datavorous/entrance-exam-dataset huggingface dataset:
$\left[\sin \left(\tan ^{-1} \frac{3}{4}\right)\right]^{2}+\left[\sin \left(\tan ^{-1} \frac{4}{3}\right)\right]^{2}=$

Options:
A 5
B 1
C -1
D 0

The correct answer is:
1

Solution:
(D)
Let $\tan ^{-1} \frac{3}{4}=\theta \Rightarrow \tan \theta=\frac{3}{4} \Rightarrow \cot \theta=\frac{4}{3}$
$\therefore \operatorname{cosec}^{2} \theta=1+\frac{16}{9}=\frac{25}{9} \Rightarrow \operatorname{cosec} \theta=\frac{5}{3} \Rightarrow \sin \theta=\frac{3}{5}$
$\therefore \sin \left(\tan ^{-1} \frac{3}{4}\right)=\sin \left(\sin ^{-1} \frac{3}{5}\right)=\frac{3}{5}$
Let $\tan ^{-1} \frac{4}{3}=\phi \Rightarrow \tan \phi=\frac{4}{3} \Rightarrow \cot \phi=\frac{3}{4}$
$\therefore \operatorname{cosec}^{2} \phi=1+\frac{9}{16}=\frac{25}{16} \Rightarrow \operatorname{cosec} \phi=\frac{5}{4} \Rightarrow \sin \phi=\frac{4}{5}$
$\therefore \sin \left(\tan ^{-1} \frac{4}{3}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=\frac{4}{5}$
Hence given expression $=\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}=\frac{9+16}{25}=1$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
sintan-145+tan-143+tan-119-tan-117=

Options:
A 12
B 12
C 32
D 1

The correct answer is:
1

Solution:
sintan-145+tan-143+tan-119-tan-117

=sintan-145-tan-117+tan-143+tan-119

Now, tan-145-tan-117=psay

⇒p=tan-145-171+45·17=tan-12339

Also, tan-143+tan-119=qsay

⇒q=tan-143+191-43·19=tan-13923

⇒q=cot-12339=π2-tan-12339

So, sinp+q=sinπ2=1

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\sinh ^{-1} 2+\sinh ^{-1} 3=x \Rightarrow \cosh x$ is equal to

Options:
A $\frac{1}{2}(3 \sqrt{5}+2 \sqrt{10})$
B $\frac{1}{2}(3 \sqrt{5}-2 \sqrt{10})$
C $\frac{1}{2}(12+2 \sqrt{50})$
D $\frac{1}{2}(12-2 \sqrt{50})$

The correct answer is:
$\frac{1}{2}(12+2 \sqrt{50})$

Solution:
$\begin{aligned} & \text { Given, } \sinh ^{-1} 2+\sinh ^{-1} 3=x \\ & \Rightarrow \quad \cosh \left(\sinh ^{-1} 2+\sinh ^{-1} 3\right)=\cosh x \\ & \Rightarrow \quad \cosh \left(\sinh ^{-1} 2\right) \cosh \left(\sinh ^{-1} 3\right) \\ & +\sinh \left(\sinh ^{-1} 2\right) \sinh \left(\sinh ^{-1} 3\right) \\ & =\cosh x \\ & \Rightarrow \cosh x=\cosh \left(\cosh ^{-1} \sqrt{1+2^2}\right) \times \\ & \cosh \left(\cosh ^{-1} \sqrt{1+3^2}\right)+2 \times 3 \\ & \Rightarrow \cosh x=(\sqrt{5} \sqrt{10}+6) \times \frac{2}{2} \\ & =\frac{1}{2}(12+2 \sqrt{50}) \\ & \end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
Sinusoidal carrier voltage of frequency $1.5 \mathrm{MHz}$ and amplitude $50 \mathrm{~V}$ is amplitude modulated by sinusoidal voltage of frequency $10 \mathrm{kHz}$ producing $50 \%$ modulation. The lower and upper side-band frequencies in $\mathrm{kHz}$ are

Options:
A 1490,1510
B 1510,1490
C $\frac{1}{1490}, \frac{1}{1510}$
D $\frac{1}{1510}, \frac{1}{1490}$

