datavorous/entrance-exam-dataset huggingface dataset:
Both in cells and extracellular fluids diabasic phosphate (HPO4) and monobasic phosphate (H2PO-4) act as acid base buffers to maintain

Options:
A K+ concentration of extracellular fluid
B Na+ concentration of extracellular fluid
C Na+ concentration of cellular fluid
D H+ concentration of cellular fluid

The correct answer is:
H+ concentration of cellular fluid

Solution:
Phosphate Buffer System has 2 main components:
• Sodium salts of dihydrogen phosphate (H2PO4¯), a weak acid
• Monohydrogen phosphate (HPO42¯), a weak base
This system is an effective acid base buffer to maintain H+ concentration of cellular fluid

Topic: Chemistry, Biomolecules, NEET
Both \( \left[\mathrm{Ni}(\mathrm{CO})_{4}\right] \) and \( \left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-} \) are diamagnetic. The types of hybridisation of \( \mathrm{Ni} \) in these complexes are and respectively.

Options:
A \( \mathrm{sp}^{3}, \mathrm{sp}^{3} \)
B \( \mathrm{sp}^{3}, \mathrm{dsp}^{2} \)
C \( \mathrm{dsp}^{2}, \mathrm{sp}^{3} \)
D \( d s p^{2}, d s p^{2} \)

The correct answer is:
\( \mathrm{sp}^{3}, \mathrm{dsp}^{2} \)

Solution:
Oxidation number of Ni in NiCO4 is zero with Ar3d84s2 electronic configuration as there are no unpaired electrons. Hence, pairing takes place to form sp3 hybridisation.

Oxidation number of Ni in NiCN42- is +2 with Ar3d84s0 electronic configuration as there are no unpaired electrons. Hence, pairing takes place to form dsp2 hybridisation.

Topic: Chemistry, Coordination Compounds, JEE Main
Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect is:

Options:
A Both form soluble bicarbonates
B Both form nitrides
C Nitrates of both Li and Mg yield NO2 and O2 on heating
D Both form basic carbonate

The correct answer is:
Both form basic carbonate

Solution:
Carbonate of magnesium is basic, sometimes appears as MgCO3.Mg(OH)2 but that of lithium is not basic.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2017 (02 Apr)
Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths λN, λA respectively. The ratio λNλA is closest to:

Options:
A 10-1
B 10-6
C 10
D 10-10

The correct answer is:
10-6

Solution:
λNλa= EaEN, where Ea and EN are energies of photons from atom and nucleus respectively. EN is of the order of MeV and Ea in few eV. So, λNλa=10-6.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2018 (16 Apr Online)
Both the roots of the given equation
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are always

Options:
A Positive
B Negative
C Real
D Imaginary

The correct answer is:
Real

Solution:
Given equation
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
can be re-written as
$3 x^2-2(a+b+c) x+(a b+b c+c a)=0$
$\begin{aligned} \Delta & =4(a+b+c)^2-3(a b+b c+c a)\left(\left(b b^2-4 a c=\Delta\right)\right. \\ & =4\left(a^2+b^2+c^2-a b-b c-a c\right) \\ & \left.=2(a-b)^2+(b-c)^2+(c-a)^2\right\} \geq 0\end{aligned}$
Hence both roots are always real.

Topic: Mathematics, Quadratic Equation, JEE Main
Bottles containing $\mathrm{C}_6 \mathrm{H}_5$ l and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{I}$ lost their original labels. They were labelled $\mathrm{A}$ and $\mathrm{B}$ for testing $\mathrm{A}$ and $\mathrm{B}$ were separately taken in test tubes and boiled with $\mathrm{NaOH}$ solution. The end solution in each tube was made acidic with dilute $\mathrm{HNO}_3$ and then some $\mathrm{AgNO}_3$ solution was added. Substance B gave a yellow precipitate. Which one of the following statements is true for this experiment?

