datavorous/entrance-exam-dataset huggingface dataset:
On adding few drops of dil \( \mathrm{HCl} \) to freshly precipitated ferric hydroxide, a red coloured solution is obtained. This
phenomenon is known as

Options:
A Peptization
B Dialysis
C Protective action
D Dissolution

The correct answer is:
Peptization

Solution:
The phenomenon of converting a freshly precipitated mass into colloidal state by the action of solute or solvent is known as peptization.

In biochemistry, dialysis is the process of separating molecules in solution by the difference in their rates of diffusion through a semi-permeable membrane, such as dialysis tubing.

Dissolution is the process by which a solute forms a solution in a solvent. The solute, in the case of solids, has its crystalline structure disintegrated as separate ions, atoms and molecules form.

Hence, option A is the correct answer.

Topic: Chemistry, Surface Chemistry, JEE Main
On addition of $1 \mathrm{~mL}$ of $10 \% \mathrm{NaCl}$ solution to 10 $\mathrm{mL}$ gold sol in the presence of $0.025 \mathrm{~g}$ of starch, the coagulation is just prevented. Starch has gold number

Options:
A
$2.5$

B
25

C
$0.25$

D
$0.025$

The correct answer is:
25

Solution:
Gold Number is the minimum amount of lyophilic colloide in miligrams (starch in given case) which prevents the flocculation of $10 \mathrm{ml}$ gold sol by the addition of $1 \mathrm{ml}$ of $10 \% \mathrm{NaCl}$ solution.
$\therefore$ Gold number $=$ miligram of starch $=0.025 \times 10^3 \mathrm{mg}$ $=25$

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2012 (19 May Online)
On addition of mineral acid to an aqueous solution of borax, the compound that is formed is:

Options:
A Pyroboric acid
B Boron hydride
C Meta boric acid
D Orthoboric acid

The correct answer is:
Orthoboric acid

Solution:
Na2B4O7+H2SO4+5H2O→Na2SO4+4H3BO3
Addition of mineral acid to an aqueous solution of borax, orthoboric acid is formed.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
On addition of one $\mathrm{ml}$ solution of $10 \% \mathrm{NaCl}$ to $10 \mathrm{~ml}$ gold sol in the presence of $0.25 \mathrm{gm}$ of starch, the coagulation is just prevented. Starch has the following gold number

Options:
A 0.025
B 0.25
C 0.5
D 250

The correct answer is:
250

Solution:
By definition gold number of starch is the amount of starch in $\mathrm{mg}$ added to 10 $\mathrm{ml}$ standard gold sol which prevents the coagulation of gold on adding $1 \mathrm{~ml}$ of $10 \% \mathrm{NaCl}$ solution. So the amount of starch is $0.25 \mathrm{~g}=250 \mathrm{mg}$. Hence gold number is 250.

Topic: Chemistry, Surface Chemistry, JEE Main
On an imaginary linear scale of temperature (called ' $\mathrm{W}$ ' scale) the freezing and boiling points of water are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$ respectively. The temperature on the new scale corresponding to $39^{\circ} \mathrm{C}$ temperature on Celsius scale will be

Options:
A $139^{\circ} \mathrm{W}$
B $78^{\circ} \mathrm{W}$
C $117^{\circ} \mathrm{W}$
D $200^{\circ} \mathrm{W}$

The correct answer is:
$117^{\circ} \mathrm{W}$

Solution:
In Celsius scale the freezing and boiling point are $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$. In the given imaginary scale the freezing and boiling points are $39^{\circ} \mathrm{W}$ and $239^{\circ} \mathrm{W}$. Hence, we can write
$$
\frac{\mathrm{C}-0}{100}=\frac{\mathrm{W}-39}{200}
$$
For $\mathrm{C}=39^{\circ}, \frac{39}{100}=\frac{\mathrm{W}-39}{200}$
Solving, $\mathrm{W}=117$

Topic: Physics, Thermal Properties of Matter, JEE Main
On an \(n \times n\) chessboard, the total number of rectangles which are not squares is 350 . Then, the number of white squares on the chessboard is .......

