datavorous/entrance-exam-dataset huggingface dataset:
What should be the value of $\mathrm{k}$ for which the equation $3 x^{2}+3 y^{2}+(k+1) z^{2}+x-y+z=0$ represents the sphere?

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

The correct answer is:
2

Solution:
Given equation is $3 x^{2}+3 y^{2}+(k+1) z^{2}+x-y+z=0$
which will represents a sphere if coeff of $x^{2}=$ coeff of $y^{2}=$ coeff of $z^{2}$ $\Rightarrow 3=\mathrm{k}+1$
$\Rightarrow \mathrm{k}=2$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 1)
What should be the value of $\mathrm{k}$ so that the system of linear equations $x-y+2 z=0, k x-y+z=0,3 x+y-3 z=0$ does not possess a unique solution?

Options:
A 0
B 3
C 4
D 5

The correct answer is:
5

Solution:
Equations are
$\begin{array}{l}
x-y+2 x=0 ...(i) \\
k x-y+z=0 ...(ii) \\
3 x+y-3 x=0 ...(iii)
\end{array}$
System of equations posses a unique solution, if
$|\mathrm{A}|=\left|\begin{array}{rrr}
1 & -1 & 2 \\
\mathrm{k} & -1 & 1 \\
3 & 1 & -3
\end{array}\right| \neq 0$
$\text { Applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\frac{1}{2} \mathrm{C}_{3}$
$|\mathbf{A}|=\left|\begin{array}{ccc}0 & 0 & 2 \\ \mathrm{k}-1 & -\frac{1}{2} & 1 \\ 4 & -\frac{1}{2} & -3\end{array}\right|=-\frac{1}{2}\left|\begin{array}{lrr}0 & 0 & 2 \\ \mathrm{k}-1 & -1 & 1 \\ 4 & -1 & -3\end{array}\right|$
$=\frac{1}{2} \times 2[(\mathrm{k}-1)(-1)-(4)(-1)], 0$
$\Rightarrow-(-k+1+4) \neq 0 \Rightarrow k-5 \neq 0 \Rightarrow k \neq 5$
Thus, the system does not posses unique solution, if $\mathrm{k}=5$

Topic: Mathematics, Determinants, NDA, NDA 2007 (Phase 1)
What should be the value of $x$ so that the matrix $\left(\begin{array}{ll}2 & 4 \\ -8 & x\end{array}\right)$
$\begin{array}{ll}\text { does not have an inverse? } & \end{array}$

Options:
A 16
B $-16$
C 8
D $-8$

The correct answer is:
$-16$

Solution:
A matrix does not have an inverse if $|\mathrm{A}|=0$
$\left|\begin{array}{cc}2 & 4 \\ -8 & x\end{array}\right|=0$
$\Rightarrow x=-16$

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
What term describes the arrangement of ions/groups around the central atom/ion in a coordination entity?

Options:
A Secondary valence
B Coordination number
C Counter ions
D Lewis base

The correct answer is:
Coordination number

Solution:
The coordination number describes the arrangement of ions/groups around the central atom/ion in a coordination entity.

Topic: Chemistry, Coordination Compounds, NEET
What term is used for ligands with one donor atom per molecule?

Options:
A Hexadentate
B Didendate
C Polydentate
D Unidentate

The correct answer is:
Unidentate

Solution:
Unidentate ligands have one donor atom per molecule and can bind through that atom.

Topic: Chemistry, Coordination Compounds, NEET
What term is used for the atom/ion to which a fixed number of ions/groups are bound in a coordination entity?

Options:
A Central atom/ion
B Secondary valence
C Counter ion
D Lewis base

The correct answer is:
Central atom/ion

Solution:
The atom/ion to which a fixed number of ions/groups are bound in a coordination entity is called the central atom/ion.

Topic: Chemistry, Coordination Compounds, NEET
What term is used to describe the spatial arrangement of atoms in the molecules of two types of crystals, according to Louis Pasteur?

