datavorous/entrance-exam-dataset huggingface dataset:
A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid then:

Options:
A Energy 4VT1r-1R is released
B Energy =3VT1r+1R is released
C Energy =3VT1r-1R is released
D Energy is neither released nor absorbed

The correct answer is:
Energy =3VT1r-1R is released

Solution:
∆U=ST∆A
Ainitial=4πr2n
Afinal=4πR2
∆A=4πr2n-4πR2
43πr3n=43πR3
n=R3r3
∆A=4πR3r3.r2-R2=4πR3r-R3R=4πR3331r-1R
∆A=3V1r-1R
∆U=3VT1r-1R

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A certain number of spherical liquid drops of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid, which one of the following statements is true for the energy $(E)$ in the process?

Options:
A $E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$ is absorbed.
B $E=4 V T\left[\frac{1}{r}-\frac{1}{R}\right]$ is released.
C $E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$ is released.
D $E=4 V T\left[\frac{1}{r}-\frac{1}{R}\right]$ is absorbed.

The correct answer is:
$E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$ is released.

Solution:
Change in surface energy is given by
$E=T(\Delta A) \quad---(1)$
The initial area is given by,
$\mathrm{A}=\left(4 \pi r^2\right) n$
The final area is given by,
$\mathrm{a}=4 \pi R^2$
Therefore, the change in area is given by,
$\begin{aligned}
& \Delta A=a-A \\
& \Rightarrow \Delta A=4 \pi\left(n r^2-R^2\right)---(2)
\end{aligned}$
Now, using volume conservation: $\left(\frac{4}{3} \pi r^2\right) n=\frac{4}{3} R^3$
$\therefore n=\frac{R^3}{r^3} \quad---(3)$
$\Delta A=4 \pi\left[\frac{R^3}{r^3} \cdot r^2-R^2\right]=4 \pi\left[\frac{R^3}{r}-\frac{R^3}{R}\right]=\left(\frac{4}{3} \pi R^3\right) 3\left[\frac{1}{r}-\frac{1}{R}\right]=3 V\left[\frac{1}{r}-\frac{1}{R}\right]$
Introducing above value in equation (1)
$E=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A certain orbital has \( \mathrm{n}=4 \) and \( \mathrm{m}_{1}=-3 \). The number of radial nodes in this orbital is . (Round off to the Nearest Integer).

Options:

The correct answer is:
0

Solution:
n=4 and mℓ=-3

Hence, ℓ value must be 3.

Now, number of radial nodes =n-ℓ-1

=4-3-1=0

Topic: Chemistry, Structure of Atom, JEE Main
A certain orbital has n=4 and ml=-3. The number of radial nodes in this orbital is _______. (Round off to the Nearest Integer).

Options:

The correct answer is:
0

Solution:
n=4 and mℓ=-3

Hence, ℓ value must be 3.

Now, number of radial nodes =n-ℓ-1

=4-3-1=0

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (17 Mar Shift 1)
A certain orbital has no angular nodes and two radial nodes. The orbital is:

Options:
A 2s
B 3s
C 3p
D 2p

The correct answer is:
3s

Solution:
l=0⇒'s' orbital
n-l-1=2
n-1=2
n=3

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A certain patient is suspected to be suffering from acquired immuno deficiency syndrome. Which diagnostic technique will you recommend for its detection?

Options:
A MRI
B Ultra Sound
C WIDAL
D ELISA

The correct answer is:
ELISA

Solution:
ELISA (Enzyme-linked Immunosorbent Assay), also known as an enzyme immunoassay (EIA), is a biochemical technique used mainly in immunology to detect the presence of an antibody or an antigen in a sample. It is a useful tool for determining serum antibody concentrations (such as with the HIV test). The ELISA was the first screning test widely used for HIV because of its high sensitivity.

