datavorous/entrance-exam-dataset huggingface dataset:
The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3OH and CH3(CH2)11OSO3-Na+ at room temperature. The correct assignment of the sketches is

Options:
A I. KCl ;
II.CH3OH ;
III.CH3CH211OSO3-Na+
B I.III.CH3CH211OSO3-Na+ ;
II.CH3OH ;
III. KCl
C I. KCl ;
II.CH3(CH2)11OSO3-Na+ ;
III.CH3OH
D I.CH3OH ;
II. KCl ;
III.CH3CH211OSO3-Na+

The correct answer is:
I.CH3OH ;
II. KCl ;
III.CH3CH211OSO3-Na+

Solution:
Impurities affect surface tension appreciably. It is observed that impurities which tend to concentrate on surface of liquids, compared to its bulk lower the surface tension.
Substances like detergents and soaps CH 3 CH 2 11 SO 3 − Na + decrease the surface tension sharply.
Those like alcohol (eg. CH3OH,C2H5OH ) lower the surface tension slightly.
Inorganic impurities present in bulk of a liquid such as KCl tend to increase the surface tension of water.

Topic: Chemistry, States of Matter, JEE Main
The quality factor of $L C R$ circuit having resistance $(R)$ and inductance $(L)$ at resonance frequency $(\omega)$ is given by

Options:
A $\frac{\omega L}{R}$
B $\frac{R}{\omega L}$
C $\left(\frac{\omega L}{R}\right)^{1 / 2}$
D $\left(\frac{\omega L}{R}\right)^2$

The correct answer is:
$\frac{\omega L}{R}$

Solution:
As we know,
Quality factor $Q=\frac{V_L}{V_R}=\frac{X_L}{R}$
since, $X_L=\omega L$ then,
$Q=\frac{\omega L}{R}$

Topic: Physics, Alternating Current, JEE Main
The quantities of heat required to raise the temperatures of two copper spheres of radii $r_{1}$ and $r_{2}\left(r_{1}=1.5 r_{2}\right)$ through $1 \mathrm{~K}$ are in the ratio of

Options:
A $\frac{27}{8}$
B $\frac{9}{4}$
C $\frac{3}{2}$
D 1

The correct answer is:
$\frac{9}{4}$

Solution:
Required heat, $Q \propto r^{2}$
where $r$ is the radius.
$$
\begin{aligned}
\therefore \quad \frac{Q_{1}}{Q_{2}} &=\frac{r_{1}^{2}}{r_{2}^{2}} \\
\frac{Q_{1}}{Q_{2}} &=\frac{(1.5)^{2}}{(1)^{2}} \\
\frac{Q_{1}}{Q_{2}} &=\frac{9}{4}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The quantities that don't change when a resistor connected to a battery is heated due to the current are

Options:
A $B$ and $C$
B D
C A
D A and D

The correct answer is:
D

Solution:
Resistance, resistivity and drift velocity varies with relaxation time which is dependent on temperature.
Number of free electrons in a conductor remain invariant even if its temperature changes.

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 2)
The quantities x=1μ0∈0,y=EB and z=lCR are defined where C-capacitance, R-Resistance, ℓ-length, E-Electric field, B-magnetic field and∈0, μ0,-free space permittivity and permeability respectively. Then:

Options:
A x, y and z have the same dimension.
B Only x and z have the same dimension
C Only x and y have the same dimension
D Only y and z have the same dimension.

The correct answer is:
x, y and z have the same dimension.

Solution:
(a) Dimensions of quantity are given bellow

μ0=T m A-1=Wb m-1 A-1=V s m-1 C-1 s=N C-2 s2ε0=C2 N-1m-2

Now the dimension of x is given by

x=1μ0∈0=1N C-2 s2C2 N-1m-2x=m s-1

(b) The dimension of electric and magnetic field is given by,

B=T=Wb m-2=V s m-2E=V m-1

Now the dimension of y is given by,

y=EB=V m-1V s m-2=m s-1

(c) The dimension of capacitance and resistance is given by,

C=C V-1R=V C-1 s

Now the dimension of z is given by

z=lCR=mC V-1V C-1 s=m s-1

All quantities are having the dimensions of velocity.