Options:
A Isomerism
B Stereochemistry
C Conformation
D Racemism

The correct answer is:
Stereochemistry

Solution:
Louis Pasteur described the spatial arrangement of atoms in the molecules of two types of crystals as a foundation of modern stereochemistry.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
What type of alcohol is produced when methanal (formaldehyde) reacts with a Grignard reagent?

Options:
A Primary alcohol
B Secondary alcohol
C Tertiary alcohol
D Quaternary alcohol

The correct answer is:
Primary alcohol

Solution:
Methanal reacts to produce a primary alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
What type of amines can be prepared using Gabriel synthesis?

Options:
A Primary amines
B Secondary amines
C Tertiary amines
D Quaternary ammonium salts

The correct answer is:
Primary amines

Solution:
Gabriel synthesis is primarily used for the preparation of primary amines.

Topic: Chemistry, Amines, NEET
What type of appendages do aquatic annelids like Nereis possess to help in swimming?

Options:
A parapodia
B tentacle
C gills
D fins

The correct answer is:
parapodia

Solution:
Aquatic annelids like Nereis possess lateral appendages called parapodia, which help in swimming.

Topic: Biology, Animal Kingdom, NEET
What type of ecological pyramid would be obtained with following data?
Secondary consumer : 120 g
Primary consumer : 60 g
Primary producer : 10 g

Options:
A Upright pyramid of numbers
B Pyramid of energy
C Inverted pyramid of biomass
D Upright pyramid of biomass

The correct answer is:
Inverted pyramid of biomass

Solution:
The inverted pyramid of biomass
Biomass is the amount of living material in an organism. The pyramids of biomass represent the relationship between different trophic levels in terms of biomass. In grassland and forest ecosystems, there is generally a gradual decrease in the biomass of organisms at successive trophic levels from the producers to the consumers. Their pyramid of biomass is upright. In an aquatic ecosystem, the producers are small organisms with the least biomass and the biomass gradually increase towards the apex of the pyramid. Thus, the pyramid of the biomass of aquatic ecosystems is inverted in shape.

Topic: Biology, Ecosystem, NEET, NEET 2018
What type of immunity is provided to the newborn infant through mother’s milk?

Options:
A Active immunity
B Cell-mediated immunity
C Humoral immune response
D Passive immunity

The correct answer is:
Passive immunity

Solution:
Mother’s milk contains abundant antibodies (IgA) and provides passive immunity to the newborn infant.

Topic: Biology, Human Health and Diseases, NEET
What type of ligand has several donor atoms in a single molecule?

Options:
A Unidentate
B Didendate
C Hexadentate
D Polydentate

The correct answer is:
Polydentate

Solution:
Polydentate ligands have several donor atoms in a single molecule.

Topic: Chemistry, Coordination Compounds, NEET
What type of ligand is ethane-1,2-diamine?

Options:
A Didendate
B Polydentate
C Hexadentate
D Unidentate

The correct answer is:
Didendate

Solution:
Ethane-1,2-diamine is a didendate ligand because it can bind through two donor atoms.

Topic: Chemistry, Coordination Compounds, NEET
What type of ligand is ethylenediaminetetraacetate ion (EDTA4–)?

Options:
A Didendate
B Polydentate
C Hexadentate
D Unidentate

The correct answer is:
Hexadentate

Solution:
EDTA4– is a hexadentate ligand because it can bind through multiple donor atoms, including two nitrogen and four oxygen atoms.

Topic: Chemistry, Coordination Compounds, NEET
What type of molecular orbitals are not symmetrical around the bond-axis in the sigma-pi nomenclature?

Options:
A Sigma (σ) molecular orbitals
B Pi (π) molecular orbitals
C Delta (Δ) molecular orbitals
D Gamma (γ) molecular orbitals

The correct answer is:
Pi (π) molecular orbitals

Solution:
Pi (π) molecular orbitals are not symmetrical around the bond-axis in the sigma-pi nomenclature.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
What type of molecular orbitals are symmetrical around the bond-axis in the sigma-pi nomenclature?

