datavorous/entrance-exam-dataset huggingface dataset:
Pb+dil. HNO3âWarmP+Qâ+H2O

incorrect statement for Q is

Options:
A Paramagnetic colourless gas
B It is oxidized to paramagnetic coloured gas by air
C It combines with Fe2(SO4)3
D It is also obtained by disproportionation of HNO2

The correct answer is:
It combines with Fe2(SO4)3

Solution:
Here only statement in option C is incorrect as it combines with FeSO4 and not it combines with Fe2(SO4)3
Pbâ+âdil . HNO 3 â Warm Pb(NO 3 ) 2 â+âNOââ+â H 2 O
paramagnetic, coloured gas
HNO2âHNO3+NOâ+H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
[ P ] â Br 2 C 2 H 4 Br 2 â NaNH 2 Q â 20 % H 2 SO 4 R â ZnHg/HCl S

The species P, Q, R and S respectively are

Options:
A ethene, ethyne, ethanal, ethane
B ethane, ethyne, ethanal, ethene
C ethene, ethyne, ethanal, ethanol
D ethyne, ethane, ethene, ethanal

The correct answer is:
ethene, ethyne, ethanal, ethane

Solution:
[ P ] â Br 2 C 2 H 4 Br 2 â NaNH 2 Q â 20 % H 2 SO 4 R â ZnHg/HCl S
As 'P' undergoes addition reaction with 1 mole of Br2 it means there is only 1 double bond
So 'P' is C2H4CH2=CH2 and the reaction is as follows
Here P, Q, R and S are ethene, ethyne, ethanal and ethane respectively

Topic: Chemistry, Hydrocarbons, JEE Main
$[P] \stackrel{\text { Br }_{2}}{\longrightarrow} C_{2} \mathrm{H}_{4}$ Br $_{2} \frac{\text { NaNH }_{2}}{\mathrm{NH}_{3}}$ ? $[Q]$
$[Q] \frac{20 \% \mathrm{H}_{2} \mathrm{SO}_{4}}{\mathrm{Hg}^{2+} \cdot \Delta}[R] \stackrel{\mathrm{Zn}-\mathrm{Hg} / \mathrm{HC}}{\longrightarrow}[S]$
The species $P, Q, R$ and $S$ respectively are

Options:
A ethene, ethyne, ethanal, ethane
B ethane, ethyne, ethanal, ethene
C ethene, ethyne, ethanal, ethanol
D ethyne, ethane, ethene, ethanal

The correct answer is:
ethene, ethyne, ethanal, ethane

Solution:
Given reactions are
$[P] \stackrel{\mathrm{Br}_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2} \frac{\mathrm{NaNH}_{2}}{\mathrm{NH}_{3}},[Q]$
$[Q] \frac{20 \% / \mathrm{H}_{2} \mathrm{SO}_{4}}{\mathrm{Hg}^{2+}, \Delta},[\mathrm{R}] \stackrel{\mathrm{Zn} \cdot \mathrm{Hg} / \mathrm{HG}}{\longrightarrow}[S]$
$[P]$ is an alkene with two carbon atoms i.e. ethene. When it reacts with Br $y$, addition reaction occurs and $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Br}_{2}$ is formed.

$\mathrm{C}_{\mathrm{H}}, \mathrm{Br}_{2}$(1, 2-dibromoethane) on reaction with $\operatorname{NaNH}_{2} / \mathrm{NH},$ gives ethyne $(\mathrm{C}, \mathrm{H}, \mathrm{y})$
Ethyne IQlin presence of $20 \% \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{Hg}^{2+} \mathrm{at} 333 \mathrm{K}$
gives ethanal.

