datavorous/entrance-exam-dataset huggingface dataset:
An inductor of inductance $\mathrm{L}=400 \mathrm{mH}$ and resistors of resistances $\mathrm{R}_1$ $=2 \Omega$ and $R_2=2 \Omega$ are connected to a battery of emf $12 \mathrm{~V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t=0$. The potential drop across $L$ as a function of time is

Options:
A
$6 e^{-5 t} V$

B
$\frac{12}{t} e^{-3 t} V$

C
$6\left(1-e^{-t / 0.2}\right) V$

D
$12 e^{-5 t} V$

The correct answer is:
$12 e^{-5 t} V$

Solution:
$$
\begin{aligned}
& \mathrm{I}_1=\frac{\mathrm{F}}{\mathrm{R}_1}=\frac{12}{2}=6 \mathrm{~A} \\
& \mathrm{E}=\mathrm{L} \frac{\mathrm{dl_{2 }}}{\mathrm{dt}}+\mathrm{R}_2 \times \mathrm{I}_2 \\
& \mathrm{I}_2=\mathrm{I}_0\left(1-\mathrm{e}^{-\mathrm{t} / \mathrm{t}_{\mathrm{c}}}\right) \Rightarrow \mathrm{I}_{\mathrm{o}}=\frac{\mathrm{E}}{\mathrm{R}_2}=\frac{12}{2}=6 \mathrm{~A} \\
& \mathrm{t}_{\mathrm{c}}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{400 \times 10^{-3}}{2}=0.2 \\
& \mathrm{I}_2=6\left(1-\mathrm{e}^{-\mathrm{t} / 0.2}\right)
\end{aligned}
$$
Potential drop across $L=E-R_2 l_2=12-2 \times 6\left(1-e^{-b t}\right)=12 e^{-5 t}$
Directions: Question numbers 28,29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle $A B C D A$, as shown in the $P-T$ diagram.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2009
An inelastic ball is dropped from a height 100 meter. If due to impact it loses 35% of its energy the ball will rise to a height of

Options:
A 35Â m
B 65 m
C 100Â m
D 135Â m

The correct answer is:
65 m

Solution:
Loss of energy =35%
Remaining energy =65%
New energy =65% of initial energy
mghâ²=65100mgh
hâ²=65100Ã100
hâ²=65 m

Topic: Physics, Work Power Energy, NEET
An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along $X$-axis at $x=1 \mathrm{~cm}, x=2 \mathrm{~cm}$, $x=4 \mathrm{~cm}, x=8 \mathrm{~cm} . . .$. and so on. In this set up if the consecutive charges have opposite sign, then the electric field in newton/coulomb at $x=0$ is :
$\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\right)$

Options:
A $12 \times 10^4$
B $24 \times 10^4$
C $36 \times 10^4$
D $48 \times 10^4$

The correct answer is:
$36 \times 10^4$

Solution:
Electric field intensity due to a point charge.
$E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$
The consecutive charges are of opposite signs.
$\therefore$ Net electric field at $x=0$, is
$E=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q}{r_1^2}-\frac{Q}{r_2^2}+\frac{Q}{r_3^2}-\frac{Q}{r_4^2} \ldots . . \infty\right]$
$=\frac{Q}{4 \pi \varepsilon_0}\left[\frac{1}{r_1^2}-\frac{1}{r_2^2}+\frac{1}{r_3^2}-\frac{1}{r_4^2}+\ldots . \infty\right\rceil$
$=9 \times 10^9 \times 5 \times 10^{-9}$
$\left[\frac{1}{\left(1 \times 10^{-2}\right)^2}-\frac{1}{\left(2 \times 10^{-2}\right)^2}+\frac{1}{\left(4 \times 10^{-2}\right)^2}+\ldots \infty\right\rfloor$
$=\frac{45}{10^{-4}}\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2}+\ldots . \infty\right]$
$=45 \times 10^4\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right]$
$\left\lceil\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right\rfloor=\frac{\frac{1}{1^2}}{1-\left(\frac{-1}{2}\right)^2}$
$\left[\right.$ Sum of infinite GP, $\left.\mathrm{S}_{\infty}=\frac{a}{1-r}\right]$
$=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$
$\therefore \quad E=45 \times 10^4 \times \frac{4}{5}=36 \times 10^4 \mathrm{~N} / \mathrm{C}$

