datavorous/entrance-exam-dataset huggingface dataset:
The ratio of momentum of an electron and $\alpha$-particle which are accelerated from rest by a potential difference of $100 \mathrm{~V}$ is

Options:
A 1
B $\sqrt{\left(2 \mathrm{~m}_{\mathrm{e}} / \mathrm{m}_{\mathrm{a}}\right)}$
C $\sqrt{\left(m_{e} / m_{\alpha}\right)}$
D $\sqrt{\left(\mathrm{m}_{\mathrm{e}} / 2 \mathrm{~m}_{\alpha}\right)}$

The correct answer is:
$\sqrt{\left(\mathrm{m}_{\mathrm{e}} / 2 \mathrm{~m}_{\alpha}\right)}$

Solution:
As we know, $\frac{1}{2} m v^{2}=e V$
\begin{aligned}
\Rightarrow \mathrm{v} &=\sqrt{\frac{2 e V}{m}} \\
\mathrm{p} &=\mathrm{mv}=\sqrt{2 m e V}
\end{aligned}
Now, $\mathrm{p}_{\mathrm{e}}=\sqrt{2 m_{e} \times e \times 100}$
and $\mathrm{p}_{\alpha}=\sqrt{4 m_{\alpha} \times e \times 100}$
$\therefore \quad \frac{p_{e}}{p_{\alpha}}=\sqrt{\frac{m_{e}}{2 m_{\alpha}}}$

Topic: Physics, Dual Nature of Matter, JEE Main
The ratio of most probable velocity to that of average velocity is

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

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

Solution:
$\frac{V_{m p}}{V_{a v}}=\frac{\sqrt{\frac{2 R T}{M}}}{\sqrt{\frac{8 R T}{\pi M}}}=\frac{\sqrt{\pi}}{2}$

Topic: Chemistry, States of Matter, JEE Main
The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are, respectively :

Options:
A 4:2:1
B 1:2:4
C 4:2:3
D 8:1:6

The correct answer is:
1:2:4

Solution:
ZSC = 8 × 18= 1

[Atoms are only at the corners]

ZBCC = 8 × 18 +1= 1 + 1=2

[Atoms are at the corners and at the body center]

ZFCC = 8 × 18 + 6 × 12= 1 + 3 = 4

[Atoms are at the corners and at the face centers]

Here, Z= effective number of atoms in the unit cell.

Therefore, ZSC​: ZBCC​: ZFCC​=1 : 2 : 4

SC:BCC:FCC=1:2:4

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The ratio of number of oxygen atoms $(\mathrm{O})$ in $16.0 \mathrm{~g}$ ozone $\left(\mathrm{O}_3\right), 28.0 \mathrm{~g}$ carbon monoxide $(\mathrm{CO})$ and $16.0$ oxygen $\left(\mathrm{O}_2\right)$ is (Atomic mass : $\mathrm{C}=12, \mathrm{O}=16$ and Avogadro's constant $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$ )

Options:
A
$3: 1: 2$

B
$1: 1: 2$

C
$3: 1: 1$

D
$1: 1: 1$

The correct answer is:
$1: 1: 1$

Solution:
$16.0 \mathrm{~g} \mathrm{O}_3=\frac{16}{48}$ mole
$$
\begin{aligned}
& =\frac{16}{48} \times 6.023 \times 10^{23} \text { molecules } \\
& =3 \times \frac{16}{48} \times 6.023 \times 10^{23} \text { atoms } \\
& =6.023 \times 10^{23} \text { atoms } \\
& 28.0 \mathrm{~g} \mathrm{CO}=\frac{28}{28} \text { mole }=1 \text { mole } \\
& =1 \times 6.023 \times 10^{23} \text { molecules } \\
& =1 \times 6.023 \times 10^{23} \text { atoms } \\
& =6.023 \times 10^{23} \text { atoms } \\
& 16.0 \mathrm{~g} \mathrm{O}_2=\frac{16}{16} \text { mole }=1 \mathrm{~mole}
\end{aligned}
$$
$=1 \times 6.023 \times 10^{23}$ molecules
$=1 \times 6.023 \times 10^{23}$ atoms
$=6.023 \times 10^{23}$ atoms
Therefore, the ratio is
$6.023 \times 10^{23}: 6.023 \times 10^{23}: 6.023 \times 10^{23}$
i.e. $1: 1: 1$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The ratio of number of oxygen atoms $(\mathrm{O})$ in $16.0 \mathrm{~g}$ ozone $\left(\mathrm{O}_3\right), 28.0 \mathrm{~g}$ carbon monoxide $(\mathrm{CO})$ and $16.0$ oxygen $\left(\mathrm{O}_2\right)$ is (Atomic mass : $\mathrm{C}=12, \mathrm{O}=16$ and Avogadro's constant $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$ )

