volume or number of moles on the product side are also decreased. Thus, the reaction is favoured by high pressure.
( Le - Chatelier 's principle)

Topic: Chemistry, Chemical Equilibrium, JEE Main
Consider the reaction equilibrium,

2SO2g+O2gâ2SO3g; âHÂ°=-198 kJ

On the basis of Le-Chatelier's principle, the condition favourable for the forward reaction is

Options:
A Lowering of temperature as well as pressure
B Increasing temperature as well as pressure
C Lowering the temperature and increasing the pressure
D Any value of temperature and pressure

The correct answer is:
Lowering the temperature and increasing the pressure

Solution:
∆n=-ve
∆H°=-ve
Takes place with decrease in number of mole or pressure, hence increase in pressure shifts equilibrium in forward side.
Takes place with evolution of heat or increase in temperature, hence decrease in temperature shifts this equilibrium in forward side.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Consider the reaction equilibrium $2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g}) ; \Delta \mathrm{H}^0=-198 \mathrm{~kJ}$. On the basis of Le Chatelier's principle, the condition favourable for the forward reaction is

Options:
A
increasing temperature as well as pressure

B
lowering the temperature and increasing the pressure

C
any value of temperature and pressure

D
lowering of temperature as well as pressure

The correct answer is:
lowering the temperature and increasing the pressure

Solution:
Due to exothermicity of reaction low or optimum temperature will be required. Since 3 moles are changing to 2 moles.
$\therefore$ High pressure will be required.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2003
Consider the reaction :
H 2 SO 3 aq + Sn aq 4 + + H 2 O l⁡ → Sn aq 2 + + HSO 4 aq - + 3 H aq +
Which of the following statements is correct ?

Options:
A Sn4+ is the oxidizing agent because it undergoes oxidation
B Sn4+ is the reducing agent because it undergoes oxidation
C H2SO3 is the reducing agent because it undergoes reduction
D H2SO3 is the reducing agent because it undergoes oxidation

The correct answer is:
H2SO3 is the reducing agent because it undergoes oxidation

Solution:
Identify oxidation states of all elements in Redox Reaction
H2SO+43+Sn4++H2O(l)→Sn2++HSO4- +6 +3H+
S is changing the oxidation state from (+4) to (+6). i.e H2SO3 is a Reducing agent because it undergoes oxidation.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2014 (19 Apr Online)
Consider the reaction: $\mathrm{N}_2+3 \mathrm{H}_2 \longrightarrow 2 \mathrm{NH}_3$ carried out at constant temperature and pressure. If $\Delta \mathrm{H}$ and $\Delta \mathrm{U}$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

Options:
A
$\Delta \mathrm{H}=0$

B
$\Delta \mathrm{H}=\Delta \mathrm{U}$

C
$\Delta \mathrm{H} < \Delta \mathrm{U}$
D
$\Delta \mathrm{H}>\Delta \mathrm{U}$

The correct answer is:
$\Delta \mathrm{H} < \Delta \mathrm{U}$

Solution:
$$
\begin{aligned}
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{nRT} \\
& \Delta \mathrm{n}=-2 \\
& \Delta \mathrm{H}=\Delta \mathrm{U}-2 \mathrm{RT} \\
& \Delta \mathrm{H} < \Delta \mathrm{U}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main
Consider the reaction: $\mathrm{N}_2+3 \mathrm{H}_2 \longrightarrow 2 \mathrm{NH}_3$ carried out at constant temperature and pressure. If $\Delta \mathrm{H}$ and $\Delta \mathrm{U}$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

Options:
A
$\Delta \mathrm{H}=0$

B
$\Delta \mathrm{H}=\Delta \mathrm{U}$

C
$\Delta \mathrm{H} < \Delta \mathrm{U}$
D
$\Delta \mathrm{H}>\Delta \mathrm{U}$

The correct answer is:
$\Delta \mathrm{H} < \Delta \mathrm{U}$

Solution:
$$
\begin{aligned}
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{nRT} \\
& \Delta \mathrm{n}=-2 \\
& \Delta \mathrm{H}=\Delta \mathrm{U}-2 \mathrm{RT} \\
& \Delta \mathrm{H} < \Delta \mathrm{U}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2005
Consider the reaction $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})$

The equilibrium constant of the above reaction is $\mathrm{K}_{\mathrm{P}}$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that

$\mathrm{P}_{\mathrm{NH}_{3}}< < \mathrm{P}_{\text {total }}$ at equilibrium)

