datavorous/entrance-exam-dataset huggingface dataset:
Unpolarised light from air incidents on the surface of a transparent medium of refractive index 1.414 such that the reflected light is completely polarised. Match the angles given in List-I with the corresponding values given in List-II.
The correct match is
Codes
$\begin{array}{llll}
A & B & C & D
\end{array}$

Options:
A (ii) (iii) (i) (iv)
B (ii) (iii) (iv) (i)
C (iv) (i) (iii) (ii)
D (iv) (iii) (i) (ii)

The correct answer is:
(iv) (iii) (i) (ii)

Solution:
Figure showing the ray diagram of refraction at a plane surface.
Given, $i_p=$ angle of polarisation
So,
$$
i_p=\tan ^{-1} \mu \Rightarrow i_p=\tan ^{-1} \sqrt{2}=\sin ^{-1} \sqrt{\frac{2}{3}}
$$
From Snell's law,
$$
\sin i_p=\mu \sin r
$$
Here,
$$
\mu=1.414=\sqrt{2} \Rightarrow \sin r=\frac{\sin i_p}{\mu}
$$
$$
\begin{aligned}
\sin r & =\frac{1}{\sqrt{2}} \frac{\sqrt{2}}{\sqrt{3}} \\
r & =\sin ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$$
angle of refraction
Now, the angle of deviation,
$$
\begin{aligned}
& \delta=i_p-r \\
& \delta=\sin ^{-1} \sqrt{\frac{2}{3}}-\sin ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$$
Angle of reflection,
$$
\begin{gathered}
\theta=90^{\circ}-r \Rightarrow r=90^{\circ}-\theta \\
\sin ^{-1} \frac{1}{\sqrt{3}}=90^{\circ}-\theta \Rightarrow \frac{1}{\sqrt{3}}=\sin \left(90^{\circ}-\theta\right) \\
\cos \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\cos ^{-1} \frac{1}{\sqrt{3}}
\end{gathered}
$$
Angle between incident and completely polarised light is given by
$$
\begin{aligned}
& \phi=i_p+\theta \\
& \phi=\sin ^{-1} \sqrt{\frac{2}{3}}+\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \sin ^{-1} \sqrt{\frac{2}{3}}
\end{aligned}
$$
Hence, the correct option is (d).

Topic: Physics, Wave Optics, JEE Main
Unpolarised light is incident from air on plane surface of a material of refractive index μ. At a particular angle of incident i it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

Options:
A i=sin-1⁡1μ
B Reflected light is polarized with its electric vector perpendicular to the plane of incidence
C Reflected light is polarized with its electric vector parallel to the plane of incident
D i=tan-1⁡1μ

The correct answer is:
Reflected light is polarized with its electric vector perpendicular to the plane of incidence

Solution:
when reflected light rays and reflected rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.
Also. Brewster's law μ=tani.

Topic: Physics, Wave Optics, JEE Main
Unpolarised light is incident on the boundary between two dielectric media, whose dielectric constants are 2.8 (medium –1) and 6.8 (medium – 2), respectively. To satisfy the condition, so that the reflected and refracted rays are perpendicular to each other, the angle of incidence should be tan-11+10θ12, the value of θ is ______.

(Given for dielectric media, μr=1)

Options:

The correct answer is:
7

Solution:
As given in the question, we can write

μ1=εrμr=2.8×1=2.8 and

μ2=εrμr=6.8×1=6.8

Now applying Snell's law and considering r=90°-i, we get

μ1 sin i=μ2 cos i

⇒tan i=μ2μ1=6.82.8

⇒tan i=2.8+42.812

⇒i=tan-11+10712

Therefore, θ=7.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Unpolarised light of intensity 32 W m−2 passes through the combination of three polaroids such that the pass axis of the last polaroids is perpendicular to that of the pass axis of first polaroids. If intensity of emerging light is 3 W m−2, then the angle between pass axis of first two polaroids is _________°.