Ethanal $[R]$ undergoes reduction in presence of $$
\text { Zn-Hg /HCl to give ethane }[S]
$$
$\mathrm{CH}_{3}\mathrm{CHO} \xrightarrow[\mathrm{HCl}]{\mathrm{Zn-Hg}} \mathrm{CH}_{3}-\mathrm{CH}_{3}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
PdFClBrI2- has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilization energy CFSE of FeCN6n-6 , respectively, are:
[Note: Ignore the pairing energy]

Options:
A 2.84 BM and -16∆0
B 5.92 BM and 0
C 1.73 BM and -2.0∆0
D 0 BM and -2.4∆0

The correct answer is:
1.73 BM and -2.0∆0

Solution:
Number of Geometrical Isomers in square planar PdFClBr2- are =3
Hence, n=3

FeCN63-
Fe3+=3d5, According to CFT, configuration is t2g221eg00; it means number of unpaired electron (n)=1

So spin-only magnetic moment
μ=nn+2=1.73 BM

CFSE=−0.4 Δ0×nt2g+0.6 Δ0×neg =-0.4∆0×5=-2.0∆0

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (09 Jan Shift 1)
P. Epithelium A is commonly found in ducts of glands and tubular parts of the nephron.
Q. Epithelium B is found in the inner surface of hollow organs like bronchioles and fallopian tubes.
R. Epithelium C provides protection against chemical and mechanical stress.

Identify the above-mentioned epithelium.

Options:
A ABCSimpleÂ cuboidalSimpleÂ columnarCompound
B ABCSimpleÂ columnarSimpleÂ cuboidalSimpleÂ squamous
C ABCSimpleÂ columnarCiliatedPseudostratified
D ABCSimple cuboidalCiliatedCompound

The correct answer is:
ABCSimple cuboidalCiliatedCompound

Solution:
The cuboidal epithelium is composed of a single layer of cube-like cells. This is commonly found in ducts of glands and tubular parts of nephrons in kidneys and its main functions are secretion and absorption. The epithelium of proximal convoluted tubule (PCT) of the nephron in the kidney has microvilli. If the columnar or cuboidal cells bear cilia on their free surface they are called the ciliated epithelium. Their function is to move particles or mucus in a specific direction over the epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and fallopian tubes. Compound or stratified epithelium is made of more than one layer (multi-layered) of cells and their main function is to provide protection against chemical and mechanical stresses. They cover the dry surface of the skin, the moist surface of the buccal cavity, pharynx, the inner lining of ducts of salivary glands, and of pancreatic ducts.

Topic: Biology, Structural Organisation in Animals, NEET
Phenol→H2SO4Na2Cr2O7X The ratio of the number of σ to π bonds in the product X is

Options:

The correct answer is:
3

Solution:
Benzoquinone (X) contains 12σ and 4π bonds. Thus, the ratio of the number of σ to π bonds is 3

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$p H$ of a $10^{-3} \mathrm{M}$ solution of hydrochloric acid will be

Options:
A 1.3
B 2.0
C 3.0
D 4.5

The correct answer is:
3.0

Solution:
$\left[H^{+}\right]=10^{-3} M, p H=-\log \left[10^{-3}\right], p H=3$

Topic: Chemistry, Ionic Equilibrium, JEE Main
pH of a 1000 cc solution is 2. It will not change if

Options:
A 100Â cc of water is added to it
B 100Â cc of 0.1Â MÂ HCl is added to it
C 100 ccâ(N/100) HCl is added to it
D 1Â cc of 0.1Â MÂ HCl is added to it.

The correct answer is:
100 ccâ(N/100) HCl is added to it

Solution:
If two solutions of same pH are mixed, no change in pH is observed. pH of water will be 7, that of 100âcc of 0.1âMâHCl and 1âcc of 0.1âMâHCl will be 1 (as H + concentration is 0.1 )
0.01âNâHCl has also pH=2.
So adding this to a 1000 cc solution of pHâ2 will not make any difference in pH

Topic: Chemistry, Ionic Equilibrium, JEE Main
pH value of 0.001MNaOH solution is

Options:

The correct answer is:
11

Solution:
Concentration of NaOH given is 0.001M

OH-=0.001=1×10-3

pOH=-logOH-

=-log1×10-3=

=3 pH=14-pOH

=14-3=11

Hence, the pH of the given solution of NaOH is 11.