Topic: Biology, Human Health and Diseases, NEET, NEET 2011 (Screening)
A certain physical quantity is calculated from the formula $\frac{\pi}{3}\left(a^2-b^2\right) h d$, where $a, b$ and $h$ are all lengths and $d$ is density. The physical quantity being calculated is

Options:
A velocity
B volume
C mass
D acceleration

The correct answer is:
mass

Solution:
$\frac{\pi}{3}\left(a^2-b^2\right)$ hd ...(1)
a, b, h has length "l" so there unit is $\mathrm{m}(\mathrm{L})$ $\mathrm{d}=$ density $=\frac{\text { Mass }}{\text { volume }}=\frac{\mathrm{kg}}{\mathrm{m}^3}=\frac{\mathrm{M}}{\mathrm{L}^3}$
$\frac{\pi}{3}$ has no dimensions
from equation (1)
Now, $\left(\mathrm{L}^2-\mathrm{L}^2\right) \frac{\mathrm{M}}{\mathrm{L}^3}(\mathrm{~L})$
$\frac{\mathrm{L}^3 \mathrm{M}}{\mathrm{L}^3}=\mathrm{M} \Rightarrow$ Mass

Topic: Physics, Units and Dimensions, JEE Main
A certain piece of silver of given mass is to be made like a wire. Which of the following combination of length $(L)$ and the area of cross-sectional $(A)$ will lead to the smallest resistance

Options:
A $L$ and $A$
B $2 L$ and $A / 2$
C $L / 2$ and $2 A$
D Any of the above, because volume of silver remains same

The correct answer is:
$L / 2$ and $2 A$

Solution:
We know,
\(\mathrm{R}=\frac{\rho \times \mathrm{L}}{\mathrm{A}} \Rightarrow \mathrm{R} \propto \frac{\mathrm{L}}{\mathrm{A}}\) (As \(\rho\) is same for all the cases)
Clearly, value of \(\frac{L}{A}\) is minimum for \(C\).

Topic: Physics, Current Electricity, JEE Main
A certain pressure 'P' is applied to 1 litre of water and 2 litre of a liquid separately. Water gets compressed to 0.01% whereas the liquid gets compressed to 0.03%. The ratio of Bulk modulus of water to that of the liquid is 3x. The value of x is ______.

Options:

The correct answer is:
1

Solution:
Bulk modulus is given by B=-ΔPΔVV

Now, in case of water, Bwater=-ΔPΔVV=-ΔP0.01100

In case of liquid, Bliquid=-ΔP0.03100

Thus, the ratio BwaterBliquid=-ΔP0.0001-ΔP0.0003=3

Hence, the value of x=1.

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A certain quantity of real gas occupies a volume of 0.15 dm3 at 100 atm and 500 K when its compressibility factor is 1.07. Its volume at 300 atm and 300 K (When its compressibility factor is 1.4) is ______×10-4dm3(Nearest integer)

Options:

The correct answer is:
392

Solution:
The compressibility factor (Z) is given as:

Z = PVnRT

n=PVZRT

Z1=1.07, P1=100atm, V1=0.15L, T1=500KZ2=1.4, P1=300atm, V2=?, T2=300K

Z1Z2=P1V1×T2T1×P2V2

1.071.4=100×0.15×300500×300×V2

V2=0.03925 dm3

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A certain radar operates at a wavelength of 50 cm. If the beat frequency between the transmitted signal and the signal reflected from aircraft âf=1 kHz, then what will be the velocity (in km h-1) of the aircraft?

Options:

The correct answer is:
900

Solution:
∆f=f2vc=2v1λ
∴v=λ∆f2=0.5×10002=250 m s-1
=900 km/hr

Topic: Physics, Waves and Sound, JEE Main
A certain radioactive nucleus emits a particle that leaves its mass unchanged, but increases its atomic number by one. Identify the particle

Options:
A Alpha particle
B Beta particle
C Gamma particle
D None of the above

The correct answer is:
Beta particle

Solution:
Beta particle emission change the atomic number but mass number remain same.
Beta negative decay increases the atomic number by 1 by decaying neutron into proton and some other particle.
Beta positive decay decreases the atomic number by 1 by decaying proton into neutron and some other particle.

