datavorous/entrance-exam-dataset huggingface dataset:
The following questions consist of two statements, one labelled as the 'Assertion $(\mathrm{A})$ ' and the other as 'Reason $(\mathrm{R})$ ' You are to examine these two statements carefully and select the answer. Assertion (A) : For a binomial distribution $\mathrm{B}(\mathrm{n}, \mathrm{p})$, Mean $>$ Variance Reason $(\mathrm{R}):$ Probability is less than or equal to 1

Options:
A Both $\mathrm{A}$ and $\mathrm{R}$ are individually tue and $\mathrm{R}$ is the correct explanation of $A$
B Both $\mathrm{A}$ and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of $A$
C A is true but $\mathrm{R}$ is false
D A is false but $\mathrm{R}$ is true

The correct answer is:
Both $\mathrm{A}$ and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of $A$

Solution:
Mean $=\mathrm{np}$ and Variance $=\mathrm{npq} < \mathrm{np}(\because \mathrm{q} < 1)$

Topic: Mathematics, Probability, JEE Main
The following reaction constitutes :
\( \mathrm{RNH}_{2}+\mathrm{S}=\mathrm{C}=\mathrm{S} \stackrel{\mathrm{HgCl}_{2}}{\stackrel{\mathrm{H}_{2} \mathrm{~S}}{\longrightarrow}} \underset{\text { Alkyl isothiocyanate }}{\mathrm{R}}-\mathrm{N}=\mathrm{C}+\mathrm{H}_{2} \mathrm{~S} \)

Options:
A Mustard oil reaction
B Test for \( 3^{\circ} \) amine
C Test for \( 2^{\circ} \) amine
D Test for \( \mathrm{CS}_{2} \)

The correct answer is:
Mustard oil reaction

Solution:
Secondary amine also reacts with carbon disulphide to form dithiocarbamic acids, but the later fails to react with mercuric chloride.
Tertiary amines do not react with carbon disulphide

Topic: Chemistry, Amines, JEE Main
The following reaction descomposes N 2 O 5 â 2NO 2 + 1 2 O 2 . At a 25 degree centigrade the rate constant of the reaction is 5Ã10-2sec-1â¡ . The initial pressure of N2O5 is 0.2âatm. If total pressure of gaseous mixture becomes 0.35âatm, then calculate the time of decomposition of N2O5.

Options:
A 238.6Â seconds
B 138.6 seconds
C 69.3Â seconds
D 89.3Â seconds

The correct answer is:
138.6 seconds

Solution:
(i) The decomposition reaction is represented by
Total pressure =p0-x+2x+x2=p0+32x=0.2 atm.
or 3x2=0.35-0.2=0.15
x=0.1 atm
Pressure of N2O5 after t seconds,
P=P0-x=0.2-0.1=0.1 atm
Now applying first order rate law
t= 2.303 k log 10 0.2 0.1
=2.3035Ã10-3log10â¡0.20.1=138.6 seconds

Topic: Chemistry, Chemical Kinetics, JEE Main
The following reaction
is known as:

Options:
A
Perkin reaction

B
Gatterman-Koch Formylation

C
Kolbe's reaction

D
Gattermann reaction

The correct answer is:
Gattermann reaction

Solution:
No Solution Available

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2014 (11 Apr Online)
The following reaction
is known as:

Options:
A
Perkin reaction

B
Gatterman-Koch Formylation

C
Kolbe's reaction

D
Gattermann reaction

The correct answer is:
Gattermann reaction

Solution:
No Solution Available

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The following reaction is known as

Options:
A Perkin reaction
B Gattermann reaction
C Kolbe reaction
D Gattermann-aldehyde reaction

The correct answer is:
Gattermann-aldehyde reaction

Solution:
In the presence of anhydrous ZnCl2, phenol form salicyladehyde. It is Gattermann-aldehyde reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The following reaction
Is known by the name:

Options:
A Friedel-Craft's reaction
B Perkin's reaction
C Acetylation reaction
D Schotten-Baumen reaction

The correct answer is:
Schotten-Baumen reaction

Solution:
The reaction is known by the name Schotten-Baumen reaction. It is a benzoylation reaction. When aniline is treated with benzoyl chloride in the presence of NaOH, benzanilide (N-benzoylaniline) is obtained.This reaction is given by primary and secondary amines.