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (27 Jun Shift 2)
P is a point inside the ÎABC. Three straight lines parallel to AB, BC and AC are drawn which divides the triangle into six regions. If the area of three triangles formed in these regions are 9,16 and 25 square units, then the area (in sq. units) of ÎABC is

Options:
A 81
B 256
C 144
D 64

The correct answer is:
144

Solution:
âµÎABC, ÎTUP, ÎPDE, ÎSPR all are similar
â´ ratio of areas =ratio of sides2
Let, PT=Î±=AS
â´ SRPT=45âSR=4Î±5
PEPT=35âPE=3Î±5
âAC=AS+SR+RC
=Î±+4Î±5+3Î±5
=12Î±5
â´arÎABCarÎPDE=AC2PE2=122Â·Î±252Â·32Î±252
Hence, arÎABC=16arÎPDE=144 sq. units

Topic: Mathematics, Straight Lines, JEE Main
\( P \) is a point on the parabola whose ordinate equals its abscissa. A normal is drawn to the parabola at \( P \) to meet it again at \( Q . \) If \( S \) is the focus of the parabola, then the product of the slopes of \( S P \) and \( S Q \) is

Options:

The correct answer is:
-1.00

Solution:
Let, Pat2,2at be a point on the parabola y2=4ax
Since, the ordinate equals its abscissa.

⇒at2=2at⇒t=2

P≡4a,4a

Equation of the normal at P4a,4a is
y+2x=2a2+a23

⇒y+2x=12a …i

y2=4ax

⇒y2=2a12a-y … [From i]

⇒y2+2ay-24a2=0

⇒y-4ay+6a=0

⇒y=4a or y=-6a

⇒S≡a,0, P≡4a,4a, Q≡9a,-6a
Slope of SP=43 and slope of SQ=-68
⇒Required product=43×-68=-1

Topic: Mathematics, Parabola, JEE Main
P is a point on the parabola whose ordinate equals its abscissa. A normal is drawn to the parabola at P to meet it again at Q. If S is the focus of the parabola, then the product of the slopes of SP and SQ is

Options:

The correct answer is:
-1

Solution:
Let, Pat2,2at be a point on the parabola y2=4ax
Since, the ordinate equals its abscissa.
âat2=2atât=2
Pâ¡4a,4a
Equation of the normal at P4a,4a is
y+2x=2a2+a23
ây+2x=12aâ¦i
y2=4ax
ây2=2a12a-y â¦ [From i]
ây2+2ay-24a2=0
ây-4ay+6a=0
ây=4a or y=-6a
âSâ¡a,0,Pâ¡4a,4a,Qâ¡9a,-6a
Slope of SP=43 and slope of SQ=-68
â Required product =43Ã-68=-1

Topic: Mathematics, Parabola, JEE Main
\( P \) is point at a distance \( r \) from the centre of a solid sphere of radius \( R \). The variation of gravitational potential at \( P(i . e, V) \) and distance \( r \) from the centre of sphere is represented by the curve.

Options:
A
B
C
D

The correct answer is:

Solution:
Let M be the mass of the sphere of radius R.

Gravitational potential at any point P due to the solid sphere can be written as,

1 At all points lies inside r<R the sphere is,

VP=-GM2R3(3R2-r2)

(2) On the sphere r=R is, VP=-GMR

3 At any point outside the sphere r>R is,

VP=-GMr

Topic: Physics, Gravitation, JEE Main
P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr, at a distance r from the nucleus. The volume of this shell is 4πr2 dr. The qualitative sketch of the dependence of P on r is

Options:
A
B
C
D

The correct answer is:

Solution:
This is correct graph for 1s.