The correct answer is:
1490,1510

Solution:
Here $f_{c}=1.5 \mathrm{MHz}=1500 \mathrm{kHz}, f_{m}=10 \mathrm{kHz}$
$\therefore$ Lower side-band frequency
$=f_{c}-f_{m}=1500 \mathrm{kHz}-10 \mathrm{kHz}=1490 \mathrm{kHz}$
Upper side-band frequency
$=f_{c}+f_{m}=1500 \mathrm{kHz}+10 \mathrm{kHz}=1510 \mathrm{kHz}$

Topic: Physics, Thermodynamics, JEE Main
Six consecutive sides of an equiangular octagon are 6, 9, 8, 7, 10, 5 in that order. The integer nearest to the sum of the remaining two sides is

Options:
A 17
B 18
C 19
D 20

The correct answer is:
18

Solution:
Let ABCDEFGH be the equiangular octagon as shown PQ=SR

⇒y2+6+92=52+10+72

⇒y=3+42

Also: PS=QR

⇒y2+x+52=92+8+72

⇒x=4+42

∴x+y=7+82=18.313

∴ Nearest integer =1818$

Topic: Mathematics, Properties of Triangles, KVPY, KVPY 2020 (SB/SX)
Six distinct numbers are chosen from the first 10 natural numbers. The probability that 6 is the third largest of those chosen numbers, is

Options:
A 27
B 521
C 1063
D 1663

The correct answer is:
27

Solution:
Total cases = C610
Favourable cases = C24·1C1·5C3
Required probability = C24·1C1·5C3C610=27

Topic: Mathematics, Probability, JEE Main
Six equal resistances are connected between points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ as shown in figure. Then net resistance will be maximum between :

Options:
A
$\mathrm{P}$ and $\mathrm{R}$

B
P and Q

C
Q and R

D
Any two points

The correct answer is:
P and Q

Solution:
Resistance between $\mathrm{P}$ and $\mathrm{Q}$
$$
r_{\mathrm{PQ}}=r_{11}\left(\frac{r}{3}+\frac{r}{2}\right)=\frac{r \times \frac{5}{6} r}{r+\frac{5}{6} r}=\frac{5}{11} r
$$
Resistance between Q and R
$$
\mathrm{r}_{\mathrm{QR}}=\frac{\mathrm{r}}{2} 11\left(\mathrm{r}+\frac{\mathrm{r}}{3}\right)=\frac{\frac{\mathrm{r}}{2} \times \frac{4}{3} \mathrm{r}}{\frac{\mathrm{r}}{2}+\frac{4}{3} \mathrm{r}}=\frac{4}{11} \mathrm{r}
$$
Resistance between $\mathrm{P}$ and $\mathrm{R}$
$$
\mathrm{r}_{\mathrm{PR}}=\frac{\mathrm{r}}{3} \mathrm{l}\left(\frac{\mathrm{r}}{2}+\mathrm{r}\right)=\frac{\frac{\mathrm{r}}{3} \times \frac{3}{2} \mathrm{r}}{\frac{\mathrm{r}}{3}+\frac{3}{2} \mathrm{r}}=\frac{3}{11} \mathrm{r}
$$
Hence, it is clear that $\mathrm{r}_{\mathrm{PQ}}$ is maximum

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (25 Apr Online)
Six fair dice are rolled. The probability that the product of the numbers appearing on top faces is prime, is

Options:
A 12164
B 126
C 164
D 12165

The correct answer is:
12164

Solution:
This can be possible if five of the dice have 1 on top faces and one die has 2, 3 or 5.
Hence, required probability =6C516536
=616512
=12164

Topic: Mathematics, Probability, JEE Main
Six identical conducting rods are joined as shown in figure. Points A and D are maintained at $200^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$ respectively. The temperature of junction B will be

Options:
A $120^{\circ} \mathrm{C}$
B $100^{\circ} \mathrm{C}$
C $140^{\circ} \mathrm{C}$
D $80^{\circ} \mathrm{C}$

The correct answer is:
$140^{\circ} \mathrm{C}$

Solution:
The equivalent electrical circuit of the arrangement is shown in figure.