Options:
A 32
B 50
C 18
D 72

The correct answer is:
18

Solution:
Number of rectangles in a chess board
(including squares) \(=\Sigma n^3\)
\(=\frac{n^2(n+1)^2}{4}\)
Number of squares in a chess board
\(=\Sigma n^2=\frac{n(n+1)(2 n+1)}{6}\)
So, Number of rectangles which are not squares
\(\begin{aligned}
&=\Sigma n^3-\Sigma n^2 \\
&=\frac{n^2(n+1)^2}{4}-n \frac{(n+1)(2 n+1)}{6}=350 \\
& \Rightarrow \quad n\left(\frac{n+1}{2}\right)\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)=350
\end{aligned}\)
\(\begin{aligned}
\Rightarrow & & n\left(\frac{n+1}{2}\right)\left(\frac{3 n^2+3 n-4 n-2}{6}\right) & =350 \\
\Rightarrow & & n(n+1)\left(3 n^2-n-2\right) & =350 \times 6 \times 2 \\
\Rightarrow & & n(n+1)\left(3 n^2-3 n+2 n-2\right) & =350 \times 6 \times 2 \\
\Rightarrow & & n(n+1)(3 n(n-1)+2(n-1) & =350 \times 6 \times 2 \\
\Rightarrow & & n(n+1)(n-1)(3 n+2) & =350 \times 6 \times 2 \\
\Rightarrow & & (n-1) n(n+1)(3 n+2) & =5 \times 6 \times 7 \times 20
\end{aligned}\)
By comparison we get
\(\therefore \quad n=6\)
Hence, total number of black and white squares
\(=n \times n=6 \times 6=36\)
Half of there are
\(\text {white }=\frac{36}{2}=18 \text { squares. }\)

Topic: Mathematics, Permutation Combination, JEE Main
On application of pressure certain crystals can produce electric signals. This phenomenon is known as

Options:
A Ferrielectricity
B Ferroelectricity
C Peizoelectricity
D Pyroelectricity

The correct answer is:
Peizoelectricity

Solution:
When polar crystal is subjected to a mechanical stress, electricity is produced - a case of piezoelectricity. Reversely, if electric field is applied, mechanical stress is developed. Piezoelectric crystal acts as a mechanical electrical transductor.

Topic: Chemistry, Solid State, JEE Main
On applying an external magnetic field, to a ferromagnetic substance domains

Options:
A align in the direction of magnetic field
B align in the direction opposite to magnetic field
C remain unaffected
D None of the above

The correct answer is:
align in the direction of magnetic field

Solution:
On applying magnetic field, domains of ferromagnetic substances align themselves in the direction of magnetic field.

Topic: Physics, Magnetic Properties of Matter, JEE Main
On applying an external magnetic field, to a ferromagnetic substance domains

Options:
A align in the direction of magnetic field
B align in the direction opposite to magnetic field
C remain unaffected
D None of the above

The correct answer is:
align in the direction of magnetic field

Solution:
On applying magnetic field, domains of ferromagnetic substances align themselves in the direction of magnetic field.

Topic: Physics, Magnetic Properties of Matter, MHT CET, MHT CET 2009
On balancing the given redox reaction

aCr2O72-+bSO32-aq+cH+aq→2aCr3+aq+bSO42-aq+c2H2Ol

The coefficients a,b,c are found to be respectively

Options:
A 8,1,3
B 1,3,8
C 3,8,1
D 1,8,3

The correct answer is:
1,3,8

Solution:
Reduction Half reaction : Cr2O72−+6e-→2Cr3+

Oxidation Half reaction :

SO32−→SO42−+2e-×3

Overall reaction :

Cr2O72−+3SO32−→2Cr3++3SO42-

To balance ‘O’ atoms, adding H2O on LHS

Cr2O72−+3SO32−→2Cr3++3SO42-+4H2O

To balance ‘H’ atoms, adding H+ on RHS

Cr2O72−+3SO32−+8H+→2Cr3++3SO42-+4H2O

Therefore, a=1, b=3, c=8.