Options:
A Sigma (σ) molecular orbitals
B Pi (π) molecular orbitals
C Delta (Δ) molecular orbitals
D Gamma (γ) molecular orbitals

The correct answer is:
Sigma (σ) molecular orbitals

Solution:
In the sigma-pi nomenclature, sigma (σ) molecular orbitals are symmetrical around the bond-axis.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
What type of sulphonamide is formed in the reaction with a secondary amine?

Options:
A N-ethyl benzene sulfonyl amide
B N,N-diethyl benzene sulphonamide
C N-methylbenzenesulphonyl amide
D N,N-dimethyl benzenesulfonamide

The correct answer is:
N,N-diethyl benzene sulphonamide

Solution:
The reaction of benzenesulfonyl chloride with a secondary amine yields N,N-diethyl benzene sulphonamide.

Topic: Chemistry, Amines, NEET
What volume of O2 measured at standard conditions will be formed by the action of 100 mL of 0.5 KMnO4 on hydrogen peroxide in an acidic solution?

Options:
A 0.12 litre
B 0.28 litre
C 0.56 litre
D 1.12 litre

The correct answer is:
0.28 litre

Solution:
Here H2O2 on reaction with KMnO4 gives O2 so
Meq. of O2=Meq of KMnO4=100×0.5
w8×1000=50 (equivalent weight of oxygen is 8)
So wO2=0.4 g
So VO2=22.4×0.432=0.28 litre.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
What was Panchanan Maheshwari's contribution to school education?

Options:
A He published the first textbooks of Biology for Higher Secondary Schools
B He encouraged students to study in general education
C He taught at Higher Secondary Schools
D He established schools in rural areas

The correct answer is:
He published the first textbooks of Biology for Higher Secondary Schools

Solution:
Panchanan made a significant contribution to school education by his leadership in bringing out the very first textbooks of Biology for Higher Secondary Schools published by NCERT in 1964.

Topic: Biology, Reproduction in Organisms, NEET
What will be the emf of the following cell at $25^{\circ} \mathrm{C}$ ?
$\mathrm{Fe} / \mathrm{Fe}^{2+}(0.001 \mathrm{M})|| \mathrm{H}^{+}(0.01 \mathrm{M}) \mid \mathrm{H}_2(\mathrm{~g})$
(1 Bar) $\mid \mathrm{Pt}(s)$
$E_{\left(\mathrm{Fe}^{2+} / \mathrm{Fe}\right)}^{\circ}=-0.44 \mathrm{~V} ; E_{\left(\mathrm{H}^{+} / \mathrm{H}_2\right)}^{\circ}=0.00 \mathrm{~V}$

Options:
A $0.44 \mathrm{~V}$
B $-0.44 \mathrm{~V}$
C $0.41 \mathrm{~V}$
D $-0.41 \mathrm{~V}$

The correct answer is:
$0.41 \mathrm{~V}$

Solution:
For the given cell representation, the cell reaction will be,
$$
\mathrm{Fe}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{H}_2(g)
$$
The standard emf of the cell will be,
$$
\begin{aligned}
E_{\text {cell }}^{\circ} & =E_{\mathrm{H}^{+} / \mathrm{H}_2}^{\circ}-E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ} \\
& =0-(-0.44)=0.44 \mathrm{~V}
\end{aligned}
$$
The Nernst equation for the cell reaction at $25^{\circ} \mathrm{C}$,
$$
E_{\text {cell }}=E_{\text {cell }}^a-\frac{0.0591}{n} \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]}
$$
$$
\begin{aligned}
& =0.44-\frac{0.0591}{2} \log \frac{[0.001]}{[0.01]^2} \\
& =0.44-0.02955(\log 10) \\
& =0.44-0.02955(1) \\
& =0.41045 \mathrm{~V}=0.41 \mathrm{~V}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, COMEDK, COMEDK 2022
Whatare the linear constraints for which the shaded area in
the above figure is the solution set?

