datavorous/entrance-exam-dataset huggingface dataset:
The rates of heat radiation from two patches of skin each of area A, on a patient's chest differ by 2%. If the patch of the lower temperature is at 300 K and emissivity of both the patches is assumed to be unity, then what is the approximate temperature (in K) of the other patch? [Ignore all the higher-order terms in approximation]

Options:

The correct answer is:
301.5

Solution:
According to Stefan's law.
E1=ÏT4 and E2=ÏT+âT4
â´ E2-E1E1= Ï[T+âT4-T4]ÏT4=4âTT
Now, E2-E1E1=2100=4âT300
â´ âT= 300Ã24Ã100=1.5 K
â´ Temperature of other patch = T+âT
= 300 +1.5 = 301.5 K

Topic: Physics, Thermal Properties of Matter, JEE Main
The rates of reaction of NaOH with:

follow the order:-

Options:
A II>I>III
B II>III>I
C $\mathrm{I}>\mathrm{III}>\mathrm{II}$
D III>II>I

The correct answer is:
$\mathrm{I}>\mathrm{III}>\mathrm{II}$

Solution:
I and II can react with NaOH but II do not react at room temperature. I and III give reaction because at O and P position electrone withdrawing group is present.

Topic: Chemistry, General Organic Chemistry, JEE Main
The ratio among most probable velocity, mean velocity and root mean square velocity is given by

Options:
A $1: 2: 3$
B $1: \sqrt{2}: \sqrt{3}$
C $\sqrt{2}: \sqrt{3}: \sqrt{8 / \pi}$
D $\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}$

The correct answer is:
$\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}$

Solution:
most probable velocity : mean velocity : Vrms
$=\sqrt{\frac{2 R T}{M}}: \sqrt{\frac{8 R T}{\pi M}}: \sqrt{\frac{3 R T}{M}}=\sqrt{2}: \sqrt{\frac{8}{\pi}}: \sqrt{3}$

Topic: Chemistry, States of Matter, JEE Main
The ratio between the most probable speed of $\mathrm{N}_2$ at $400 \mathrm{~K}$ and $\mathrm{CO}$ at $800 \mathrm{~K}$ is (molar mass of $\mathrm{N}_2=28 \mathrm{~g} \mathrm{~mol}^{-1}$, molar mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Options:
A 0.75
B 0.25
C 0.707
D 1.414

The correct answer is:
0.707

Solution:
Most probable speed $=\sqrt{\frac{2 R T}{M}}$
Molar mass of $\mathrm{N}_2=28 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{CO}=28 \mathrm{~g} \mathrm{~mol}^{-1}$
$\mathrm{N}_2$ at $400 \mathrm{~K}$
$\mathrm{CO}$ at $800 \mathrm{~K}$
Ratio between the most probable speed of
$$
\begin{aligned}
& \begin{aligned}
\mathrm{N}_2 \text { and } \mathrm{CO} & =\frac{\sqrt{\frac{2 R T_1}{M_2}}}{\frac{\sqrt{2 R T_2}}{M_2}}=\frac{\sqrt{\frac{2 R \times 400}{28}}}{\sqrt{\frac{2 R \times 800}{28}}} \\
& =\sqrt{\frac{400}{800}}=\sqrt{\frac{1}{2}}=\sqrt{0.5}
\end{aligned} \\
& \text { Remaining ratio } r_{\left(\mathrm{N}_2\right)}: r_{(\mathrm{CO})}=1: 0.707
\end{aligned}
$$

Topic: Chemistry, States of Matter, TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
The ratio for the speed of the electron in the 3rd orbit of He+ to the speed of the electron in the 3rd orbit of hydrogen atom will be :

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

The correct answer is:
2:1

Solution:
The speed of electron is given by v=2.18×106×Zn m s-1, where, Z is atomic number and n is number of orbit.