Benzoylation of aniline is an example of Schotten-Baumen reaction.

Topic: Chemistry, Amines, JEE Main
The following reaction is performed at 298 K.
2NOg+O2g⇌2NO2g
The standard free energy of the formation of NOg is 86.6 kJ mol-1 at 298 K. What is the standard free energy of the formation of NO2g at 298 K? (KP=1.6×1012)

Options:
A 0.5[2×86,600-R298ln⁡(1.6×1012)]
B R298lnâ¡1.6Ã1012-86,600
C 86,600+R298lnâ¡(1.6Ã1012)
D Â 86,600-lnâ¡(1.6Ã1012)R(298)

The correct answer is:
0.5[2×86,600-R298ln⁡(1.6×1012)]

Solution:
Δ G reac ∘ = - 2.303 RT log K P
= - RT ln K P
=-R298ln 1.6×1012
ΔGreac∘=2ΔGfNO2∘-2ΔGfNO∘
⇒-R298ln 1.6×1012=2ΔGfNO2∘-2×86.6×103
⇒2ΔGfNO2∘=-R298ln1.6×1012+2×86,600
Δ G f NO 2 ∘ = 86,600 - R 2 9 8 2 ln 1.6 × 1 0 1 2
= 0.5 2 × 86,600 - R 2 9 8 ln 1.6 × 1 0 1 2

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2015 (04 Apr)
The following reaction occurs during rusting of iron

2H + â+â2eâ+â 1 2 O 2 â H 2 O;â E o =â+1.23âV

Fe 2+ â+â2eâFeâ(s);â E o =â0.44âV

Calculate magnitude of ÎG o (kJ) for the net process

Feâ(s)â+â 2H + â+â 1 2 â O 2 â Fe 2+ â+â H 2 O

Options:

The correct answer is:
322

Solution:
Fe(s)→ Fe 2+  +  2e − ;  ΔG 1 o
2H + â+â 2e â â+â 1 2 â O 2 â H 2 O(l);â ÎG 2 o
Fe(s)â+â 2H + â+â 1 2 â O 2 â Fe 2+ â+â H 2 O;â ÎG 3 o
Applying, ÎG 1 o â+â ÎG 2 o â=â ÎG 3 o
ÎG 3 o â=â â2FÃ0.44 â+â â2FâÃâ1.23
ÎG 3 o â=â â2âÃâ96500âÃâ0.44 â+â â2âÃâ96500âÃâ1.23
ÎG 3 o â=ââ322310âJ
â´â ÎG 3 o =â322âkJ

Topic: Chemistry, Electrochemistry, JEE Main
The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal:

Fe2O3s+3COg â2Fe l+3CO2g

Using the Le Chatelier's principle, predict which one of the following will not disturb the equilibrium?

Options:
A Addition of CO2
B Removal of CO2
C Addition of Fe2O3
D Removal of CO

The correct answer is:
Addition of Fe2O3

Solution:
It is heterogeneous equilibrium Fe is liquid Fe2O3 is solid whose active mass is taken to be unity so, it will not be considered in kp, kc,& kx expresssion.
So, it will not disturb the equilibrium

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2017 (09 Apr Online)
The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal:
\[
\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Fe}(\mathrm{l})+3 \mathrm{CO}_{2}(\mathrm{~g})
\]
Using the Le Chatelier's principle, predict which one of the following will not disturb the equilibrium?