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The quantity $\frac{\mathrm{PV}}{\mathrm{KT}}$ represents $(\mathrm{K}=$ Boltzmann constant)

Options:
A number of moles of gas.
B kinetic energy of gas
C mass of the gas
D number of molecules of gas in one mole

The correct answer is:
number of molecules of gas in one mole

Solution:
$\mathrm{PV}=\mathrm{nRT}$ $\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}$ Number of molecules $=\mathrm{N}_{\mathrm{A}} \frac{\mathrm{PV}}{\mathrm{RT}}$ $=\frac{\mathrm{PV}}{\left(\mathrm{R} / \mathrm{N}_{\mathrm{A}}\right) \mathrm{T}}=\frac{\mathrm{PV}}{\mathrm{k}_{\mathrm{B}} \mathrm{T}}$
where, $\mathrm{k}_{\mathrm{B}}=$ Boltzmann constant.

Topic: Physics, Kinetic Theory of Gases, JEE Main
The quantity of electricity in Faraday needed to reduce 1 mol of Cr2O72- to Cr3+ is

Options:

The correct answer is:
6

Solution:
1 faraday of charge is the magnitude of charge of one mole of electrons.

According to the given reaction:
Cr2O72-+ 6e-⟶2Cr3+
Reduction of 1 mol of Cr2O72- to Cr3+ requires 6 moles of electrons. Hence, charge required =6 Faraday

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The quantity of heat required to change the unit mass of a solid substance, from solid-state to liquid state, while the temperature remains constant, is known as

Options:
A latent heat
B sublimation
C hoar frost
D latent heat of fusion

The correct answer is:
latent heat of fusion

Solution:
The heat required to convert a solid into a liquid without any change in temperature is called the latent heat of fusion.

Topic: Physics, Thermal Properties of Matter, NEET
The quantity, when measured from different inertial reference frames, remains the same, among the following is

Options:
A Force
B Velocity
C Displacement
D Kinetic energy

The correct answer is:
Kinetic energy

Solution:
Kinetic energy being a scalar quantity, hence measured from different inertial frame gives the same value, while the other three being vector quantities their values vary.

Topic: Physics, Laws of Motion, NEET
The quantity which changes with temperature is:

Options:
A Molarity
B Mass percentage
C Molality
D Mole fraction

The correct answer is:
Molarity

Solution:
Molarity=No. of Moles of solute Volume of solution in L

Since volume depends on temperature, molarity will change upon change in temperature.

As the temperature rises, the volume of the solution increases. When the temperature increases the molarity of the solution will decrease as temperature depends on volume.

Hence option A is the answer.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The quantity X= Îµ 0 LV t where Îµ 0 is the permittivity of free space, L is length, V is the potential difference and t is time. The dimensions of X are the same as that of

Options:
A Resistance
B Charge
C Voltage
D Current

The correct answer is:
Current

Solution:
[ ε 0 L ] = [C]
∴ X= ε 0 LV t = C×V t = Q t = current

Topic: Physics, Units and Dimensions, JEE Main
The Quantity $X=\frac{\varepsilon_0 L V}{t}$ is the permittivity of free space, is length, is potential difference and is time. The dimensions of are same as that of

Options:
A Resistance
B Charge
C Voltage
D Current

The correct answer is:
Current

Solution:
$\left[\varepsilon_0 L\right]=[C] \therefore X=\frac{\varepsilon_0 L V}{t}=\frac{C \times V}{t}=\frac{Q}{t}=$ current
or
Dimensional formula for \(\epsilon_0=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right]\)
Dimensional formula for \(\mathrm{L}=[\mathrm{L}]\)
Dimensional formula for \(\Delta \mathrm{V}=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]\)
Dimensional formula for \(\Delta t=\left[\mathrm{T}^1\right]\)
Therefore, dimensional formula for X is:
\(\frac{\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{I}^2\right][\mathrm{L}]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{I}^{-1}\right]}{\left[\mathrm{T}^1\right]}=[\mathrm{I}]\)
Therefore, its the same as that of current.