Topic: Chemistry, Redox Reactions, JEE Main
On breaking a cubic solid (edge = 1 m) into fine cubic particles of edge 1 μm , surface becomes n×100000 . The value of n is?

Options:

The correct answer is:
10.00

Solution:
For a cube of dimension (1mÃ1mÃ1m) , the surface area
=(1Ã1)Ã6 faces
=6m2Ã1Î¼m10-6m2
=6Ã1012(Î¼m)2
The volume = 1  m 3  ×  1μm 10 −6 m 3
=1018(Î¼m)3
For 1018 cubes of dimension (1Î¼mÃ1Î¼mÃ1Î¼m) , the total surface area
=1018Ã1Î¼mÃ1Î¼mÃ6 faces
=6Ã1018(Î¼m)2
On breaking, the surface area becomes
=6Ã1018(Î¼m)26Ã1012(Î¼m)2 times
= 1000000 times
Hence n = 10

Topic: Chemistry, Solid State, JEE Main
On celcius scale the temperature of body increases by $40^{\circ} \mathrm{C}$. The increase in temperature on Fahrenheit scale is :

Options:
A $68^{\circ} \mathrm{F}$
B $75^{\circ} \mathrm{F}$
C $72^{\circ} \mathrm{F}$
D $70^{\circ} \mathrm{F}$

The correct answer is:
$72^{\circ} \mathrm{F}$

Solution:
We know that per ${ }^{\circ} \mathrm{C}$ change is equivalent to $1.8^{\circ}$ change in ${ }^{\circ} \mathrm{F}$.
$\therefore 40^{\circ}$ change on celcius scale will corresponds to $72^{\circ}$ change on Fahrenheit scale.
Hence option (3)

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2024 (04 Apr Shift 1)
On combustion of Li, Na and K in excess of air, the major oxides formed, respectively, are:

Options:
A Li2O2, Na2O and K2O2
B Li2O, Na2O2 and KO2
C Li2O, Na2O and K2O2
D Li2O, Na2O2 and K2O

The correct answer is:
Li2O, Na2O2 and KO2

Solution:
On heating in excess air Li form oxide sodium form peroxide while K, Rb, Cs form superoxide.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2020 (04 Sep Shift 1)
On complete combustion 0.30 g of an organic compound gave 0.20 g of carbon dioxide and 0.10 g of water. The percentage of carbon in the given organic compound is____(Nearest Integer)

Options:

The correct answer is:
18

Solution:
CxHyOz+x+y4-z2O2→xCO2+y2H2O0.3 g 0.2 g 0.1 g

CO2nH2On=xy2=0.2440.118=911

x=9y22

Now, CxHyO2nCO2n=1x

⇒0.312x+y+16z×440.2=1x

66x = 12 x + y + 16 z 54x = y + 16 z

54×9y22 – y = 16 z

z = 29y22

CxHyOz =C9y22HyO29y22= C9H22O29

% of C = 12×912 ×9 +22 +29× 16×100 = 18.18%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (26 Jun Shift 1)
On complete combustion, 0.492 g of an organic compound gave 0.792 g of CO2. The % of carbon in the organic compound is (Nearest integer)

Options:

The correct answer is:
44

Solution:
44 grams of CO2 contain 12 grams of carbon.

Hence, one gram of carbon dioxide contain 1244 gram of carbon.

0.792 gram of carbon dioxide contain 1244 × 0.792 gram of carbon

Percentage of carbon = Mass of carbonMass of organic compound × 100

= 1244 × 0.7920.492 ×100

= 43.90%

So, answer is 44.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (31 Jan Shift 1)
On complete combustion, 0.500 g of an organic compound gave 0.150 of carbon dioxide. Determine the percentage composition of carbon in the compound.