Options:
A Addition of \( \mathrm{CO}_{2} \)
B Removal of \( \mathrm{CO}_{2} \)
C Addition of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \)
D Removal of CO

The correct answer is:
Addition of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \)

Solution:
It is heterogeneous equilibrium Fe is liquid Fe2O3 is solid whose active mass is taken to be unity so, it will not be considered in kp, kc,& kx expresssion.
So, it will not disturb the equilibrium

Topic: Chemistry, Chemical Equilibrium, JEE Main
The following reaction occurs in the Blast furnance where iron ore is reduced to iron metal
\(\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{Fe}_{(\mathrm{l})}+3 \mathrm{CO}_{2(\mathrm{~g})}\)
Using the Le-chatelier's principle, predict which one of the following will not disturb the equilibrium.

Options:
A Addition of \(\mathrm{CO}_2\)
B Removal of \(\mathrm{CO}_2\)
C Addition of \(\mathrm{Fe}_2 \mathrm{O}_3\)
D Removal of CO

The correct answer is:
Addition of \(\mathrm{Fe}_2 \mathrm{O}_3\)

Solution:
When solid added no effect on equilibrium.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The following reaction produces:

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction of carbonyl compounds with phosphorous ylide Ph3P=CH2 results in the formation of an alkene. This is known as the Wittig reaction.

The phosphorous ylide is known as the Wittig reagent (organophosphorous reagent). The ylide has a carbon which acts as a nucleophile and helps in forming an alkene.

Mechanism of the reaction is as follows:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The following reactions show the H2O2 behaviour in A and B reactions as
(1) PbS(s)+4H2O2aq→PbSO4s+4H2Ol
(2) HOCl+H2O2→H3O++Cl-+O2

Options:
A oxidising in acidic medium and reducing in basic medium
B reducing in acidic medium and oxidising in basic medium
C oxidising in acidic medium and reducing in acidic medium
D reducing in acidic medium and oxidising in acidic medium

The correct answer is:
oxidising in acidic medium and reducing in acidic medium

Solution:
Hydrogen peroxide oxidies lead sulphide to lead sulphate and itself gets reduced. Thus, H2O2 acts as an oxidising agent in acidic medium.
Hydrogen peroxide reduces hypochlorous acid to chloride ion in acidic medium and itself gets oxidised.
Thus, H2O2 acts as a reducing agent in acidic medium.

Topic: Chemistry, s Block Elements, JEE Main
The following reactions take place in the blast furnace in the preparation of impure iron. Identify the reaction pertaining to the formation of the slag.

Options:
A $\mathrm{Fe}_2 \mathrm{O}_3(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(l)+3 \mathrm{CO}_2(g)$
B $\mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_2(\mathrm{~g})$
C $\mathrm{CaO}(s)+\mathrm{SiO}_2(s) \longrightarrow \mathrm{CaSiO}_3(s)$
D $2 \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})$

The correct answer is:
$\mathrm{CaO}(s)+\mathrm{SiO}_2(s) \longrightarrow \mathrm{CaSiO}_3(s)$

Solution:
A slag is an easily fusible material which is formed when gangue still present in the roasted or the calcined ore combines with the flux. For example, in the metallurgy of iron, $\mathrm{CaO}$ (flux) combines with silica gangue to form easily fusible calcium silicate $\left(\mathrm{CaSiO}_3\right)$ slag.
$$
\mathrm{CaO}+\mathrm{SiO}_2 \longrightarrow \mathrm{CaSiO}_3 \text { (slag) }
$$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET, NEET 2011 (Mains)
The following results were obtained during kinetic studies of the reaction.
2A+B→ product