Options:
A $\quad x-y \geq 1 ; x+2 y \leq 8 ; x+y \geq 1 ; x, y \geq 0$
B $\quad x-y \leq 1 ; x+2 y \geq 8 ; x+y \leq 1 ; x, y \geq 0$
C $x-y \leq 1 ; x+2 y \leq 8 ; x+y \geq 1 ; x, y \geq 0$
D $x-y \leq 1 ; x+2 y \leq 8 ; x+y \leq 1 ; x, y \geq 0$

The correct answer is:
$x-y \leq 1 ; x+2 y \leq 8 ; x+y \geq 1 ; x, y \geq 0$

Solution:
The linear constraints for which the shaded region in the given figure is the solution set, are given by:
$x-y \leq 1, x+2 y \leq 8, x+y \geq 1, x, y \geq 0$

Topic: Mathematics, Straight Lines, NDA, NDA 2007 (Phase 2)
When 3-methylbutant 2-0l is treated with $\mathrm{HBr}$, the following reaction takes place:
Give a mechanism for this reaction.
(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

Options:

None

Solution:
Protonation of the given alcohol followed by loss of water gives a $2^{\circ}$ carbocation(I), which being unstable rearranges by 1 , 2-hydride shift to form the more stable $3^{\circ}$ carbocation (II). Nucleophilic attack by Br ion on this carbocation (II) gives the final product.

Topic: Chemistry, Alcohols Phenols and Ethers
When $6 \cdot 0 \mathrm{~g}$ of graphite reacts with dihydrogen to give methane gas, $37 \cdot 4 \mathrm{~kJ}$ of heat is liberated. What is standard enthalpy of formation of $\mathrm{CH}_{4_{(g)}}$ ?

Options:
A $112 \cdot 2 \mathrm{~kJ} \mathrm{~mol} \cdot 1$
B $-74 \cdot 8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-37 \cdot 4 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-112 \cdot 2 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-74 \cdot 8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
(A)
$\mathrm{C}_{(8)}+2 \mathrm{H}_{2(8)} \longrightarrow \mathrm{CH}_{4(\mathrm{~g})}$
For $6.0 \mathrm{~g}$ of graphite, $\Delta \mathrm{H}=-37.4 \mathrm{~kJ}$
$\therefore$ For $12.0 \mathrm{~g}$ of graphite, $\quad \Delta \mathrm{H}^{0}=\frac{-37.4 \times 12}{6}=-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Topic: Chemistry, Thermodynamics (C), MHT CET, MHT CET 2020 (19 Oct Shift 2)
When $a$ and $b$ are eliminated from the equation $x y=a e^{x}+$ $b e^{-x}$, the resulting differential equation is of $\quad$

Options:
A first order and first degree
B first order and second degree
C second order and first degree
D second order and second degree

The correct answer is:
second order and first degree

Solution:
Given equation is $x y=a e^{x}+b e^{-x}$
Differentiate both side, w.r.t. $^{\circ} x$
$$
x \frac{d y}{d x}+y=a e^{x}-b e^{-x}
$$
Again differentiate both side w.r.t ' $x$ '
$$
\begin{aligned}
& x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+\frac{d y}{d x}=a e^{x}+b e^{-x} \\
\Rightarrow & x \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=x y(\text { from }(1))
\end{aligned}
$$
Hence, this is the differential equation of second order and first degree.

Topic: Mathematics, Differential Equations, NDA, NDA 2009 (Phase 2)
When a beam of light is passed through a colloidal sol, what is observed?

Options:
A The light is not visible.
B Tyndall Effect The path of the beam becomes visible due to scattering of light by the colloidal particles.
C The light changes color.
D The light becomes more intense.

The correct answer is:
Tyndall Effect The path of the beam becomes visible due to scattering of light by the colloidal particles.

Solution:
Correct Option is : (B)
Tyndall Effect The path of the beam becomes visible due to scattering of light by the colloidal particles.