Options:
A
$3: 1: 2$

B
$1: 1: 2$

C
$3: 1: 1$

D
$1: 1: 1$

The correct answer is:
$1: 1: 1$

Solution:
$16.0 \mathrm{~g} \mathrm{O}_3=\frac{16}{48}$ mole
$$
\begin{aligned}
& =\frac{16}{48} \times 6.023 \times 10^{23} \text { molecules } \\
& =3 \times \frac{16}{48} \times 6.023 \times 10^{23} \text { atoms } \\
& =6.023 \times 10^{23} \text { atoms } \\
& 28.0 \mathrm{~g} \mathrm{CO}=\frac{28}{28} \text { mole }=1 \text { mole } \\
& =1 \times 6.023 \times 10^{23} \text { molecules } \\
& =1 \times 6.023 \times 10^{23} \text { atoms } \\
& =6.023 \times 10^{23} \text { atoms } \\
& 16.0 \mathrm{~g} \mathrm{O}_2=\frac{16}{16} \text { mole }=1 \mathrm{~mole}
\end{aligned}
$$
$=1 \times 6.023 \times 10^{23}$ molecules
$=1 \times 6.023 \times 10^{23}$ atoms
$=6.023 \times 10^{23}$ atoms
Therefore, the ratio is
$6.023 \times 10^{23}: 6.023 \times 10^{23}: 6.023 \times 10^{23}$
i.e. $1: 1: 1$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2012 (07 May Online)
The ratio of number of water molecules in Mohr's salt and potash alum is ×10-1. (Integer answer)

Options:

The correct answer is:
5

Solution:
Mohr's salt =FeSO4NH42SO4·6H2O
Potash alum =KAlSO42·12H2O
So water molecule ratio =612=12=0.5=5×10-1

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2021 (26 Aug Shift 1)
The ratio of powers of two motors is 3xx+1, that are capable of raising 300 kg water in 5 minutes and 50 kg water in 2 minutes respectively from a well of 100 m deep. The value of x will be

Options:
A 16
B 2
C 2.4
D 4

The correct answer is:
16

Solution:
The work done in raising the water is W=mgh.

Power is P=Wt

Let the powers of both the motors be P1 & P2 respectively.

Hence, the ratio of the powers is,

P1P2=300×g×1005×6050×g×1002×60=6025⇒3xx+1=2.4

⇒3x=2.4x+2.4

⇒x=16

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The ratio of root mean square velocity to average velocity of gas molecules at a particular temperature is

Options:
A $1.086: 1$
B $1: 1.086$
C $2: 1.086$
D $1.086: 2$

The correct answer is:
$1.086: 1$

Solution:
$V_{max}=\sqrt{\frac{3 R T}{M}}, V_{a v}=\sqrt{\frac{8 R T}{\pi M}} ; \frac{V_{m x}}{V_{a v}}=\sqrt{\frac{3 \pi}{8}}$
$=\sqrt{\frac{66}{56}} \Rightarrow \frac{1.086}{1}$

Topic: Chemistry, States of Matter, JEE Main
The ratio of secondary and primary turns of step-up transformer is $4: 1 .$ If a current of 4 A is applied to the primary, the induced current in secondary will be

Options:
A $8 \mathrm{~A}$
B $2 \mathrm{~A}$
C $1 \mathrm{~A}$
D $0.5 \mathrm{~A}$