Topic: Physics, Nuclear Physics, JEE Main
A certain reaction A→B follows the given concentration (molarity)-time graph. Which of the following statement is true?

Options:
A The reaction is of second order with respect to A
B The rate for this reaction at 40 second will be approximately 3.5×10–3Ms–1
C The rate for this reaction at 80 second will be 1.75×10-3M s–1
D The [B] will be 0.25 M at 60 second

The correct answer is:
The rate for this reaction at 40 second will be approximately 3.5×10–3Ms–1

Solution:
Time required for concentration to change from 0.4to 0.2is same as the time required for the concentration to change from 0.2to 0.1, it means t 1/2 is independent of initial concentration, i.e. it is a first order reaction.
And from the graph t 1/2 =20 s
∴ K= 0.693 20 s −1
∴Rate=k[A]
At 40second [A]=0.1
∴ rate= 0.693 20
=3.46× 10 −3   Ms −1
=3.5× 10 −3 s −1

Topic: Chemistry, Chemical Kinetics, JEE Main
A certain reaction occurs in two steps as
(i) 2SO2g+2NO2g→2SO3g+2NOg
(ii) 2NOg+O2g→2NO2g
in the reaction,

Options:
A NO2g is intermediate
B NOg intermediate
C NOg is catalyst
D O2g is intermediate

The correct answer is:
NOg intermediate

Solution:
Here NO2 is used as a catalyst and it is Lead chamber process for the manufacture of H2SO4 .

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A certain sample of gas has a volume of 0.2 lithe measured at $1 \mathrm{~atm}$. pressure and $0^{\circ} \mathrm{C}$. At the same pressure but at $273^{\circ} \mathrm{C}$, its volume will be

Options:
A 0.4 litres
B 0.8 litres
C 27.8 litres
D 55.6 litres

The correct answer is:
0.4 litres

Solution:
$\frac{V_1}{V_2}=\frac{T_1}{T_2} \because V_2=\frac{T_2}{T_1} V_1=\frac{546^{\circ} \mathrm{K}}{273^{\circ} \mathrm{K}} \times 0.2 \mathrm{~L}=0.4 \mathrm{~L}$.

Topic: Chemistry, States of Matter, JEE Main
A certain series R-C circuit is formed using resistance R, a capacitor without dielectric having a capacitance C = 2 F and a battery of emf E = 3 V. The circuit is completed and it is allowed to attain the steady state. After this, at t = 0 half the thickness of the capacitor is filled with a dielectric constant k = 2 as shown in the figure. The system is again allowed to attain a steady state. What will be the heat generated (in joule) in the capacitor between t = 0 and t = â ?

Options:
A 3
B 5
C 2
D 6

The correct answer is:
3

Solution:
Initial charge on capacitor = CE
Initial potential energy of capacitor = CE2/2
Now, C = Îµ 0 A/d
And new capacitance C â² = Îµ 0 A d/2 , in series with = Îµ 0 K A d/2
â C ' = 2 â 0 A d in series with 4 â 0 A d
= Îµ 0 A d 2 Ã 4 2 + 4 = 4 Îµ 0 A 3 d = 4 3 C
â New charge = 4 3 CE
New energy = 1 2 Ã 4 3 CE 2 = 2 3 CE 2
Now W battery = Î H + Î U
â E 4 3 CE - CE = Î H + 2 3 CE 2 - 1 2 CE 2
â Î H = 1 6 CE 2 = 3

Topic: Physics, Electrostatics, JEE Main
A certain species of ionised atoms produce emission line spectrum according to the Bohr model. A group of lines in the spectrum is forming a series in which the shortest wavelength is \( 22.79 \mathrm{~nm} \) and the longest wavelength is \( 41.02 \mathrm{~nm} \). The atomic number of atom is \( Z \) is

Options:
A \( 2 \)
B \( 3 \)
C \( 4 \)
D \( 5 \)