Topic: Chemistry, Surface Chemistry, NEET
When a bimetallic strip is heated, it

Options:
A does not bend at all
B gets twisted in the from of an helix
C bends in the form of an arc with the more expandable metal outside
D bends in the form of an arc with the more expandable metal inside

The correct answer is:
bends in the form of an arc with the more expandable metal outside

Solution:
A bimetallic strip on being heated bends in from of an arc with more expandable metal (A) outside (as shown)

Topic: Physics, Thermal Properties of Matter, NEET
When a card is drawn from a well shuffled pack of cards. what is the probability of getting a Queen?

Options:
A $2 / 13$
B $1 / 13$
C $1 / 26$
D $1 / 52$

The correct answer is:
$1 / 13$

Solution:
Since, $n(S)=52$ and $n(E)=4$
$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{4}{52}=\frac{1}{13}$

Topic: Mathematics, Probability, NDA, NDA 2009 (Phase 1)
When a current of $4 \mathrm{~A}$ changes to $8 \mathrm{~A}$ in $0.6 \mathrm{~s}$ in a primary coil it induces an e. m. f. of $50 \mathrm{mV}$ in secondary coil. The mutual inductance between two coils is

Options:
A $1.2 \mathrm{mH}$
B $3.33 \mathrm{mH}$
C $7.5 \mathrm{mH}$
D $10.5 \mathrm{mH}$

The correct answer is:
$7.5 \mathrm{mH}$

Solution:
Change in current in primary coil,
$\mathrm{di}_1=(8-4) \mathrm{A}=4 \mathrm{~A}$
Time taken, $\mathrm{dt}=0.6 \mathrm{~s}$
Induced e. m. f in secondary coil,
$\mathrm{E}_2=50 \times 10^{-3} \mathrm{~V}$
So, mutual inductance is,
$M=\frac{E_2}{\mathrm{di}_1 / \mathrm{dt}}=\frac{\mathrm{E}_2 \mathrm{dt}}{\mathrm{di}_1}=\frac{50 \times 10^{-3} \mathrm{~V} \times 0.6 \mathrm{~s}}{4 \mathrm{~A}}=7.5 \times 10^{-3} \mathrm{H}=7.5 \mathrm{mH}$

Topic: Physics, Electromagnetic Induction, MHT CET, MHT CET 2022 (07 Aug Shift 2)
When a Daniel cell is connected in the secondary circuit of a potentiometer, the balancing length is found to be 540 cm. If the balancing length becomes 500 cm when the cell is short-circuited with 1 Î©, the internal resistance of the cell is

Options:
A 0.08 Ω
B 0.04 Ω
C 1.08 Ω
D 1.45 Ω

The correct answer is:
0.08 Ω

Solution:
Internal resistance,
r=Rl1-l2l2=1540-500500=40500=0.08 Ω

Topic: Physics, Current Electricity, NEET
When a meristematic tissue called cambium is present inside a vascular bundle, the bundle is said to be

Options:
A Conjoint
B Open
C Closed
D Collateral

The correct answer is:
Open

Solution:
Vascular bundles having cambium in between xylem and phloem are called open bundles. Due to cambium, secondary growth is possible. Vascular bundles which lack cambium are called closed bundles. Open bundles occur in dicots, and closed bundles occur in monocots.

Topic: Biology, Anatomy of Flowering Plants, NEET
When a picture drawn on paper is seen through a slab of a transparent material of thickness 5 cm, it appears to be raised by 1.5 cm. The critical angle at the boundary of this transparent material and air is -

Options:
A sin-123
B sin-157
C sin-1611
D sin-1710

The correct answer is:
sin-1710

Solution:
s=1–1μt
1-st=1μ
μ=11–st=11–32×5=107
⇒θc=sin-1⁡1μ
⇒θc=sin-1710

Topic: Physics, Ray Optics, NEET
When a string is divided into three segments of lengths l1, l2 and l3 the fundamental frequencies of these three segments are Î½1, Î½2 and Î½3 respectively. The original fundamental frequency Î½ of the string is