Topic: Physics, Electrostatics, JEE Main
An infinite number of masses are placed on a frictionless table and they are connected via massless strings. Their masses follow the sequence, $m, \frac{m}{2}, \frac{m}{6}, \ldots \ldots \ldots . . . \frac{m}{n !}, \ldots \ldots . .$. and they are further connected to a mass $m$ that hangs over a massless pulley. The acceleration of the hanging mass is

Options:
A $\frac{g}{e-1}$
B $\frac{g}{e+1}$
C $\frac{g}{e}$
D $\frac{g}{2 e}$

The correct answer is:
$\frac{g}{e}$

Solution:
The given situation is shown in the following figure
Effective mass of the system of $n$ masses placed on the table,
$M=m+\frac{m}{2}+\frac{m}{6}+\ldots+\frac{m}{n !}$
$=m\left(1+\frac{1}{2}+\frac{1}{6}+\ldots+\frac{1}{n !}\right)$
$=m\left(\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots+\frac{1}{n !}\right)$
$=m\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots+\frac{1}{n !}-1\right)$
$M=m(e-1)$
$\left[\because e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\ldots+\frac{1}{n !}\right]$
The given system of masses can be redrawn as
If $a$ be the acceleration of system of masses, then according to Newton's second law of motion,
$T=M a \Rightarrow T=m(e-1) a$ $\ldots(i)$
and $m g-T=m a \Rightarrow T=m g-m a$
$T=m(g-a)$ $\ldots(ii)$
From Eq. (i) and Eq. (ii), we get
$m(e-1) a=m(g-a) \Rightarrow a=g / e$

Topic: Physics, Laws of Motion, JEE Main
An infinite number of point charges, each carrying 1 μC charge, are placed along the y-axis at y=1 m,2 m,4 m,8 m ... .

The total force on a 1 C point charge, placed at the origin, is x×103 N. The value of x, to the nearest integer, is ___ .

[Take 14πϵ0=9×109 N m2 C-2]

Options:

The correct answer is:
12

Solution:
F=k1 C1 μC1+122+142+182+…

=9×10311-14=12×103 N

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (18 Mar Shift 2)
An infinite number of straight wires each carrying current I are equally placed as shown in the figure. Adjacent wires have current in the opposite direction. The net magnetic field at point P is

Options:
A μ0I4πl⁡n23ak^
B μ0I4πln43ak^
C μ0I4πln43a-k^
D Zero

The correct answer is:
μ0I4πln43ak^

Solution:
B1=μ04πIasinα+sinβ=μ04πI×2sin30∘acos30∘
B → = B → = B 1 k ^ + B 2 k ^ + B 3 k ^ - B 4 k ^ + .....
=μ04πI×2sin30∘acos30∘k^-μ04πI×2sin30∘2acos30∘k^+μ04πI×2sin30∘3acos30∘k^+.....
=μ04π2I3ak^-μ04π2I3×2ak^+μ04π2I3×3ak^+.....
=μ04π2I3a1-12+13-14+...k^=μ02I4π3aloge1+1k^=μ02I4π3aloge2k^
=μ0I4π3aloge4k^

Topic: Physics, Magnetic Effects of Current, JEE Main
An infinite thin non-conducting sheet has uniform surface charge density Ï. Find the potential difference VA-VB between points A and B as shown in the figure. The line AB makes an angle of 37Â° with the normal to the sheet. sinâ¡37Â°=35

Options:
A 2σd5ε0
B -2σd5ε0
C 310σdε0
D -3σd10ε0

The correct answer is:
2σd5ε0

Solution:
Vf-Vi=-E→.S→ (for uniform field)
VB-VA=-σ2ε0dcos⁡37°=-2σd5ε0
⇒VA-VB=2σd5ε0

Topic: Physics, Electrostatics, JEE Main
An infinite wire placed along z-axis, has current I 1 in positive z-direction. A conducting rod placed in xy plane parallel to y-axis has current I 2 in positive y-direction. The ends of the rod subtend angles of +30o and -60o at the origin with positive x-direction. The rod is at a distance 'a' from the origin. Find net force on the rod.