Now, for third orbit, ratio of speed of electron is vHvHe+=ZHZHe+=12

⇒vHe+vH=21

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The ratio \(\frac{K_P}{K_C}\) for the reaction : \(\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}\) is :

Options:
A \(\frac{1}{\sqrt{\mathrm{RT}}}\)
B \((\mathrm{RT})^{1 / 2}\)
C \(\mathrm{RT}\)
D 1

The correct answer is:
\(\frac{1}{\sqrt{\mathrm{RT}}}\)

Solution:
$\begin{aligned} & \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_2(\mathrm{~g}) \\ & \Delta \mathrm{n}_{\mathrm{g}}=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2} \\ & \frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{K}_{\mathrm{C}}}=(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}=\frac{1}{\sqrt{\mathrm{RT}}}\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2024 (06 Apr Shift 2)
The ratio $\frac{K_p}{K_c}$ for the reaction
$\mathrm{CO}(g)+\frac{1}{2} O_2(g) \rightleftharpoons C O_2(g)$ is:

Options:
A
$\frac{1}{\sqrt{R T}}$

B
$(R T)^{1 / 2}$

C
$\mathrm{RT}$

D
$1$

The correct answer is:
$\frac{1}{\sqrt{R T}}$

Solution:
$\mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$
For the reaction
$\begin{aligned}
& \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(g) \leftrightharpoons \mathrm{CO}_2(\mathrm{~g}) \\
& \Delta n_g=1-\left(1+\frac{1}{2}\right)=-\frac{1}{2}
\end{aligned}$
$\therefore K_{\mathrm{P}}=\frac{\mathrm{K}_{\mathrm{C}}}{\sqrt{\mathrm{RT}}} ; \quad \frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{C}}}=\frac{1}{\sqrt{R T}}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2013 (25 Apr Online)
The ratio Kp to Kc of a reaction is 24.63 L atm molâ1 at 27Â°C. If heat of reaction at constant pressure is 98.8 kcal, what is the heat of reaction (in kcal) at constant volume?

Options:

The correct answer is:
98.20

Solution:
KpKc=RTÎng

RTÎng=24.63 Latmmol-1
0.0821 Ã 3 0 0 Î n g = 24.63
â´ Î n g = 1
Î H = Î E + Î n g RT
â´ Î E = Î H - RT = 98.8 - 2 Ã 1 0 - 3 Ã 3 0 0 = 98.2 kcal

Topic: Chemistry, Thermodynamics (C), JEE Main
The ratio maximum wavelength to minimum wavelength in Lyman series is :

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

The correct answer is:
$\frac{4}{3}$

Solution:
For maximum wavelenth \(\lambda_{\max }\)
\(\begin{aligned}
& n_1=1, n_2=2 \\
& \frac{1}{\lambda_{\max }}=R\left(\frac{1}{(1)^2}-\frac{1}{(2)^2}\right) \\
& \lambda_{\max }=\frac{4}{3 \pi}
\end{aligned}\)
For minimum wavelength \(\lambda_{\text {min }}\)
\(\begin{aligned}
& n_1=1, n_2=\infty \\
& \frac{1}{\lambda_{\min }}=R \\
& \lambda_{\min }=\frac{1}{R} \\
& \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{4}{3 R \frac{1}{R}}=\left(\frac{4}{3}\right)
\end{aligned}\)

Topic: Physics, Atomic Physics, JEE Main
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to

Options:
A the speed of light in vacuum
B reciprocal of speed of light in vacuum
C the ratio of magnetic permeability to the electric susceptibility of vacuum
D unity

The correct answer is:
reciprocal of speed of light in vacuum

Solution:
As $E=c B$
So required ratio $\frac{E}{B}=c$
$$
\frac{B}{E}=\frac{1}{c}
$$
Therefore, the ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to reciprocal of speed of light.