Options:

The correct answer is:
30

Solution:
The given data is

I0=32 W m-2

If=3 W m-2

Thus, by substituting the value, we get

If=I02cos2θ×cos290°-θ ...(i)

⇒3=16cos2θ·sin2θ

⇒cosθ·sinθ=34⇒sin2θ=32⇒θ=30°

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Unpolarized light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam

Options:
A cos-123
B cos-132
C cos-113
D cos-112

The correct answer is:
cos-113

Solution:
I'=I2cos2Î¸=I6

â cosÎ¸=13

Topic: Physics, Wave Optics, JEE Main
Unpolarized light of intensity 32 W m-2 passes through three polarizers arranged such that the transmission axes of the first and the last polarizer are at right angles. If the intensity of emerging light is 3 W m-2, then what is the angle (in degree) between the transmission axes of the first two polarizers?

Options:

The correct answer is:
30

Solution:
Since unpolarized light is passing through the first polarizer, hence the intensity of light after crossing the first polarizer will be

I1=12I0=16 W m-2

Let us assume that the angle between the transmission axis of the first and second polarizer is θ, then from Malus law we can find out the intensity of light after it crosses the second polarizer.

I2=I1cos2θ=16cos2θ

Similarly, the intensity of light after crossing the third polarizer is

I3=I2cos290°-θ=16cos2θsin2θ

⇒I3=16cos2θsin2θ=3

⇒4cos2θsin2θ=34

⇒sin22θ=34

θ=30°

Topic: Physics, Wave Optics, JEE Main
Unpolarized light of intensity 32 W m-2 passes through three polarizers such that transmission axes of the first and second polarizer make an angle 30Â° with each other and the transmission axis of the last polarizer is crossed with that of the first. The intensity of the final emerging light will be

Options:
A 32 W m-2
B 3 W m-2
C 8 W m-2
D 4 W m-2

The correct answer is:
3 W m-2

Solution:
I0=32
For 1st polaroid
I1=I02
For 2nd polaroid
I2=I1cos2â¡Î¸
I2=I02Ãcos2â¡30Â°
I2=I02Ã34
I2=38I0
As last crosses with 1st polarizer, hence angle between 2nd & last polariser is 60Â°
I3=I2cos2â¡60Â°
I3=I2Ã14
I3=38I0Ã14
I3=3I032
I3=332Ã32
I3=3 W m-2

Topic: Physics, Wave Optics, JEE Main
Unpolarized light of intensity I0 is incident on surface of a block of glass at Brewster's angle. In that case, which one of the following statements is true?

Options:
A transmitted light is partially polarized with intensity I02
B transmitted light is completely polarized with intensity less than I02
C reflected light is partially polarized with intensity I02
D reflected light is completely polarized with intensity less than I02

The correct answer is:
reflected light is completely polarized with intensity less than I02

Solution:
By polarizations due to reflection at a polarising angle reflected light is completely polarise and the intensity is less than average intensity I02.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2015 (11 Apr Online)
Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I2 . If a third polarizer C is placed between A and B the intensity of emergent light is reduced to I3 . The angle between the polarizers A and C is θ , then

Options:
A cos⁡θ=2314
B cos⁡θ=1314
C cos⁡θ=1312
D cos⁡θ=2312

The correct answer is:
cos⁡θ=2314

Solution:
A and B have same alignment of transmission axis.
Let's assume c is introduced at Î¸ angle
I2cos2â¡Î¸Ãcos2â¡Î¸=I3
cos4â¡Î¸=23
â cosâ¡Î¸=2314

Topic: Physics, Wave Optics, JEE Main
Unpolarized light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I2 . If a third the polarizer C is placed between A and B the intensity of emergent light is reduced to I3 . The angle between the polarizers A and C is θ , then