The correct answer is:
$1 \mathrm{~A}$

Solution:
$\frac{\mathrm{I}_{\mathrm{S}}}{\mathrm{I}_{\mathrm{P}}}=\frac{\mathrm{N}_{\mathrm{P}}}{\mathrm{N}_{\mathrm{S}}}=\frac{1}{4} \Rightarrow \mathrm{I}_{\mathrm{S}}=\frac{1}{4} \times 4=1 \mathrm{~A}$

Topic: Physics, Electromagnetic Induction, JEE Main
The ratio of sigma and π bonds present in pyrophosphoric acid is

Options:

The correct answer is:
6

Solution:
To determine the ratio of sigma and pi bonds in pyrophosphoric acid, we first need to understand the molecular structure of the compound.

Pyrophosphoric acid has the chemical formula H4P2O7. The molecule contains two phosphorus atoms, seven oxygen atoms, and four hydrogen atoms.

The structure of pyrophosphoric acid is

It contains total 12 sigma bonds and 2 pi bonds. So
No. of σ bonds No. of π bonds =122=6:1

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (08 Apr Shift 2)
The ratio of specific heats CPCV in terms of degree of freedom f is given by :

Options:
A 1+f3
B 1+2f
C 1+f2
D 1+1f

The correct answer is:
1+2f

Solution:
From Mayer's relation CP=CV+R and CV=f2R

Ratio of specific heat,

γ=CPCV=1+Rf2R=1+2f

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The ratio of specific heats of a gas is $\gamma$. The change in internal energy of one mole of the gas, when the volume changes from $V$ to $2 V$ at constant pressure $p$ is

Options:
A $\frac{\gamma-1}{p V}$
B $p V$
C $\frac{p V}{\gamma-1}$
D $\frac{p V}{\gamma}$

The correct answer is:
$\frac{p V}{\gamma-1}$

Solution:
Change in internal energy,
$d U=d Q-d W$
At constant pressure,
$\begin{aligned}
d U & =C_p d T-p d V \\
& =C_p d T-R d T \\
& =\left(C_p-R\right) d T=C_V d T \\
& =\frac{R}{\gamma-1} d T \\
& =\frac{R}{\gamma-1} \times \frac{p V}{R} \\
& =\frac{p V}{\gamma-1}
\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
The ratio of speed of electrons in the first excited state of hydrogen atom to the speed of light in vacuum is [Given, Planck's constant $=6.625 \times 10^{-34} \mathrm{Js}$ and permittivity of free space is $8.85 \times 10^{-12} \mathrm{Fm}^{-1}$ ]

Options:
A $5 \times 10^{-3}$
B $7.3 \times 10^{-3}$
C $3.6 \times 10^{-3}$
D $36.5 \times 10^{-3}$

The correct answer is:
$3.6 \times 10^{-3}$

Solution:
Velocity of electron in the nth orbit of an atom is given by $v=\frac{2 \pi k e^2}{h} \times \frac{Z}{n}$
where, $\frac{2 \pi k e^2}{h}$ is a constant quantity and $\mathrm{Z}$ is atomic number.
Hence, speed of electron in first excited state
(n = 2) is
$\begin{aligned}
v & =2.188 \times 10^6 \times \frac{1}{2} \quad(Z=1, \text { for hydrogen atom }) \\
& =1.094 \times 10^6 \mathrm{~ms}^{-1}
\end{aligned}$
Ratio of speed of electron in lst excited state to
$\begin{aligned}
\text {the speed of light in vacuum } & =\frac{1.094 \times 10^6}{3 \times 10^8} \\
& =0.364 \times 10^{-2} \\
& =3.6 \times 10^{-3}
\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is:

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

The correct answer is:
$4: 1$

Solution:
Given $\mathrm{M}_{\mathrm{H}_2}=2 ; \mathrm{M}_{\mathrm{O}_2}=32$
Speed of sound, $\mathrm{v}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}}$
$\Rightarrow \quad v \propto \frac{1}{\sqrt{M}}$
$\therefore \quad \frac{\mathrm{v}_{\mathrm{H}_2}}{\mathrm{v}_{\mathrm{O}_2}}=\sqrt{\frac{\mathrm{M}_{\mathrm{O}_2}}{\mathrm{M}_{\mathrm{H}_2}}}=\sqrt{\frac{32}{2}}=4: 1$

