datavorous/entrance-exam-dataset huggingface dataset:
The total number of molccules with zero dipole moment among CH4,BF3,H2O,HF,NH3,CO2 and SO2 is ______.

Options:

The correct answer is:
3

Solution:
Dipole moment:- The polarity of a covalent bond can be conveniently measured in terms of a physical quantity called dipole moment. The dipole moment is the vector sum of all the bond moments. The molecules with perfect geometries with zero lone pairs have zero dipole moment.

Amongst the given compounds,

CO2, BF3, CH4 have zero dipole moments.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule

Options:

The correct answer is:
08

Solution:
2s and 2p atomic orbitals combine, forming σ (sigma) and π (pi) molecular orbitals. σ orbitals result from head-on overlap, while π orbitals arise from sideways overlap. This molecular orbital theory explains the bonding in molecules

2s orbital will form, two molecular orbitals σ2 s and σ*2 s.
2p orbital will form six molecular orbitals σ2pz and σ*2pz. π2px,π2py and π*2px,π*2py

Hence the total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule formed is 8.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The total number of moles of neutrons present in 108 mL H 2 O l are

Options:

The correct answer is:
48

Solution:
â´ density of H 2 O( l )= 1 g/mL
1 mL contains =1 g H 2 O
108 mL contains =108 g H 2 O
Number of moles of H 2 O l = 1 0 8 1 8 = 6
H has no neutron, Number of neutrons in O = 1 6 - 8 = 8
Number of moles of neutrons in 1 mole of H 2 O=0+8=8
Number of moles of neutrons in 6 mole of H 2 O = 6 Ã 8 = 4 8

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The total number of monohalogenated organic products in the following (including stereoisomers) reaction is

A(Simplest opticallyactive alkene) →(ii) X2/∆(i) H2/Ni/∆

Options:

The correct answer is:
8

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The total number of negative charge in the tetrapeptide, Gly-Glu-Asp-Tyr, at pH12.5 will be_______ (Integer answer)

Options:

The correct answer is:
4

Solution:
All acidic groups loose a proton to form negative charge at pH = 12.5.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1,2,3,4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is

Options:

The correct answer is:
32

Solution:
We need three digits numbers.

Since 1+2+3+4+5=15

So, number of possible triplets for multiple of 15 is 1×2×2

so =4×⌊3+4×3-1×2×⌊2=32

Alternate:

divisible by 5=4×3=12

Divisible by 3

12→3, 4, 5→3!=615→2, 3, 4→3!=624→1, 3, 5→3!=642→1, 2, 3→3!=6 24

Divisible by 15

1,4,5; 4,1,5; 3,4,5; 4,3,5

Required No. =24+12-4=32

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The total number of octahedral void (s) per atom present in a cubic close packed structure is :

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
In CCP there are four atoms per unit cell octahedral voids are
12Edge centres14+1Body centre=4 O.V. [Contribution at edge centre is14 ]

Four octahedral voids are present in a unit cell i.e. four atoms.

Therefore, per atom one octahedral void will be present.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2014 (19 Apr Online)
The total number of optically active compounds formed in the following reaction is:

Options:
A
Zero

B
Six

C
Four

D
Two

The correct answer is:
Four

Solution:
The total number of optically active compounds formed is four. The product has two chiral $\mathrm{C}$ atoms. Thus, it has $2^n=2^2=4$ stereoisomers.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The total number of orbitals associated with the principal quantum number 5 is:

Options:
A 20
B 25
C 10
D 5

The correct answer is:
25

Solution:
Number of electrons in a shell of quantum number n=2n2.
Since, 2 electrons are stored in 1 orbital,
Number of orbitals in a shell =n2=52=25.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2016 (09 Apr Online)
The total number of permutations of the letters of the word "BANANA" is

Options:
A $60$
B $120$
C $720$
D $24$

The correct answer is:
$60$

Solution:
Total no. of permutations $=\frac{6!}{3!2!}=60$.