Options:
A F = Î¼ 0 4 Ï I 1 I 2 â n 3 - k ^
B F = μ 0 4 π 2I 1 I 2 â n 3 - k &Hat;
C F = μ 0 4 π I 1 I 2 â n 3 k &Hat;
D F = μ 0 4 π 2 I 1 I 2 â n 3 k &Hat;

The correct answer is:
F = μ 0 4 π I 1 I 2 ℓ n 3 - k ^

Solution:
x a = tan Î¸
x = a tan Î¸ 1
d x = a sec 2 Î¸ d Î¸
B = Î¼ 0 4 Ï 2 I 1 r Â· Â· Â· 2
dF â = I 2 dx B sin Î¸ - k ^ Â· Â· Â· 3
From (1),(2) & (3)
dF â = I 2 asec 2 Î¸ d Î¸ . Î¼ 0 4 Ï 2 I 1 a sec Î¸ . sin Î¸ = Î¼ 0 4 Ï 2 I 1 I 2 sin Î¸ sec Î¸ d Î¸ ( â k ^ )
F â = Î¼ o 4Ï 2 I 1 I 2 ââââ â« â30 â 60 â tanÎ¸dÎ¸= Î¼ o 4Ï 2 I 1 I 2 [ â log e cosÎ¸ ] â 30 â 60 â ( â k ^ ) F â = Î¼ o 4Ï 2 I 1 I 2 [ â log e cos 60 â + log e cos( â 30 â ) ]= Î¼ o 4Ï 2 I 1 I 2 [ log e 3 2 Ã2 ] ( â k ^ ) F â = Î¼ o 4Ï I 1 I 2 ln3( â k ^ )

Topic: Physics, Magnetic Effects of Current, JEE Main
An infinite wire placed along z-axis, has current \( \mathrm{I}_{1} \) in positive z-direction. A conducting rod placed in xy plane parallel to y- axis has current \( \mathrm{I}_{2} \) in positive y-direction. The ends of the rod subtend angles of \( +30^{\circ} \) and \( -60^{\circ} \) at the origin with positive \( \mathrm{x}- \) direction. The rod is at a distance 'a' from the origin. Find net force on the rod.

Options:
A \( \mathrm{F}=\frac{\mu_{0}}{4 \pi} \mathrm{I}_{1} \mathrm{I}_{2} \ell \ln 3(-\widehat{\mathrm{k}}) \)
B \( \mathrm{F}=\frac{\mu_{0}}{4 \pi} 2 \mathrm{I}_{1} \mathrm{I}_{2} \ell \mathrm{n} 3(-\widehat{\mathrm{k}}) \)
C \( \mathrm{F}=\frac{\mu_{0}}{4 \pi} \mathrm{I}_{1} \mathrm{I}_{2} \ell \mathrm{n} 3(\widehat{\mathrm{k}}) \)
D \( \mathrm{F}=\frac{\mu_{0}}{4 \pi} 2 \mathrm{I}_{1} \mathrm{I}_{2} \ell \mathrm{n} 3(\widehat{\mathrm{k}}) \)

The correct answer is:
\( \mathrm{F}=\frac{\mu_{0}}{4 \pi} \mathrm{I}_{1} \mathrm{I}_{2} \ell \ln 3(-\widehat{\mathrm{k}}) \)

Solution:
x a = tan θ
x = a tan θ 1
d x = a sec 2 θ d θ
B = μ 0 4 π 2 I 1 r · · · 2
dF → = I 2 dx B sin θ - k ^ · · · 3
From (1), (2) & (3)
dF→=I2asec2θdθμ04π2I1asecθsinθ (−k^)