Options:
A $\frac{3^{3 / 2} \mathrm{~K}_{\mathrm{P}}^{1 / 2} \mathrm{P}^{2}}{16}$
B $\frac{\mathrm{K}_{\mathrm{P}}^{1 / 2 \mathrm{P}^{2}}}{16}$
C $\frac{\mathrm{K}_{\mathrm{P}}^{1 / 2 \mathrm{P}^{2}}}{4}$
D $\frac{3^{3 / 2} \mathrm{~K}_{\mathrm{P}}^{1 / 2} \mathrm{P}^{2}}{4}$

The correct answer is:
$\frac{3^{3 / 2} \mathrm{~K}_{\mathrm{P}}^{1 / 2} \mathrm{P}^{2}}{16}$

Solution:

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Consider the reaction,
$$
R \mathrm{CHO}+\mathrm{NH}_2 \mathrm{NH}_2 \longrightarrow \mathrm{RCH}=\mathrm{N}-\mathrm{NH}_2
$$
What sort of reaction is it?

Options:
A Electrophilic addition elimination reaction
B Free radical addition elimination reaction
C Electrophilic substitution elimination reaction
D Nucleophilic addition elimination reaction

The correct answer is:
Nucleophilic addition elimination reaction

Solution:
In this reaction
acts as nucleophilic
species which adds on carbon. Therefore this is
the case of nucleophilic addition reaction. Simultaneously, a mole of $\mathrm{H}_2 \mathrm{O}$ also eliminated. Therefore, overall this reaction is a nucleophilic addition elimination reaction.

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2012 (Mains)
Consider the reaction sequence below:

X is :

Options:
A
B
C
D

The correct answer is:

Solution:
In the given sequence, the succinic anhydride in the presence of AlCl3 reacts with methoxy benzene to give the following product.

In the second step of the sequence, the compound A undergoes Clemmenson's reduction, resulting in the formation of:

Overall reaction would be.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2016 (10 Apr Online)
Consider the reaction sequence given below :

Which of the following statements is true:

Options:
A Changing the base from OH⊖ to OR⊖ will have no effect on reaction (2)
B Changing the concentration of base will have no effect on reaction (1)
C Doubling the concentration of base will double the rate of both the reactions
D Changing the concentration of base will have no effect on reaction (2)

The correct answer is:
Changing the concentration of base will have no effect on reaction (1)

Solution:
First reaction is SN1 in which rate does not depend on conc. of nucleophile.

Second reaction is E2 reaction in which rate depends on conc. of base.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Consider the reactions
(i) $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br} \stackrel{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}{\longrightarrow}\left(\mathrm{CH}_3\right)_2 \mathrm{CH} -\mathrm{CH}_2 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{HBr}$
(ii) $\left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br} \stackrel{\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}}{\longrightarrow}\left(\mathrm{CH}_3\right)_2 \mathrm{CH} -\mathrm{CH}_2 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{Br}^{-}$
The mechanisms of reactions (i) and (ii) are respectively

Options:
A $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$
B $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 1$
C $\mathrm{S}_{\mathrm{N}} 2$ and $\mathrm{S}_{\mathrm{N}} 2$
D $\mathrm{S}_{\mathrm{N}} 2$ and $\mathrm{S}_{\mathrm{N}} 1$

The correct answer is:
$\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$

Solution:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ being a weaker nucleophile, when used as a solvent in case of hindered $1^{\circ}$ halide, favours $\mathrm{S}_{\mathrm{N}} 1$ mechanism while $\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}$being a strong nucleophile in this reaction favours $\mathrm{S}_{\mathrm{N}} 2$ mechanism.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2011 (Mains)
Consider the regions $\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}) \mid \mathrm{x}^{2}+\mathrm{y}^{2} \leq 100\right\}$ and $=\{(\mathrm{x}, \mathrm{y}) \mid \sin (\mathrm{x}+\mathrm{y})>0\}$ in the plane. Then the area of the region $\mathrm{A} \cap \mathrm{B}$ is-

Options:
A $10 \pi$
B 100
C $100 \pi$
D $50 \pi$

The correct answer is:
$50 \pi$

Solution:
$\quad \mathrm{x}^{2}+\mathrm{y}^{2} \leq 100 \rightarrow$ inside of a circle
$$
\begin{array}{l}
\sin (x+y)>0 \\
x+y \in(0, \pi) \cup(2 \pi, 3 \pi) \ldots \ldots \\
x+y=c \rightarrow \text { equation of a line }
\end{array}
$$
required area $=$ shaded region $=\frac{1}{2} \pi(10)^{2}=50 \pi$

Topic: Mathematics, Area Under Curves, JEE Main
Consider the set $A=\{1,2,3, \ldots, 30\}$. The number of ways in which one can choose three distinct numbers from Aso that the product of the chosennumbers is divisible by 9 is