Topic: Mathematics, Permutation Combination, JEE Main
The total number of possible isomers for squareplanar $\left[\mathrm{Pt}(\mathrm{Cl})\left(\mathrm{NO}_2\right)\left(\mathrm{NO}_3\right)(\mathrm{SCN})\right]^{2-}$ is:

Options:
A
16

B
12

C
8

D
24

The correct answer is:
12

Solution:
The square planar complex of the type $[\mathrm{Mabcd}]^{\mathrm{n} \pm}$, where all four ligands are different, has 3 geometrical isomers. But if one of the ligands is ambidentate, then $2 \times$ $3=6$ geometrical isomers are possible. But if two ligands are ambidentate, then $4 \times 3=$ 12 geometrical isomers are possible.
In the given example, $\mathrm{NO}_2^{-}$and $\mathrm{SCN}^{-}$are ambidentate ligands.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The total number of possible isomers for the complex compound CuNH34[PtCl4]

Options:
A 4
B 5
C 6
D 3

The correct answer is:
4

Solution:
1. CuNH34[PtCl4]
2. CuNH33Cl [PtCl3(NH3)]
3*. CuNH32Cl2 PtCl2NH32 - cis isomer
4*. CuNH32Cl2PtCl2NH32 - trans isomer
5. CuNH3Cl3 [Pt(Cl)(NH3)3]
6, PtNH34 [CuCl4]
* These isomers would not exist as charge on the cation & anion would be zero.

Topic: Chemistry, Coordination Compounds, JEE Main
The total number of progeny obtained from a dihybrid cross is 1280 in the F2 generation. How many of them are recombinant?

Options:
A 240
B 360
C 480
D 720

The correct answer is:
480

Solution:
In the F2 generation of a dihybrid cross, normally 16 progeny will be formed. 10 of the progeny shows the parental phenotypes and the remaining 6 progeny shows the new phenotypes involving recombination. Accordingly, if in the F2 generation 1280 progenies are formed, 800 progeny shows parental phenotypes and 480 progenies are recombinants.

Topic: Biology, Principles of Inheritance and Variation, NEET
The total number of reagents from those given below, that can convert nitrobenzene into aniline is __________ . (Integer answer)
I. Sn-HCl
II. Sn-NH4OH
III. Fe-HCl
IV. Zn-HCl
V. H2-Pd
VI. H2- Raney Nickel

Options:

The correct answer is:
5

Solution:
Nitrobenzene converted into Aniline by -Sn+HCl, Fe+HCl, H2/Pd, H2/ Raney Ni.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The total number of 'Sigma' and Pi bonds in 2-formylhex-4-enoic acid is

Options:

The correct answer is:
22

Solution:
A single bond is a result of the axial overlap of bonding orbitals. Hence, it contributes a sigma bond. A multiple bond (double or triple bond) contain one sigma bond and the remaining are pi bonds.

The total number of sigma and pi bond in

2-Formyl hex-4-enoic acid is 22.

The compound can be represented as,

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The total number of 'sigma' and 'Pi' bonds in 2-oxohex-4-ynoic acid is ______

Options:

The correct answer is:
18

Solution:
$\begin{aligned} & \text { Number of } \sigma \text {-bonds }=14 \\ & \begin{aligned} \text { Number of } \pi \text {-bonds } & =4 \\ & =18\end{aligned}\end{aligned}$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The total number of sigma bonds in the structure of P4O10 is

Options:

The correct answer is:
16

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The total number of six digit numbers, formed using the digits 4,5,9 only and divisible by 6, is _____ .

Options:

The correct answer is:
81

Solution:
Given,

We have to form 6 digit number using the digits 4, 5 & 9 and for number should be divisible by 6 which means that the addition of number should be divisible by 3 and unit place should be even, so we will make following cases:

Case 1 The number is 444444 which can be arranged in 1 way,

Case 2 when the number is formed using the digits 4,4,4,5,9 and last place is fixed with 4, that can be arranged in 5!3!=20 ways

Case 3 when number is formed using the digits 4,4,5,5,5 and last place is fixed with 4, that can be arranged in 5!3!2!=10 ways

Case 4 when number is formed using the digits 4,4,9,9,9 and last place is fixed by 4, that can be arranged in 5!3!2!=10 ways