Options:
A cos⁡θ=2314
B cos⁡θ=1314
C cos⁡θ=1312
D cos⁡θ=2312

The correct answer is:
cos⁡θ=2314

Solution:
A and B have the same alignment of the transmission axis. Let's assume c is introduced at θ angle, ⇒I2cos2θ×cos2θ=I3⇒cos4θ=23
⇒cosθ=2314.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2018 (16 Apr Online)
Unpolarized light of intensity I passes through an ideal polariser A. Another identical polariser B is placed behind A. The intensity of light beyond B is found to be I2. Now another identical polariser C is placed between A and B. The intensity beyond B is now found to be I8 . The angle between polariser A and C is

Options:
A 60°
B 0°
C 30°
D 45°

The correct answer is:
45°

Solution:
Transformation axis A and B are parallel. If the transformation axis of C is making angle θ with A, it will make angle 90−θ with B. Therefore,
⇒I2cos2θcos2(90−θ)=I8⇒14sin22θ=14
⇒2θ=π2⇒θ=π4

Topic: Physics, Wave Optics, JEE Main, JEE Main 2018 (08 Apr)
Upon dipping a copper rod, the aqueous solution of the salt that can turn blue is:-

Options:
A CaNO32
B MgNO32
C ZnNO32
D AgNO3

The correct answer is:
AgNO3

Solution:
2AgNO3+Cu⟶Ag+CuNO32
Metal can reduce that metal cation which is placed below it in reactivity series.

Topic: Chemistry, Electrochemistry, JEE Main
Upon long standing, concentrated HNO3

Options:
A remains colourless, but gives out NO
B turns yellow brown due to formation of NO2
C turns yellow brown due to the formation of N2O4
D remains colourless, but gives N2O

The correct answer is:
turns yellow brown due to formation of NO2

Solution:
When nitric acid is left for long standing, its color changes due to decomposition reaction.
The reaction in which one reactant yields two or more products, it is called a decomposition reaction. Due to decomposition, concentrated nitric acid (HNO3) gets broken down to nitrogen dioxide, oxygen, and water. Due to the formation of nitrogen dioxide, it turns yellow-brown. The chemical reaction involved in the reaction is as follows:
4  HNO 3 →4  NO 2  +  O 2  + 2  H 2 O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Uracil is base present in RNA with the following structure. % of N in uracil is _______.

Options:

The correct answer is:
25

Solution:
The molecular formula of uracil is C4H4 N2O2

The molar mass of uracil = 112 g/mol

Mass % of N=28112×100 = 25 %

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Uranium has two isotopes of masses 235 and 238 wits. If both of them are present in wranium hexafluoride gas, find the percentage ratio of difference in rms velocities of two isotopes to the rms velocity of heavier isotope.

Options:
A 1.64
B 0.064
C 0.64
D 6.4

The correct answer is:
0.64

Solution:
Given,
Mass of first uranium isotope $=238$ units
Mass of second uranium isotope $=235$ units
We know that,
$$
\begin{aligned}
& v_{\text {rms }} & =\sqrt{\frac{\gamma R T}{M}} \\
\Rightarrow & v_{\text {rms }} & \propto \frac{1}{\sqrt{M}}
\end{aligned}
$$
So, the percentage ratio of difference in rms velocity of isotopes is given by
$$
\begin{aligned}
& \frac{V_{\mathrm{rms}_2}-V_{\mathrm{rms}_1}}{V_{\mathrm{rms}_2}} \times 100 \\
& =\frac{\sqrt{M_1}-\sqrt{M_2}}{\sqrt{M_1}} \times 100 \\
& =\frac{\sqrt{238}-\sqrt{235}}{\sqrt{238}} \times 100
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{15.427-15.329}{15.427} \times 100=\frac{0.097}{15.427} \times 100 \\
& =6.34 \times 10^{-3} \times 100=0.634 \approx 0.64
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
Urea (NH2-CO-NH2) needs to be dissolved in 100g of water, in order to decrease the vapour pressure of water by 25%? What will be the molality of the solution?