Topic: Physics, Waves and Sound, JEE Main
The ratio of speed of sound in hydrogen gas to the speed of sound in oxygen gas at the same temperature is:

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

The correct answer is:
4:1

Solution:
The formula to calculate the speed of sound in hydrogen gas is given by

vH2=γRTMH2 ...1

The formula to calculate the speed of sound in oxygen is given by

vO2=γRTMO2 ...2

Divide equation (1) by (2) and simplify to obtain the required ratio.

vH2vO2=MO2MH2 ...3

Substitute the values of the known parameters into equation (3) to calculate the required ratio.

vH2vO2=322= 4:1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The ratio of spin-only magnetic moment values μeffCr(CN)63-/μeffCrH2O63+ is _______

Options:

The correct answer is:
1

Solution:
In case of Cr3+ ion , the valence shell electronic configuration is same whether the ligand is strong field or weak field.

The spin only magnetic moment , μeff=nn+2BM

The valence shell electronic configuration of Cr3+ is 3d3, hence, it has three unpaired electrons.

μeff of Cr(CN)6-3=15 B.M.

μeff of CrH2O6+3=15 B.M.

Ratio = 1

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The ratio of surface tensions of mercury and water is given to be 7.5, while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135Â° and 0Â°, respectively. If it is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius r2, then the ratio r1r2 is close to

Options:
A 3/5
B 2/3
C 4/5
D 2/5

The correct answer is:
2/5

Solution:
|hHg|=|hwater|

2S1cosθ1ρ1r1g=2S2cosθ2ρ2r2g

Therefore, the ratio of the radii of the respective tubes is,

r1r2=ρ2ρ1×S1cosθ1S2cosθ2

r1r2=113.6×7.5×12

r1r2≈25

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The ratio of the accelerations for a solid sphere (mass 'm' and radius 'R') rolling down an incline of angle ' $\theta$ ' without slipping and slipping down the incline without rolling is:

Options:
A 5: 7
B 2: 3
C 2: 5
D 7: 5

The correct answer is:
5: 7

Solution:
For solid sphere rolling without slipping on inclined plane, acceleration
$$
a_{1}=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}
$$
For solid sphere slipping on inclined plane without rolling, acceleration
$$
a_{2}=g \sin \theta
$$
Therefore required ratio $=\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}$
$$
=\frac{1}{1+\frac{K^{2}}{R^{2}}}=\frac{1}{1+\frac{2}{5}}=\frac{5}{7}
$$

Topic: Physics, Rotational Motion, JEE Main
The ratio of the angular speed of minutes hand and hour hand of a watch is:

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

The correct answer is:
$12: 1$

Solution:
Time taken by minute hand to rotate by $2 \pi$ radian angle is 60 minute $=3600$ s
$\Longrightarrow$ Angular speed of minute hand $w_m=\frac{2 \pi}{3600}=\frac{\pi}{1800} \mathrm{rad} / \mathrm{s}$
Time taken by hour hand to rotate by $2 \pi$ radian angle is $12 \mathrm{hrs}=43200 \mathrm{~s}$
$\begin{aligned}
& \Longrightarrow \text { Angular speed of } \\
& \therefore \frac{w_m}{w_h}=\frac{\frac{\pi}{1800}}{\frac{\pi}{21600}}=12 \\
& \Longrightarrow w_m: w_h=12: 1
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
The ratio of the circumradius and inradius of an equilateral triangle is

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

The correct answer is:
$1: 2$

Solution:
$\frac{r}{R}=\frac{a \cos A+b \cos B+c \cos C}{a+b+c}$
In equilateral triangle $\mathrm{A}=\mathrm{B}=\mathrm{C}=60^{\circ}$
$=\frac{(a+b+c) \cos 60^{\circ}}{a+b+c}\frac{1}{2}$

Topic: Mathematics, Properties of Triangles, JEE Main
The ratio of the coefficient of the middle term in the expansion of (1+x)20 and the sum of the coefficients of two middle terms in expansion of (1+x)19 is

.