Temperature difference between the end points $\mathrm{A}$ and $\mathrm{D}=200-20=180^{\circ} \mathrm{C}$ As the resistances for the three parts are equal, the temperature difference must be distribuited equally in the three parts $(=180 /$ ) $3=60^{\circ} \mathrm{C}$ )
$\therefore$ Temperature of $\mathrm{B}=200^{\circ} \mathrm{C}-60^{\circ}=140^{\circ} \mathrm{C}$.

Topic: Physics, Thermal Properties of Matter, VITEEE, VITEEE 2017
Six identical conducting rods are joined as shown in the figure. Points A and D are maintained at 200 Â°C and 20 Â°C respectively. The temperature of junction B will be -

Options:
A 120 °C
B 100 °C
C 140 °C
D 80 °C

The correct answer is:
140 °C

Solution:
Electrical circuit,
dqdt1=dQdt2
ka(200-θ)l=kA(θ-20)2l
400-2θ=θ-20 ;
θ=4203=140 °C

Topic: Physics, Thermal Properties of Matter, NEET
Six identical conducting rods are joined as shown in the figure, Points A and D are maintained at 200 Â°C and 20 Â°C respectively. The temperature of the junction B will be

Options:
A 120 °C
B 100 °C
C 140 °C
D 80 °C

The correct answer is:
140 °C

Solution:
The equivalent electrical circuit of the arrangement is shown in the figure.
The temperature difference between the endpoints A and D=200−20=180 °C
As the resistances for the three parts are equal, the temperature difference must be distributed equally in the three parts (=180/3=60 °C)
∴ Temperature of B=200 °C−60 °C =140 °C.

Topic: Physics, Thermal Properties of Matter, NEET
Six married couples are sitting in a room. The number of ways in which four persons can be selected, such that there is exactly one married couple among the four, is

Options:
A 276
B 600
C 840
D 240

The correct answer is:
240

Solution:
Select one couple in C61 ways.
Remaining 5 couples have 5 men and 5 women.
Select two men or two women or one man and one women so that they are not couple.
Two men can be selected in C25 ways
Two women can be selected in C25 ways
One man and one woman such that they are not a couple can be selected in C15ÃC14 ways
Hence, the total number of ways =C61C52+C52+C51C41=610+10+20=240

Topic: Mathematics, Permutation Combination, JEE Main
Six molecules of an ideal gas have velocities \(1,3,5,5,6\) and \(5 \mathrm{~m} / \mathrm{s}\) respectively. At any given temperature, if \(\overline{V}\) and \(V_{\mathrm{rms}}\) represent average and rms speed of the molecules, then

Options:
A \(\overline{V}=5 \mathrm{~m} / \mathrm{s}\)
B \(V_{r m s} > \overline{V}\)
C \(V_{\text {rms }}^2 < \overline{V}^2\)
D \(V_{r m s}=\overline{V}\)

The correct answer is:
\(V_{r m s} > \overline{V}\)

Solution:
Hint : \(\overline{V}=\frac{1+3+5+5+6+5}{6}=\frac{25}{6}=4.16\)
\(\begin{aligned}
& \overline{V}_{\mathrm{ms}}=\sqrt{\frac{1+9+25 \times 3+36}{6}}=\sqrt{\frac{121}{6}}=\frac{11}{\sqrt{6}}=4.48 \\
& \mathrm{~V}_{\mathrm{rms}} > \overline{V}
\end{aligned}\)

Topic: Physics, Kinetic Theory of Gases, JEE Main
Six molecules speed \( 2 \) unit, \( 5 \) unit, \( 3 \) unit, \( 6 \) unit, \( 3 \) unit and \( 5 \) unit respectively. The rms speed is

Options:
A \( 4 \) unit
B \( 1.7 \) unit
C \( 4.2 \) unit
D \( 5 \) unit

The correct answer is:
\( 4.2 \) unit

Solution:
We have,

vrms=v12+v22+...+vn2n

=4+25+9+36+9+256

=1086=18

=32=3×1.414=4.242 unit.

Topic: Physics, Kinetic Theory of Gases, JEE Main
Six persons A, B, C, D, E and F are to be seated at a circular table facing towards the centre. Then the number of ways that can be done if A must have either E or F on his immediate right and E must have either F or D on his immediate right, is

Options:
A 18
B 30
C 12
D 24

The correct answer is:
18

Solution:
There are two cases,
I is to the immediate right of A.