Options:
A
A and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{I}$

B
$B$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{I}$

C
Addition of $\mathrm{HNO}_3$ was unnecessary

D
A was $\mathrm{C}_6 \mathrm{H}_5 \mathrm{I}$

The correct answer is:
A was $\mathrm{C}_6 \mathrm{H}_5 \mathrm{I}$

Solution:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{I}$ will not respond to silver nitrate test because $\mathrm{C}$-I bond has a partial double bond character.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2003
Bouveault-Blanc reduction reaction involves:

Options:
A Reduction of an acyl halide with \( \mathrm{H}_{2} / \mathrm{Pd} \).
B Reduction of an anhydride with \( \mathrm{LiAlH}_{4} \)
C Reduction of an ester with \( \mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \).
D Reduction of a carbonyl compound with \( \mathrm{Na} / \mathrm{Hg} \) and \( \mathrm{HCl} \)

The correct answer is:
Reduction of an ester with \( \mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \).

Solution:
Bouveault-Blanc reduction involves reduction of following compounds into alcohol.

R−C≡N into primary alcohol by reduction with Na/C2H5OH

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Bouveault-Blanc reduction reaction involves:

Options:
A Reduction of an acyl halide with H2/Pd.
B Reduction of an anhydride with LiAlH4
C Reduction of an ester with Na/C2H5OH .
D Reduction of a carbonyl compound with Na/Hg and HCl.

The correct answer is:
Reduction of an ester with Na/C2H5OH .

Solution:
Bouveault-Blanc reduction involves reduction of following compounds into alcohol.

R−C≡N into primary alcohol by reduction with Na/C2H5OH

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2016 (09 Apr Online)
Bouveault-Blanc reduction reaction involves:

Options:
A Reduction of an acyl halide with H2/Pd.
B Reduction of an anhydride with LiAlH4
C Reduction of an ester with Na/C2H5OH .
D Reduction of a carbonyl compound with Na/Hg and HCl.

The correct answer is:
Reduction of an ester with Na/C2H5OH .

Solution:
Bouveault-Blanc reduction involves reduction of following compounds into alcohol.

R−C≡N into primary alcohol by reduction with Na/C2H5OH

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Bouveault-Blanc reduction reaction involves:

Options:
A Reduction of an acyl halide with H2/Pd.
B Reduction of an anhydride with LiAlH4
C Reduction of an ester with Na/C2H5OH .
D Reduction of a carbonyl compound with Na/Hg and HCl

The correct answer is:
Reduction of an ester with Na/C2H5OH .

Solution:
Bouveault-Blanc reduction involves reduction of an pure impure carbonyl
into alcohol and R−C≡N into R− CH 2 − NH 2 by Na/ C 2 H 5 OH

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Bowman's capsule is found in :

Options:
A glomerulus
B uriniferous tubule
C nephron
D Malpighian capsule

The correct answer is:
Malpighian capsule

Solution:
Human kidney possesses about one million nephrons. Each nephron in made up of two main parts i.e., Malpighian body and renal tubule.
Malpighian body comprises a cup shaped Bowman's capsule and a meshwork of blood capillaries, the glomerulus.

Topic: Biology, Excretory Products and their Elimination, JIPMER, JIPMER 2006
Box 1 contains 30 cards numbered 1 to 30 and Box 2 contains 20 cards numbered 31 to 50. A box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box 1 is

Options:
A 23
B 817
C 417
D 25

The correct answer is:
817

Solution:
PB1=12=PB2

Box B1 have 20 non-prime numbers 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30

⇒PNP/B1=2030

Box B1 have 15 non-prime numbers 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50

⇒PNP/B2=1520

Probability of non-prime=PNP.

PNP=PB1.PNP/B1+PB2.PNP/B2

=12·2030+12·1520

PB1/NP=12·203012·2030+12·1520=817

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Box \( A \) contains \( 2 \) block and \( 3 \) red balls while box \( B \) contains \( 3 \) black and \( 4 \) red balls. Out of these two boxes one is selected
at random and the probability of choosing box \( A \) is double that of box \( B \). If a red ball is drawn from the selected box, then
the probability that it has come from box \( B \) is

Options:
A \( \frac{21}{41} \)
B \( \frac{10}{31} \)
C \( \frac{12}{31} \)
D \( \frac{13}{41} \)

The correct answer is:
\( \frac{10}{31} \)

Solution:
PA=2B⇒ PB=13 PA=23

Required Probability =13·4723·35+13·47=1031.