Options:

The correct answer is:
1

Solution:
In 1+xn, if n is even, then the middle term is n2+1th term.

So, Coefficient of middle term in (1+x)20=C1020

In 1+xn, if n is odd, then the middle term is n+12th & n+32th term.

So, Sum of Coefficient of two middle terms in

(1+x)19=C919+C1019

So required ratio =C1020C919+C1019=C1020C1020=1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The ratio of the coefficient of volume expansion of a glass container to that of a viscous liquid kept inside the container is $1: 4$. What fraction of the inner volume of the container should the liquid occupy so that the volume of the remaining vacant space will be same at all temperatures ?

Options:
A
$2: 5$

B
$1: 4$

C
$1: 64$

D
$1: 8$

The correct answer is:
$1: 4$

Solution:
When there is no change in liquid level in vessel then $\gamma_{\text {real }}^{\prime}=\gamma^{\prime}$ vessel Change in volume in liquid relative to vessel
$$
\Delta \mathrm{V}_{\mathrm{app}}=\mathrm{V} \gamma_{\text {app }}^{\prime} \Delta \theta=\mathrm{V}\left(\gamma_{\text {real }}^{\prime}-\gamma_{\text {vessel }}^{\prime}\right)
$$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2013 (23 Apr Online)
The ratio of the coefficient of x15 to the term independent of x in the expansion of x2+2x15 is

Options:
A 1:8
B 1:12
C 1:16
D 1:32

The correct answer is:
1:32

Solution:
Tr+1=15Crx215-r2xr
=15Crx30-3r2r
For coefficient of x15, r=5 and for term independent of x, r=10.
Required ratio =15C52515C10210=125=132

Topic: Mathematics, Binomial Theorem, JEE Main
The ratio of the coefficient of $x^{15}$ to the term independent of $x$ in the expansion of $\left(x^2+\frac{2}{x}\right)^{15}$ is:

Options:
A
$7: 16$

B
$7: 64$

C
$1: 4$

D
$1: 32$

The correct answer is:
$1: 32$

Solution:
$$
\begin{aligned}
& \mathrm{T}_{r+1}={ }^{15} \mathrm{C}_r\left(x^2\right)^{15-r} \cdot\left(2 x^{-1}\right)^r \\
& ={ }^{15} \mathrm{C}_r \times(2)^r \times x^{30-3 r}
\end{aligned}
$$
For independent term, $30-3 r=0 \Rightarrow r=10$
Hence the term independent of $x$,
$$
\mathrm{T}_{11}={ }^{15} \mathrm{C}_{10} \times(2)^{10}
$$
For term involve $x^{15}, 30-3 r=15 \Rightarrow r=5$
Hence coefficient of $x^{15}={ }^{15} \mathrm{C}_5 \times(2)^5$
Required ratio
$$
\begin{aligned}
& =\frac{{ }^{15} \mathrm{C}_5 \times(2)^5}{{ }^{15} \mathrm{C}_{10} \times(2)^{10}}=\frac{\frac{15 !}{10 ! 5 !}}{\frac{15 !}{5 ! 10 !} \times(2)^5} \\
& =1: 32 \\
&
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2013 (09 Apr Online)
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy:
(Assume mp=me×1849)

Options:
A 1:43
B 1:30
C 1:62
D 2:43

The correct answer is:
1:43

Solution:
The de-Broglie wavelength is given by

λ=hp=h2mK (As, p2=2mK).

It is given that

(K)e=(K)p

Thus, the ratio of the wavelengths of the proton and electron is

λpλe=2Kme2Kmp=memp=11849=143.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The ratio of the density of oxygen nucleus O816 and helium nucleus He24 is

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

The correct answer is:
1:1

Solution:
The radius of a nucleus, R is related to its atomic number A as:

R=R0A13

where R0 is the nuclear radius of hydrogen.