Topic: Physics, Electromagnetic Waves, JEE Main
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to

Options:
A the speed of light in vacuum
B reciprocal of speed of light in vacuum
C the ratio of magnetic permeability to the electric susceptibility of vacuum
D unity

The correct answer is:
reciprocal of speed of light in vacuum

Solution:
As $E=c B$
So required ratio $\frac{E}{B}=c$
$$
\frac{B}{E}=\frac{1}{c}
$$
Therefore, the ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to reciprocal of speed of light.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2012 (Mains)
The ratio of amplitudes of two coherent waves in Young's double-slit experiment is A1A2=13. What is the ratio of maximum and minimum intensities of fringes?

Options:

The correct answer is:
04.00

Solution:
ImaxImin=I1+I22(I1-I2)2=A1+3A1A1-3A12

ImaxImin=A1+3A1A1-3A12=422=4

Topic: Physics, Wave Optics, JEE Main
The ratio of angular speeds of minute hand and hour hand of a watch is

Options:
A $1: 12$
B $6: 1$
C $12: 1$
D $1: 6$

The correct answer is:
$12: 1$

Solution:
$\omega_{\min }=\frac{2 \pi}{60} \frac{\mathrm{rad}}{\min }$
and $\omega_{\mathrm{hr}}=\frac{2 \pi}{12 \times 60} \frac{\mathrm{rad}}{\mathrm{min}}$
$$
\begin{aligned}
\therefore \frac{\omega_{\min }}{\omega_{\mathrm{hr}}} &=\frac{2 \pi / 60}{2 \pi / 12 \times 60} \\
&=\frac{12}{1}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
The ratio of area covered by second orbital to the first orbital is

Options:
A $1: 2$
B $1: 16$
C $8: 1$
D $16: 1$

The correct answer is:
$16: 1$

Solution:
$r_n \propto n^2: A_n \propto n^4$
$\frac{A_2}{A_1}=\frac{n_2^4}{n_1^4}=\frac{2^4}{1^4}=\frac{16}{1}=16: 1$

Topic: Chemistry, Structure of Atom, JEE Main
The ratio of average electric energy density and total average energy density of electromagnetic wave is:

Options:
A 2
B 1
C 3
D 12

The correct answer is:
12

Solution:
Average electric energy density is stated as uE=12ε0E2

Total average energy density U= 12ε0E2+12μ0B2

=12ε0E2+12μ0E2c2 (As B=Ec)

=12ε0E2+12μ0E2ε0μ0=12ε0E2+12E2ε0=ε0E2

Hence, the ratio, uEU=12.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The ratio of C14C12 in a piece of wood is 18 part that of atmosphere. If half life of C14 is 5730 years, the age of wood sample is ..... years.

Options:

The correct answer is:
17190

Solution:
Given half life of C14= 5730 years.

λt=lnC14/C12atmosphere C14/C12wood sample

As per the question,

C14/C12woodC14/C12atmosphere=18

So, λt=ln8

ln 2t1/2t=ln 8

t=3×t1/2=3×5730=17190 years

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The ratio of close-packed atoms to tetrahedral holes in cubic close packing is

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

The correct answer is:
$1: 2$

Solution:
Every constituent has two tetrahedral voids. In $c c p$ lattice atoms
$=8 \times \frac{1}{8}+6 \times \frac{1}{2}=4$
$\therefore$ Tetrahedral void $=4 \times 2=8$,
Thus ratio $=4: 8:: 1: 2$.

Topic: Chemistry, Solid State, JEE Main
The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is:
(c= speed of electromagnetic waves)

Options:
A 1 :1
B 1 :c
C 1 :c2
D c :1

The correct answer is:
1 :1

Solution:
​​​In EMW, electric field and magnetic field have same energy density and same intensities.

Topic: Physics, Electromagnetic Waves, JEE Main
The ratio of de-Broglie wavelength of an α-particle and a proton accelerated from rest by the same potential is 1m , the value of m is:

Options:
A 4
B 16
C 8
D 2

The correct answer is:
8

Solution:
As we know, de Broglie wavelength in terms of energy of particle is given by, λ=12mE.

Energy gained by proton, E=qV.