Options:
A \(8.18 \%\)
B \(81.8 \%\)
C \(0.81 \%\)
D \(81 \%\)

The correct answer is:
\(8.18 \%\)

Solution:
The required equation is:
Percentage of carbon \(=12 * \mathrm{~m} 2 * 100 / 44 * \mathrm{~m}\)
In this case, \(m=0.500\)
\(\mathrm{m} 2=0.150\)
Substituting in the equation \(\rightarrow\) Percentage of carbon \(=12 * 0.150 * 100 / 44 * 0.500\)
\(=8.18 \%\)
Therefore, the percentage of carbon present in the organic compound is \(8.18 \%\).

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
On complete combustion of 0.492 g of an organic compound containing C,H and O,0.7938 g of CO2 and 0.4428 g of H2O was produced. The % composition of oxygen in the compound is____. (Nearest Integer)

Options:

The correct answer is:
46

Solution:
Organic compound +O2⟶ CO2g + H2O

0.7938 gram 0.4428 gram

Weight of the organic compound=0.492 gram

Number of moles of CO2=Number of moles of C

=0.793844Weight of carbon=0.79384412=0.216g

Weight of hydrogen=0.4428182=0.0491g

Weight of oxygen=0.492-0.216-0.0491=0.2269g

% of O=0.22490.492×100=46.11≈46

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (28 Jul Shift 1)
On complete hydrogenation, natural rubber produces

Options:
A ethylene-propylene copolymer
B vulcanised rubber
C polypropylene
D polybutylene

The correct answer is:
ethylene-propylene copolymer

Solution:
nCH2=C-CH| CH3 =CH2→(-CH2-C=CH| -CH3 CH2-)nNatural rubber→H2/Pt-CH2-CH-| CH3CH2-CH2-n

Topic: Chemistry, Polymers, JEE Main
On complete reaction of FeCl3 with oxalic acid in aqueous solution containing KOH, resulted in the formation of product A. The secondary valency of Fe in the product A is ______ .

(Round off to the Nearest Integer).

Options:

The correct answer is:
6

Solution:
Fe3++3 K++3C2O42-→K3FeC2O43(A)

Secondary valency of Fe in 'A' is 6 .

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (17 Mar Shift 2)
On differentiating tan â1 [ 1+ x 2 â1 x ] with respect to x, the result would be

Options:
A

1 2 . 1 1+ x 2

B

1 1+ x 2

C

2 1+ x 2

D

1 2 . 1 1+2x

The correct answer is:

1 2 . 1 1+ x 2

Solution:
Let us assume that,

y= tan â1 [ 1+ x 2 â1 x ]

Put : x=tanÎ¸

â´âââââ1+x2â1x=1+tan2Î¸â1tanÎ¸=secÎ¸â1tanÎ¸=1âcosÎ¸sinÎ¸=2sin2Î¸22sinÎ¸2cosÎ¸2=tanÎ¸2

On differentiating with respect to x:

â´ââây= tan â1 ( tan Î¸ 2 )= 1 2 Î¸= 1 2 tan â1 x ââââ dy dx = 1 2 . 1 1+ x 2 .

Topic: Mathematics, Continuity and Differentiability, JEE Main
On dissolving moderate amount of sodium metal in liquid $\mathrm{NH}_3$ at low temperature, which one of the following does not occur?

Options:
A Blue coloured solution is obtained.
B $\mathrm{Na}^{+}$ions are formed in the solution.
C Liquid $\mathrm{NH}_3$ becomes good conductor of electricity.
D Liquid ammonia remains diamagnetic.

The correct answer is:
Liquid ammonia remains diamagnetic.

Solution:
The alkali metals dissolve in liquid ammonia without evolution of hydrogen. The colour of the dilute solution is blue and are paramagnetic in nature. $M \rightarrow M^{+}$(in liquid ammonia) $+e^{-}$(ammoniated) $M+(x+y) \mathrm{NH}_3 \rightarrow\left[M\left(\mathrm{NH}_3\right)_x\right]^{+}+e^{-}\left(\mathrm{NH}_3\right)_y$
solvated electron

Topic: Chemistry, s Block Elements, AIIMS, AIIMS 2003
On dissolving moderate amount of sodium metal in liquid $\mathrm{NH}_3$ at low temperature, which one of the following does not occur

Options:
A Blue coloured solution is obtained
B $\mathrm{Na}^{+}$ions are formed in the solution
C Liquid $N \mathrm{NH}_3$ becomes good conductor of electricity
D Liquid ammonia remains diamagnetic

The correct answer is:
Liquid ammonia remains diamagnetic

Solution:
Due to free electron liquid ammonia becomes paramagnetic.