Topic: Mathematics, Probability, JEE Main
Boyle's law is applicable for an

Options:
A
adiabatic process

B
isothermal process

C
isobaric process
(
D
isochoric process

The correct answer is:
isothermal process

Solution:
Boyle's law is applicable at constant temperature and temperature remains constant in isothermal process.
For ideal gas, $p V=n R T=$ constant
So, $p V=$ constant (at constant temperature)
$$
p \propto \frac{1}{V}
$$
So, this process can be called as isothermal process.

Topic: Physics, Kinetic Theory of Gases
$\mathrm{Br}^{-}$is converted into $\mathrm{Br}_2$ by using

Options:
A $\mathrm{Cl}_2$
B Conc. $\mathrm{HCl}$
C $\mathrm{HBr}$
D $\mathrm{H}_2 \mathrm{~S}$

The correct answer is:
$\mathrm{Cl}_2$

Solution:
$\mathrm{Cl}_2+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_2$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Bredig's arc method cannot be used to prepare colloidal solution of which of the following?

Options:
A Pt
B \( \mathrm{Fe} \)
C \( \mathrm{Ag} \)
D \( \mathrm{Au} \)

The correct answer is:
\( \mathrm{Fe} \)

Solution:
Bredig's arc method cannot be used to prepare colloidal solution of iron(Fe) because at high temperature water react with iron and forms steam. This method can be used to prepare the colloidal solutions of metals such as copper, silver gold and platinum.

Topic: Chemistry, Surface Chemistry, JEE Main
Briefly describe the valence bond theory of covalent bond formation by taking an example of hydrogen. How can you interpret energy changes taking place in the formation of dihydrogen?

Options:

None

Solution:
The valence bond theory was put forward by Heitler and London in 1927. It was later improved and developed by L. Pauling and J.C. Slater in 1931. The valence bond theory is based on the knowledge of atomic orbitals and electronic configurations of elements, overlap criteria of atomic orbitals and stability of molecule.
The main points of valence bond theory are
(i) Atoms do not lose their identity even after the formation of the molecule.
(ii) The bond is formed due to the interaction of only the valence electrons as the two atoms come close to each other. The inner electrons do not participate in the bond formation.
(iii) During the formation of bond, only the valence electrons from each bonded atom lose their identity. The other electrons remain unaffected.
(iv) The stability of bond is accounted by the fact that the formation of bond is accompanied by release of energy. The molecule has minimum energy at a certain distance between the atoms known as internuclear distance. Larger the decrease in energy, stronger will be the bond formed.

Topic: Chemistry, Chemical Bonding and Molecular Structure
Broadcasting antennas are generally

Options:
A
vertical type

B
both vertical and horizontal type

C
omni directional type

D
horizontal type

The correct answer is:
vertical type

Solution:
Broadcasting antennas are generally vertical type.

Topic: Physics, Communication System, JEE Main, JEE Main 2012 (07 May Online)
Bromination of cyclohexene under conditions given below yields:

Options:
A
B
C
D

The correct answer is:

Solution:
In presence of u.v. light allylic C-H- bond undergoes bromination, i.e., free radical mechanism.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2016 (10 Apr Online)
Bromination of PhCOMe in acetic medium produces mainly

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, WBJEE, WBJEE 2013
Bromine in excess is dropped to a 0.01âM SO2. All of SO2 is oxidized to H2SO4 and the excess Br2 is removed by flushing with gaseous N2. Determine the pH of the resulting solution assuming Ka1 of H2SO4 very large & K a2 = 10 â2 . Take the value of log (3.24) = 0.51. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
HSO 4 − 0.01-x aq ⇌ H + 0.03+x aq + SO 4 2+ x ;  Ka 2 = 10 −2
Ka 2 = 10 −2 = x 0.03+x 0.01−x
x=2.36× 10 −3 =0.00236
H + Total =0.03+0.00236
=0.03236=3.24+0.00236
pH=−log3.240× 10 −2
=2−log3.24=1.49

Topic: Chemistry, Ionic Equilibrium, JEE Main
Bromine in excess is dropped to a 0.01âM SO2. All of SO2 is oxidized to H2SO4 and the excess Br2 is removed by flushing with gaseous N2. Determine the pH of the resulting solution assuming Ka1 of H2SO4 very large & K a2 = 10 â2 . Take the value of log (3.24) = 0.51.