Topic: Chemistry, Structure of Atom, JEE Main
P is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
$p K_a$ of a weak acid is defined as

Options:
A $\log_10 K_a$
B $\frac{1}{\log _{10} K_a}$
C $\log_{10} \frac{1}{K_a}$
D $-\log_{10} \frac{1}{K_a}$

The correct answer is:
$\log_{10} \frac{1}{K_a}$

Solution:
$p K_a=\log _{10} \frac{1}{K_a}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
P on treatment with Br2/FeBr3 in CCl4 produced a single isomer C8H7O2Br wile heating P with sodalime gives toluene. The compound P is:

Options:
A
B
C
D

The correct answer is:

Solution:
Isomer C8H7O2Br on heating with sodalime gives toluene,hence it is an acid.Moreover since it has Degree of unsaturation 5 therefore it is an aromatic carboxylic acid.Since Carboxylic acid on heating gives toluene ,therefore it is having -CH3 group attached to the benzene ring.

Now it forms single isomer on halogenation, therefore -CH3 group must be attached at para position with respect to -COOH group.

Hence correct option is 1.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2020 (04 Sep Shift 1)
$(p+q)^2=p^2+2 p q+q^2=1$, represents an equation used in:

Options:
A
population genetics

B
Mendelian genetics

C
biometrics

D
molecular genetics

The correct answer is:
population genetics

Solution:
$(p+q)^2=p^2+2 p q+q^2=1$, represents an equation used in population genetics.
It is a mathematical representation of 'Hardy-Weinberg principle'. This principle states that allele frequencies in a population are stable and is constant from generation to generation, i.e the gene pool remains a constant.

Topic: Biology, Evolution
$P, Q$ and $R$ are three coplanar forces acting at a point and are in equilibrium. Given $P=1.931 .8$ $\mathrm{kg} \omega t, \sin \theta_1=0.9659$, the value of $R$ is (in $\mathrm{kg} w t$ )

Options:
A 0.9659
B 2
C 1
D $\frac{1}{2}$

The correct answer is:
1

Solution:
$\frac{P}{\sin \theta_1}=\frac{Q}{\sin \theta_2}=\frac{R}{\sin 150^{\circ}}$
$\Rightarrow \frac{1.93}{\sin \theta_1}=\frac{R}{\sin 150^{\circ}}$
$\Rightarrow R=\frac{1.93 \times \sin 150^{\circ}}{\sin \theta_1}=\frac{1.93 \times 0.5}{0.9659}=1$

Topic: Physics, Mathematics in Physics, JEE Main
$\sim(p \vee(\sim q))$ is equal to

Options:
A $\sim p \vee q$
B $(\sim p) \wedge q$
C $\sim p \vee \sim p$
D $\sim p \wedge \sim q$

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

Solution:
$\sim(p \vee(\sim q)) \equiv \sim p \wedge \sim(\sim q) \equiv(\sim p) \wedge q$

Topic: Mathematics, Mathematical Reasoning, JEE Main
∼p∨q∨∼p∧q is equivalent to

Options:
A p
B ∼p
C q
D ∼q

The correct answer is:
∼p

Solution:
~p∨q∨~p∧q≡~p∧~q∨~p∧q (By De morgan's Law)
≡∼p∧∼q∨q (By complement laws)
≡~p∧t (By Identity Laws)

≡~p

Topic: Mathematics, Mathematical Reasoning, JEE Main
$\sim(p \vee q) \vee(\sim p \wedge q)$ is logically equivalent to

Options:
A $\sim p$
B $p$
C $q$
D $\sim q$

The correct answer is:
$\sim p$

Solution:
$\sim(p \vee q) \vee(\sim p \vee q))$
$\equiv(\sim p \wedge \sim q) \vee(\sim p \wedge q)$
$\equiv \sim p \wedge(\sim q \vee q) \equiv \sim p$

Topic: Mathematics, Mathematical Reasoning, JEE Main
$P, Q, R$ and $S$ are four points with the position vectors $3 \mathbf{i}-4 \mathbf{j}+5 \mathbf{k},-4 \mathbf{i}+5 \mathbf{j}+\mathbf{k}$ and $-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$, respectively. Then, the line $P Q$ meets the line $R S$ at the point

Options:
A $3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$
B $-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$
C $-\mathbf{i}+4 \mathbf{j}+\mathbf{k}$
D $\mathbf{i}+\mathbf{j}+\mathbf{k}$