Thus, the density of a nucleus is given by:

d=μA43πR0A133=3μ4πR03=Constant

(Here, μ is atomic mass unit)

Hence, the ratio of nuclear density of oxygen to helium nucleus is 1:1

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The ratio of the energies of the hydrogen atom in its first to second excited states is

Options:
A $1 / 4$
B $4 / 9$
C $9 / 4$
D 4

The correct answer is:
$9 / 4$

Solution:
Ist excited state corresponds to $\mathrm{n}=2$ 2nd excited state corresponds to $\mathrm{n}=3$
As $E \propto \frac{1}{n^{2}}$
$$
\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{n}_{2}^{2}}{\mathrm{n}_{1}^{2}}=\frac{3^{2}}{2^{2}}=\frac{9}{4}
$$

Topic: Physics, Atomic Physics, JEE Main
The ratio of the energy required to raise a satellite upto a height \( h \) above the earth of radius \( R \) to that the kinetic energy of the satellite into that orbit is

Options:
A \( \mathrm{R}: \mathrm{h} \)
B \( h: R \)
C \( \mathrm{R}: 2 \mathrm{~h} \)
D \( 2 \mathrm{~h}: \mathrm{R} \)

The correct answer is:
\( 2 \mathrm{~h}: \mathrm{R} \)

Solution:
Energy required to raise the satellite to a height h from surface of earth is given by,
U = - GmM R + h - - GmM R = GMmh R R + h
Kinetic energy of satellite is given by,
K = 1 2 mv 0 2 = 1 2 m GM R + h
∴ U K = 2 h R

Topic: Physics, Gravitation, JEE Main
The ratio of the equivalent resistance of the network (shown in figure) between the points a and b when switch is open and switch is closed is x: 8. The value of x is

Options:

The correct answer is:
9

Solution:
When switch is closed

R1=2R3+2R3=4R3

When open R2=3R·3R3R+3R=3R2

R1R2=4R3×23R=89=8x

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The ratio of the fifth term from the beginning to the fifth term from the end in the expansion of 2 4 + 1 3 4 n is 6:1. If n=20Î», then the value of Î» is

Options:

The correct answer is:
2

Solution:
T 5 T n−5+2 =  n C 4 2 1 4 n−4 ⋅ 3 − 1 4 4 n C n−4 2 1 4 4 ⋅ 3 − 1 4 n−4
T 5 T n−5+2 =  2 n−4 4 ⋅ 3 n−4 4 6 = 6
⇒ 6 n−4 4 = 6 3 2
⇒n=10= 20 λ
⇒λ=2

Topic: Mathematics, Binomial Theorem, JEE Main
The ratio of the height above the surface of earth to the depth below the surface of earth, for gravitational accelerations to be the same (assuming small heights), is

Options:
A 0.25
B 0.5
C 1.0
D 1.25

The correct answer is:
0.5

Solution:
gh=g1-2hR ...1

gd=g1-dR ...2

On equating 1 & 2,

⇒ 1-2hR=1-dR

⇒ 2hR=dR

∴ hd=12=0.5

Topic: Physics, Gravitation, JEE Main
The ratio of the length of the subnormal to the square of the length of the subtangent at any point $P$ on the curve $y^2=(2 x+1)^3$ is

Options:
A 27
B $\frac{1}{9}$
C 9
D $\frac{8}{27}$

The correct answer is:
27

Solution:
Given equation of curve is
Let $P\left(x_1, y_1\right)$ be any point on Eq. (i) So,
$$
y_1^2=\left(2 x_1+1\right)^3
$$
Now, differentiate Eq. (i) w.r.t. $x$, we get
$$
\begin{aligned}
& \qquad 2 y \frac{d y}{d x}=3(2 x+1)^2 \times 2 \Rightarrow \frac{d y}{d x}=\frac{3(2 x+1)^2}{y} \\
& \begin{aligned}
\therefore\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=m=\frac{3\left(2 x_1+1\right)^2}{y_1} \\
\mathrm{ST}=\text { Length of subtangent } \\
\quad=\frac{y_1}{m}=y_1 \times \frac{y_1}{3\left(2 x_1+1\right)^2}=\frac{y_1^2}{3\left(2 x_1+1\right)^2}
\end{aligned} \\
& \mathrm{SN}=\text { length of subnormal }=y_1 m=y_1\left[\frac{3\left(2 x_1+1\right)^2}{y_1}\right] \\
& =3\left(2 x_1+1\right)^2
\end{aligned}
$$
According to the question, $\frac{\mathrm{SN}}{(\mathrm{ST})^2}=\frac{3\left(2 x_1+1\right)^2}{\left[\frac{y_1^2}{3\left(2 x_1+1\right)^2}\right]^2}$
$$
=\frac{3\left(2 x_1+1\right)^2}{1} \times \frac{9\left(2 x_1+1\right)^4}{y_1^4}
$$
= 27