Options:
A 18.52
B 62.45
C 28.52
D 35.64

The correct answer is:
18.52

Solution:
p0-pSp0=xSolutes
=WSolutesMSoluteWSolutesMSolutes+WSolventMSolvent
P0-PSP0=wAmAÃmBWB (A stands for solute and B for solvent)
x-xÃ75100xÃ75100=w60Ã18100
[Let the vapour pressure of pure water, P0=x ]
0.25x0.75x=w60Ã18100
â´w=111.118
Molality= 111.11 60 Ã 1000 100 =18.52

Topic: Chemistry, Solutions, JEE Main
Urea on slow heating gives

Options:
A $\mathrm{NH}_2 \mathrm{CON} \cdot \mathrm{HNO}_2$
B $\mathrm{NH}_2 \mathrm{CONHCONH}_2$
C $\mathrm{HCNO}$
D $\mathrm{NH}_2 \mathrm{CONH}_2 \cdot \mathrm{HNO}_3$

The correct answer is:
$\mathrm{NH}_2 \mathrm{CONHCONH}_2$

Solution:
$2 \mathrm{NH}_2 \mathrm{CONH}_2 \rightarrow \mathrm{NH}_2 \mathrm{CONHCONH}_2+\mathrm{NH}_3$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Ureters act as urogenital ducts in

Options:
A human males
B human females
C frog's both males and females
D frog's males

The correct answer is:
frog's males

Solution:
In male frog, vasa efferential finally communicates with the urogenital duct that comes out of the kidneys and open into cloaca. The cloaca is a small, median chamber that is used to pass faecal matter, urine and sperms to the exterior.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2011 (Mains)
Uric acid is the chief nitrogenous component of the excretory products of -

Options:
A Frog
B Man
C Earthworm
D Cockroach

The correct answer is:
Cockroach

Solution:
No Solution Available

Topic: Biology, Structural Organisation in Animals, NEET, NEET 2009 (Mains)
Use of anti-histamines and steroids give a quick relief from -

Options:
A Headache
B Allergy
C Nausea
D Cough

The correct answer is:
Allergy

Solution:
No Solution Available

Topic: Biology, Human Health and Diseases, NEET, NEET 2009 (Mains)
Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t=0 it was 1600 counts per second and t=8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t=6 seconds is close to:

Options:
A 400
B 200
C 150
D 360

The correct answer is:
200

Solution:
After 8 seconds count rate changes from A1=1600 per second to A2=100 per second.
A 2 = A 1 16 = A 1 2 4
∴8s in 4 half life
⇒1 half life =2 s
t=6s is 3 half life.
∴ Count rate after 3 second
A=A123=16008=200

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Using column chromatography, mixture of two compounds ‘A’ and ‘B’ was separated. ‘A’ eluted first, this indicates ‘B’ has

Options:
A low Rf, stronger adsorption
B high Rf, weaker adsorption
C high Rf, stronger adsorption
D low Rf, weaker adsorption

The correct answer is:
low Rf, stronger adsorption

Solution:
In chromatography, the retardation factor (Rf) is the fraction of an analyte in the mobile phase of a chromatographic system. In planar chromatography in particular, the retardation factor Rf is defined as the ratio of the distance travelled by the centre of a spot to the distance travelled by the solvent front. The more polar spot travels slower, and the less polar spot travels faster. Rf values, on the other hand, are directly related to the rate of movement. The fastest moving spot has the highest Rf value, least polar (fastest moving), and the spot with the lowest Rf value is the most polar (slowest moving). Hence, the compound B is strongly adsorbed and has low Rf value.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Using dimensional analysis the resistivity in terms of fundamental constants $\mathrm{h}, \mathrm{m}_{\mathrm{e}}, \mathrm{c}, \mathrm{e}, \varepsilon_{0}$ can be expressed as