Experiment	A in mol L-1	B in mol L-1	Initial rate of reaction in mol L-1 min -1

I	0.10	0.20	6.93×10-3

II	0.10	0.25	6.93×10-3

III	0.20	0.30	1.386×10-2

The time (in minutes) required to consume half of A is

Options:
A 100
B 10
C 5
D 1

The correct answer is:
10

Solution:
Let's assume x and y are the order of the reaction with respect to A and B, respectively.
From equation 1 and 2,
0.10.1x0.20.25y=6.93×10-36.93×10-3⇒y=0
From equation 1 and 3 and put the value y=0,
0.10.2x0.20.30=6.93×10-31.386×10-2 ∴y=0
12x=12⇒x=1
The rate law will be R=K[A]1[B]0
Hence, the reaction is of the first order.
K=RA=6.93×10-30.1=6.93×10-2
Here t12 for the first-order reaction is given by, t12=0.693K=0.693(6.93×10-2)=10

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The following series $L-C-R$ circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like

Options:
A purely resistive circuit
B series R-L circuit
C series R-C circuit
D series L-C circuit with R = 0

The correct answer is:
series R-L circuit

Solution:
Impedance, $Z=\sqrt{\left(X_L \sim X_C\right)^2+R^2}$
or $\quad Z=\sqrt{\left(\omega L \sim \frac{1}{\omega C}\right)^2+R^2}$
Inductive reactance
$\begin{aligned}
X_L & =\omega L=70 \times 10^3 \times 100 \times 10^{-6} \\
& =7 \Omega
\end{aligned}$
Capacitance reactance
$\begin{aligned}
X_C & =\frac{1}{\omega C}=\frac{1}{70 \times 10^3 \times 1 \times 10^{-6}} \\
& =\frac{1}{7 \times 10^{-2}}=\frac{10^2}{7}=\frac{100}{7}
\end{aligned}$
As $\quad X_C>X_L$
Hence, circuit behave like as $R-C$ circuit.

Topic: Physics, Alternating Current, JEE Main
The following sets of quantum numbers represent four electrons in an atom.
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$
(iv) $n=3, l=1$
The sequence representing increasing order of energy, is

Options:
A
(iii) < (i) < (iv) < (ii)
B
(iv) < (ii) < (iii) < (i)
C
(i) < (iii) < (ii) < (iv)
D
(ii) < (iv) < (i) < (iii)

The correct answer is:
(iv) < (ii) < (iii) < (i)

Solution:
(i) $4 p$
(ii) $4 \mathrm{~s}$
(iii) $3 d$
(iv) $3 p$
According to Bohr Bury's $(n+\ell)$ rule, increasing order of energy will be (iv) < (ii) $ < $ (iii) $ < $ (i).
Note: If the two orbitals have same value of $(n+\ell)$ then the orbital with lower value of $n$ will be filled first.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2012 (26 May Online)
The following statements are given about hydrogen atom.
A. The wavelengths of the spectral lines of Lyman series are greater than the wavelength of the second spectral line of Balmer series.
B. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths.

Options:
A $A$ is false, $B$ is true
B $A$ is true, $B$ is false
C $A$ is false, $B$ is false
D $A$ is true, $B$ is true

The correct answer is:
$A$ is false, $B$ is true

Solution:
The wavelengths of a spectral lines of Lyman series are smaller than the wavelength of the second spectral line of Balmer series therefore statement $A$ is false.
Whereas, statement $B$ is true.

Topic: Physics, Atomic Physics, JEE Main
The following statements concern elements in the periodic table. Which of the following is true?

Options:
A For group 15 elements, the stability of +5 oxidation state increases down the group.
B Elements of group 16 have lower ionization enthalpy values compared to those of group 15 in the corresponding periods.
C The group 13 elements are all metals.
D All the elements in group 17 are gases.

The correct answer is:
Elements of group 16 have lower ionization enthalpy values compared to those of group 15 in the corresponding periods.