Topic: Physics, Units and Dimensions, JEE Main
The quantum number of four electrons are given below:
I. n=4, l=2, ml=–2, ms=–1/2
II. n=3, l=2, ml=1, ms=+1/2
III. n=4, l=1, ml=0, ms=+1/2
IV. n=3, l=1, ml=1, ms=–1/2
The correct order of their increasing energies will be:

Options:
A I<III<II<IV
B IV<II<III<I
C I<II<III<II
D IV<III<II<I

The correct answer is:
IV<II<III<I

Solution:
Higher the value of n+l higher will be energy.
Hence, 3p4<3d5<4p5<4d7
If n+l same, then higher the value of, higher will be energy.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The quantum numbers for the outermost electron of an element are given below as $n=2, l=0, m=0, s=+\frac{1}{2}$. The atom is

Options:
A Lithium
B Beryllium
C Hydrogen
D Boron

The correct answer is:
Lithium

Solution:
The electronic configuration must be $1 s^2 2 s^1$. Hence, the element is lithium $(z=3)$.

Topic: Chemistry, Structure of Atom, JEE Main
The quarter disc of radius R (see figure) has a uniform surface charge density σ . Find electric potential at a point O,O,Z .

Options:
A σ8ϵ01-ZR2+Z2
B σ4ϵ0R2+Z2+Z
C σ8ϵ0R2+Z2-Z
D σ8ϵ01+ZR2+Z2

The correct answer is:
σ8ϵ0R2+Z2-Z

Solution:
V=∫0Rkαπrdrαr2+z2=σ8ϵ0∫0Rrdrr2+z2
t2=r2+z2
tdt=rdr
V=σ8ϵ0∫z22+R2dt=σ8ϵ0R2+z2-z

Topic: Physics, Electrostatics, JEE Main
The quotient when $1+x^{2}+x^{4}+x^{6}+\ldots \ldots+x^{34}$ is divided by $1+x+x^{2}+x^{3}+\ldots \ldots+x^{17}$ is

Options:
A $x^{17}-x^{15}+x^{13}-x^{11} \ldots \ldots+x$
B $x^{17}+x^{15}+x^{13}+x^{11}+\ldots \ldots+x$
C $x^{17}+x^{16}+x^{15}+x^{14}+\ldots \ldots+1$
D $x^{17}-x^{16}+x^{15}-x^{14}+\ldots \ldots-1$

The correct answer is:
$x^{17}-x^{16}+x^{15}-x^{14}+\ldots \ldots-1$

Solution:
Do in this way option (D)

Topic: Mathematics, Quadratic Equation, JEE Main
The 'rad' is the correct unit used to report the measurement of

Options:
A
the rate of decay of radioactive source

B
the ability of a beam of gamma ray photons to produce ions in a target

C
the energy delivered by radiation to a target.

D
the biological effect of radiation

The correct answer is:
the biological effect of radiation

Solution:
No Solution Available

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2006
The radial wave equation for hydrogen atom is
ψ=11641aο32x⁡-1x2⁡-8x⁡+12e-x2
where x=2rao , aο= radius of first bohr's orbit the minimum and maximum position of radial nodes from nucleus are:

Options:
A aο,3ao
B 0.5aο,3aο
C 0.5aο,aο
D 0.5aο,4aο

The correct answer is:
0.5aο,3aο

Solution:
For node Ï2=0 , So also Î¨=0
So for: Ï=11641aÎ¿32xâ¡-1x2â¡-8xâ¡+12e-x2
xâ¡-1x2â¡-8xâ¡+12 should be equal to zero
xâ¡-1x2â¡-8xâ¡+12=0 , solve the equation
xâ¡-1x2â¡-8xâ¡+12
X=1, 2, 6
i.e.r=0.5aÎ¿,aÎ¿,3aÎ¿;
âµx=2raÎ¿
So minimum 0.5aÎ¿
maximum 3aÎ¿

Topic: Chemistry, Structure of Atom, JEE Main
The radiation corresponding to 3 → 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 1 0 - 4 T . If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to :

Options:
A 1.8 eV
B 1.1 eV
C 0.8 eV
D 1.6 eV

The correct answer is:
1.1 eV

Solution:
We know that
R = m υ Bq
∴ υ = Bqr m
∴ KE = 1 2 m v 2
= B 2 q 2 R 2 2m
= 0.79 eV
Now from photoelectric equation
E = Ï + KE
∴ Ï = E - KE
= 1.89 - 0.79
= 1.1 eV