=μ04π2I1I2sinθsecθdθ(−k^)

F=μ02πI1I2∫-60°30°sinθsecθdθ (−k^)

F=μ02πI1I2lncosθ30°-60° (k^)
⇒F=μ04πI1I2ℓn3-k^

Topic: Physics, Magnetic Effects of Current, JEE Main
An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is a and distance of its centre from the wire is d d≫a. If the loop applies a force F on the wire then:

Options:
A F∝ad2
B F=0
C F∝a2d3
D F∝ad

The correct answer is:
F∝ad2

Solution:
U=-P→.B-
F=-∂U∂r=PdBdr
F=i2πa2×ddrμ0i12πr
F=i1i2π0a221r2
F∝ar2
As r=d
F∝a2d2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (09 Jan Shift 1)
An infinitely long hollow conducting cylinder with inner radius R2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, B→ as a function of the radial distance r from the axis, is best represented by:

Options:
A
B
C
D

The correct answer is:

Solution:
r<R2 , B=0
r>R , B=μ0I2πr
R2<r<R , B=μ0I2πrr2-R22R2-R22

Topic: Physics, Magnetic Effects of Current, JEE Main
An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, B as a function of the radial distance r from the axis, is best represented by

Options:
A
B
C
D

The correct answer is:

Solution:
The required cylinder is shown in figure
Let j= current density, then
For, r<R2

B=0
For, R2â¤râ¤R

BÃ2Ïr=Î¼0jÏr2-R24
B=Î¼0j2r-R24r
For, r>R

BÃ2Ïr=Î¼0jÏR2-R24
B=Î¼0j2â2R24r
âBâ1/r [this is an equation for rectangular hyperbola.]

Topic: Physics, Magnetic Effects of Current, JEE Main
An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is at a distance u>f from the mirror. Its image will have a length

Options:
A f 2 u - f
B u f u - f
C f 2 u + f
D u f u + f

The correct answer is:
f 2 u - f

Solution:
1 v + 1 u = 1 f â let v 1 = distance of near end of image
1 v 1 + 1 - u = 1 - f â 1 v 1 = 1 u + 1 - f â 1 v 1 = f - u u f
v 1 = u f f - u
â1v2+1â=1-fâv2=-f ( v2=the distance of last end of image )
Length of Imageâv1-v2=uff-u--fâv1-v2=uf-uf+f2u-f
v 1 - v 2 = f 2 u - f

Topic: Physics, Ray Optics, JEE Main
An infinitely long solid cylinder of radius R with uniform volume charge density ρ has a spherical cavity of radius R 2 with its centre on the axis of the cylinder as shown in the figure. The magnitude of the electric field at a point P which is at a distance 2R from the axis of the cylinder is given by 23ρR6Kε0. What is the value of K ?

Options:

The correct answer is:
16.00

Solution:
The electric field at the point P due to the entire cylinder is

E1=λ2πε0(2R)=ρπR24πε0R=ρR4ε0

The electric field at the P due to the spherical cavity is
E2=14πε0ρ·43πR38(2R)2=ρR(24×4)ε0

Enet=E1-E2

=ρR4ε0-ρR96ε0

=ρR4ε01-124

=2396ρRε0=236×16ρRε0

∴ K=16

Topic: Physics, Electrostatics, JEE Main
An infinitely long straight conductor is bent into the shape as shown below. It carries a current of I ampere and the radius of the circular loop is R metre. Then, the magnitude of magnetic induction at the centre of the circular loop is

Options:
A $\frac{\mu_0 I}{2 \pi R}$
B $\frac{\mu_0 n I}{2 R}$
C $\frac{\mu_0 I}{2 \pi R}(\pi+1)$
D $\frac{\mu_0 I}{2 \pi R}(\pi-1)$