Solution:
Group 16 elements have less ionization potential than Group 15 elements due to half-filled configurations of group 15 elements. So need more amount of energy to remove an electron. In the halogen family iodine is solid and bromine is liquid. In the boron family boron is non metal.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2016 (10 Apr Online)
The following statements relate to the adsorption of gases on a solid surface. Identify the incorrect statement among them:

Options:
A Entropy of adsorption is negative
B Enthalpy of adsorption is negative
C On adsorption decrease in surface energy appears as heat
D On adsorption, the residual forces on the surface are increased

The correct answer is:
On adsorption, the residual forces on the surface are increased

Solution:
Adsorption is spontaneous process ∆G is -ve. During adsorption randomness of adsorbate molecules reduced and ∆Stotal is +ve.
∆G=∆H-T∆S
∆H=∆G+T∆S
∆H is highly -ve and residual forces on surface are decreased.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2015 (10 Apr Online)
The following system of equation $3 x-2 y+z=0, \quad \lambda x-14 y+15 z=0, \quad x+2 y-3 z=0$ has a solution other than $x=y=z=0$ for $\lambda$ equal to

Options:
A 1
B 2
C 3
D 5

The correct answer is:
5

Solution:
The system of equations has infinitely many (non-trivial) solution, if $\Delta=0$ i.e., if
$\left|\begin{array}{ccc}3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & -3\end{array}\right|=0$
$\Rightarrow 3(42-30)-\lambda(6-2)+1(-30+14)=0 \Rightarrow \lambda=5$

Topic: Mathematics, Determinants, JEE Main
The following system of equations 5x-7y+3z=3,5x+y+3z=7 and 5x+3y+2z=5 is

Options:
A Consistent with trivial solution
B Consistent with a unique non trivial solution
C Consistent with infinite solutions
D Inconsistent with no solution

The correct answer is:
Consistent with a unique non trivial solution

Solution:
∆=5-73513532=5-7+7-5+310
=-40≠0
Hence, system of equation has non trivial unique solution as 0,0,0 does not satisfy system of equation.

Topic: Mathematics, Determinants, JEE Main
The following system of linear equations
2x+3y+2z=9
3x+2y+2z=9
x-y+4z=8

Options:
A has infinitely many solutions
B has a unique solution
C has a solution α,β,γ satisfying α+β2+γ3=12
D does not have any solution

The correct answer is:
has a unique solution

Solution:
2x+3y+2z=9 ...1

3x+2y+2 z=9 ...2

x-y+4z=8 ...3

1-2⇒-x+y=0⇒x-y=0

from 3 4z=8⇒z=2

from 1 2x+3y=5

⇒x=y=1

∴ system has unique solution.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The following system of linear equations
7x+6y-2z=0
3x+4y+2z=0
x-2y-6z=0, has

Options:
A infinitely many solutions, x,y,z satisfying y=2z
B no solution
C infinitely many solutions, x,y,z satisfying x=2z
D only the trivial solution

The correct answer is:
infinitely many solutions, x,y,z satisfying x=2z

Solution:
76-23421-2-6

=7-20-6-20-2-10

=-140+120+20=0

So infinite non-trivial solution exist
Now equation (1)+3 equation (3)
⇒10x-20z=0
⇒x=2z

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The following two reactions are known
$$
\begin{aligned}
& \mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{~g}), \\
& \Delta \mathrm{H}=-26.8 \mathrm{~kJ} \\
& \mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \\
& \Delta \mathrm{H}=-16.5 \mathrm{~kJ}
\end{aligned}
$$
The value of $\Delta \mathrm{H}$ for the following reaction $\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$ is

Options:
A $+10.3 \mathrm{~kJ}$
B $-43.3 \mathrm{~kJ}$
C $-10.3 \mathrm{~kJ}$
D $+6.2 \mathrm{~kJ}$