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2014 (06 Apr)
The radiation corresponding to the transitions of electrons from n=2 to n=1 of hydrogen atoms, fall on He+ ions which are in n=1 and n=2 states. The possible transition of helium ions as they absorb energy from the radiation is

Options:
A n=2→n=3
B n=1→n=4
C n=2→n=5
D n=2ân=4

The correct answer is:
n=2→n=4

Solution:
E=13.6×1×112-122=10.2 eV
For He+ in n=2,
10.2 eV=13.6×4122-1n2
⇒n=4

Topic: Physics, Atomic Physics, JEE Main
The radiation emitted by the surface of the Sun emits maximum power at a wavelength of about 500 nm. Assuming the Sun to be a blackbody emitter. If its surface temperature K is given by Î±âÎ²Ã10Î³ then fill the value of Î±+Î²+Î³. (Wienâs constant is given by 2.898 mm K )

Options:

The correct answer is:
16

Solution:
λ=500 nm, T=bλ
T=2.898 mm500 nm=5.8×103

Topic: Physics, Thermal Properties of Matter, JEE Main
The radiation emitted, when an electron jumps from n=3 to n=2 orbit in a hydrogen atom, falls on a metal to produce photo electrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1320 T in a radius of 10-3 m. Find the work function of the metal

Options:
A 1.03 eV
B 1.89 eV
C 0.86 eV
D 2.03 eV

The correct answer is:
1.03 eV

Solution:
E3-E2=13.6122-132
=13.6Ã536=1.89 eV
Photoelectron with KEmax are moving on circular path.
r=mvqB
mv=qBr
P=qBr=1.6Ã10-19Ã13200Ã10-3
=12Ã10-24=5Ã10-25 kg m s-1
Energy of photoelectron =KEmax
=p22m=25Ã10-502Ã9.1Ã10-31Ã1.6Ã10-19
=0.86 eV
Now, using Einstein equation
hÎ½=Ï+KEmax
1.89=0.86+ÏâÏ=1.03 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The radiation emitted, when an electron jumps from n=3 to n=2 orbit in a hydrogen atom, falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1320 T in a radius of 10-3 m. Find the work function of the metal

Options:
A 1.03 eV
B 1.89 eV
C 0.86 eV
D 2.03 eV

The correct answer is:
1.03 eV

Solution:
E3-E2=13.6122-132
=13.6Ã536=1.89 eV
Photoelectron with KEmax is moving on a circular path.
r=mvqB
mv=qBr
P=qBr=1.6Ã10-19Ã13200Ã10-3
=12Ã10-24=5Ã10-25 kg m s-1
The energy of photoelectron =KEmax
=p22m=25Ã10-502Ã9.1Ã10-31Ã1.6Ã10-19
=0.86 eV
Now, using the Einstein equation
hÎ½=Ï+KEmax
1.89=0.86+ÏâÏ=1.03 eV

Topic: Physics, Dual Nature of Matter, JEE Main
The radical axis of circles $x^2+y^2+5 x+4 y-5=0$ and $x^2+y^2-3 x+5 y-6=0$ is

Options:
A $8 y-x+1=0$
B $8 x-y+1=0$
C $8 x-8 y+1=0$
D $y-8 x+1=0$

The correct answer is:
$8 x-y+1=0$

Solution:
Let $S_1 \equiv x^2+y^2+5 x+4 y-5=0$
and $S_2=x^2+y^3-3 x+5 y-6=0$
The radical axis is $\quad S_1-S_3=0$
$\begin{array}{rrr}\Rightarrow & \left(x^2+y^2+5 x+4 y-5\right) \\ & -\left(x^2+y^2-3 x+5 y-6\right)=0 \\ \Rightarrow & 5 x+3 x+4 y-5 y-5+6=0 \\ \Rightarrow & 8 x-y+1=0\end{array}$

Topic: Mathematics, Circle, JEE Main
The radical centre of the circles $x^2+y^2-16 x+60=0, x^2+y^2-12 x+27=0, \quad x^2+y^2-12 y+8=0$ is