The correct answer is:
\( 4 \)

Solution:
Let n be the lowest energy level of the given series of lines, then maximum wavelength is given by
hcλmax=13.6×Z21n2-1n+12
hcλmin=13.6×Z21n2-1∞2
Where λmax=41.02 nm and λmin=22.79 nm.
Solving above equations, we get n=2, which corresponds to Balmer series.
Z=4

Topic: Physics, Atomic Physics, JEE Main
A certain stellar body has radius 50Rs and temperature 2Ts and is at a distance of 2×1010 AU from the earth. Here, AU refers to the earth-sun distance and Rs and Ts refer to the sun's radius and temperature, respectively. Take, both star and sun to be ideal black bodies. The ratio of the power received on earth from the stellar body as compared to that received from the sun is close to

Options:
A 4×10-20
B 2×10-6
C 10-8
D 10-16

The correct answer is:
10-16

Solution:
Intensity of radiation of sun on earth or solar constant,

S1=P4πR02=4πRs2.σ.Ts44πR02

=σRs2R02.Ts4

where, Rs= radius of sun,

Ro= radius of orbit

σ= Stefan-Boltzmann constant and Ts= temperature of sun.

Now, intensity of radiation received from steller body on earth's surface,

S2=σ50Rs22×1010R02.2Ts4

⇒S2=2500×164×1020×σ.Rs2R02.Ts4

⇒S2=10-16S1 (from Eq. (i))

⇒S2S1=10-16

Topic: Physics, Thermal Properties of Matter, JEE Main
A chain is held on a frictionless table with 1nth of its length hanging over the edge. If the chain has a length L and a mass M, how much work is required to pull the hanging part back on the table?

Options:
A MgL2n2
B MgLn2
C MgL4n3
D MgL3n-1

The correct answer is:
MgL2n2

Solution:
If m is the mass per unit length of the chain, the mass of the chain of length y will be yÃm=ym and the force acting on it due to gravity will be myÃg=mgy(assuming that y is the length of the chain hanging over the edge). So, the work done in pulling the dy length of the chain on the table
dW=F-dy [as y is decreasing]
i.e., dW=mgy(-dy) [asF=mgy]
So, the work done in pulling the hanging portion on the table :
W = - â« L/n 0 mgydy = MgL 2 2n 2
or W = MgL/2n 2 as M = mL

Topic: Physics, Work Power Energy, JEE Main
A chain is placed on a rough table, partially hanging, as shown in the figure. The coefficient of static friction between the chain and the table is Î¼=0.4. If the maximum possible length of the hanging part is x and the length of the chain is L, then what is the value of Lx?

Options:

The correct answer is:
3.5

Solution:
If x= length suspended
m1=λx (mass of hanging part)
m2=λ(L-x) (mass of the part kept on the table)
Then for balance,
m1g=μsm2g
λx=0.4λ(L-x)
x=0.4L-0.4x
1.4x=0.4L
Lx=1.40.4=3.5

Topic: Physics, Laws of Motion, JEE Main
A chain (mass M) is hanging from a wooden structure as shown in the figure A. It is slowly taken to a state as shown in the figure B. Calculate the work done by gravity in the process (assume that after reaching the ground, the chain does not form a heap)

Options:
A Mg 4 a + π b 4 a 2 + b 2 + a b π
B Mg b 4 a + π b b - a π
C 2 Mg 2 a + π b 4 a 2 - b 2 + ab π
D Mg b 4 a + π b b + a π

The correct answer is:
Mg b 4 a + π b b + a π

Solution:
Work done by gravity = – (Change in P.E.)
⇒ ui = ulinear + ucurved
= Mass per unit length = M a + π b 4
W gravity = u i - u f = Mgb 4 a + π b b + a π

Topic: Physics, Work Power Energy, JEE Main
A chain of mass m forming a circle of radius R is slipped on a smooth round cone with half-angle θ . Find the tension in the chain if it rotates with a constant angular velocity ω about a vertical axis coinciding with the symmetry axis of the cone.