The correct answer is:
$\frac{\mu_0 I}{2 \pi R}(\pi-1)$

Solution:
Magnetic field due to long wire at $O$ point
$B_1=\frac{\mu_0}{2 \pi}\left(\frac{I}{R}\right)$
(upward direction)
Magnetic field due to loop at $O$ point
$\begin{aligned}
& B_2=\frac{\mu_0}{4 \pi} \cdot \frac{I \cdot 2 \pi R}{R^2} \\
& B_2=\frac{\mu_0}{2} \cdot \frac{I}{R} \quad \text { (in upward direction) }
\end{aligned}$
Hence, resultant magnetic field at centre $O$
$\begin{aligned}
B & =B_1+B_2 \\
B & =\frac{\mu_0 I}{2 \pi \cdot R}(\pi+1) \mathrm{T}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
An infinitely long straight wire carrying current I, one side opened rectangular loop and a conductor C with a sliding connector are located in the same plane, as shown in the figure. The connector has length l. and resistance R. It slides to the right with a velocity v. The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation r, between the connector and the straight wire is

Options:
A μ04πIvℓRr
B μ0πIvℓRr
C 2μ0πIvℓRr
D μ02πIvℓRr

The correct answer is:
μ02πIvℓRr

Solution:
The induced current in the wire will be due to the motional EMF developed in the moving resistor and that back emf developed at that point will be equal to VBL

V=velocity of the conductor

B=perpendicular magnetic field

L=length of conductor

We know magnetic field due to a long wire is

B=μ0i2πr

Hence the induced current across wire will be :

I=eR

I=μ02πIvℓRr

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2020 (05 Sep Shift 2)
An infinitely long wire carrying current I is along Y axis such that its one end is at point A(0, b) while the wire extends upto + â . The magnitude of magnetic field strength at point (a, 0).

Options:
A Î¼0I4Ïa1+ba2Â +Â b2
B Î¼0I4Ïa1-ba2 + b2
C μ0I4πa1-aa2 + b2
D None of these

The correct answer is:
Î¼0I4Ïa1-ba2 + b2

Solution:
B=Î¼0I4Ï ia sinâ¡90o+sinâ¡-Î¸=Î¼04Ïia 1-sinâ¡Î¸
=Î¼04Ïia 1-ba2+b2

Topic: Physics, Magnetic Effects of Current, JEE Main
An inflated rubber balloon contains one mole of an ideal gas, has a pressure $p$, volume $V$ and temperature $T$. If the temperature rises to $1.1 \mathrm{~T}$, and the volume is increased to $1.05 \mathrm{~V}$, the final pressure will be

Options:
A
$1.1 p$

B
$\mathrm{p}$

C
less than $p$

D
between $p$ and $1.1$

The correct answer is:
between $p$ and $1.1$

Solution:
As we know that an ideal gas equation, $p V=n R T$
where, $n=$ Number of moles, $p=$ Pressure, $V=$ Volume, $R=$ Gas constant,
$T=$ Temperature
$$
n=\frac{p V}{R T}
$$
If $n, R$ are constant for the system or as number of moles of the gas remains fixed, hence, we can write
$$
\frac{p V}{T}=\text { constant }
$$
$$
\text { or } \begin{aligned}
\frac{p_1 V_1}{R T_1}=\frac{p_2 V_2}{R T_2} \\
p_2 &=\left(p_1 V_1\right)\left(\frac{T_2}{V_2 T_1}\right) \\
&=\frac{(p)(V)(1.1 T)}{(1.05) V(T)} \quad\left[p_1=p, V_2=1.05 \mathrm{~V}\right.\\
&\left.=p \times\left(\frac{1.1}{1.05}\right) \quad \text { and } T_2=1.1 T\right] \\
&=p(1.0476)
\end{aligned}
$$
So, final pressure $p_2$ will lies between $p$ and $1.1 p$.