Options:
A $(13,33 / 4)$
B $(33 / 4,-13)$
C $(33 / 4,13)$
D None of these

The correct answer is:
None of these

Solution:
$S_1 \equiv x^2+y^2-16 x+60=0...(i)$
$S_2 \equiv x^2+y^2-12 x+27=0...(ii)$
$S_3 \equiv x^2+y^2-12 y+8=0...(iii)$
The radical axis of circle (i) and circle $(i)$ is $s_1-s_2=0 \Rightarrow-4 x+33=0...(iv)$
the radical axis of circle (ii) and circle (iii) is $S_2-S_3=0$
$\Rightarrow-12+12 y+19=0...(v)$
Solving (iv) and (v), we get the radical centre
$\left(\frac{33}{4}, \frac{20}{3}\right)$

Topic: Mathematics, Circle, JEE Main
The radical centre of three circles described on the three sides of a triangle as diameter is

Options:
A The orthocentre
B The circumcentre
C The incentre of the triangle
D The centroid

The correct answer is:
The orthocentre

Solution:
Let us consider a triangle as shown in fig.
Equations of the circles with $A B, B C$ and $C A$ as diameters are $S_1 \equiv(x+a)(x-a)+y^2=0$
$\begin{aligned} & \qquad S_2 \equiv(x-a)(x-\alpha)+y(y-\beta)=0 \\ & \text { and } S_3 \equiv(x+a)(x-\alpha)+y(y-\beta)=0 \\ & \text { i.e., } S_1 \equiv x^2+y^2-a^2=0\end{aligned}$
$S_2 \equiv x^2+y^2-(a+\alpha) x-\beta y+a \alpha=0$ and $S_3 \equiv x^2+y^2-(\alpha-a) x-\beta y-a \alpha=0$
$\therefore$ Radical axis of $S_2$ and $S_3$ is $S_3-S_2=0$
i.e., $2 a x-2 a \alpha=0$
$\Rightarrow 2 a(x-\alpha)=0, \text { as } a \neq 0, x=\alpha$
But $x=\alpha$ is the orthogonal through $\mathrm{C}$. Similarly other radical axes will be orthogonals through $A$ and $B$. Hence radical ceptre will be the orthocentre.

Topic: Mathematics, Circle, JEE Main
The radical which mainly causes ozone depletion in the presence of UV radiations is :

Options:
A Cl·
B NO·
C O·H
D CH3·

The correct answer is:
Cl·

Solution:
Chlorofluorocarbons are chemicals which contain carbon, chlorine and fluorine. Ozone layer breaks when these CFC's are released into the atmosphere.So. Chlorine is one of the radical which belongs to CFC's family, responsible for break down.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The radii and Young's moduli of two uniform wires $A$ and $B$ are in the ratio $2: 1$ and $1: 2$ respectively. Both wires are subjected to the same longitudinal force. If the increase in length of the wire $A$ is one percent, the percentage increase in length of the wire $B$ is

Options:
A 1.0
B 1.5
C 2.0
D 3.0

The correct answer is:
2.0

Solution:
Increase in length, $\Delta l=\frac{F l}{A Y}$
or $\Delta l=\frac{F l}{\pi r^2 y}$
Percent increase in length,
$\Delta x=\frac{\Delta l}{l} \times 100=\frac{F}{\pi r^2 y}$
Here, same longitudinal force is applied.
So,
$\begin{aligned}
& \frac{\Delta x_1}{\Delta x_2}=\left(\frac{r_2}{r_1}\right)^2 \cdot\left(\frac{y_2}{y_1}\right) \\
& \frac{1}{\Delta x_2}=\left(\frac{1}{2}\right)^2 \cdot\left(\frac{2}{1}\right)=\frac{1}{4} \times \frac{2}{1} \\
& \frac{1}{\Delta x_2}=\frac{1}{2} \\
& \Delta x_2=1 \times 2=2 \%
\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The radii of $F, F^{-}, O$ and $O^{-2}$ are in the order of

Options:
A $\mathrm{O}^{2-} \gt \mathrm{F}^{-} \gt \mathrm{O} \gt \mathrm{F}$
B $O^{2-} \gt F^{-} \gt F\gt0$
C $F^{-} \gt O^{2-} \gt F\gt0$
D $O^{2-} \gt O\gtF^{-} \gt F$

The correct answer is:
$\mathrm{O}^{2-} \gt \mathrm{F}^{-} \gt \mathrm{O} \gt \mathrm{F}$