Topic: Mathematics, Application of Derivatives, JEE Main
The ratio of the lengths of two wires $A$ and $B$ of same material is $1: 2$ and the ratio of their diameter is $2: 1$. They are stretched by the same force, then the ratio of increase in length will be

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

The correct answer is:
$1: 8$

Solution:
$l=\frac{F L}{A Y} \Rightarrow l \propto \frac{L}{d^2}=\frac{l_1}{l_2}=\frac{L_1}{L_2} \times\left(\frac{d_2}{d_1}\right)^2=\frac{1}{2} \times\left(\frac{1}{2}\right)^2=\frac{1}{8}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the 5th excited state of a hydrogen atom is :

Options:
A 4
B 14
C 12
D 1

The correct answer is:
12

Solution:
As we know, KE=12|PE| for any orbit.

∴KE|PE|=12

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The ratio of the mass percentages of 'C & H' and 'C & O' of a saturated acyclic organic compound 'X' are 4:1 and 3:4 respectively. Then, the moles of oxygen gas required for complete combustion of two moles of organic compound 'X' is_______

Options:

The correct answer is:
5

Solution:
Mass ratio of C : H is 4 : 1⇒12 : 3

& C : O is 3 : 4 ⇒ 12 : 16

so mass mole mole ratioC 12 1 1H 3 3 3O 16 1 1

Empirical formula ⇒CH3O

as compound is saturated acyclic so molecular formula is C2H6O2.

C2H6O2+52O2(g)⟶2CO2(g)+3H2O(g)
2mole 5mole

so required moles of O2 is ⇒5

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The ratio of the masses of methane and ethane in a gas mixture is 4:5 . The ratio of number of their molecules in the mixture is

Options:
A 4:5
B 3:2
C 2:3
D 5:4

The correct answer is:
3:2

Solution:
nCH4nC2H6=mCH4MC2H6mCH4MCH4=3:2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The ratio of the nuclear radius, of an atom with mass number A and 2 4 He is 1413. What is the value of A?

Options:

The correct answer is:
56

Solution:
From the relation r ∝ A 1 / 3 ,
we have, r 2 r 1 = A 2 A 1 1 / 3
or A 2 4 1 / 3 = 1 4 1 / 3
Therefore, A2 = 56

Topic: Physics, Nuclear Physics, JEE Main
The ratio of the numerical values of the average velocity and average speed of a body is always

Options:
A Unity
B Unity or less
C Unity or more
D Less than unity

The correct answer is:
Unity or less

Solution:
$\frac{\mid \text { Average velocity } \mid}{\mid \text { Average speed } \mid}=\frac{\mid \text { displaceme nt } \mid}{\mid \text { distance } \mid} \leq 1$
because displacement will either be equal or less than distance. It can never be greater than distance.

Topic: Physics, Motion In One Dimension, JEE Main
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is:

Options:
A $2: 3$
B $2: 1$
C $\sqrt{5}: \sqrt{6}$
D $1: \sqrt{2}$

The correct answer is:
$\sqrt{5}: \sqrt{6}$

Solution:
The radius of gyration of disc about a tangential axis in the plane of disc is $k_1$
$$
\therefore \quad k_1=\frac{\sqrt{5}}{2} R
$$
And radius of gyration of circular ring of same radius about tangential axis is given by:
$$
\begin{aligned}
k_2 & =\frac{\sqrt{3}}{2} \mathrm{R} \\
\therefore \quad \frac{k_1}{k_2} & =\frac{\sqrt{5}}{2} R \times \frac{\sqrt{2}}{\sqrt{3} R} \\
& =\frac{\sqrt{5}}{\sqrt{6}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main