Number of ways E can fill=3 ways

E cannot take sit on the left of A as there will be no space.

Number of ways D can fill=1 ways

The remaining seats can be filled in 2 ways.

The total numbers of ways are 3×1×2=6

The required number of ways are 12+6=18

Topic: Mathematics, Permutation Combination, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
Six text books numbered $1,2,3,4,5$ and 6 are arranged at random. What is the probability that the text books 2 and 3 will occupy consecutive places ?

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{4}$
D $\frac{1}{6}$

The correct answer is:
$\frac{1}{3}$

Solution:
Total number of possible arrangements $\mathrm{n}(\mathrm{s})=6 !$. Since 2 and 3 occupy consecutive places, so, they are grouped together. So, there will be $5 !$ such arrangements. But 2 and 3 can be arranged in themselves in $2 !$ ways
Required Probability $=\frac{5 ! \times 2 !}{6 !}=\frac{2}{6}=\frac{1}{3}$

Topic: Mathematics, Probability, JEE Main
Six vectors $\vec{a}$ through $\vec{f}$ have the magnitudes and directions indicated in the figure. Which of the following statements is true?

Options:
A $\vec{b}+\vec{c}=\vec{f}$
B $\vec{d}+\vec{c}=\vec{f}$
C $\vec{d}+\vec{e}=\vec{f}$
D $\vec{b}+\vec{e}=\vec{f}$

The correct answer is:
$\vec{d}+\vec{e}=\vec{f}$

Solution:
If two non-zero vectors are represented by the two adjacent sides of a parallelogram, then the resultant is given by the diagonal of the parallelogram passing through the point of intersection of the two vectors
$$
\therefore \quad \overrightarrow{\mathrm{d}}+\overrightarrow{\mathrm{e}}=\overrightarrow{\mathrm{f}}
$$

Topic: Physics, Mathematics in Physics, JEE Main
Sketch the curve for $\quad y=\left.|| x\right|^2-2|x|-3 \mid$.

Options:
A

B

C

D none of these

The correct answer is:

Solution:
As we know the graph for $y=x^2-2 x-3$, shown as;
(i) $y=x^2-2 x-3 \rightarrow y=|x|^2-2|x|-3$.
(ii) $y=|x|^2-2|x|-3 \rightarrow y=\left.|| x\right|^2-2|x|-3 \mid$
Clearly, above figure is differentiable for all $x \in R-\{-3,0,3\}$.

Topic: Mathematics, Functions, JEE Main
Sketch the curve $|x|+|y|=1$.

Options:
A

B

C

D

The correct answer is:

Solution:
As the graph for $y=1-x$ is;
(i) $y=1-x \longrightarrow y=1-|x|$.
(ii) $y=1-|x| \longrightarrow|y|=1-|x|$.
Clearly above figure represents a square.

Topic: Mathematics, Functions, JEE Main
Sketch the graph for $\quad y=\left|e^{-|x|}-\frac{1}{2}\right|$.

Options:
A

B

C

D

The correct answer is:

Solution:
As we know the graph for $y=e^{-x}$.
(i) $y=e^{-x} \longrightarrow y=e^{-x}-\frac{1}{2}$.
(ii) $y=e^{-x}-\frac{1}{2} \longrightarrow y=e^{-|x|}-\frac{1}{2}$.
(iii) $y=\left|e^{-|x|}-\frac{1}{2}\right|$

Topic: Mathematics, Functions, JEE Main
Sky looks blue due to

Options:
A Dispersion effect
B Reflection
C Transmission
D Scattering

The correct answer is:
Scattering

Solution:
The sky looks blue due to the scattering of light by the molecules of air as blue light is scattered more than the red light due to its shorter wavelength.

Topic: Chemistry, Surface Chemistry, JEE Main
Slope of a line passing through P(2, 3) and intersecting the line x+y=7 at a distance of 4 units from P, is

Options:
A 7-17+1
B 1–71+7
C 5-15+1
D 1–51+5

The correct answer is:
1–71+7

Solution:
Given, the point P2, 3 on the line x+y=7.