The correct answer is:
$+6.2 \mathrm{~kJ}$

Solution:
Given,
(I) $\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{~g})$;
$\Delta \mathrm{H}=-26.8 \mathrm{~kJ}$
(II) $\mathrm{FeO}$ (s) $+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$;
$\Delta \mathrm{H}=-16.5 \mathrm{~kJ}$
On multiplying Eq (II) with 2, we get
(III) $2 \mathrm{FeO}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+2 \mathrm{CO}_2(\mathrm{~g})$;
$\Delta \mathrm{H}=-33 \mathrm{~kJ}$
On subtracting Eq (III) from I, we get
$$
\begin{array}{r}
\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) ; \\
\Delta \mathrm{H}=-26.8-(-33) \\
=+6.2 \mathrm{~kJ}
\end{array}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2010 (Mains)
The foot of perpendicular from the origin O to a plane P which meets the co-ordinate axes at the point A, B,C is (2,a,4),a∈N. If the volume of the tetrahedron OABC is 144 unit3, then which of the following points is NOT on P?

Options:
A (0,4,4)
B (3,0,4)
C (0,6,3)
D (2,2,4)

The correct answer is:
(3,0,4)

Solution:
Plotting the diagram of the given data we get,

As (2,a,4) is the foot of perpendicular from origin,

So, direction ratio of normal vector will be (2-0,a-0,4-0)≡(2,a,4)

So, equation of plane P will be,

2(x-2)+a(y-a)+4(z-4)=0

⇒2x+ay+4z=20+a2

⇒x20+a22+y20+a2a+z20+a24=1

Now Volume of tetrahedron OABC is given by,

V=16·OA·OB·OC

⇒1620+a2220+a2a20+a24=144

⇒20+a2123=4a

Let a=2k3

⇒20+4k6123=8k3

⇒20+4k612=2k

⇒k6+5=6k

⇒k=1⇒a=2

So, equation of plane 2x+2y+4z=24

⇒x+y+2z=12

Clearly all three points (0,4,4),(0,6,3),(2,2,4) satisfy the equation of plane but (3,0,4) does not satisfy plane

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The foot of perpendicular of the point 2,0,5 on the line x+12=y-15=z+1-1 is α,β,γ. Then. Which of the following is NOT correct?

Options:
A αβγ=415
B αβ=-8
C βγ=-5
D γα=58

The correct answer is:
βγ=-5

Solution:
Given line is

L:x+12=y-15=z+1-1=λ (let)

Let foot of perpendicular is P2λ-1,5λ+1,-λ-1

PA→=3-2λi^-5λ+1j^+6+λk^

Direction ratio of line is

b→=2i^+5j^-k^

Now, PA→⊥L, so

PA→·b→=0

⇒23-2λ-55λ+1-6+λ=0

⇒6-4λ-25λ-5-6-λ=0

⇒-30λ-5=0

⇒λ=-16

So,

P2λ-1,5λ+1,-λ-1≡Pα,β,γ

Now,

α=2-16-1=-43⇒ α=-43

β=5-16+1=16⇒β=16

γ=-λ-1=16-1⇒ γ=-56

So,

βγ=-15≠-5

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The foot of the normal from the point (4,3) to a circle is (2,1) and a diameter of the circle has the equation 2x-y-2=0, then the equation of the circle is

Options:
A x 2 + y 2 −4y+2=0
B x 2 + y 2 −4y+1=0
C x 2 + y 2 â2xâ1=0
D x 2 + y 2 −2x+1=0

The correct answer is:
x 2 + y 2 −2x−1=0

Solution:
Equation of the diameter of the circle is given as
2x-y-2=0 ...(i)

If P(4,3) and N(2,1) are the given points, then
slope of PN=3-14-2=1
Equation of normal through PN is

y-1=(x-2)
x-y-1=0 ...(ii)
solving (i) and (ii), we get, the centre as (1, 0)
Hence, the equation of the circle is
(x-1)2+y2=(2 - 1)2 + 1
x2+y2-2x-1=0

Topic: Mathematics, Circle, JEE Main
The foot of the perpendicular drawn from the origin, on the line, $3 x+y=\lambda(\lambda \neq 0)$ is $P$. If the line meets $x$-axis at $A$ and $y$-axis at $B$, then the ratio $B P$ $: P A$ is