Topic: Physics, Kinetic Theory of Gases
An initially parallel cylindrical beam travels in a medium of refractive index $\mu(I)=\mu_0+\mu_2 I$, where $\mu_0$ and $\mu_2$ are positive constants and $\mathrm{I}$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
The speed of light in the medium is

Options:
A
minimum on the axis of the beam

B
the same everywhere in the beam

C
directly proportional to the intensity I

D
maximum on the axis of the beam

The correct answer is:
minimum on the axis of the beam

Solution:
No Solution Available

Topic: Physics, Ray Optics, JEE Main, JEE Main 2010
An initially parallel cylindrical beam travels in a medium of refractive index $\mu(I)=\mu_0+\mu_2 I$, where $\mu_0$ and $\mu_2$ are positive constants and $\mathrm{I}$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
The initial shape of the wave front of the beam is

Options:
A
convex

B
concave

C
convex near the axis and concave near the periphery

D
planar

The correct answer is:
planar

Solution:
For a parallel cylindrical beam, wavefront will be planar.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2010
An initially parallel cylindrical beam travels in a medium of refractive index $\mu(I)=\mu_0+\mu_2 I$, where $\mu_0$ and $\mu_2$ are positive constants and $\mathrm{I}$ is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.
As the beam enters the medium, it will

Options:
A
diverge

B
converge

C
diverge near the axis and converge near the periphery

D
travel as a cylindrical beam

The correct answer is:
converge

Solution:
As intensity is maximum at axis,
$\therefore \quad \mu$ will be maximum and speed will be minimum on the axis of the beam.
$\therefore \quad$ beam will converge.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2010
An inorganic compound A made of two most occurring elements in the earth’s crust and it is used in building construction when made to react with carbon forms a poisonous gas B which is diatomic molecule. Compounds A and B are

Options:
A SiO2,CO2
B SiO2,CO
C SiO2,N2
D CaO,CO2

The correct answer is:
SiO2,CO

Solution:
SiO2 is a compound of oxygen and silicon most abundant elements of earth’s crust and is used in construction.
SiO2+2C→Si+2CO (Poisonous gas and diatomic molecule)

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
An inorganic Compound 'X' on treatment with concentrated H2SO4 produces brown fumes and gives dark brown ring with FeSO4 in presence of concentrated H2SO4. Also Compound 'X' gives precipitate 'Y', when its solution in dilute HCl is treated with H2S gas. The precipitate 'Y' on treatment with concentrated HNO3 followed by excess of NH4OH further gives deep blue coloured solution, Compound 'X' is:

Options:
A CoNO32
B PbNO22
C CuNO32
D PbNO32

The correct answer is:
CuNO32

Solution:
∴X→CuNO32

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (20 Jul Shift 1)
An inorganic substance liberates oxygen on heating and turns and acidic solution of Kl brown and reduces acidified KMnO4 solution. The substance is

Options:
A HgO
B H2O2
C KNO3
D PbNO32

The correct answer is:
H2O2

Solution:
It is H2O2 and the reactions are as follows
2H2O2→Heat2H2O+O2
2KI+H2O2→Acidic2KOH+I2Brown

Topic: Chemistry, s Block Elements, JEE Main
An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is $1 / 3$. If the line joining the centre of the hemispherical surface to the insect makes an angle $\alpha$ with the vertical, the maximum possible value of $\alpha$ so that the insect does not slip is given by

Options:
A
$\cot \alpha=3$

B
$\sec \alpha=3$

C
$\operatorname{cosec} \alpha=3$

D
$\cos \alpha=3$

The correct answer is:
$\cot \alpha=3$

Solution:
The insect crawls up the bowl upto a certain height $h$ only till the component of its weight along the bowl is balanced by limiting frictional force.
For limiting condition at point $\mathrm{A}$ $R=m g \cos \alpha$
$F_1=m g \sin \alpha$
Dividing eq. (ii) by (i)
$\tan \alpha=\frac{1}{\cot \alpha}=\frac{F_1}{R}=\mu\left[\right.$ As $\left.F_1=\mu R\right]$ $\Rightarrow \quad \tan \alpha=\mu=\frac{1}{3}\left[\because \mu=\frac{1}{3}(\right.$ Given $\left.]\right)$ $\therefore \quad \cot \alpha=3$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (12 May Online)
An insect is at the bottom of a hemispherical ditch of radius 1m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then his :g=10 m s-2