Options:

The correct answer is:
1.49

Solution:
HSO 4 − 0.01-x aq ⇌ H + 0.03+x aq + SO 4 2+ x ;  Ka 2 = 10 −2
Ka 2 = 10 −2 = x 0.03+x 0.01−x
x=2.36× 10 −3 =0.00236
H + Total =0.03+0.00236
=0.03236=3.24+0.00236
pH=−log3.240× 10 −2
=2−log3.24=1.49

Topic: Chemistry, Ionic Equilibrium, JEE Main
Bromine is liberated when an aqueous solution of potassium bromide is treated with

Options:
A $\mathrm{Cl}_2$
B $I_2$
C Dilute $\mathrm{H}_2 \mathrm{SO}_4$
D $\mathrm{SO}_2$

The correct answer is:
$\mathrm{Cl}_2$

Solution:
$\mathrm{Cl}_2+2 \mathrm{KBr} \rightarrow 2 \mathrm{KCl}+\mathrm{Br}_2$
A more electronegative halogen can displaces less electronegative halogen

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Buffer solutions have constant acidity and alkalinity because

Options:
A these give unionised acid or base on reaction with added acid or alkali
B acids and alkalies in these solutions are shielded from attack by other ions
C they have large excess of $\mathrm{H}^{+}$or $\mathrm{OH}^{-}$ions
D they have fixed value of $\mathrm{pH}$

The correct answer is:
these give unionised acid or base on reaction with added acid or alkali

Solution:
If small amount of an acid or alkali is added to a buffer solution, it converts them into unionised acid or base. Thus, its $\mathrm{pH}$ remains unaffected or in other words its acidity/alkalinity remains constant. e.g.,
$$
\begin{aligned}
& \mathrm{H}_3 \mathrm{O}^{+}+A^{-} \rightleftharpoons \mathrm{H}_2 \mathrm{O}+\mathrm{H} A \\
& { }^{-} \mathrm{OH}+\mathrm{H} A \longrightarrow \mathrm{H}_2 \mathrm{O}+A^{-}
\end{aligned}
$$
If acid is added, it reacts with $A^{-}$to form undissociated HA. Similarly, if base/alkali is added, $\mathrm{OH}^{-}$combines with $\mathrm{H} A$ to give $\mathrm{H}_2 \mathrm{O}$ and $A^{-}$and thus, maintains the acidity/ alkalinity of buffer solution.

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2012 (Screening)
Bulk of carbon dioxide $\left(\mathrm{CO}_2\right)$ released from body tissues into the blood is present as

Options:
A bicarbonate in blood plasma and RBCs
B free $\mathrm{CO}_2$ in blood plasma
C $70 \%$ carbamino-haemoglobin and $30 \%$ as bicarbonate
D carbamino-haemoglobin in RBCs

The correct answer is:
bicarbonate in blood plasma and RBCs

Solution:
$70-75 \%$ of $\mathrm{CO}_2$ is transported as sodium bicarbonate $\left(\mathrm{NaHCO}_3\right)$ by plasma, and as potassium bicarbonate $\left(\mathrm{KHCO}_3\right)$ by RBCs.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2011 (Mains)
Buna-S can be represented as:

Options:
A
B
C
D

The correct answer is:

Solution:
Buna-S is a copolymer of 1,3-butadiene and styrene. It is prepared by copolymerisation of 1,3 butadiene and styrene along with sodium. In this process peroxide is used as a catalyst at 5∘C therefore the formed product is also known as cold rubber.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Buna-N synthetic rubber is a copolymer of :

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Polymers, JEE Main, JEE Main 2009
Bundle of His is a network of:

Options:
A muscle fibres distributed throughout the heart walls
B muscle fibres found only in the ventricle wall
C nerve fibres distributed in ventricles
D nerve fibres found throughout the heart

The correct answer is:
muscle fibres found only in the ventricle wall

Solution:
Bundle of His is a network of muscle fibres, found in the walls of ventricles. They originate from AV node and further give rise to Purkinje fibres. This helps in maintaining the rhythmic contraction of the ventricles.
Related Theory
Bundle of His is a collection of specialized heart muscle cells for electrical conduction while Purkinje fibres are thin filaments originating from the left posterior fascicle and left anterior fascicles that help in distributing impulses to the ventricular muscles.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2003
But-2-ene on reaction with alkaline KMnO4 at elevated temperature followed by acidification will give:

Options:
A 2 molecules of
B
C 2 Molecules of
D One molecular of and one molecular of

The correct answer is:
2 molecules of

Solution:
The reaction will be;

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Butane can be prepared from ethyl chloride using the following synthesis

The reaction is

Options:
A Wurtz reaction
B Kolbe's electrolytic method
C Sandmayer's reaction
D Corey-House synthesis

The correct answer is:
Corey-House synthesis

Solution:
Here an alkyl halide is first converted to lithium dialkyl cuperate, R 2 CuLi , which on reaction with alkyl halide gives an alkane. This method is also suitable for the preparation of unsymmetrical alkane of the type R − R ′ , where R ≠ R ′ .

Topic: Chemistry, Hydrocarbons, JEE Main
Butene- 1 may be converted to butane by reaction with

Options:
A
$\mathrm{Sn}-\mathrm{HCI}$

B
$\mathrm{Zn}-\mathrm{Hg}$

C
$\mathrm{Pd} / \mathrm{H}_2$

D
$\mathrm{Zn}-\mathrm{HCI}$

The correct answer is:
$\mathrm{Pd} / \mathrm{H}_2$

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Butene- 1 may be converted to butane by reaction with

Options:
A
$\mathrm{Sn}-\mathrm{HCI}$

B
$\mathrm{Zn}-\mathrm{Hg}$

C
$\mathrm{Pd} / \mathrm{H}_2$

D
$\mathrm{Zn}-\mathrm{HCI}$

The correct answer is:
$\mathrm{Pd} / \mathrm{H}_2$

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2003
∝ -butylene when subjected to hydroboration oxidation reaction, yields

Options:
A iso-butyl alcohol
B sec-butyl alcohol
C n -butyl alcohol
D tert-butyl alcohol

The correct answer is:
n -butyl alcohol

Solution:
CH2=CH-CH2-CH3+H2O→iB2H6,THFHO-CH2CH2CH2CH3n-Butylalcohol

Topic: Chemistry, Hydrocarbons, JEE Main
By a cell a current of $0.9 \mathrm{~A}$ flows through $2 \mathrm{ohm}$ resistor and $0.3 \mathrm{~A}$ through $7 \mathrm{ohm}$ resistor. The internal resistance of the cell is

Options:
A $0.5 \Omega$
B $1.0 \Omega$
C $1.2 \Omega$
D $2.0 \Omega$

The correct answer is:
$0.5 \Omega$

Solution:
$\begin{aligned} & 0.9(2+r)=0.3(7+r) \Rightarrow 6+3 r=7+r \Rightarrow r \\ & =0.5 \Omega\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
By adding a strong acid to the buffer solution, the $p H$ of the buffer solution

Options:
A Remain constant
B Increases
C Decreases
D Becomes zero

The correct answer is:
Remain constant

Solution:
The $\mathrm{pH}$ of buffer solution never changed.

Topic: Chemistry, Ionic Equilibrium, JEE Main
By adding inert gas at a constant volume, which of the following equilibrium will not be affected?

Options:
A H2(g)+I2(g)⇄2Hl(g)
B 3H2(g)+N2(g)⇄2NH3(g)
C PCl5(g)⇄PCl3(g)+Cl2(g)
D All the above

The correct answer is:
All the above

Solution:
Adding inert gas at constant volume does not affect any equilibrium whether Δn≥0 or Δn≤0

Topic: Chemistry, Chemical Equilibrium, JEE Main
By adding soluble impurity in a liquid, angle of contact

Options:
A decreases
B increases
C remains unchanged
D first increases and then decreases

The correct answer is:
decreases

Solution:
\begin{array}{|l|l|l|} \hline Std.12 & Ch-2 & Subtopic-2.4 \end{array}

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2023 (14 May Shift 1)
By corpuscular theory of light, the phenomenon which can be explained is

Options:
A Refraction
B Interference
C Diffraction
D Polarisation

The correct answer is:
Refraction

Solution:
Corpuscular theory explains refraction of light.