Options:
A 1590
B 1505
C 1110
D 1025

The correct answer is:
1590

Solution:
We define the following subsets of set A :
$\mathrm{X}=9,18,27 \quad$ (divisible by 9) Number of elements
$=3$
$\mathrm{Y}=3,6,12,15,21,24,30 \quad \quad($ divisible by 3 but not by 9 ) Numher if
elements $=7$
$\mathrm{Z}=$ all the remaining elements of set A (not divisible by 3 ) Number of elements $=20$
Number of ways of choosing 3 distinct numbers such that their product is NOT divisible by 9 :
(I) Choose any 3 numbers from subset $\mathrm{Z}={ }^{20} \mathrm{C}_{3}=1140$
(II) Choose any 2 numbers from subset $\mathrm{Z}$ and any 1 number from subset $\mathrm{Y}=$ ${ }^{20} \mathrm{C}_{2} \times{ }^{7} \mathrm{C}_{1}=1330$
Total number of ways of choosing 3 distinct numbers such that their product is NOT divisible by $9=1140+1330=2470$
Total number of ways of choosing 3 distinct numbers from set $\mathrm{A}={ }^{30} \mathrm{C}_{3}=$ 4060
So, Number of ways of choosing 3 distinct numbers such that their product is divisible by $9=4060-2470=1590$

Topic: Mathematics, Permutation Combination, KVPY, KVPY 2017 (5 Nov SA)
Consider the set An of points x, y such that 0≤x≤n,0y≤n where n, x, y are integers. Let Sn be the set of all lines passing through at least two distinct points from An. Suppose we choose a line l at random from Sn. Let Pn be the probability that l is tangent to the circle x2+y2=n21+1-1n2 Then, the limit limn→∞Pn is

Options:
A 0
B 1
C 1π
D 12

The correct answer is:
0

Solution:
Equation of line passing through two points x1,y1 and x2,y2 is

y-y1=y2-y1x2-x1x-x1

⇒x2-x1y+y1-y2x+y1x1-x2+x1y2-y1=0

⇒ax+by+c=0, where a,b,c∈I

a=y1-y2,b=x2-x1

c∈yx1-x2+x1y2-y1

Perpendicular distance from origin to line

ax+by+c=0 is ca2+b2 and c2a2+b2 is rational.

Case I n is not perfect square and square f radius =n211-1n2= rational

⇒r2≠c2a2+b2

⇒ax+by+c=0 never be tangent to give n circle

⇒limn→∞Pn=0

Case II If n is perfect square. In this case number of tangents passing through two points from given set are few, but total number of lines are in much quantity when n approaches to infinite.
⇒limn→∞Pn=0

Topic: Mathematics, Straight Lines, JEE Main
Consider the set A of all matrices of order $3 \times 3$ with entries 0 or 1 only. Let $\mathrm{B}$ be the subset of A consisting of all matrices whose determinant is $1 .$ Let $\mathrm{C}$ be the subset of A consisting of all matrices whose determinant is $-1$. Then which one of the following is correct?

Options:
A $\mathrm{C}$ is empty
B $\mathrm{B}$ has as many elements as $\mathrm{C}$
C $A=B \cup C$
D $\mathrm{B}$ has thrice as many elements as $\mathrm{C}$

The correct answer is:
$\mathrm{B}$ has as many elements as $\mathrm{C}$

Solution:
determinant of $\mathrm{B}=1$
Let ' $\mathrm{B}$ ' be identity matrix
So, $B=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$
If we interchange any 2 rows, determinant will be $-1$
Let, $\mathrm{C}=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right|,|\mathrm{C}|=-1$
Here, number of elements in $\mathrm{B}$ and $\mathrm{C}$ are equal.

Topic: Mathematics, Determinants, JEE Main
Consider the set of all lines px+qy+r=0 such that 3p+2q+4r=0. Which one of the following statements is true?

Options:
A The lines are not concurrent.
B The lines are concurrent at the point 34,12 .
C The lines are all parallel.
D Each line passes through the origin.

The correct answer is:
The lines are concurrent at the point 34,12 .

Solution:
Given condition is 3p+2q+4r=0
⇒34p+12q+r=0 .......1
And given family of line is px+qy+r=0 ...(ii)
From (i) we can say that equation (ii) always passes through 34,12 .
Hence all lines are concurrent at 34,12 .