Options:
A 0.20 m
B 0.45 m
C 0.60 m
D 0.80 m

The correct answer is:
0.20 m

Solution:
mgsinθ=μmgcosθ

tanθ=μ=0.75

⇒θ=37

cosθ=R-hR

cos37=1-h1⇒h=0.2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (06 Sep Shift 1)
An insect starts from the origin in the argand plane and goes 4 km (N 45° E) then it moves 3 km N 45°W and then takes an angular movement of π3 about origin in the anticlockwise direction. The final position of the insect is

Options:
A 4-3iei5π6
B 4+3iei5π6
C 4-3iei3π4
D 4+3iei7π12

The correct answer is:
4+3iei7π12

Solution:
Travelling 4 km N 45°E it reaches at z1=4eiπ4
Travelling 3 km N 45°W it reaches at z2
So, 0-z1z2-z1=43i⇒z2=4+3ieiπ4
After an angular movement ofπ3 in the anticlockwise direction it reaches at z3
∴z3=z2eiπ3=4+3iei7π12

Topic: Mathematics, Complex Number, JEE Main
An installation consists of an electric motor which drives a water pump to lift 75 L of water per second to a height of 5 m, where water is disbursed at negligible speed. If the motor consumes a power of 5 kW, then what is the efficiency (%) of the installation? [g=10 m s-2]

Options:

The correct answer is:
75

Solution:
Power consumed by the motor

Pmotor=5000 W

Power used in lifting the water

Pwater=mght=75×10×51=3750 W

The efficiency of the motor is

η=PwaterPmotor×100%

η=37505000×100%=75%

Topic: Physics, Work Power Energy, JEE Main
An instantaneous displacement of a simple harmonic oscillator is $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 4) .$ Its speed will be maximum at time

Options:
A $\pi / 4 \omega$
B $\pi / 2 \omega$
C $\pi / \omega$
D $2 \pi \omega$

The correct answer is:
$\pi / 4 \omega$

Solution:
Velocity, $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{A} \omega \sin (\omega \mathrm{t}+\pi / 4)$

Velocity will be maximum, when $\omega t+\pi / 4=\pi / 2$ or $\omega t=\pi / 2-\pi / 4=\pi / 4$

or $t=\pi / 4 \omega$

Topic: Physics, Oscillations, JEE Main
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $P_1$ and temperature $T_1$. The other chamber has volume $V_2$ and contains ideal gas at pressure $P_2$ and temperature $T_2$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Options:
A
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}$

B
$\frac{P_1 V_1 T_1+P_2 V_2 T_2}{P_1 V_1+P_2 V_2}$

C
$\frac{P_1 V_1 T_2+P_2 V_2 T_1}{P_1 V_1+P_2 V_2}$

D
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_1+P_2 V_2 T_2}$

The correct answer is:
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}$

Solution:
$$
\begin{aligned}
& U=U_1+U_2 \\
& T=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{\left(P_1 V_1 T_2+P_2 V_2 T_1\right)}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_1$ and contains ideal gas at pressure $P_1$ and temperature $T_1$. The other chamber has volume $V_2$ and contains ideal gas at pressure $P_2$ and temperature $T_2$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

Options:
A
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}$

B
$\frac{P_1 V_1 T_1+P_2 V_2 T_2}{P_1 V_1+P_2 V_2}$

C
$\frac{P_1 V_1 T_2+P_2 V_2 T_1}{P_1 V_1+P_2 V_2}$

D
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_1+P_2 V_2 T_2}$

The correct answer is:
$\frac{T_1 T_2\left(P_1 V_1+P_2 V_2\right)}{P_1 V_1 T_2+P_2 V_2 T_1}$

Solution:
$$
\begin{aligned}
& U=U_1+U_2 \\
& T=\frac{\left(P_1 V_1+P_2 V_2\right) T_1 T_2}{\left(P_1 V_1 T_2+P_2 V_2 T_1\right)}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2008