Topic: Physics, Wave Optics, JEE Main
By diluting a weak electrolyte , specific conductivity K c and equivalent conductivity (λc) change as-

Options:
A Both increase
B K c increases, λc decreases
C K c decreases, λc increases
D Both decrease

The correct answer is:
K c decreases, λc increases

Solution:
When the solution of a weak electrolyte is diluted, the volume of the solution increases, hence equivalent conductivity (Î»c) increases. However, during this process, the number of current carrying particles per cm2 decreases, hence specific conductivity K c decreases.

Topic: Chemistry, Electrochemistry, JEE Main
By distilling glycol with fuming sulphuric acid, which of following is obtained

Options:
A Glycerol
B Pinacol
C Dioxan
D Ethylene oxide

The correct answer is:
Dioxan

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecules in a container of fixed volume is increased from $5 \times 10^4 \mathrm{~cm} / \mathrm{s}$ to $10 \times 10^4 \mathrm{~cm} / \mathrm{s}$ ?

Options:
A
Two

B
Three

C
Six

D
Four

The correct answer is:
Four

Solution:
r.m.s. velocity $\mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}}$
i.e., $\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}$
$\frac{5 \times 10^4}{10 \times 10^4}=\frac{1}{2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}$
$\therefore \mathrm{T}_2=4 \mathrm{T}_1$

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2013 (09 Apr Online)
By increasing the temperature, the specific resistance of a conductor and a semiconductor

Options:
A
increases for both

B
decreases for both

C
increases, decreases

D
decreases, increases

The correct answer is:
increases, decreases

Solution:
Use $\mathrm{R}_{\mathrm{t}}=\mathrm{R}_0\left(\frac{\mathrm{T}}{273}\right)$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2002
By inserting a capillary tube upto a depth $l$ in water, the water rises to a height $h$. If the lower end of the capillary is closed inside water and the capillary is taken out and closed end opened, to what height the water will remain in the tube

Options:
A Zero
B $I+h$
C $2 h$
D $h$

The correct answer is:
$h$

Solution:
The water rises to height $h$ due to capilarity

Topic: Physics, Mechanical Properties of Fluids, JEE Main
By keeping moment of inertia of a body constant, if we double the time period, then angular momentum of body

Options:
A Remains constant
B Becomes half
C Doubles
D quadruples

The correct answer is:
Becomes half

Solution:
Angular of the body is given by
L=Iω
or L=I×2πT or L∝1T
⇒ L1L2=T2T1
LL2=2TT
(As, T2=2T)
So, L2=L2
Thus, on doubling the time period, angular momentum of body becomes half.

Topic: Physics, Rotational Motion, NEET
By passing excess $\mathrm{Cl}_{2}(\mathrm{g})$ in boiling toluene, which one of the following compounds is exclusively formed?

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
By Simpson's rule, the value of $\int_{1}^{2} \frac{d x}{x}$ dividing the interval (1, 2) into four equal parts, is

Options:
A $0.6932$
B $0.6753$
C $0.6692$
D $7.1324$

The correct answer is:
$0.6932$

Solution:
$\quad \begin{aligned} h=\frac{2-1}{4} &=\frac{1}{4} \\ \text { Now, } x_{0} &=1, x_{1}=1+\frac{1}{4}, x_{2}=1+2 \times \frac{1}{4}, \\ x_{3} &=1+3 \times \frac{1}{4}, x_{4}=1+4 \times \frac{1}{4} \end{aligned}$
$$
\text { ie, } \begin{aligned}
x_{0} &=1, x_{1}=1.25, x_{2}=1.5, x_{3} \\
&=1.75, x_{4}=2 \\
\Rightarrow \quad y_{0} &=1, y_{1}=0.8, y_{2}=0.667, y_{3} \\
&=0.571, y_{4}=0.5
\end{aligned}
$$
$\therefore$ Using Simpson's $\frac{1}{3}$ rd rule
$$
\begin{aligned}
\int_{1}^{2} \frac{d x}{x}=& \frac{1}{12}[(1+0.5)+4(0.8+0.571)\\
&+2(0.667)] \\
=& \frac{1}{12}[1.5+5.484+1.334] \\
=& \frac{1}{12}[8.318]=0.6932
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main