Topic: Chemistry, s Block Elements, JEE Main
On electrolysis of water, a total of 1 mole of gases is evolved. The amount of water decomposed is

Options:
A 1 mol
B 2 mol
C 13 mol
D 23 mol

The correct answer is:
23 mol

Solution:
∵ Only one mole of total gases are evolved and H2O decomposes as

H2O→decompositionH2+12O2

i.e. total moles =1+12=32 moles by one mole of H2O

Now,

∵ 32 moles of gases are evolved by =1 mole of H2O

∴ 1 mole of gases are evolved by =13/2=23

Hence, 1 mole of total gases are evolved by 23 mole of H2O.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
On going from right to left in a period in the periodic table the electronegativity of the elements

Options:
A Increases
B Decreases
C Remain unchanged
D Decreases first then increases

The correct answer is:
Decreases

Solution:
Decrease as atomic size increases.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
On going left to right in a period, in transition metals, their atomic volumes

Options:
A Decrease
B Increase
C Remain same
D None of these of correct

The correct answer is:
Decrease

Solution:
Decrease

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
On heating a mixture of NaCl, K2Cr2O7 and conc. H2SO4 which of the following is formed ?

Options:
A CrO2Cl
B CrO2Cl2
C CrOCl2
D NaClO2

The correct answer is:
CrO2Cl2

Solution:
4 NaCl + 3 H 2 SO 4 + K 2 Cr 2 O 7 → CrO 2 Cl 2 Chromyl Chloride (deep red colour vapour)

Topic: Chemistry, Practical Chemistry, JEE Main
On heating a mixture of one mole each \( \mathrm{Li}_{2} \mathrm{CO}_{3}, \mathrm{MgCO}_{3} \) and \( \mathrm{Na}_{2} \mathrm{CO}_{3} \), the volume of \( \mathrm{CO}_{2} \) evolved under \( \mathrm{STP} \) condition
will be

Options:
A \( 67.2 \mathrm{~L} \)
B \( 33.6 \mathrm{~L} \)
C \( 44.8 \mathrm{~L} \)
D \( 22.4 \mathrm{~L} \)

The correct answer is:
\( 44.8 \mathrm{~L} \)

Solution:
Li2CO3â and MgCO3â are thermally unstable and they decompose on heating. On the other hand, sodium carbonate is thermally stable, which does not decompose on heating.

Li2CO3→Li2O+CO2 1 mole

MgCO3→MgO+CO2 1 mole

Na2CO3 does not decompose.

So, the total of 2 moles of CO2 gas is evolved by heating the given mixture.

The volume of 1 mole of CO2 gas at STP=22.4 L

The volume of 2 moles of CO2 gas =2×22.4 L=44.8 L

Topic: Chemistry, s Block Elements, JEE Main
On heating a mixture of sodium benzoate and sodalime, the following is obtained

Options:
A Toluene
B Phenol
C Benzene
D Benzoic acid

The correct answer is:
Benzene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
On heating acetamide in presence of $\mathrm{P}_2 \mathrm{O}_5$, which of the following is formed?

Options:
A Ammonium acetate
B Acetonitrile
C $\mathrm{NH}_3$
D Methylamines

The correct answer is:
Acetonitrile

Solution:
Explanation: On heating acetamide in presence of $\mathrm{P}_2 \mathrm{O}_5$, acetonitrile is formed.

Topic: Chemistry, Amines, JEE Main
On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is :

Options:
A An alkanol
B An alkanediol
C An alkyl cyanide
D An alkyl isocyanide

The correct answer is:
An alkyl isocyanide

Solution:
Reaction of aliphatic primary amine with chloroform and enhanolic potassium hydroxide leads to the formation of isocyanides/ carbylamines. This reaction is called carbylamine reaction.