Let, the point of intersection of the two lines be Q

Then, Q can be taken as Q≡(α, 7-α)

Given, PQ=4 units and the distance between the points x1, y1 & x2, y2 is x1-x22+y1-y22

⇒(α-2)2+7-α-32=4

⇒(α-2)2+4-α2=4

⇒(α-2)2+4-α2=42

⇒α2-4α+4+16-8α+α2=16

⇒2α2-12α+4=0

⇒α2-6α+2=0

⇒α=--6±-62-4×1×22×2

⇒α=6±36-84

⇒α=6±274

⇒α=3±7

The slope of a line joining the points x1, y1 & x2, y2 is y2-y1x2-x1.

If we take α=3-7, then Q≡3-7, 4+7

And, the slope of PQ=m=4+7-33-7-2=1+71-7

If we take α=3+7, then Q≡3+7, 4-7

And, the slope of PQ=m=4-7-33+7-2=1-71+7.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (09 Apr Shift 1)
'Smack' is a drug obtained from the

Options:
A
latex of Papaver somniferum

B
leaves of Cannabis sativa

C
flowers of Dhatura pinata

D
fruits of Erythroxyl coca

The correct answer is:
latex of Papaver somniferum

Solution:
Heroin, chemically called as diacetyl morphine that is a white, odourless, bitter, crystalline compound. It is obtained by acetylation of morphine and extracted from the latex of poppy plant (Papaver somniferum), but Leaves of Cannabis sativa commonly called bhang produces connabinoids. Active chemical of Dhatura flower is tropane alkaloids mainly scopolamines, hyoscyamine and atropine.

Topic: Biology, Human Health and Diseases
Small water droplets of radius $0.01 \mathrm{~mm}$ are formed in the upper atmosphere and falling with a terminal velocity of $10 \mathrm{~cm} / \mathrm{s}$. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be _____$\mathrm{cm} / \mathrm{s}$.

Options:

The correct answer is:
40

Solution:
$\mathrm{m}=$ mass of small drop
$\mathrm{M}=$ mass of bigger drop
$\begin{aligned} & V_t=\frac{2}{9} \frac{R^2(\rho-\sigma) g}{\eta} \\ & 8 \propto m=M \\ & 8 r^3=R^3 \Rightarrow R=2 R\end{aligned}$
as $\mathrm{V}_{\mathrm{t}} \times \mathrm{R}^2 \quad \because$ Radius double so $\mathrm{V}_{\mathrm{t}}$ becomes 4 time
$\therefore 4 \times 10=40 \mathrm{~cm} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Smallest division on the main scale of given Vernier calipers is 0.5 mm . Vernier scale has 25 divisions and these coincide with 24 main scale divisions. The least count of Vernier callipers is

Options:
A 0.001 cm
B 0.002 cm
C 0.01 cm
D 0.02 cm

The correct answer is:
0.002 cm

Solution:
25âVSD=24âMSD
1 VSD=2425 MSD .......(i)
The least count of Vernier calipers is LC=1âMSD-1âVSD
=1 MSD-2425 MSD=125 MSD (using (i))
As 1âMSD=0.5âmm (given)
â´ LC=125(0.5âmm)=0.02âmm=0.002âcm

Topic: Physics, Experimental Physics, JEE Main
Smoke is an example of:

Options:
A
Solid dispersed in solid

B
Gas dispersed in liquid

C
Solid dispersed in gas

D
Gas dispersed in solid

The correct answer is:
Solid dispersed in gas

Solution:
Smoke is an example of solid particles dispersed in gas.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2013 (23 Apr Online)
$\mathrm{Sn}^{++}$loses two electrons in a reaction. What will be the oxidation number of tin after the reaction

Options:
A $+2$
B Zero
C $+4$
D $-2$

The correct answer is:
$+4$

Solution:
$\operatorname{Sn}^{2+} \Rightarrow \operatorname{Sn}^{4+} +2 e$

Topic: Chemistry, Redox Reactions, JEE Main
$\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{~S} \longrightarrow 3 \mathrm{~S}+2 \mathrm{H}_{2} \mathrm{O}$. This equation represents preparation of sulphur sol by