Options:
A Râ¡ Ï 2 + g cot Î¸ mâ¡ 2 Ï
B R&ApplyFunction; ω 2 - g cot θ m&ApplyFunction; 2 π
C R&ApplyFunction; ω 2 + g cot θ m&ApplyFunction; π
D R&ApplyFunction; ω - g cot θ m&ApplyFunction; 2 π

The correct answer is:
R⁡ ω 2 + g cot θ m⁡ 2 π

Solution:
Every element of chain moves on a circular path. So, every element of the chain experiences centripetal force. We consider a small element Îm on the chain making angle ÎÎ¸ at the centre
ÎN sin Î¸ = Îmg ...... (i)
Net force towards centre is
2 T sin Î Î¸ 2 - Î N cos Î¸ = Î mR Ï 2
Î Nâ¡ cos Î¸ = 2 Tâ¡ sin Î Î¸ 2 - Î mR Ï 2 .......... (ii)
From eqn.(i) and (ii) we get
cos Î¸ = 2 T sin Î Î¸ 2 - Î mR Ï 2 Î mg
But Î Î¸ is very small.
â´ sin Î Î¸ 2 = Î Î¸ 2
âµ cot Î¸ = T Î Î¸ - Î mR Ï 2 Î mg
Here Îm makes an angle ÎÎ¸ at the centre. The total mass of chain is m.
â´ The mass of chain per unit angle is
Î» = mâ¡ 2 Ï
â´ Î mâ¡ = Î» Î Î¸ = mâ¡ 2 Ï Î Î¸ = mâ¡ Î Î¸ 2 Ï
â´ cotââÎ¸= TÎÎ¸â mÎÎ¸ 2Ï R Ï 2 m 2Ï ÎÎ¸ g
â´ Tâ¡ = Râ¡ Ï 2 + g cot Î¸ mâ¡ 2 Ï

Topic: Physics, Rotational Motion, JEE Main
A change in the amount of yolk and its distribution in the egg will affect

Options:
A formation of zygote
B pattern of cleavage
C number of blastomeres produced
D fertilization

The correct answer is:
pattern of cleavage

Solution:
The mode of cleavage is determined by the amount of yolk and its distribution.
(i) Holoblastic-The cleavage, in which the segmentation lines pass through the entire egg, dividing it completely, eg, frog, human egg, etc.
(ii) Meroblastic-The lines of segmentation do not completely pass through the egg and remain confined to a part of the egg, eg, Insects, birds, reptiles.

Topic: Biology, Human Reproduction, NEET, NEET 2009 (Screening)
A change in the amount of yolk and its distribution in the egg will effect :

Options:
A Fertilization
B Formation of zygote
C Pattern of cleavage
D Number of blastomeres produced

The correct answer is:
Pattern of cleavage

Solution:
No Solution Available

Topic: Biology, Human Reproduction, NEET, NEET 2009 (Mains)
A change of 0.04 volts takes place between the base and the emitter when an input signal is connected to the common emitter transistor amplifier. As a result, 20 μA change takes place in the base current and a change of 2 mA takes place in the collector current. The input resistance and AC current gain are

Options:
A 1 kΩ &100
B 2 kΩ &100
C 2 kΩ &1000
D 1 kΩ & 200

The correct answer is:
2 kΩ &100

Solution:
Given: ΔIb=20 μA ΔIc=2 mA

ΔVBE=0.04 V

Input resistance Rinput =VBEΔIb=0.0420×10-6=2 kΩ

A.C. current gain β=ΔIcΔIb=2×10-320×10-6=100

Topic: Physics, Semiconductors, JEE Main
A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. The values of Î± and Î² are

Options:
A 0.99, 90
B 0.96, 79
C 0.97, 99
D 0.99, 79

The correct answer is:
0.99, 79

Solution:
Given that, change in emitter current, ∆IE= 8 mA
and change in collector current, ∆IC=7.9 mA
We know that,
α=∆IC∆IE
⇒=7.98⇒ α≈0.99
Also, we know that
β=α1-α
⇒=7.981-7.98=7.98-7.9
Or β=7.90.1 = 79
Hence, the required answer is α=0.99 and β=79.