R - NH 2 → KOH(alc) CHCl 3 R - N ≡ C
This is the confirmatory test for primary amines. A foul smelling isocyanide is formed.
R≠−CH3

Topic: Chemistry, Amines, JEE Main, JEE Main 2014 (06 Apr)
On heating aqueous solution of benzene diazonium chloride, which is formed

Options:
A Benzene
B Chlorobenzene
C Phenol
D Aniline

The correct answer is:
Phenol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
On heating $\mathrm{CHCl}_3$ with aq. $\mathrm{NaOH}$, the product is

Options:
A $\mathrm{CH}_3 \mathrm{COONa}$
B $\mathrm{HCOONa}$
C Sodium oxalate
D $\mathrm{CH}_3 \mathrm{OH}$

The correct answer is:
$\mathrm{HCOONa}$

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
On heating compound A gives a gas B which is constituent of air. This gas when treated with H2 in the presence of a catalyst gives another gas C which is basic in nature. A should not be :

Options:
A NaN3
B PbNO32
C NH42Cr2O7
D NH4NO2

The correct answer is:
PbNO32

Solution:
PbNO32⟶ΔPbO+2NO2

N2 gas is constituent of air and it produces ammonia gas (basic nature) after reacting with H2 gas.

NH4NO2, (NH4)2Cr2O7 and NaN3 give nitrogen gas on heating but lead nitrate doesn't give nitrogen gas so it will be answer.

NH4NO2⟶ΔN2+2H2O

NH42Cr2O7⟶ΔN2+Cr2O3+H2O

NaN3⟶300°C3N2+2Na

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2020 (02 Sep Shift 1)
On heating glucose with Fehling's solution we get a precipitate whose colour is

Options:
A Yellow
B Red
C Black
D White

The correct answer is:
Red

Solution:
Glucose + Fehling solution $\rightarrow$ Gluconic acid $+\underset{\text { (Red ppt) }}{\mathrm{Cu}_2 \mathrm{O}}$

Topic: Chemistry, Biomolecules, JEE Main
On heating, LiNO3 gives how many compounds among the following?
Li2O,N2,O2,LiNO2,NO2

Options:

The correct answer is:
3

Solution:
Decomposition of Lithium Nitrate occurs via the following reaction:
2LiNO3⟶ΔLi2O+2NO2+12O2

Hence three products Li2O,NO2 and O2

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (29 Jan Shift 2)
On heating potassium permanganate, one of the following compound is not obtained:

Options:
A \( \mathrm{O}_{2} \)
B \( \mathrm{MnO} \)
C \( \mathrm{MnO}_{2} \)
D \( \mathrm{K}_{2} \mathrm{MnO}_{4} \)

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

Solution:
Heating of potassium permanganate salt at \( 513 \mathrm{~K} \)
\( 2 \mathrm{KMnO}_{4}^{-}->\mathrm{K}_{2} \mathrm{MnO}_{4}+\mathrm{MnO}_{2}+\mathrm{O}_{2} \)
Hence, on heating of potassium permanganate, MnO will not be obtain.

Topic: Chemistry, d and f Block Elements, JEE Main
On heating Potassium permanganate the product is obtained is/are: -

Options:
A Manganese dioxide
B Potassium manganite
C Oxygen
D All of these

The correct answer is:
All of these

Solution:
2KMnO4→ΔK2MnO4+MnO2+O2

Topic: Chemistry, d and f Block Elements, JEE Main
On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detatch is :
(density of water is ρ w )

Options:
A R 2 2Ï w g 3 T â
B R 2 ρ w g 6 T
C R 2 ρ w g T
D R 2 3 ρ w g T

The correct answer is:
R 2 2Ï w g 3 T â

Solution:
When the bubble gets detached,
Buoyant force = force due to surface tension
∫ T × dl sin θ = 4 3 π R 3 ρ w g
⇒ T × 2 π r × r R = 4 3 π R 3 ρ w g
⇒ r 2 = 2 R 4 ρ w g 3T