Options:
A
$9: 1$

B
$1: 3$

C
$1: 9$

D
$3: 1$

The correct answer is:
$3: 1$

Solution:
Equation of the line, which is perpendicular to the line, $3 x+y=\lambda(\lambda \neq 0)$ and passing through origin, is given by
$$
\frac{x-0}{3}=\frac{y-0}{1}=r
$$
For foot of perpendicular
$$
r=\frac{-((3 \times 0)+(1 \times 0)-\lambda)}{3^2+1^2}=\frac{\lambda}{10}
$$
So, foot of perpendicular $P=\left(\frac{3 \lambda}{10}, \frac{\lambda}{10}\right)$
Given the line meets $\mathrm{X}$-axis at $A=\left(\frac{\lambda}{3}, 0\right)$ and meets $Y$-axis at $B=(0, \lambda)$
So,
$$
\begin{aligned}
&B P=\sqrt{\left(\frac{3 \lambda}{10}\right)^2+\left(\frac{\lambda}{10}-\lambda\right)^2} \Rightarrow B P=\sqrt{\frac{9 \lambda^2}{100}+\frac{81 \lambda^2}{100}} \\
&\Rightarrow B P=\sqrt{\frac{90 \lambda^2}{100}} \\
&\text { Now, } P A=\sqrt{\left(\frac{\lambda}{3}-\frac{3 \lambda}{10}\right)^2+\left(0-\frac{\lambda}{10}\right)^2} \\
&\Rightarrow P A \sqrt{\frac{\lambda^2}{900}+\frac{\lambda^2}{100}} \Rightarrow P A=\sqrt{\frac{10 \lambda^2}{900}}
\end{aligned}
$$
Therefore $B P: P A=3: 1$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The foot of the perpendicular drawn from the point 4, 2, 3 to the line joining the points 1, -2, 3 and 1, 1, 0 lies on the plane

Options:
A 2 x+y-z=1
B x-y-2 z=1
C x-2 y+z=1
D x+2 y-z=1

The correct answer is:
2 x+y-z=1

Solution:
Equation of a line through given points 1, -2, 3 and 1, 1, 0 will be

x-10=y+23=z-3-3=λ.

Direction ratio of line is =3j^-3k^.

A point on this line can be taken as 1, 3λ-2, -3λ+3.

Let this point is foot of perpendicular from point 4, 2, 3.

Direction ratio of line joining these points is =3i^+4-3λj^+3λk^.

Both lines are perpendicular to each other.

⇒Dot product of directions ratio will be zero.

⇒3j^-3k^·3i^+4-3λj^+3λk^=0

⇒0+34-3λ+-33λ=0

⇒18λ=12 ⇒λ=23

Replace value of λ for foot of perpendicular.

⇒Foot of perpendicular is 1, 0, 1.

This point satisfies the plane 2x+y-z=1.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is

Options:
A $(-2,3,4)$
B $(2,-1,3)$
C $(2,3,-1)$
D $(3,2,-1)$

The correct answer is:
$(2,3,-1)$

Solution:
Let $N$ be the foot of the perpendicular from the point $P(0,2,3)$ on the given line
$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=r \quad$ (say) ... (i)
Any point on the line (i) is
$(5 r-3,2 r+1,3 r-4)$
If this point is $N$, The direction ratio's of $N P$ are
$5 r-3,2 r+1-2,3 r-4-3$
i.e.
$5 r-3,2 r-1,3 r-7$
Since, $N P$ is perpendicular to the given line, then
$\begin{aligned} & \qquad 5(5 r-3)+2(2 r-1)+3(3 r-7)=0 \\ & \Rightarrow \quad 38 r-38=0 \Rightarrow r=1 \\ & \text { The point } N \text { is } \quad(5-3,2+1,3-4) \\ & \text { i.e., } \quad(2,3,-1) .\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main