Topic: Physics, Semiconductors, JEE Main
A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. The values of α and β are

Options:
A 0.99, 90
B 0.96, 79
C 0.97, 99
D 0.99, 79

The correct answer is:
0.99, 79

Solution:
Given that, change in emitter current, ∆IE=8 mA and change in collector current, ∆IC=7.9 mA.
We know that,
α=∆IC∆IE
⇒α=7.98
⇒α≈0.99
Also, we know that
β=α1-α
⇒β=7.981-7.98=7.98-7.9
⇒β=7.90.1=79
Hence, the required answer is α=0.99 and β=79.

Topic: Physics, Semiconductors, JEE Main
A charge is uniformly distributed inside a spherical body of radius r1=2r0 having a concentric cavity of radius r2=r0 (Ï is charge density inside the sphere). The potential of a point P at a distance 3r02 from the centre is

Options:
A 7ρr026ε0
B 101ρr0272ε0
C 17ρr0272ε0
D none of these

The correct answer is:
101ρr0272ε0

Solution:
Electric field at a distance r, r0<r<2r0 from the center is given by,
E=14πε0.qenclosedr2
=14πε0.ρ.43π(r3−r03)r2
=ρ3ε0.(r−r03r2)
Potential at the outer surface,
Vs=14πε0.Q2r0
=14πε0.ρ.43π((2r0)3-r03)2r0
=7ρ6ε0. r02
∴ If V is the required potential at r=3r02,
V−Vs=−∫2r03r02E. dr
V=Vs−ρ3ε0∫2r03r02(r−r03r2)dr
=7ρ6ε0r02−ρ3ε0(r22+r03r)|2r03r02
=7ρ6ε0r02−ρ3ε0[12(9x024−4r02)−(2r033r0−r032r0)]
=101ρr0372ε0

Topic: Physics, Electrostatics, JEE Main
A charge of $1 \mu \mathrm{C}$ is divided into two parts such that their charges are in the ratio of $2: 3$. These two charges are kept at a distance $1 \mathrm{~m}$ apart in vacuum. Then, the electric force between them $($ in $N)$ is

Options:
A $0.216$
B $0.00216$
C $0.0216$
D $2.16$

The correct answer is:
$0.00216$

Solution:
Ratio of charges $=2: 3$
$$
\therefore q_1=\frac{2}{5} \times 1 \mu \mathrm{C} \text { and } q_2=\frac{3}{5} \times 1 \mu \mathrm{C}
$$
Electrostatic force between the two charges
$$
\begin{aligned}
F & =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \\
& =\frac{9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}}{5 \times 5 \times(1)^2} \\
& =2.16 \times 10^{-3} \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
A charge of 4.0 μC is moving with a velocity of 4.0×106 m s-1 along the positive y-axis under a magnetic field B of strength (2k^) T. The force acting on the charge is xi^ N. The value of x is ______.

Options:

The correct answer is:
32

Solution:
Given:

q=4 μC,v→=4×106 j^ m s-1 and B→=2 k^ T

Force acting on charged particle, F→=q(v→×B→)

=4×10-64×106 j^×2 k^

=4×10-6×8×106i^

⇒F→=32 i^ N

⇒x=32

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A charge of 4μC is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be:

Options:
A 1μC and 3μC
B 2μC and 2μC
C 0 and 4μC
D 1.5μC and 2.5μC

The correct answer is:
2μC and 2μC

Solution:
If after division, one of the charge is q the other charge will be 4-q. The force between them can be written as,

F=Kq4-qd2

For the force between them to be maximum,

dFdq=0⇒Kd24-2q=0

⇒q=2 μC

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A charge of total amount $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R(R$ $>r$ ) such that the surface charge densities on the two spheres are equal. The electric potential at the common centre is