Options:
A ν=ν1+ν2+ν3
B ν=ν1+ν2+ν3
C 1Î½=1Î½1+1Î½2+1Î½3
D 1ν=1ν1+1ν2+1ν3

The correct answer is:
1ν=1ν1+1ν2+1ν3

Solution:
Let l be the length of the string.
The fundamental frequency is given by
ν=12lT⁡μ
orν∝1l⁡ ( ∵ T and μ are constant )
orν∝kl⁡ ( where k is a constant )
Here, l1=kν1, l2=kν2, l3=kν3 and l=kν
But l=l1+l2+l3
∴ 1ν=1ν1+1ν2+1ν3

Topic: Physics, Waves and Sound, NEET
When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of these three segments are $v_1$, $v_2$ and $v_3$ respectively. The original fundamental frequency $(v)$ of the string is

Options:
A $\sqrt{v}=\sqrt{v_1}+\sqrt{v_2}+\sqrt{v_3}$
B $v=v_1+v_2+v_3$
C $\frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}$
D $\frac{1}{\sqrt{v}}=\frac{1}{\sqrt{v_1}}+\frac{1}{\sqrt{v_2}}+\frac{1}{\sqrt{v_3}}$

The correct answer is:
$\frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}$

Solution:
The fundamental frequency of string
$\begin{aligned}
v & =\frac{1}{2 l} \sqrt{\frac{T}{m}} \\
\therefore v_1 l_1 & =v_2 l_2=v_2 l_3=k
\end{aligned}$
From Eq. (i)
$l_1=\frac{k}{v_1}, l_2=\frac{k}{v_2}, l_3=\frac{k}{v_3}$
Original length
$\begin{aligned}
l & =\frac{k}{v} \\
l & =l_1+l_2+l_3 \\
\frac{k}{v} & =\frac{k}{v_1}+\frac{k}{v_2}+\frac{k}{v_3} \\
\frac{1}{v} & =\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}
\end{aligned}$
Here,
$\begin{aligned}
& l=l_1+l_2+l_3 \\
& \frac{k}{v}=\frac{k}{v_1}+\frac{k}{v_2}+\frac{k}{v_3} \\
& \frac{1}{v}=\frac{1}{v_1}+\frac{1}{v_2}+\frac{1}{v_3}
\end{aligned}$

Topic: Physics, Waves and Sound, NEET, NEET 2012 (Screening)
When a string of length ' $l$ ' is divided into three segments of length $l_1, l_2$ and $l_3$. The fundamental frequencies of three segments are $\mathrm{n}_1, \mathrm{n}_2$ and $\mathrm{n}_3$ respectively. The original fundamental frequency ' $n$ ' of the string is

Options:
A $\mathrm{n}=\mathrm{n}_1+\mathrm{n}_2+\mathrm{n}_3$
B $\sqrt{\mathrm{n}}=\sqrt{\mathrm{n}_1}+\sqrt{\mathrm{n}_2}+\sqrt{\mathrm{n}_3}$
C $\frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_1}+\frac{1}{\mathrm{n}_2}+\frac{1}{\mathrm{n}_3}$
D $\frac{1}{\sqrt{\mathrm{n}}}=\frac{1}{\sqrt{\mathrm{n}_1}}+\frac{1}{\sqrt{\mathrm{n}_2}}+\frac{1}{\sqrt{\mathrm{n}_3}}$

The correct answer is:
$\frac{1}{\mathrm{n}}=\frac{1}{\mathrm{n}_1}+\frac{1}{\mathrm{n}_2}+\frac{1}{\mathrm{n}_3}$

Solution:
The fundamental frequency of a string is given by
Given: $l=l_1+l_2+l_3$... (i)
$\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
$\Rightarrow \mathrm{n} \propto \frac{1}{l}$ or $\mathrm{n} l=\mathrm{k}$
$\therefore \quad l_1=\frac{\mathrm{k}}{\mathrm{n}_1}, l_2=\frac{\mathrm{k}}{\mathrm{n}_2}$ and $l_3=\frac{\mathrm{k}}{\mathrm{n} 3}$... (ii)
$\therefore \quad$ Original length $l=\frac{\mathrm{k}}{\mathrm{n}}$... (iii)
Putting eq (ii) and (iii) into eq (i)
$\frac{\mathrm{k}}{\mathrm{n}}=\frac{\mathrm{k}}{\mathrm{n}_1}+\frac{\mathrm{k}}{\mathrm{n}_2}+\frac{\mathrm{k}}{\mathrm{n}_3}$
$\therefore \quad \frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (09 May Shift 1)
When a tetraploid individual is crossed with another individual it produces triploid embryo, 5n endosperm and 4n fruit. In this cross, the male individual is:

Options:
A Diploid
B Tetraploid
C 2n or 4n
D None of these

The correct answer is:
Diploid

Solution:
When Tetraploid female is crossed with diploid male then
Diploid male 2n - Microspore mother cell 2n - Pollen grain (n) -Male gamete n
Tetraploid female 4n - Megaspore mother cell (4n)- Megaspore 2n - Embryo sac 2n
Embryo = Male gamete + Female gamete =n+2n=3n
Endosperm = Male gamete + Secondary nucleus =n+4n=5n
Fruit is developed by ovary which is sporophytic in nature. 4n ploidy is present in ovary so fruit has also 4n ploidy.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
When alkyl halide is boiled with large excess of alcoholic ammonia it forms

Options:
A primary amine
B tertiary amine
C secondary amine
D quaternary ammonium salt

The correct answer is:
primary amine

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, MHT CET, MHT CET 2020 (12 Oct Shift 1)
When an aldehyde reacts with a Grignard reagent, what type of alcohol is typically formed?

Options:
A Primary alcohol
B Secondary alcohol
C Tertiary alcohol
D Quaternary alcohol

The correct answer is:
Secondary alcohol

Solution:
Aldehydes react to form secondary alcohols.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
When an open pipe is closed from one end then the third overtone of the closed pipe is higher in frequency by 150 Hz than the second overtone of an open pipe. The fundamental frequency of the open-end pipe will be

Options:
A 75 Hz

B 150 Hz
C 225 Hz
D 300 Hz

The correct answer is:
300 Hz

Solution:
Let f0=v02L be the fundamental frequency of the open pipe
â´ is second overtone is 3f0=3v02L
Let f=v04L be the fundamental frequency of close pipe.
And third overtone of the organ pipe closed at one end only is, f3closed =74 v0L
As given 7v04L-3v02L=150
74-64 v0L=150
â´ v04L=150
v02L = 300 Hz

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2016
When an open pipe is closed from one end then the third overtone of the closed pipe is higher in frequency by 150 Hz than the second overtone of an open pipe. The fundamental frequency of the open-end pipe will be

Options:
A 75 Hz

B 150 Hz
C 225 Hz
D 300 Hz

The correct answer is:
300 Hz

Solution:
Let f0=v02L be the fundamental frequency of the open pipe
â´ is second overtone is 3f0=3v02L
Let f=v04L be the fundamental frequency of close pipe.
And third overtone of the organ pipe closed at one end only is, f3closed =74 v0L
As given 7v04L-3v02L=150
74-64 v0L=150
â´ v04L=150
v02L = 300 Hz

Topic: Physics, Wave Optics, NEET
When Butan - 2 -ol is dehydrated using sulphuric acid, the concentration of acid and temperature needed respectively is

Options:
A 600/0 Conc. and 373 K
B $200 / 0$ conc. and $373 \mathrm{~K}$
C $200 / 0$ conc. and $363 \mathrm{~K}$
D $950 / 0$ conc. and $373 \mathrm{~K}$

The correct answer is:
600/0 Conc. and 373 K

Solution:
When Butan-2-ol is dehydrated using sulphuric acid, the concentration of acid and temperature needed respectively is $\mathrm{6 0 \%}$ cone. and $373 \mathrm{~K}$.
Secondary alcohol $\left(2^{\circ}\right)$ is dehydrated by heating with $60 \% \mathrm{H}_{2} \mathrm{SO}_{4}$ at $373 \mathrm{~K}$.

Topic: Chemistry, Alcohols Phenols and Ethers, MHT CET, MHT CET 2020 (14 Oct Shift 1)