Options:
A hydrolysis
B oxidation
C reduction
D double decomposition

The correct answer is:
oxidation

Solution:
The reaction, $\mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{~S} \stackrel{\text { Oxidation }}{\longrightarrow} 3 \mathrm{~S}$ (sol.) $+2 \mathrm{H}_{2} \mathrm{O}$, represent the preparation of sulphur sol by oxidation.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
SO2Cl2 on reaction with excess of water results into acidic mixture SO2Cl2+2H2O→H2SO4+2HCl 16 moles of NaOH is required for the complete neutralisation of the resultant acidic mixture. The number of moles of SO2Cl2 used is

Options:
A 16
B 8
C 4
D 2

The correct answer is:
4

Solution:
SO2Cl2+2H2O→2HCl+H2SO4

Let x moles of SO2Cl2 are required,

∴ nH2SO4=x,nHCl=2x

⇒nH+=4x

Moles of H+ formed =4x

Moles of OH- from 16 moles of NaOH=16

4x=16 (for complete neutralisation)

x=4

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (25 Jul Shift 1)
SO 2 is considered as an air pollutant because

Options:
A Its concentration increases with temperature increase of atmosphere.
B It is used as an insecticide which is air pollutant
C It reacts with O 2 and H 2 O to produce acid rain
D It is a strong oxidant and oxidizes other components of atmosphere

The correct answer is:
It reacts with O 2 and H 2 O to produce acid rain

Solution:
SO 2 is considered as air pollutant because it reacts with O 2 and H 2 O to produce acid rain. Oxides from nitrogen and sulphur, released into the atmosphere from thermal power plants, industries and automobiles are the main sources of acid rain. These oxides on oxidation followed by hydrolysis give sulphuric acid and nitric acid that get dissolved in rain water and are responsible for the acidity of rain.

Topic: Chemistry, Environmental Chemistry, JEE Main
Soaps are

Options:
A sodium and potassium salts of higher fatty acids
B sodium salts of sulphonic acids
C high molecular mass alcohols
D More than one of the above

The correct answer is:
sodium and potassium salts of higher fatty acids

Solution:
Soaps are water-soluble sodium or potassium salts of fatty acids. Soaps are made from fats and oils, or their fatty acids, by treating them chemically with a strong alkali.
$\mathrm{C}_{12} \mathrm{H}_{35} \mathrm{COONa}$ is a good example of soap.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Soaps do not work in hard water containing calcium and magnesium ions because

Options:
A Ca2+ and Mg2+ ions form insoluble calcium and magnesium salts in the form of scum
B Na+ and K+ present in soap react with Ca2+ and Mg+ and hinder cleansing process
C a large amount of soap is to be used in presence of Ca+ and Mg+
D scum formed by combination of Na+ , Ca2+ and Mg2+ stick to the cloth and are not removed on agitation

The correct answer is:
Ca2+ and Mg2+ ions form insoluble calcium and magnesium salts in the form of scum

Solution:
Ca2+ and Mg2+ ions form insoluble calcium and magnesium soaps respectively when sodium or potassium soaps are dissolved in hard water.
2C17H35COONasoap+CaCl2→2NaCl+(C17H35COO)2CaInsoluble calciumstearate soap
These insoluble soaps separate as scum in water and are useless as cleansing agent.

Topic: Chemistry, s Block Elements, JEE Main
Sodium acetate can be converted to ethane by

Options:
A Heating with $\mathrm{LiAlH}_4$
B Electrolysing its aqueous solution
C Heating with soda lime
D Heating with calcium acetate

The correct answer is:
Electrolysing its aqueous solution

Solution:
$\underset{\text { Sodiumacetate }}{2 \mathrm{CH}_3 \mathrm{COONa}+2 \mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Electolysis }}}$ $\mathrm{CH}_3-\mathrm{CH}_3+2 \mathrm{CO}_2+2 \mathrm{NaOH}+\mathrm{H}_2$

Topic: Chemistry, Hydrocarbons, JEE Main
Sodium and copper have work functions $2.3 \mathrm{eV}$ and $4.5 \mathrm{eV}$ respectively. Then the ratio of the wave lengths is nearest to

Options:
A
1 : 2

B
4 : 1

C
2 : 1

D
1 : 4

The correct answer is:
2 : 1

Solution:
$\mathrm{hc} / \lambda_0=\mathrm{W}_0 ; \frac{\left(\lambda_0\right)_1}{\left(\lambda_0\right)_2}=\frac{\left(\mathrm{W}_0\right)_2}{\left(\mathrm{~W}_0\right)_1}=\frac{4.5}{2.3}=2: 1$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2002
Sodium and potassium react with water much more vigorously than lithium because:

Options:
A Sodium and potassium have high values of hydration energy as compared to that of lithium.
B Sodium and potassium have higher melting point than that of lithium.
C Sodium and potassium have lower melting point than that of lithium.
D Sodium and potassium have lower hydration energy than that of lithium.