Case 5 when number is formed using the digits 4,5,5,9,9 and last place is fixed with 4, that can be arranged in 5!2!2!=30 ways

Case 6 when number is formed using the digits 5,9,9,9,9 and last place is fixed with 4, that can be arranged in 5!4!=5 ways

Case 7 when number is formed using the digits 5,5,5,5,9 and last place is fixed with 4, that can be arranged in 5!4!=5 ways

Now adding all the cases we get, 1+20+10+10+30+5+5=81 ways.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The total number of solution(s) of the equation 2x+3tan⁡x=5π2 in x∈0,2π is/are equal to

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
2x+3tan⁡x=5π2⇒tan⁡x=-23x+5π6
y=tan⁡x and y=5π6-2x3
Both the graphs meet exactly three times in 0,2π.
Thus, there are 3 solutions.

Topic: Mathematics, Trigonometric Equations, JEE Main
The total number of solutions of the equation sin⁡xtan⁡4x=cos⁡x for all x∈0,π are

Options:

The correct answer is:
5

Solution:
tan⁡4x=cot⁡x=tan⁡π2-x
4x=nπ+π2-x
5x=nπ+π2=2n+1π2
⇒x=2n+1π10
x=π10,3π10,5π10,7π10,9π10
Hence, 5 solutions in 0,π

Topic: Mathematics, Trigonometric Equations, JEE Main
The total number of species from the following in which one unpaired electron is present, is _______
\(\mathrm{N}_2, \mathrm{O}_2, \mathrm{C}_2^{-}, \mathrm{O}_2^{-}, \mathrm{O}_2^{2-}, \mathrm{H}_2^{+}, \mathrm{CN}^{-}, \mathrm{He}_2^{+}\)

Options:

The correct answer is:
4

Solution:
One unpaired $\mathrm{e}^{-}$is present in : $\mathrm{C}_2^{-} ; \mathrm{O}_2^{-} ; \mathrm{H}_2^{+} ; \mathrm{He}_2^{+}$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The total number of stereoisomers that can exist for M is

Options:

The correct answer is:
2

Solution:
Compound-M has '2' stereocenters.
Total number of stereoisomers = 2 (the given compound and its mirror image)

Topic: Chemistry, General Organic Chemistry, JEE Main
The total number of three-digit numbers, divisible by 3, which can be formed using the digits 1, 3, 5, 8, if repetition of digits is allowed, is

Options:
A 21
B 20
C 22
D 18

The correct answer is:
22

Solution:
Given,

Three-digit number to be formed which are divisible by 3 using the digits 1,3,5,8 and repetition is allowed,

Now taking case 1, where all digits are same, we get

1,1,1, 3,3,3, 5,5,5, 8,8,8→4 ways

Now taking case 2, where 2-digit are same and one is distinct, we get

5,5,8→3!2!=3 ways, 8,8,5→ 3 ways

Now taking case 3, where all are distinct, we get

1,3,5→6 ways, 1,8,3→6 ways

So, total ways will be 4+3+3+6+6=22

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The total number of three-digit numbers, with one digit repeated exactly two times, is ______.

Options:

The correct answer is:
243

Solution:
For three digit number we have following cases,

Case I: When digit zero is repeated

Number of numbers =C19=9

Case II: When digit zero is present but not repeated

Number of numbers =C19·2!=18

Case III: When digit zero is not present

Number of numbers =9·8·3!2!=216

Total number of numbers =243

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The total number of turns and cross-section area in a solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to:

Options:
A L
B 1L2
C 1L
D L2

The correct answer is:
1L

Solution:
Self-Inductance =μ0n2πr2L (n = number of turns per unit length)
=μ0NL2πr2L
=μ0N2πr2L
Self-Inductance ∝1L

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The total number of unpaired electrons present in $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$ is

Options:

The correct answer is:
1

Solution:
CoNH36Cl2
Co2+:[Ar] 3d74 s04p0
For strong field ligand and d7 configuration: t2g2,2,2 eg1,0
Total 1 unpaired electron is present.
CoNH36Cl3
Co3+:[Ar]3 d64 s04p0
The hybridisation of the complex is d2sp3 hybridisation.
NH3 acts as strong field ligand, because Δ0>P.E..
So here all electrons becomes paired.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The total number of unpaired electrons present in the complex K3Cr(oxalate)3 is______.