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Consider the set P=n1n,n∈N, then the largest element of set P is

Options:
A 212
B 313
C e1e
D 717

The correct answer is:
313

Solution:
Let f x = x 1 x ⇒ f ′ x = x 1 x 1−lnx x 2
At x=e, f ′ x =0 and sign change of derivative is from positive to negative, hence x=e is a point of local maxima of fx.
Now 2<e<3, hence f2 or f3 is the greatest element of set P

Since 23<32⇒212<313
So, f3=313 is the largest element of set P

Topic: Mathematics, Application of Derivatives, JEE Main
Consider the situation given in figure. The wire \( A B \) is slid on the fixed rails with a constant
velocity. If the wire \( A B \) is replaced by a semicircular wire, the magnitude of the induced current
will

Options:
A decrease
B increase
C increase or decrease depending on whether the semicircle bulges towards the resistance
or away from it
D remain same

The correct answer is:
remain same

Solution:
(D)
The induced emf e \( =l(\vec{v} \times \vec{B})=I(v B \sin \theta)=B v \mid \)
Induced current, \( \mathrm{i}=\frac{e}{R} \)
If the straight wire is replaced by a semicircular wire, resistance \( R \) remains unchanged.
Hence, current remains same.

Topic: Physics, Electromagnetic Induction, KCET, KCET 2019
Consider the situation shown in the figure. Here Î¸ is the maximum angle for which the light suffers total internal reflection at the vertical surface. Then sinÎ¸ will be:-

Options:

The correct answer is:
00.75

Solution:
The critical angle for this case is
Î¸ â³ = sin -1 1 1 . 2 5 = sin -1 4 5
or, sin Î¸ â³ = 4 5
Since Î¸ â³ = Ï 2 - Î¸ â² , we have sin Î¸ â² = cos Î¸ â³ = 3 / 5 . From Snell's law,
sin Î¸ sin Î¸ â² = 1 . 2 5
or, sin Î¸ = 1 . 2 5 Ã sin Î¸ â²
= 1 . 2 5 Ã 3 5 = 3 4
or, Î¸ = sin -1 3 4 .
If Î¸ â³ is greater than the critical angle, Î¸ will be smaller than this value. Thus, the maximum value of Î¸ , for which total reflection takes place at the vertical surface, is sin â1 (3/4).

Topic: Physics, Ray Optics, JEE Main
Consider the situation shown in the figure. The wall is smooth but the surfaces of blocks A and B in contact are rough. The friction on A due to B in equilibrium

Options:
A Is upward
B Is downward
C Is zero
D The system cannot remain in equilibrium for any value of F

The correct answer is:
Is upward

Solution:
Block 'A' cannot be in equilibrium. Due to Fsinθ component (upwards) on block 'B', Block 'A' moves downwards relative to block 'B'. So, friction on block 'B' downwords.

Topic: Physics, Laws of Motion, NEET
Consider the situation shown in the figure. The wire $A B$ is sliding on the fixed rails with a constant velocity. If the wire $A B$ is replaced by semicircular wire, the magnitude of the induced current will

Options:
A increase
B remain the same
C decrease
D increase or decrease depending whether the semicircle bulges towards the resistance or away from it

The correct answer is:
remain the same

Solution:
Magnitude of induced current in the wire $A B$ is given as
$$
I=\frac{B v l \sin \theta}{R}
$$
Here,
$$
\begin{aligned}
& \theta=90^{\circ} \\
& I=\frac{B v l}{R}
\end{aligned}
$$
$$
\therefore
$$
When wire $A B$ is replaced by semicircular wire, then neither $l$ nor $\theta$ are changed. $B, v$ and $R$ are already constant. Hence, induced current will remains same.

Topic: Physics, Electromagnetic Induction, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 1)
Consider the situation shown in the figure. The wire AB is sliding on the fixed rails with constant velocity v. If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will

Options:
A increase
B remain the same
C decrease
D increase or decrease depending on whether the semicircle bulges towards the resistance or away from it

The correct answer is:
remain the same

Solution:
As we have to consider the length perpendicular to velocity vector and magnetic field vector independent of shape and length of wire, the perpendicular length between two wires is constant hence value of induced current will remain constant.

Topic: Physics, Electromagnetic Induction, NEET
Consider the spectral line resulting from the transition n=2ân=1 in the atoms and ions given below. The shortest wavelength is produced by

Options:
A hydrogen atom
B deuterium atom
C singly ionized helium
D doubly ionized lithium

The correct answer is:
doubly ionized lithium

Solution:
Shortest wavelength will correspond to maximum energy. As the value of the atomic number (Z) increases, the magnitude of energy in different energy states gets increased. Value of Z is maximum for doubly ionised lithium atom (Z=3) among the given elements. Hence, the wavelength corresponding to this will be least.