Options:
A $\frac{\mathrm{h}}{\varepsilon_{0} \mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$
B $\frac{\varepsilon_{0} \mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}{\mathrm{~h}}$
C $\frac{\mathrm{h}^{2}}{\mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$
D $\frac{\mathrm{m}_{\mathrm{e}} \varepsilon_{0}}{\mathrm{ce}^{2}}$

The correct answer is:
$\frac{\mathrm{h}^{2}}{\mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$

Solution:
$\rho=\frac{E}{J}$
$E=\left[M L T^{-3} I^{-1}\right]$
$\mathrm{J}=\left[\mathrm{I} \mathrm{L}^{-2}\right]$
$\mathrm{h}=\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-1}$
$\mathrm{~m}_{\mathrm{e}}=\mathrm{M}$
$\mathrm{c}=\mathrm{LT}^{-1}$
$\varepsilon_{0}=\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{I}^{2}$
$\therefore \rho=\frac{\mathrm{h}^{2}}{\mathrm{~m}_{\mathrm{e}} \mathrm{ce}^{2}}$

Topic: Physics, Units and Dimensions, JEE Main
Using electrolytic method, the cost of production of 5L of oxygen at STP , is Rs X , the cost of production of same volume of hydrogen at STP , will be

Options:
A 2X
B X2
C 8X
D X8

The correct answer is:
X2

Solution:
During electrolysis of water oxygen is produced at anode and hydrogen at cathode according to reaction, 2 H 2 O→2 H 2 + O 2 H 2 and O2 are produced in 2:1

Topic: Chemistry, Electrochemistry, JEE Main
Using equipartition of energy, the specific heat (in J kg-1 K-1 ) of Aluminium at high temperature can be estimated to be (atomic weight of Aluminium =27)

Options:
A 25
B 1850
C 410
D 925

The correct answer is:
925

Solution:
Solid atom possesses 6 degrees of freedom due to vibrations at high temperature.
Cv=12fR=6R2=3R for 1 mole
∴ Specific heat capacity in J kg-1 K-1=CvM
=3RM=3×8.31427×1000
≈925

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2015 (11 Apr Online)
Using MO theory predict which of the following species has the shortest bond length ?

Options:
A
$\mathrm{O}_2^{2+}$

B
$\mathrm{O}_2^{+}$

C
$\mathrm{O}_2^{-}$

D
$\mathrm{O}_2^{2-}$

The correct answer is:
$\mathrm{O}_2^{2+}$

Solution:
Bond length $\alpha \frac{1}{\text { bond order }}$
Bond order $=\frac{\text { no..of bonding } \overline{\mathrm{e}}-\text { no.of antibonding } \overline{\mathrm{e}}}{2}$
Bond orders of $\mathrm{O}_2^{+}, \mathrm{O}_2^{-}, \mathrm{O}_2^{-2}$ and $\mathrm{O}_2^{+2}$ are respectively $2.5,1.5,1$ and 3 .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2009
Using MO theory predict which of the following species has the shortest bond length?

Options:
A O2+
B O2-
C O22-
D O22+

The correct answer is:
O22+

Solution:
Bond length ∝1bond order
Bond length =no. of bonding ē-no. of antibonding ē2
Bond orders of O2+, O2-, O22- and O22+ are respectively 2.5, 1.5, 1 and 3
Hence O22+ is shortest in bond length as have max .bond order.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Using molecular orbital theory predict the correct decreasing bond order of the following species
1.O2+ 2.O2- 3.O22- 4.O22+

Options:
A 1 > 2 > 3 > 4
B 4 > 1 > 2 > 3
C 4 > 1 > 3 > 2
D 1 > 4 > 2 > 3

The correct answer is:
4 > 1 > 2 > 3

Solution:
Bond order =no.of bonding eÂ¯-no. of anti bonding eÂ¯2
Bond orders of O2+,O2-,O22-andO22+ are respectively 2.5, 1.5, 1 and 3
Hence O22+ has max bond order is
O22+>O2+>O2->O22-