Options:

The correct answer is:
3

Solution:
K3Cr( oxalate )3
Chromium is in +3 oxidation state.

It has stable t2g 3 configuration

Number of unpaired electrons in Cr+3 will be 3.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

Options:
A 75
B 150
C 210
D 243

The correct answer is:
150

Solution:
$\because \quad$ Each person gets at least one ball.

$\therefore \quad 3$ Persons can have 5 balls as follow.

\begin{array}{|c|c|c|}

\hline Person & No. of balls & No. of balls \\

\hline I & 1 & 1 \\

\hline II & 1 & 2 \\

\hline III & 3 & 2 \\

\hline

\end{array}

The number of ways to distribute balls 1,1,3 in first to three persons

$={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1} \times{ }^{3} \mathrm{C}_{3}$

Also 3 , persons having 1,1 and 3 balls can be arranged in $\frac{3 !}{2 !}$ ways.

$\therefore \quad$ Total no. of ways to distribute $1,1,3$ balls to the three persons

$={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1} \times{ }^{3} \mathrm{C}_{3} \times \frac{3 !}{2 !}=60$

Similarly, total no. of ways to distribute $1,2,2$ balls to

three persons $={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2} \times{ }^{2} \mathrm{C}_{2} \times \frac{3 !}{2 !}=90$

$\therefore \quad$ The required number of ways $=60+90=150$

Topic: Mathematics, Permutation Combination, JEE Main
The total number of ways in which 5 balls of different colours can be distributed among 3 persons such that each person gets at least one ball is

Options:
A 75
B 150
C 210
D 243

The correct answer is:
150

Solution:
We know that total number of ways of distributing n distinct things among m persons such that each person gets at least one thing
= m n â C m 1 . ( mâ1 ) n + C m 2 ( mâ2 ) n â C m 3 ( mâ3 ) n +.....
So the required total number of ways
= 3 5 â C 3 1 .2 5 + C 3 2 .1 5 .
=243â96+3
=150

Topic: Mathematics, Permutation Combination, JEE Main
The total number of words (with or without meaning) that can be formed out of the letters of the word "DISTRIBUTION" taken four at a time, is equal to ______.

Options:

The correct answer is:
3734

Solution:
Given,

Word 'DISTRIBUTION'

Now, we have III, TT, D, S, R, B, U, O, N

Number of words with 3 alike letter and 1 different will be,

=8C1×4!3!=32

Number of words with 2 alike and other 2 alike will be,

=4!2!2!=6

Number of words with 2 alike and 2 different will be,

=2C1×8C2×4!2!=672

Number of words with all different letters will be,

=9C4×4!=3024

Therefore, the total number of ways will be,

=3024+672+6+32

=3734

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The total pressure exerted by a number of non-reacting gases is equal to the sum of the partial pressures of the gases under the same conditions is known as

Options:
A Boyle's law
B Charle's law
C Avogadro's law
D Dalton's law

The correct answer is:
Dalton's law

Solution:
Dalton's law
The correct answer is Dalton's law.
CONCEPT:
- Boyle's law states that at a constant temperature, the volume of a given mass of gas is inversely proportional to its pressure.
- $\mathbf{P}_1 \mathbf{V}_1=\mathbf{P}_2 \mathbf{V}_2$
- Graham's law of diffusion states that at a constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density.
- $\frac{r_A}{r_B}=\sqrt{\frac{d_A}{d_B}}$ or $\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}$
- Where $r_A$ and $r_B$ are rates of diffusions of $A$ and $B, d_A$ and $d_B$ are densities and $M_A$ and $M_B$ are molecular masses of $A$ and $B$ respectively
- Charles law: This law states that at constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature.
- $\frac{V 1}{T 1}=\frac{V 2}{T 2}$
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a number of non-reacting gases in a mixture is equal to the sum of partial pressures of the gases in the mixture.
- In a mixture, the pressure exerted by individual gas is called partial pressure.
- $\mathrm{p}_{\text {total }}=\mathrm{p}_1+\mathrm{p}_2+\mathrm{p}_3+\ldots$. (at constant $\mathrm{T}$ and $\mathrm{V}$), where $\mathrm{p}_{\text {total }}$ is the total pressure exerted by the mixture of gases and $p_1, p_2, p_3$ are the partial pressures of gases.