The correct answer is:
$-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}$

Solution:
Let the coordinates of four points $P, Q, R$ and $S$ be $(3,-4,5), \quad(0,0,4), \quad(-4,5,1) \quad$ and $\quad(-3,4,3)$ respectively.
Now, equation of line $P Q$ is
$$
\begin{aligned}
& \frac{x-3}{0-3}=\frac{y+4}{0+4}=\frac{z-5}{4-5} \\
\Rightarrow \quad & \frac{x-3}{-3}=\frac{y+4}{4}=\frac{z-5}{-1}=r_1
\end{aligned}
$$
Equation of line $R S$ is
$$
\frac{x+4}{-3+4}=\frac{y-5}{4-5}=\frac{z-1}{3-1}
$$
$$
\Rightarrow \quad \frac{x+4}{1}=\frac{y-5}{-1}=\frac{z-1}{2}=r_2
$$
Let $\left(-3 r_1+3,4 r_1-4,-r_1+5\right)$ and $\left(r_2-4,-r_2+5\right.$, $\left.2 r_2+1\right)$ be the points on line (i) and (ii), respectively. Since, both lines intersect at a common point, then
$$
\begin{aligned}
& & -3 r_1+3 & =r_2-4 \\
\Rightarrow & & 3 r_1+r_2 & =7 \\
\text { and } & & -r_2+5 & =4 r_1-4 \\
\Rightarrow & & 4 r_1+r_2 & =9
\end{aligned}
$$
On subtracting Eq. (iv) from Eq. (iii), we get
$$
r_1=2
$$
On putting the value of $r_1$ in Eq. (iii), we get
$$
3(2)+r_2=7 \Rightarrow r_2=1
$$
So, required point of intersection is $(-3,4,3)$ i.e.,
$$
-3 \mathbf{i}+4 \mathbf{j}+3 \mathbf{k}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
PQR is a triangular park with PQ=PR=200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P,Q and R are respectively, 45°, 30° and 30°, then the height of the tower (in m) is:

Options:
A 502
B 100
C 50
D 1003

The correct answer is:
100

Solution:
Let the height of the tower MN be h

In ∆QMN

tan30°=MNQM

∴QM=3 h=MR …(1){∵M is the point QM=MR}

Now in ΔMNP

∵∠MPN=∠NMP=45°

∴MN=PM(∵ In a triangle sides opposite to equal angles are

equal...2 )

∴MN=PM=h

In ∆PMQ

By using Equation 1

PM=(200)2-(3 h)2 {By using Pythagoras theorem}

∴ From 2

h=(200)2-(3h)2

⇒h=(40000)-3 h2

⇒h2=40000-3 h2 {By squaring both the sides}

⇒4 h2=40000

⇒h2=10000 m

⇒h=100 m

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2018 (08 Apr)
PQR is a triangular park with PQ=PR=200 m. A TV tower stands at the mid-point of QR. If the angles of elevation of the top of the tower from P,Q and R are 45o, 30o and 30o respectively, then the height of the tower (in meters) is

Options:

The correct answer is:
100

Solution:
In the figure PQR is triangular park with PQ=PR=200.

TV tower of height h stands at midpoint M of QR.

In triangle TMR, MR=hcot30o

In triangle TMP, PM=TM=h

In triangle PMR, PR2=PM2+MR2

∴2002=h2+3h2

∴h=100 m

Topic: Mathematics, Heights and Distances, JEE Main
p∧r⇔p∧~q is equivalent to ~p when r is