Solution:
Atomic radius decreases on going from left to right in a period. Thus size of $O \gt F$. As $O^{2-}$ and $\mathrm{F}^{-}$are isoelectronic, therefore, size of $\mathrm{O}^{2-} \gt \mathrm{F}^{-}$.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The radii of Na+and Cl- ions are 95 pm and 181 pm respectively. The edge length of NaCl unit cell is

Options:
A 276 pm
B 138 pm
C 552 pm
D 415 pm

The correct answer is:
552 pm

Solution:
NaCl has fcc structure.
In fcc lattice with ions in octahedral voids
r + â+â r â â=â a 2
Where, a = edge length
r + â=â95âpm, â r â â=â181âpm
Edge length =â 2r + â+â 2r â
=(2Ã95+2Ã181)âpm
=â190â+â362â=â552âpm

Topic: Chemistry, Solid State, JEE Main
The radii of two mercury drops are $R_1$ and $R_2$. Under isothermal conditions, a single drop of radius $R$ is formed from them. The relations between $R, R_1$ and $R_2$ is

Options:
A $R^2=R_1^2+R_2^2$
B $R=R_1+R_2$
C $R=\frac{R_1+R_2}{2}$
D $R^3=R_1^3+R_2^3$

The correct answer is:
$R^3=R_1^3+R_2^3$

Solution:
Total volume remains same,
$\begin{aligned}
& \frac{4}{3} \pi R^3=\frac{4}{3} \pi R_1^3+\frac{4}{3} \pi R_2^3 \\
& R^3=R_1^3+R_2^3
\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2022 (11 Aug Shift 1)
The radii of two metallic spheres \(P\) and \(Q\) are \(r_1\) and \(r_2\) respectively. They are given the same charge. If \(r_1 > r_2\) Then on connecting them with a thin wire, the charge will flow:

Options:
A From \(P\) to \(Q\)
B From \(Q\) to \(P\)
C Neither the charge will flow from \(P\) to \(Q\) nor from \(Q\) to \(P\)
D The information is incomplete

The correct answer is:
The information is incomplete

Solution:
Since charge flows from high potential to lower potential.
If positive charge is given, then \(\mathrm{V}_1 < \mathrm{V}_2\) as \(\mathrm{r}_1 > \mathrm{r}_2\)
So positive charge flows from \(\mathrm{Q} \rightarrow \mathrm{P}\)
If negative charge is given, then \(V_1 > V_2\)
So negative charge flows form \(\mathrm{P} \rightarrow \mathrm{Q}\).
Since it is not given that whether the charge given is positive or negative, hence the information is incomplete.

Topic: Physics, Capacitance, JEE Main
The radii of two planets A and B are R and 4R and their densities are ρ and ρ3 respectively. The ratio of acceleration due to gravity at their surfaces gA:gB will be

Options:
A 4:3
B 1:16
C 3:16
D 3:4

The correct answer is:
3:4

Solution:
Acceleration due to gravity is written as g=GMr2=G×43×π×r3×ρr2=4πG3ρr

Given:

rA=RrB=4RρA=ρρB=ρ3

Hence, the ratio is

gAgB=R×ρ4R×ρ3=34.

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The radii of two planets are R1 and R2 and their densities are Ï1 and Ï2 respectively. If g1 and g2 represent the acceleration due to gravity at their respective surfaces, then g1g2 is

Options:
A ρ1R22ρ2R12
B ρ1R12ρ2R22
C ρ2R1ρ1R2
D ρ1R1ρ2R2

The correct answer is:
ρ1R1ρ2R2

Solution:
The value of g at surface
g= Gm e R e 2 = G R e 2 . 4Ï 3 â R e 3 Ï
g=43ÏGRÏ
So, g1g2=43ÏGR1Ï143ÏGR2Ï2â g1g2=R1Ï1R2Ï2

Topic: Physics, Gravitation, JEE Main
The radii of two soap bubbles are $\mathbf{r}_1$ and $\mathbf{r}_2$. In isothermal conditions, two meet together in yaccums. Then the radius of the resultant bubble is given by

Options:
A $R=\left(r_1+r_2\right) / 2$
B $R=r_1\left(r_1 r_2+r_2\right)$
C $R^2=r_1^2+r_2^2$
D $R=r_1+r_2$