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (06 Apr)
On heating water, bubbles beings formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius \( R \) and making a circular contact of radius \( r \) with the bottom of the vessel. If \( r< < R \) and the surface tension of water is \( T \), value of \( r \) just before bubbles detach is (density of water is \( \rho \) )

Options:
A \( R^{2} \sqrt{\frac{2 \rho_{w} g}{3 T}} \)
B \( R^{2} \sqrt{\frac{\rho_{w} g}{6 T}} \)
C \( R^{2} \sqrt{\frac{\rho_{w} g}{T}} \)
D \( R^{2} \sqrt{\frac{3 \rho_{w} g}{T}} \)

The correct answer is:
\( R^{2} \sqrt{\frac{2 \rho_{w} g}{3 T}} \)

Solution:
Forces acting on the bubble are:

a Buoyant force in upward direction

B=ρwVg

⇒B=ρw43πR3g …1,

where, ρw=density of water.

b Force due to surface tension.

If we consider an elemental length dl at bottom contact, then force due to surface tension is dF=Tdl along the tangent to surface & perpendicular to dl.

Clearly dFcosθ components on two diametrically elements cancel out & dFsinθ components add up.

∴ Net force due to surface tensions on the bubble

F=∫dFsinθ=∫02πrTdlrR

⇒F=TrR2πr=2πTr2R …2

When the bubble is on the verge of detaching then

B=F

⇒ρw43πR3g=2πTr2R

⇒r=R223ρwgT

Topic: Physics, Mechanical Properties of Fluids, JEE Main
On hydrolysis $\mathrm{Me}_2 \mathrm{SiCl}_2$ will produce

Options:
A $(\mathrm{Me})_2 \mathrm{Si}(\mathrm{OH})_2$
B $(\mathrm{Me})_2 \mathrm{Si}=\mathrm{O}$
C $-\left[-\mathrm{O}-(\mathrm{Me})_2 \mathrm{Si}-\mathrm{O}-\right]_n-$
D $\mathrm{Me}_2 \mathrm{SiCl}(\mathrm{OH})$

The correct answer is:
$-\left[-\mathrm{O}-(\mathrm{Me})_2 \mathrm{Si}-\mathrm{O}-\right]_n-$

Solution:
The hydrolysis of dimethyldichlorosilane involves the replacement of chlorine atoms with hydroxyl groups from water, resulting in the formation of dimethyldihydroxysilane.

Topic: Chemistry, Coordination Compounds, JEE Main
On increasing the reverse bias to a large value in a P-N junction diode, current

Options:
A increase slowly.
B remains fixed.
C suddenly increases.
D decreases slowly.

The correct answer is:
suddenly increases.

Solution:
When the reverse voltage increases beyond the breakdown voltage, the current suddenly increases.

Topic: Physics, Semiconductors, MHT CET, MHT CET 2023 (13 May Shift 1)
On increasing the temperature of a conductor, its resistance increases because the

Options:
A relaxation time increases
B electron density decreases
C relaxation time decreases
D relaxation time remains constant

The correct answer is:
relaxation time decreases

Solution:
Resistance of a conductor, $R=\frac{m}{n e^2 \tau} \frac{l}{A}$ where the symbols have their usual meaning. As the temperature increases, the relaxation time $\tau$ decreases.

Topic: Physics, Current Electricity, AIIMS, AIIMS 2016
On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ . How much was the resistance on the left slot before interchanging the resistances?

Options:
A 910Â Î©
B 990Â Î©
C 505Â Î©
D 550 Î©

The correct answer is:
550 Î©

Solution:
ab=l100-l
a+b=1000
b a = l−10 110−l
100−l l = l−10 110−l
l 2 −10l=11000+ l 2
−100 l=110 l
200l=11000
l= 110 2 =55
55 45 = a b
⇒a=550 Ω     b=450 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (08 Apr)
on mercuration-demercuration produces the major product:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2014 (12 Apr Online)