Options:
A p
B ~p
C q
D ~q

The correct answer is:
q

Solution:
Given p∧r⇔p∧~q≡~p

Assuming all option one by one,

Taking r=p

p	q	~p	~q	p∧p	p∧~q	p∧r⇔p∧~q

T	T	F	F	T	F	F

T	F	F	T	T	T	T

F	T	T	F	F	F	T

F	F	T	T	F	F	T

Now taking r=~p

p	q	~p	~q	p∧~p	p∧~q	p∧r⇔p∧~q

T	T	F	F	F	F	T

T	F	F	T	F	T	F

F	T	T	F	F	F	T

F	F	T	T	F	F	T

Taking r=q

p	q	~p	~q	p∧q	p∧~q	p∧r⇔p∧~q

T	T	F	F	T	F	F

T	F	F	T	F	T	F

F	T	T	F	F	F	T

F	F	T	T	F	F	T

So clear, p∧r⇔p∧~q≡~p

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2022 (27 Jul Shift 1)
P-V diagram of a diatomic gas is a straight line parallel to P-axis. The molar heat capacity of the gas in the process will be

Options:
A 4R
B 2.5R
C 3R
D 4R3

The correct answer is:
2.5R

Solution:
The process described is isochoric
∴ molar heat capacity CV=52R=2.5R

Topic: Physics, Thermodynamics, JEE Main
P - V diagram of an ideal gas is shown. The gas undergoes from initial state A to final state B such that initial and final volumes are same. Select the correct alternative for given process AB .

Options:
A Work done by gas is positive
B Work done by gas is negative
C Temperature of gas increases continuously
D Process is isochoric

The correct answer is:
Work done by gas is negative

Solution:
Volume does not remain constant throughout the process AB. As T=PVnR, temperature decreases initially as both P and V decreases. By area under curve, net work done is negative.

Topic: Physics, Thermodynamics, JEE Main
P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should correspond respectively to

Options:
A Heand O 2
B O 2 and He
C Heand Ar
D O 2 and N 2

The correct answer is:
O 2 and He

Solution:
In adiabatic process, slope of P-V graph is given as, dPdv=-γPV.
For a given P-V graph, slope∝-γ

From the given graph,
slope 2 > slope 1
∴ γ 2 > γ 1
Therefore, plot 1 should correspond to O 2 γ = 1.4 and plot 2 should correspond to He γ = 1.67 .

Topic: Physics, Thermodynamics, JEE Main
Palisade parenchyma is absent in leaves of -

Options:
A Gram
B Sorghum
C Mustard
D Soybean

The correct answer is:
Sorghum

Solution:
No Solution Available

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2009 (Mains)
Pancreatic juice/enzyme digests

Options:
A vitamins and fats only
B vitamins, starch and fat only
C triglycerides, proteins and nucleic acids only
D nucleic acids, proteins, fat and starch.

The correct answer is:
nucleic acids, proteins, fat and starch.

Solution:
The exocrine part of the pancreas consists of rounded lobules (acini) that secrete an alkaline pancreatic juice with $\mathrm{pH}$ 8.4. The pancreatic juice is carried by the main pancreatic duct into the duodenum through the hepatopancreatic ampulla. The accessory pancreatic duct directly pours the pancreatic juice into the duodenum. The pancreatic juice contains sodium bicarbonate, three proenzymes; trypsinogen, chymotrypsinogen and procarboxypeptidase and some enzymes such as elastase, pancreatic $\alpha$-amylase, DNase, RNase and pancreatic lipase. The pancreatic juice helps in the digestion of starch, proteins, fats and nucleic acids.

Topic: Biology, Digestion and Absorption, AIIMS, AIIMS 2019 (25 May)
Paragraph:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation
$\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question:
Equations of the sides $Q R, R P$ are

Options:
A
$y=\frac{2}{\sqrt{3}} x+1, y=-\frac{2}{\sqrt{3}} x-1$

B
$y=\frac{1}{\sqrt{3}} x, y=0$

C
$y=\frac{\sqrt{3}}{2} x+1, y=-\frac{\sqrt{3}}{2} x-1$

D
$y=\sqrt{3} x, y=0$

The correct answer is:
$y=\sqrt{3} x, y=0$

Solution:
Clearly, point $E$ and $F$ satisfy the equations given in option (d).