Topic: Chemistry, States of Matter, JEE Main
The Total pressure observed by mixing two liquid A and B is 350 mm Hg when their mole fractions are 0.7 and 0.3 respectively. The Total pressure becomes410 mm Hg if the mole fractions are changed to0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is ______ mm Hg. (Nearest integer) Consider the liquids and solutions behave ideally

Options:

The correct answer is:
314

Solution:
Let Vapour pressure of pure A be PAo

Let Vapour pressure of pure B be PBo

Using Raoult's law and Dalton's law:

When XA = 0.7 & XB = 0.3Ps = 350⇒ PAo ×0.7 + PBo ×0.3 = 350 ....(i)When XA = 0.2 & XB = 0.8 Ps = 410⇒ PAo ×0.7 + PBo ×0.3 = 350 ....(ii)Solving (i) and (ii)PAo = 314 mm HgPBo = 434 mm Hg

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The total pressure of a mixture of non-reacting gases X(0.6 g) and Y(0.45 g) in a vessel is 740 mm of Hg. The partial pressure of the gas X is mm of Hg. (Nearest Integer)
(Given : molar mass X=20 and Y=45 g mol-1 )

Options:

The correct answer is:
555

Solution:
Total pressure (PTotal) = 740 mm of Hg

Number of moles of gas X = 0.620

Number of moles of gas Y = 0.4545

Total number of moles = 0.03 + 0.01 = 0.04 mole

The partial pressure of the gas X = Mole fraction × Total pressure

X = 0.030.04 × 740 = 555

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The total work done in the following PV curve is

Options:
A 150-25π2 L-atm
B 150+25π2 L-atm
C -25π2 L-atm
D 150 L-atm

The correct answer is:
150-25π2 L-atm

Solution:
Total work done = area of rectangle of 15 atm height and 10 L width- area of half of circle
1520-10-π×522
=150-25π2

Topic: Chemistry, Thermodynamics (C), JEE Main
The toy-top (initially at rest) shown in the figure, has a moment of inertia 4Ã10-4 kg m2 and it is free to rotate about the stationary axis AA'. The string wrapped around the peg of the toy-top is pulled outwards without letting it slip and simultaneously maintaining a constant tension of 5.625 N. If Ï is the angular speed (in rad s-1) of the top after 80 cm of string has been pulled off the peg, then mark your answer as Ï100. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
Work done
=F×Δr=5.625×0.8 m=4.5 J and
work done =ΔK= 1 2 Iω r 2 − 1 2 Iω 1 2
∴4.5= 1 2 ×4× 10 −4 × ω r 2 −0
... ( ∵ top starts from rest, ω i =0 )
ω f =150 rad/s

Topic: Physics, Rotational Motion, JEE Main
The toy-top (initially at rest) shown in the figure, has a moment of inertia 4Ã10-4 kg m2 and it is free to rotate about the stationary axis AA'. The string wrapped around the peg of the toy-top is pulled outwards without letting it slip and simultaneously maintaining a constant tension of 5.625 N. If Ï is the angular speed (in rad s-1) of the top after 80 cm of string has been pulled off the peg, then mark your answer as Ï100.