The correct answer is:
$R^2=r_1^2+r_2^2$

Solution:
The correct option is $\mathbf{3}$
$r^2=r_1{ }^2+r_2^2$
In isothermal Condition.
$\Delta P_1 V_1+\Delta P_2 V_2=\Delta P V$
$\Delta \mathrm{P}_1 \frac{4}{3} \pi r_1^3+\Delta \mathrm{P}_2 \frac{4}{3} \pi r_2^3=\Delta \mathrm{P} \frac{4}{3} \pi r^3$
$\Delta P_1 r_1^3+\Delta P_2 r_2^3=\Delta P^3[$ where ' $r$ ' is the radius of new bubble]
$\frac{4 T}{r_1}\left(r_1^3\right)+\frac{4 T}{r_2}\left(r_2^3\right)=\frac{4 T}{r}\left(r^3\right)$
$\Rightarrow r_1^2+r_2^2=r^2$
$\Rightarrow r=\sqrt{r_1^2+r_2^2}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The radioactive isotope 90 234 Th undergoes two successive Î² -decay followed by Î± decay. The atomic number and the mass number respectively of the resulting isotope are

Options:
A 92 and 237
B 94 and 230
C 90 and 230
D 92 and 230

The correct answer is:
90 and 230

Solution:
9 0 Th 2 3 4 â¶ - 2 Î² 9 2 Î§ 2 3 4 â¶ - Î± 9 0 Th 2 3 0
Elimination of 1 Î± and 2 Î² particles give isotope.

Topic: Chemistry, Chemical Kinetics, JEE Main
The radioactive isotope X with a half-life of 109 years decays to Y which is stable. A sample of rocks was found to contain both the elements X and Y in the ratio 1:7. If initially, the quantity of Y in the rock was zero, then the age of the rocks is

Options:
A 2×109 years
B 3Ã109 years
C 6×109 years
D 7×109 years

The correct answer is:
3Ã109 years

Solution:
NN0=11+7=18
NN0=12n=18
â´ n=3
t=nT=3Ã109 years

Topic: Physics, Nuclear Physics, JEE Main
The radioactivity isotope X with a half-life of 109 year decays to Y which is stable. A sample of rocks was found to contain both the elements X and Y in the ratio 1:7. If initially the no. of nuclei of Y were zero then the age of rock is

Options:
A 2×109yr
B 3×109yr
C 6×109yr
D 7×19yr

The correct answer is:
3×109yr

Solution:
In radioactive decay N=No2nNN0=11+7=18
NN0=12n=18
â´ n=3
t=nT=3Ã109 yr

Topic: Physics, Nuclear Physics, JEE Main
The radio-isotope used for treatment of thyroid disorders is

Options:
A $\mathrm{Na}-24$
B $\mathrm{P}-32$
C Co-60
D I-131

The correct answer is:
I-131

Solution:
I-131 radio-isotope is used for thyroid disorders.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The radionucleide ${ }_{90}^{234} \mathrm{Th}$ undergoes two successive $\beta$-decays followed by one $\alpha$-decay. The atomic number and the mass number respectively of the resulting radionucleide are

Options:
A
94 and 230

B
90 and 230

C
92 and 230

D
92 and 234

The correct answer is:
90 and 230

Solution:
${ }_{90}^{234} \mathrm{Th} \stackrel{-\beta}{\longrightarrow}{ }_{91}^{234} \mathrm{X} \stackrel{-\beta}{\longrightarrow}{ }_{92}^{234} \mathrm{Th}--\alpha{ }_{90}^{230} \mathrm{Th}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The radionucleide ${ }_{90}^{234} \mathrm{Th}$ undergoes two successive $\beta$-decays followed by one $\alpha$-decay. The atomic number and the mass number respectively of the resulting radionucleide are

Options:
A
94 and 230

B
90 and 230

C
92 and 230

D
92 and 234

The correct answer is:
90 and 230

Solution:
${ }_{90}^{234} \mathrm{Th} \stackrel{-\beta}{\longrightarrow}{ }_{91}^{234} \mathrm{X} \stackrel{-\beta}{\longrightarrow}{ }_{92}^{234} \mathrm{Th}--\alpha{ }_{90}^{230} \mathrm{Th}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2003