Therefore,

λαλp=h2mαqαVh2mpqpV

⇒λαλp=18

⇒m=8

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The ratio of electric field and potential (E/V) at mid-point of electric dipole of length \( l \), for which separation is

Options:
A \( 1 / l \)
B \( l \)
C \( 2 / l \)
D None of these

The correct answer is:
None of these

Solution:
Let AB be a dipole of length l. Hence O be the midpoint of the dipole.
Then the electric field at point O, due to +q charge
E1=14πε0ql22
⟹E1=14πε04ql2 (along OA)
and the electric field at point O, due to -q charge
E2=14πε0ql22
⟹E2=14πε04ql2 (along OA)
Hence, the total electric field at point O
E=E1+E2
E=2.14πε04ql2
⟹ E=14πε08ql2 ….i
Now, potential at O, due to +q charge
⟹V1=14πε02ql
and potential at point O, due to –q charge
⟹ V2=-14πε02ql
Hence, total potential at point O,
V=0 ......................(ii)
On dividing Eq. (i) by Eq. (ii), we get
EV=∞

Topic: Physics, Electrostatics, JEE Main
The ratio of energy required to raise a satellite to a height ' $h$ ' above the earth's surface to that required to put it into the orbit at the same height is ( $\mathrm{R}=$ radius of earth)

Options:
A $\frac{2 \mathrm{~h}}{\mathrm{R}}$
B $\frac{\mathrm{h}}{\mathrm{R}}$
C $\frac{\mathrm{R}}{\mathrm{h}}$
D $\frac{\mathrm{R}}{2 \mathrm{~h}}$

The correct answer is:
$\frac{2 \mathrm{~h}}{\mathrm{R}}$

Solution:
The formula for the energy required to raise a satellite to height $\mathrm{h}$ is
$\mathrm{E}_1=\Delta \mathrm{U}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}=\frac{\mathrm{mghR}}{\mathrm{R}+\mathrm{h}}$
The formula for the energy required to set the satellite in orbit is
$\mathrm{E}_2=\frac{-\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}+\frac{\mathrm{GMm}}{\mathrm{R}}$
$=\operatorname{mgR}\left[1-\frac{1}{2\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)}\right]\left(\because \mathrm{GM}=\mathrm{gR}^2\right)$
$\begin{aligned} \therefore \quad E_2 & =\frac{m g R\left(\frac{2 h}{R}+1\right)}{2\left(1+\frac{h}{R}\right)} \\ \therefore \quad \frac{E_1}{E_2} & =\frac{m g h}{1+\frac{h}{R}} \times \frac{2\left(1+\frac{h}{R}\right)}{m g R} \\ & =\frac{2 h}{R} \quad\left(\because h \ll R \Rightarrow 1+\frac{2 h}{R} \approx 0\right)\end{aligned}$

Topic: Physics, Gravitation, JEE Main
The ratio of escape velocity of a planet to the escape velocity of earth will be:-

Given: Mass of the planet is 16 times mass of earth and radius of the planet is 4 times the radius of earth.

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

The correct answer is:
2 : 1

Solution:
The formula to calculate the escape velocity is given by

ve=2GMR ...1

For the Earth, it is given by

veE=2GMERE...2

and, for the planet, it is given by

veP=2GMPRP...3

Divide equation (2) by equation (3) and simplify to obtain the required ratio.

veEveP=MEMP×RPRE ...4

Substitute the values of the known parameters into equation (4) to obtain the required ratio.

veEveP=ME16ME×4RERE=12⇒veP:veE= 2:1

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The ratio of heat dissipated per second through the resistance $5 \Omega$ and $10 \Omega$ in the circuit given below is :

Options:
A $1: 2$
B $2: 1$
C $4: 1$
D $1: 1$

The correct answer is:
$2: 1$

Solution:
$\begin{aligned} & \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{10}{5}=\frac{2}{1} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{i}_1^2 \mathrm{R}_1}{\mathrm{i}_2^2 \mathrm{R}_2}=\left(\frac{2}{1}\right)^2 \times \frac{5}{10}=\frac{2}{1}\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The ratio of intensities at two points P and Q on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are π3 and π2, respectively are