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Using monochromatic light of wavelength λ , an experimentalist sets up the Young's double slit experiment in three ways as shown.
If she observes that y⁡ = β ′ , the wavelength of light used is :

Options:
A 520 nm
B 580 nm
C 540 nm
D 560 nm

The correct answer is:
540 nm

Solution:
Î² = Î» D d .
â´ Î² 1 = Î» 2 D d.
Now shift = Î¼ - 1 t D d
y= Î² 1
Î¼ - 1 t D d = Î» . 2 D d
Î» = Î¼ - 1 t 2 = 1.6 - 1 Ã 1.8 2
= 0.6 Ã 1.8 2 Î¼ m
= 0.3 Ã 1.8 Î¼ m
= 0.54 Î¼ m
= 540 nm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2014 (09 Apr Online)
Using screw gauge of pitch 0.1cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as,

Options:
A 2.121cm
B 2.124cm
C 2.125cm
D 2.123cm

The correct answer is:
2.124cm

Solution:
Thickness = M.S. Reading + Circular Scale Reading (L.C.)

Here least count is 0.002 cm per division.

From all the four options, the second answer is divisible by this least count. Hence option 2 is correct.

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Using the data provided, find the value of equilibrium constant for the following reaction at 298 K and 1 atm pressure. NO(g)+12O2( g)⇌NO2( g)
ΔfH0NO(g)=90.4kJ.mol−1

ΔfH0NO2(g)=32.48kJ⋅mol−1

ΔS∘@298K=−70.8J⋅K−1⋅mol−1

antilog⁡(0.50)=3162

Options:
A 3.162×104
B 3.162×10-4
C 3.162×106
D 3.162×107

The correct answer is:
3.162×106

Solution:
The given reaction is

NO(g)+12O2( g)⇌NO2( g)

The enthalpy of the above reaction given by the difference of enthalpies of reactant side and product side.

Therefore,

∆cH = ∆fHNO2-∆fHNO+12∆fHO2∆cH = 32.8-90.4+120∆cH = -57.6 KJmol-1

Also, we know that Gibb's free energy can be determined by

∆G = ∆H-T∆S∆G = 57600 Jmol-1-298×-70.8 JK-1mol-1∆G =- 36501.5 Jmol-1

Again, Using relation of Gibb's free energy with equilibrium constant i.e.

∆G = -2.303×8.31×298 logk =-36501.5 Jmol-1K= 3.162 ×106

Topic: Chemistry, Chemical Equilibrium, JEE Main
Using the following data
Mass of hydrogen atom = 1.00783 u
Mass of neutron = 1.00867 u
Mass of nitrogen atom ( 7N14)= 14.00307 u
The calculated value of the binding energy of the nucleus of the nitrogen atom ( 7N14) is close to

Options:
A 56  MeV
B 98  MeV
C 104ââMeV
D 112  MeV

The correct answer is:
104  MeV

Solution:
The binding energy of nucleus may be defined as the energy equivalent to the mass defect of the nucleus.
If âm is mass defect than according to Einstein's mass energy relation.
Binding Energy
=â mc2 = Zmp + A-Zmn-Mc2
=7 Ã 1.00783 + 7 Ã 1.00867-14.00307c2
or BE=0.1124 Ã 931.5 MeV
or BE=104.6 MeV

Topic: Physics, Nuclear Physics, JEE Main
Using the Gibbs energy change, ÎGÂ°=+63.3 kJ , for the following reaction,
Ag2CO3â2Ag+(aq)+CO32-(aq)
the Ksp of Ag2CO3(s) in water at 25oC is
(R=8.314 J K-1mol-1) Given:â 10 0.91 â8

Options:
A 7.9×10-2
B 8.0×10-12
C 2.9×10-3
D 3.2×10-26

The correct answer is:
8.0×10-12

Solution:
ΔG o =−2.303 RT log  K sp
63.3Ã1000=-2.303Ã8.314Ã298logâ¡Ksp
logâ¡Ksp=-11.09
Ksp=10-11.09=8Ã10-12