Topic: Mathematics, Straight Lines, JEE Main
Paragraph:
A circle $C$ of radius 1 is inscribed in an equilateral $\triangle P Q R$. The points of contact of $C$ with the sides $P Q, Q R, R P$ are $D, E, F$ respectively. The line $P Q$ is given by the equation
$\sqrt{3} x+y-6=0$ and the point $D$ is $\left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)$. Further, it is given that the origin and the centre of $C$ are on the same side of the line $P Q$.Question:
The equation of circle $C$ is

Options:
A
$(x-2 \sqrt{3})^2+(y-1)^2=1$

B
$(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1$

C
$(x-\sqrt{3})^2+(y+1)^2=1$

D
$(x-\sqrt{3})^2+(y-1)^2=1$

The correct answer is:
$(x-\sqrt{3})^2+(y-1)^2=1$

Solution:
Let centre of circle $C$ be $(h, k)$.
Then, $\left|\frac{\sqrt{3} h+k-6}{\sqrt{3+1}}\right|=1$
$$
\Rightarrow \quad \sqrt{3} h+k-6=2,-2
$$
$$
\Rightarrow \quad \sqrt{3} h+k=4
$$
(Rejecting 2 because origin and centre of $C$ are on the same side of $P Q$ ) The point $(\sqrt{3}, 1)$ satisfies Eq. (i).
$\therefore$ Equation of circle $C$ is $(x-\sqrt{3})^2+(y-1)^2=1$.

Topic: Mathematics, Circle, JEE Main
Paragraph:
A person Jack stands at left end of a railway car sitting on a railway track. There is no frictional force between the track and the railway car.
Person + Railway track Mass $=\mathrm{M}$
Length of the railway car $\quad \mathrm{L}$
Jack throws a ball of mass $m$ with velocity $u$ towards the right end of the railway car where it collides elastically with the wall and travel back to the other end and strike the wall inelastically and coming to rest. $\mathrm{i}$ is the unit vector across right end.Question:

Options:
A $\mathrm{P} \rightarrow \mathrm{C}, \mathrm{Q} \rightarrow \mathrm{A}, \mathrm{R} \rightarrow \mathrm{B}, \mathrm{S} \rightarrow \mathrm{A}$
B $\mathrm{P} \rightarrow \mathrm{A}, \mathrm{Q}-\gt\mathrm{C}, \mathrm{R}-\gt\mathrm{B}, \mathrm{S}-\gt\mathrm{A}$
C $\mathrm{P} \rightarrow \mathrm{C}, \mathrm{Q}-\gt\mathrm{A}, \mathrm{R} \rightarrow \mathrm{D}, \mathrm{S}-\gt\mathrm{B}$
D $\mathrm{P}-\gt\mathrm{D}, \mathrm{Q}-\gt\mathrm{A}, \mathrm{R}-\gt\mathrm{B}, \mathrm{S}-\gt\mathrm{C}$

The correct answer is:
$\mathrm{P} \rightarrow \mathrm{C}, \mathrm{Q} \rightarrow \mathrm{A}, \mathrm{R} \rightarrow \mathrm{B}, \mathrm{S} \rightarrow \mathrm{A}$

Solution:
Taking Railway car + jack + ball as the system, all the forces are internal, So linear momentum will remain conserved throughout the motion and on every collision
$M \mathbf{v}_1+m \mathbf{v}_2=0--(1)$
Now Velocity of the railway car before first collision
$M \mathbf{v}+m u \mathbf{i}=0$
Or
$\mathbf{v}=-\frac{-m}{M} u \mathbf{i}$
Now Velocity of the railway car after first collision
The effect of first collision will be simply to reverse the both velocity vector
$\mathbf{v}=\frac{m u \mathbf{i}}{M}$
Now Velocity of the railway car after second collision.
Since ball comes to rest, so as per equation (1)
Velocity of railway car $=0$
Since no external force acting, Center of mass will remain at rest.
$\begin{aligned} & \mathrm{P} \rightarrow \mathrm{C} \\ & \mathrm{Q} \rightarrow \mathrm{A} \\ & \mathrm{R} \rightarrow \mathrm{B} \\ & \mathrm{S} \rightarrow \mathrm{A}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main