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

The correct answer is:
3 : 2

Solution:
The formula to calculate the intensity of the interference pattern in Young's double slit experiment can be written as

I=4I0cos2ϕ2 ...1

Hence, the ratio of the required intensities can be calculated as follows-

I1I2=4I0cos2π64I0cos2π4=3412=32

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is

Options:
A 1 : 2 : 3
B 1 : 4 : 9
C 1Â :2Ï2Â :3Ï2
D 1:49Ï2 :425Ï2

The correct answer is:
1:49Ï2 :425Ï2

Solution:
I=I0sinαα2and α=0,3π2,5π2, ………, for maxima
The ratio of Intensities of successive maxima is
1 :23π2 :25π2 or 1 : 49π2 :425π2

Topic: Physics, Wave Optics, JEE Main
The ratio of kinetic energy at the mean position to potential energy at A2 of a particle performing SHM is

Options:
A 2:1
B 4:1
C 8:1
D 1:1

The correct answer is:
4:1

Solution:
Kinetic energy K=12mÏ2(A2-y2)
At mean position y=0
K=12mÏ2(A2)
Potential energy U=12mÏ2y2 ...(i)
U at y=A2
U=12 mA24 Ï2 ...(ii)
Dividing Eq. (i) by Eq. (ii), we get
KEU=41

Topic: Physics, Oscillations, JEE Main
The ratio of lengths of two rods $A$ and $B$ of same material is 1:2 and the ratio of their radii is $2: 1$, then the ratio of modulus of rigidity of $A$ and $B$ will be

Options:
A $4: 1$
B $16: 1$
C $8: 1$
D $1: 1$

The correct answer is:
$1: 1$

Solution:
Modulus of rigidity is the property of material.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is $x$. When both the current and radius is doubled the ratio will be

Options:
A $x / 8$
B $x / 4$
C $x / 2$
D $2 x$

The correct answer is:
$x / 8$

Solution:
$$
\begin{aligned}
& \text { Hints: } \mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 a} \quad \quad \mathrm{M}=\mathrm{I}\left(\pi a^2\right) \\
& \frac{\mathrm{B}}{\mathrm{M}}=\frac{\mu_0 \mathrm{I}}{2 a} \times \frac{1}{\mathrm{I} \pi a^2}=\frac{\mu_0}{2 \pi a^3}=x
\end{aligned}
$$
Again, Ratio $=\frac{\mu_0}{2 \pi(2 a)^3}=\frac{1}{8}\left(\frac{\mu_0}{2 \pi a^3}\right)=\frac{x}{8}$

Topic: Physics, Magnetic Effects of Current, JEE Main
The ratio of magnetic field at the centre of a current carrying coil of radius r to the magnetic field at distance r from the centre of coil on its axis is x:1. The value of x is _____.

Options:

The correct answer is:
8

Solution:
The formula to calculate the magnetic field at any axial point situated at a distance x from the centre of a circular current carrying conductor of radius r can be written as

B=μ0Ir22r2+x23/2 ...1

Substitute 0 for x in equation (1) to calculate the magnetic field Bc at the centre of the conductor.

Bc=μ0Ir22r2+023/2= μ0Ir22r3=μ0I2r ...2

Substitute r for x in equation (1) to calculate the magnetic field Br at the given distance from the centre of the conductor.

Br=μ0Ir22r2+r23/2= μ0Ir242r3=μ0I42r ...3

Divide equation (2) by equation (3) to calculate the required ratio of the magnetic fields.