Options:
A
$\frac{1}{4 \pi \varepsilon_0} \frac{(R-r) Q}{\left(R^2+r^2\right)}$

B
$\frac{1}{4 \pi \varepsilon_0} \frac{(R+r) Q}{2\left(R^2+r^2\right)}$

C
$\left.\frac{1}{4 \pi \varepsilon_0} \frac{(R+r) Q}{\left(R^2+r^2\right.}\right)$

D
$\frac{1}{4 \pi \varepsilon_0} \frac{(R-r) Q}{2\left(R^2+r^2\right)}$

The correct answer is:
$\left.\frac{1}{4 \pi \varepsilon_0} \frac{(R+r) Q}{\left(R^2+r^2\right.}\right)$

Solution:
Let $q_1$ and $q_2$ be charge on two spheres of radius ' $r$ ' and ' $R$ ' respectively As, $q_1+q_2=\mathrm{Q}$ and $\sigma_1=\sigma_2 \quad$ [Surface charge density are equal]
$$
\therefore \frac{q_1}{r \pi r^2}=\frac{q_2}{4 \pi R^2}
$$
So, $q_1=\frac{Q r^2}{R^2+r^2}$ and $q_2=\frac{Q R^2}{R^2+r^2}$
Now, potential, $V=\frac{1}{4 \pi \varepsilon_0}\left[\frac{q_1}{r}+\frac{q_2}{R}\right]$
$$
\begin{aligned}
& =\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q r}{R^2+r^2}+\frac{Q R}{R^2+r^2}\right] \\
& =\frac{Q(R+r)}{R^2+r^2} \frac{1}{4 \pi \varepsilon_0}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2012 (19 May Online)
A charge particle is moving in a uniform magnetic field 2i^+3j^ T. If it has an acceleration of αi^-4j^ m s-2, then the value of α will be

Options:
A 3
B 6
C 12
D 2

The correct answer is:
6

Solution:
The magnetic force on a moving charged particle is given by F→=qv→×B→. Therefore, we know that the magnetic force is perpendicular to the magnetic field. Therefore, the resulting acceleration will also be perpendicular to the magnetic field. As the scalar product of two mutually perpendicular vectors is zero,

⇒a→·B→=0 ⇒αi^-4j^·2i^+3j^=0

⇒2α-12=0⇒α=6

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (26 Jul Shift 1)
A charge particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charge particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charge particle?

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

The correct answer is:
2:1

Solution:
Radius of a charged particle moving in a magnetic field perpendicular to its velocity is given by, R=mvqB.

mv can be written in terms of kinetic energy as mv=2mKE. Therefore,

R=mvqB=2mKEqB

⇒R∝KE⇒R2R1=K2K1=41=21

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be

Options:
A helical path with the axis perpendicular to the direction of magnetic field B
B helical path with the axis along magnetic field B
C circular path
D straight along the direction of magnetic field B

The correct answer is:
helical path with the axis along magnetic field B

Solution:
The formula to calculate the radius of the path followed by the charged particle under an external magnetic field can be written as

r=mvBq ...1

The perpendicular component of velocity results in circular motion, while the parallel component results in linear motion. Hence, the path of the charged particle is helical in nature, the axis of which lies along the direction of the magnetic field.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (08 Apr Shift 1)
A charge particle of 2 μC accelerated by a potential difference of 100 V enters a region of uniform magnetic field of magnitude 4 mT at right angle to the direction of field. The charge particle completes semicircle of radius 3 cm inside magnetic field. The mass of the charge particle is ______ ×10-18 kg.

Options:

The correct answer is:
144

Solution:
As we know, radius of the circular path is given by,

r=mvqB=2mKqB=2×m×qVqB

⇒r2=2mVqB2

⇒m=r2qB22V=3×10-22×2×10-6×4×10-322×100=144×10-18 kg

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (01 Feb Shift 1)