Topic: Chemistry, Ionic Equilibrium, JEE Main
Using the given figure, the ratio of \(\mathrm{R}_f\) values of sample \(\mathrm{A}\) and sample \(\mathrm{C}\) is \(x \times 10^{-2}\). Value of \(x\) is ______

Options:

The correct answer is:
50

Solution:
$\begin{aligned} & \mathrm{R}_{\mathrm{f}} \text { of } \mathrm{A}=\frac{5}{12.5} \quad \mathrm{R}_{\mathrm{f}} \text { of } \mathrm{C}=\frac{10}{12.5} \\ & \text { Ratio }=\frac{\mathrm{R}_{\mathrm{f}(\mathrm{A})}}{\mathrm{R}_{\mathrm{f}(\mathrm{C})}}=\frac{1}{2}=0.5 \text { or } 50 \times 10^{-2}\end{aligned}$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Using the provided information in the following paper chromatogram:
Paper chromatography for compounds A and B.
The calculated \( \mathrm{R}_{\mathrm{f}} \) value of \( \mathrm{A} \) is
\( \times 10^{-1} \).

Options:

The correct answer is:
4

Solution:
Rf= Distance travelled by compound Distance travelled by solvent

On chromatogram distance travelled by compound is →2 cm

Distance travelled by solvent=5 cm

So Rf=25=4×10-1=0.4

Topic: Chemistry, General Organic Chemistry, JEE Main
Using the provided information in the following paper chromatogram:

The calculated Rf value of A is _____×10-1.

Options:

The correct answer is:
4

Solution:
Rf = Distance travelled by compound Distance travelled by solvent

On chromatogram distance travelled by compound is →2 cm

Distance travelled by solvent=5 cm

So, Rf=25=4×10-1=0.4

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (25 Feb Shift 1)
Using the rules for significant figures, the correct answer for the expression 0.02858×0.1120.5702 will be:

Options:
A 0.005613
B 0.00561
C 0.0056
D 0.006

The correct answer is:
0.00561

Solution:
In multiplication and division, the result must be reported with no more significant figures as in the measurement with few significant figures

0.02858×0.1120.5702=0.000561

(Or) Reported answer should not be more precise than the least precise term in calculations, so there should be three significant figures in reported answer.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Using the standard electrode potentials given below, predict if the reaction between the following is feasible.
\[
\begin{array}{l}
\mathrm{E}_{\mathrm{Cu}(\mathrm{red})}^{\circ}=0.34 \mathrm{~V} \\
\mathrm{E}_{\mathrm{Ag}(\mathrm{red})}^{\circ}=0.80 \mathrm{~V}
\end{array}
\]
\( \mathrm{Ag}^{+}(\mathrm{aq}), \mathrm{Cu}(\mathrm{s}) \) Report the value of \( \mathrm{E}_{\text {cell }}^{\mathrm{o}} \)

Options:
A \( +0.46 \mathrm{~V} \)
B \( +0.56 \mathrm{~V} \)
C \( +0.66 \mathrm{~V} \)
D \( +0.76 \mathrm{~V} \)

The correct answer is:
\( +0.46 \mathrm{~V} \)

Solution:
The electrode with less SRP will become anode.

Oxidation half reaction;
Cu s ⟶ Cu 2 + aq + 2 e - ; E o = - 0 . 3 4 V
Reduction half reaction;
2 Ag + aq + 2 e - ⟶ 2 Ag s ; E o = + 0 . 8 0 V
Overall reaction,
Cu s + 2 Ag + aq ⟶ Cu 2 + aq + 2 Ag ; E o = + 0 . 4 6 V
Positive emf indicates that the reaction is feasible.

Topic: Chemistry, Electrochemistry, JEE Main