BcBr=μ0I2rμ0I42r= 22=8

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The ratio of magnetic fields due to a bar magnet at the two axial points P1 and P2 which are separated from each other by 10 cm is 25:2. P1 Point is situated at 10 cm from the centre of the magnet. (Points P1 and P2 are on the same side of magnet and distance of P1 and P2 from the centre of magnet is much greater than the distance of two ends of the magnet from the centre). Magnetic length of the bar magnet is

Options:
A 5 cm

B 10 cm

C 15 cm

D 20 cm

The correct answer is:
10 cm

Solution:
B 1 B 2 = 25 2
Î¼ 0 4Ï M d 1 ( d 1 2 â l 2 ) 2 Î¼ 0 4Ï M d 2 ( d 2 2 â l 2 ) 2 = 25 2
d 1 d 2 Ã ( d 2 2 â l 2 ) 2 ( d 1 2 â l 2 ) 2 = 25 2
d 1 =10 cm, d 2 =20 cm
10 20 Ã ( 20 2 â l 2 10 2 â l 2 ) 2 = 25 2
400â l 2 =5( 100â l 2 )
4 l 2 =100â l 2 =25
l=5 cm

Topic: Physics, Magnetic Properties of Matter, JEE Main
The ratio of mass densities of nuclei of 40Ca and 16O is close to:

Options:
A 1
B 0.1
C 2
D 5

The correct answer is:
1

Solution:
Nucleus density is independent of the mass number. So, the ratio of density will be close to 1.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The ratio of mass percent of C and H of an organic compound CXHYOZ is 6 : 1. If one molecule of the above compound CXHYOZ contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O. The empirical formula of the compound CXHYOZ is

Options:
A C2H4O3
B C3H6O3
C C2H4O
D C3H4O2

The correct answer is:
C2H4O3

Solution:
Mass of C: Mass of H=6 : 1.
Or, for every atom of C (12 u), we have 2 atoms of H (1u × 2=2u)
Thus, the ratio of atoms of C : H=x : y=1 : 2
Thus, the formula is: CxH2xOz
Combustion of CxH2x:
CxH2x+(3x/2)O2→xCO2+xH2O
Hence, the combustion of CxH2x requires (3x/2) oxygen molecules.

Hence, the original molecule contains (3x/4) molecules =(3x/2) atoms.
Or, z=3x/2

Empirical Formula=CxH2xO3x/2=C2xH4xO3x=C2H4O3

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2018 (08 Apr)
The ratio of mass percent of C and H of an organic compound CXHYOZ is 6 : 1. If one molecule of the above compound CXHYOZ contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and H2O . The empirical formula of compound CXHYOZ is

Options:
A C2H4O3
B C3H6O3
C C2H4O
D C3H4O2

The correct answer is:
C2H4O3

Solution:
Ratio of mass % of C and H in CXHYOZ is 6 : 1.
Therefore, Ratio of mole % of C and H in CXHYOZ will be 1 : 2.
Therefore x : y = 1 : 2, which is possible in options 1, 2 and 3.
Now oxygen required to burn CXHY
CXHY+x+y4O2→xCO2+y2H2O
Now z is half of oxygen atoms required to burn CXHY
So Z=2X+Y22=X+Y4
Now putting values of x and y from the given options
Option A, x=2,y=4
Z=2+44=3
Option B, x=3,y=6
Z=3+64=4.5
Therefore correct option is 1 C2H4O3

Topic: Chemistry, Hydrocarbons, JEE Main
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecules is:

Options:
A 1 : 4
B 7 : 32
C 1 : 8
D 3 : 16

The correct answer is:
7 : 32

Solution:
Let the mass of oxygen be x gm
∴ Mass of nitrogen = 4x gm
Number of molecules = 6.023 x 1023 x number of moles
∴ Ratio = 6.023 × 1 0 2 3 × no. of moles of oxygen 6.023 × 1 0 2 3 × no. of moles of nitrogen
= m O 2 M O 2 m N 2 M N 2 = x 3 2 4 x 2 8
= 7 3 2 = 7 : 32

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2014 (06 Apr)
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s-1. At, t=0 the displacement is 5 m. What is the maximum acceleration? The initial phase is π4 .