The correct answer is:
Sodium and potassium have lower melting point than that of lithium.

Solution:
When sodium and potassium react with water, the heat evolved causes them to melt, giving a larger area of contact with water, lithium on the other hand, does not melt under these conditions and thus reacts more slowly.

Topic: Chemistry, s Block Elements, JEE Main
Sodium benzoate âÎNaOH+CaOXâ310-320KCl2,FeCl3YâÎconc.âH2SO4Z(Major Product). Gram molecular weight of the major product(in grams) is

Options:

The correct answer is:
192.5

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Sodium borate on reaction with conc. $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ gives a compound $\mathrm{A}$, which burns with a green edged flame. The compound A is:
A. $\mathrm{H}_2 \mathrm{~B}_4 \mathrm{O}_7$
B. $\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{~B}_4 \mathrm{O}_7$
C. $\mathrm{H}_3 \mathrm{BO}_3$
D. $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}\right)_3 \mathrm{~B}$

Options:
A $\left(\mathrm{C}_2 \mathrm{H}_5\right)_2 \mathrm{~B}_4 \mathrm{O}_7$
B $\mathrm{H}_3 \mathrm{BO}_3$
C $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}\right)_3 \mathrm{~B}$
D $\mathrm{H}_2 \mathrm{~B}_4 \mathrm{O}_7$

The correct answer is:
$\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}\right)_3 \mathrm{~B}$

Solution:
When borate is heated with ethanol and conc. $\mathrm{H}_2 \mathrm{SO}_4$, triethyl borate is produced and these vapours will burn with a green edged flame on ignition.
$2 \mathrm{Na}_3 \mathrm{BO}_3+3 \mathrm{H}_2 \mathrm{SO}_4$ (conc.) $) \rightarrow 3 \mathrm{Na}_2 \mathrm{SO}_4(\mathrm{aq})+2 \mathrm{H}_3 \mathrm{~B}$
$\mathrm{H}_3 \mathrm{BO}_3+3 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightarrow \mathrm{BO}_3\left(\mathrm{C}_2 \mathrm{H}_5\right)_3+3 \mathrm{H}_2 \mathrm{O}$
Triethyl Borate
Thus, the green edged flame is due to Triethyl Borate.

Topic: Chemistry, Practical Chemistry, JEE Main
Sodium Carbonate cannot be used in place of $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$ for the identification of $\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}$ and $\mathrm{Sr}^{2+}$ ions (in group $\mathrm{V}$ ) during mixture analysis because :

Options:
A
$\mathrm{Mg}^{2+}$ ions will also be precipitated.

B
Concentration of $\mathrm{CO}_3{ }^{2-}$ ions is very low.

C
Sodium ions will react with acid radicals.

D
$\mathrm{Na}^{+}$ions will interfere with the detection of $\mathrm{Ca}^{2+}, \mathrm{Ba}^{2+}, \mathrm{Sr}^{2+}$ ions.

The correct answer is:
$\mathrm{Mg}^{2+}$ ions will also be precipitated.

Solution:
If $\mathrm{Na}_2 \mathrm{CO}_3$ is used in place of $\left(\mathrm{NH}_4\right)_2 \mathrm{CO}_3$. It will precipitate group $\mathrm{V}$ radicals as well as magnesium radicals. The reason for this is the high ionization of $\mathrm{Na}_2 \mathrm{CO}_3$ in water into $\mathrm{Na}^{+}$and $\mathrm{CO}_3^{--}$. Now the higher concentration of $\mathrm{CO}_3^{--}$is available which exceeds the solubility product of group $\mathrm{V}$ radicals as well as that of magnesium radicals.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2013 (09 Apr Online)