Topic: Physics, Atomic Physics, JEE Main
Consider the statement: "For an integer n, if n3-1 is even, then n is odd". The contrapositive statement of this statement is:

Options:
A For an integer n, if n is even, then n3-1 is odd.
B For an integer n, if n3-1 is not even, then n is not odd.
C For an integer n, if n is even, then n3-1 is even.
D For an integer n, if n is odd, then n3-1 is even.

The correct answer is:
For an integer n, if n is even, then n3-1 is odd.

Solution:
P:n3-1 is even, q:n is odd

contrapositive of p→q=~q→~p

⇒ "if n is not odd then n3-1 is not even"

⇒For an integer n if n is even, then n3-1 is odd.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Consider the statement p : If a cylinder is right circular cylinder then its volume is 13πr2h.
Contrapositive of the statement p is

Options:
A If the volume of a cylinder is 13πr2h then it is not right circular cylinder
B If the volume of a cylinder is not 13πr2h then it is not right circular cylinder
C If a cylinder is not right circular cylinder then its volume is not 13πr2h
D Volume can’t be 13πr2h

The correct answer is:
If the volume of a cylinder is not 13πr2h then it is not right circular cylinder

Solution:
Contrapositive of “p→q” is “~q→~p”

Hence, the contrapositive of the given statement is: If the volume of a cylinder is not 13πr2h then it is not right circular cylinder

Topic: Mathematics, Mathematical Reasoning, JEE Main
Consider the statement p: If a hexagon is regular then all its sides and angles are equal. The contrapositive of statement p is

Options:
A If all sides and angles of a hexagon are not equal then it is not regular
B If all sides and angles of a hexagon are equal then it is regular
C If all sides and angles of a hexagon are not equal then it is regular
D If all sides and angles of a hexagon are equal then it is not regular

The correct answer is:
If all sides and angles of a hexagon are not equal then it is not regular

Solution:
Contrapositive of “If p then q” is “If ~q then ~p” hence contrapositive is
“If all sides and angles of a hexagon are not equal then it is not regular”

Topic: Mathematics, Mathematical Reasoning, JEE Main
Consider the statement p: If slope of a straight line is 1 then it is equally inclined to both the axes.

Then, the contrapositive of the statement p is

Options:
A If a straight line is equally inclined to both the axes then its slope is 1
B If a straight line is equally inclined to both the axes then its slope is not 1
C If a straight line is not equally inclined to both the axes then its slope is not 1
D If a straight line is not equally inclined to both the axes then its slope is 1

The correct answer is:
If a straight line is not equally inclined to both the axes then its slope is not 1

Solution:
Contrapositive of statement “if p then q” is “if ~q then ~p.
Hence contrapositive is
"If a straight line is not equally inclined to both the axes then its slope is not 1"

Topic: Mathematics, Mathematical Reasoning, JEE Main
Consider the statement: "Pn:n2-n+41 is prime". Then which one of the following is true?

Options:
A P3 is false but P5 is true
B Both P3 and P5 are false
C Both P3 and P5 are true
D P5 is false but P3 is true

The correct answer is:
Both P3 and P5 are true

Solution:
Given Pn=n2-n+41

P3=9-3+41=47

P5=25-5+41=61

Hence P3 and P5 both are prime.

Hence, both are true.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Consider the statement "The match will be played only if the weather is good and ground is not wet". Select the correct negation from the following:

Options:
A The match will not be played and weather is not good and ground is wet.
B If the match will not be played, then either weather is not good or ground is wet.
C The match will be played and weather is not good or ground is wet.
D The match will not be played or weather is good and ground is not wet.

The correct answer is:
The match will be played and weather is not good or ground is wet.

Solution:
Let p: weather is good
and q: ground is not wet

Given statement "The match will be played only if the weather is good and ground is not wet" can be written as p∧q

Now, negation of p∧q is ~p∧q≡~p∨~q

which is equivalent to "The match will be played and weather is not good or ground is wet".

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Consider the statements given below regarding contraception and answer as directed thereafter
(a) Medical Termination of Pregnancy (MTP) during first trimester is generally safe
(b) Generally chances of conception are nil until mother breast-feeds the infant upto two years
(c) Intrauterine devices like copper-T are effective contraceptives
(d) Contraception pills may be taken upto one week after coitus to prevent conception Which two of the above statements are correct?

Options:
A a, b
B b, c
C $\mathrm{c}, \mathrm{d}$
D a, c

The correct answer is:
a, c

Solution:
During intense lactational period the chances of conception are almost nil, maximum upto six months and not two years.
After unwanted sexual intercourse, conception can be avoided by taking contraceptive pills within 72 hrs.

Topic: Biology, Reproductive Health, NEET, NEET 2008 (Mains)