Options:
A 500 m s-2
B 7502 m s-2
C 750 m s-2
D 5002 m s-2

The correct answer is:
5002 m s-2

Solution:
In a simple harmonic motion,

Maximum acceleration, fmax=ω2A
Maximum velocity, vmax=Aω.
fmaxvmax=ω2AAω=ω
ω=10
Displacement in SHM is x=Asin⁡ωt+π4
At t=0
5=Asin⁡π4
A=52.
Max acceleration = ω2A
=100×52
=5002 m s-2

Topic: Physics, Oscillations, JEE Main, JEE Main 2017 (08 Apr Online)
The ratio of minimum to maximum wavelengths in Balmer series of H-atom is

Options:
A 5:9
B 5:36
C 1:4
D 3:4

The correct answer is:
5:9

Solution:
We know that
1λmin=R14-1∞=R4
1λmax=R14-19=5R36⇒λminλmax=4R365R=59

Topic: Chemistry, Structure of Atom, JEE Main
The ratio of minimum wavelength of Lyman and Balmer series will be

Options:
A $1.25$
B $0.25$
C 5
D 10

The correct answer is:
$0.25$

Solution:
We know that, wavelength of hydrogen spectrum is given by
$$
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)
$$
For minimum wavelength of Lyman series,
$$
\begin{aligned}
& n_{1}=1 \text { and } n_{2}=\infty \\
\therefore \quad & \frac{1}{\left(\lambda_{L}\right)_{\min }}=R\left(\frac{1}{(1)^{2}}-\frac{1}{\infty}\right)=R \\
\Rightarrow \quad\left(\lambda_{2}\right)_{\min }=\frac{1}{R}
\end{aligned}
$$
For minimum wavelength of Balmer series,
$n_{1}=2$ and $n_{2}=\infty$
$\therefore \quad \frac{1}{\left(\lambda_{B}\right)_{\min }}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)$
$=R\left(\frac{1}{4}-0\right)=\frac{R}{4}$
$\Rightarrow \quad\left(\lambda_{B}\right)_{\min }=\frac{4}{R}$
$\therefore \quad \frac{\left(\lambda_{L}\right)_{\min }}{\left(\lambda_{B}\right)_{\min }}=\frac{\frac{1}{R}}{\frac{4}{R}}=\frac{1}{4}=0.25$

Topic: Physics, Atomic Physics, JEE Main
The ratio of minimum wavelengths of Lyman and Balmer series will be____

Options:
A $1.25$
B $0.25$
C $5$
D $10$

The correct answer is:
$0.25$

Solution:
The expression for wavelength is written as $\frac{1}{\lambda}=R Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$
The last line is a line of shortest wavelength or highest energy. When $n_2=\infty$, we get the last line wavelength.
$\frac{1}{\lambda}=R Z^2\left[\frac{1}{\mathrm{n}_1^2}\right]$
$\lambda_{\text {Series limit }}=\frac{\mathrm{n}_1^2}{\mathrm{RZ}}$
For Lyman series, $\mathrm{n}_1=1$ and for Balmer series $\mathrm{n}_1=2$.
Therefore, $\frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\frac{n_{1 \text { Lyman }}^2}{n_{1 \text { Balmer }}^2}=\frac{1^2}{2^2}=\frac{1}{4}=0.25$

Topic: Physics, Atomic Physics, JEE Main
The ratio of minimum wavelengths of Lyman and Balmer series will be

Options:
A 1.25
B 0.25
C 5
D 10

The correct answer is:
0.25

Solution:
The series end of Lyman series corresponds to transition from ni=∞ to
nf=1, corresponding to the wavelength
1λminL = R11-1∞ = R
⇒ (λmin)L =1R = 912 â« ...(i)
For last line of Balmer series
1(λmin)B=R122-1∞2=R4
⇒ (λmin)B =4R= 3 6 4 8 â« ...(ii)
Dividing Eq.(i) by Eq. (ii) .we get
(λmin)L(λmin)B= 0.25

Topic: Physics, Atomic Physics, JEE Main