Options:

The correct answer is:
1.50

Solution:
Work done
=F×Δr=5.625×0.8 m=4.5 J and
work done =ΔK= 1 2 Iω r 2 − 1 2 Iω 1 2
∴4.5= 1 2 ×4× 10 −4 × ω r 2 −0
... ( ∵ top starts from rest, ω i =0 )
ω f =150 rad/s

Topic: Physics, Rotational Motion, JEE Main
The trajectory of a projectile in a vertical plane is y=αx-βx2, where α and β are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by

Options:
A

tan-1α,4α2β

B

tan-1βα,α2β

C

tan-1β,α22β

D

tan-1α,α24β

The correct answer is:

tan-1α,α24β

Solution:
y=αx-βx2
comparing with trajectory equation

y=xtanθ-12gx2u2cos2θ

tanθ=α⇒θ=tan-1α

β=12gu2cos2θ

u2=g2βcos2θ

Maximum height : H

H=u2sin2θ2 g=g2βcos2θsin2θ2 g

H=tan2θ4β=α24β

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The trajectory of a projectile in a vertical plane is \( y=a x-b x^{2} \), where \( a, b \) are constants, and \( x \) and \( y \) are respectively the horizontal and vertical distances of the projectile from the point of projection. Find the maximum height \( H_{\max } \) attained by the projectile and the angle of projection from the horizontal.

Options:
A \( H_{\max }=\frac{a^{2}}{4 b} \)
\( =\tan ^{-1} a \)
B \(H_{\max } =\frac{a}{2 b} \)
\( =\tan ^{-1} b \)
C \( H_{\max }=\frac{a^{2}}{2 b} \)
\( \theta=\tan ^{-1}\left(\sqrt{a^{2}+1}\right) \)
D \( H_{\max }=\frac{a}{4 b^{2}} \)
\( \theta=\tan ^{-1} a \)

The correct answer is:
\( H_{\max }=\frac{a^{2}}{4 b} \)
\( =\tan ^{-1} a \)

Solution:
Trajectory of the projectile in a vertical plane -
y=ax-bx2
Comparing it with equation of a projectile
y=xtanθ-gx22u2cos2θ

⇒ tanθ=a ....(i)

⇒ θ=tan-1(a)

and g2u2cos2θ=b .....(ii)

From (ii) g2b2cos2θ=ga2+12b ....(iii)

∴cosθ=1a2+1

Maximum height attained, Hmax=u2sin2θ2a .....(iv)

From (ii) g2b=u2cos2θ
⇒g2b=u2-u2sin2θ
⇒u2sin2θ=ga2+12b-g2b=ga22b ......(v)
From (iv) and (v) Hmax=ga22b×2g=a24b

Alternatively
dydx=a-2bx=0 (For Maximum Height)
⇒x=a2h
Substituting the value of x in y=ax-bx2 to find max height
Hmax=aa2b-ba24b2=a22b-a24b=a24b

Topic: Physics, Motion In Two Dimensions, JEE Main
The trajectory of a projectile in vertical plane in y=ax-bx2, where a and b are constant and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal is:

Options:
A a24b,tan-1(b)
B a2b,tan-1(2b)
C a24b,tan-1(a)
D 2a2b,tan-1(a)

The correct answer is:
a24b,tan-1(a)

Solution:
Let the angle of projection be Î¸
y = ax - bx 2
To get maxima, dy dx = 0
dy dx = a - b . 2x = 0
â x = a 2b
y max = a 2 2 b - ba 2 4 b 2 = a 2 4b
Angle of projections can be obtained by slope dy dx at x = 0
â Slope at x = 0 is a
â tan Î¸ = a
â Î¸ = tan -1 a

Topic: Physics, Motion In Two Dimensions, JEE Main
The trajectory of a projectile near the surface of the earth is given as y=2x-9x2. If it were launched at an angle θ0 with speed v0 then g=10 m s-2:

Options:
A θ0=cos-1⁡15 and v0=53 ms-1
B θ0=cos-1⁡25 and v0=35 ms-1
C θ0=sin-1⁡15 and v0=53 ms-1
D θ0=sin-1⁡25 and v0=35 ms-1

The correct answer is:
θ0=cos-1⁡15 and v0=53 ms-1

Solution:
y=2x-9x2
Comparing above equation with general equation of trajectory of projectile y=xtan⁡θ1-xR
tan θ=2
⇒sin θ=25 or cos θ=15
θ=sin-125 or θ=cos-115
And R = 2/9
R=v02sin⁡2θg=29
After substituting the value for g and sin 2θ. We get v0=35 m/s

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (12 Apr Shift 1)