datavorous/entrance-exam-dataset huggingface dataset:
Three distinct points A, B and C are given in the 2 - dimensional coordinate plane such that the ratio of the distance of any one of them from the point $(1,0)$ to the distance from the point $(-1,0)$ is equal to $\frac{1}{3}$. Then the circumcentre of the triangle $A B C$ is at the point

Options:
A
$(0,0)$

B
$\left(\frac{5}{4}, 0\right)$

C
$\left(\frac{5}{2}, 0\right)$

D
$\left(\frac{5}{3}, 0\right)$

The correct answer is:
$\left(\frac{5}{4}, 0\right)$

Solution:
$$
\begin{aligned}
& P=(1,0) ; Q(-1,0) \\
& \text { Let } A=(x, y) \\
& \frac{A P}{A Q}=\frac{B P}{B Q}=\frac{C P}{C Q}=\frac{1}{3} \\
& \Rightarrow 3 A P=A Q \Rightarrow 9 A P^2=A Q^2 \Rightarrow 9(x-1)^2+9 y^2=(x+1)^2+y^2 \\
& \Rightarrow 9 x^2-18 x+9+9 y^2=x^2+2 x+1+y^2 \Rightarrow 8 x^2-20 x+8 y^2+8=0 \\
& \Rightarrow x^2+y^2-\frac{5}{2} x+1=0
\end{aligned}
$$
$\therefore$ A lies on the circle
Similarly B, C are also lies on the same circle
$\therefore$ Circumcentre of $\mathrm{ABC}=$ Centre of Circle $(1)=\left(\frac{5}{4}, 0\right)$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2009
Three elements X,Y and Z are in the 3rd period of the periodic table. The oxides ofX,Y and Z, respectively, are basic, amphoteric and acidic. The correct order of the atomic numbers of X, Y and Z is:

Options:
A Z < Y < X
B X< Y < Z
C X < Z < Y
D Y < X < Z

The correct answer is:
X< Y < Z

Solution:
On moving left to right in a period. Acidic character of oxides is increase. 3rd period element oxides. Na2O ⏟Basic MgO Al2O3⏟Amphoteric SiO2 P2O5 Cl2O7⏟Acidic
(i) Acidic character↑
(i) Atomic No↑so X have minimum Atomic No
& Z have maxima Atomic No So correct order is X<Y<Z

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Three elements X, Y, and Z have atomic numbers 19, 37, and 55 respectively. Then the correct statements(s) is/are

Options:
A their ionization potential would increase with increasing atomic numbers
B Y would have an ionization potential between those of X and Z
C Y would have the highest ionization potential
D Z would have the highest ionization potential

The correct answer is:
Y would have an ionization potential between those of X and Z

Solution:
The elements X(19), Y(37) and Z(55) are the elements of same group (IA). As atomic number, increases, the ionization potential decrease down the group. Since, the position of Y is in between X and Z. Thus, the ionization potential of Y will also be in between X and Z.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Three Faraday of electricity is passed through aqueous solutions of AgNO3,NiSO4 and CrCl3 kept in three vessels using inert electrodes. The ratio in mol in which the metals Ag, Ni and Cr will be deposited is

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

The correct answer is:
6:3:2

Solution:
(i) Ag+aq1 mol = 1F3 F+e-âAgs1 mol3 mol
(ii) Ni2+aq1 mol = 2F3 F+2e-âNis1 mol32 mol
(iii) Cr 3+ aq 1 mol = 3F +3 e − → Cr s 1 mol
The required ratio of moles of Ag, Ni and Cr is
3 mol Ag: 3/2 mol Ni : 1 mol Cr or 6 mol Ag : 3 mol Ni : 2 mol Cr.

Topic: Chemistry, Electrochemistry, JEE Main
Three faradays of electricity are passed through molten $\mathrm{Al}_2 \mathrm{O}_3$, aqueous solution of $\mathrm{CuSO}_4$ and molten $\mathrm{NaCl}$ taken in different electrolytic cells. The amount of $\mathrm{Al}, \mathrm{Cu}$ and $\mathrm{Na}$ deposited at the cathodes will be in the ratio of]

Options:
A 1 mole : 2 mole : 3 mole
B 3 mole : 2 mole : 1 mole
C 1 mole : 1.5 mole : 3 mole
D 1.5 mole : 2 mole : 3 mole

The correct answer is:
1 mole : 1.5 mole : 3 mole

Solution:
At cathode : $\mathrm{Al}^{3+}+3 e^{-} \rightarrow A l$
$E_{A l}=\frac{\text { Atomic mass }}{3}$
At cathode : $\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}$
$E_{C_l}=\frac{\text { Atomic mass }}{2}$
At cathode : $\mathrm{Na}^{+}+e^{-} \rightarrow \mathrm{Na}$
$E_{\mathrm{Na}}=\frac{\text { Atomic mass }}{1}$
For the passage of 3 faraday; mole atoms of deposited $=1$
mole atoms of $C u$ deposited $=\frac{1 \times 3}{2}=1.5$
mole atoms of $\mathrm{Na}$ deposited $=1 \times 3=3$

Topic: Chemistry, Electrochemistry, JEE Main
Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$-direction, the magnitude of the minimum additional force needed is

Options:
A $0.5 \mathrm{~N}$
B $1.5 \mathrm{~N}$
C $\frac{\sqrt{3}}{4} \mathrm{~N}$
D $\sqrt{3} \mathrm{~N}$

The correct answer is:
$0.5 \mathrm{~N}$

Solution:
Minimum additional force needed
$\begin{aligned}
& F=-\left(F_{\text {resultant }}\right)_x \\
& F_{\text {resultant }}=[(4-2)(\cos 30 \hat{\mathbf{j}}-\sin 30 \hat{\mathbf{i}}) \\
= & {\left[2\left(\frac{\sqrt{3}}{2} \hat{\mathbf{j}}-\frac{1}{2} \hat{\mathbf{i}}\right)+\left(\frac{1}{2} \hat{\mathbf{i}}+\frac{\sqrt{3}}{2} \hat{\mathbf{j}}\right)\right] } \\
= & {\left[\left(\sqrt{3}+\frac{\sqrt{3}}{2}\right) \hat{\mathbf{j}}+\sin 60 \hat{\mathbf{j}}\right) } \\
= & {\left.\left[-\frac{1}{2} \hat{\mathbf{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathbf{i}}+\frac{\hat{\mathbf{i}}}{2}\right)\right] } \\
= & -\frac{\hat{\mathbf{i}}}{2}+\frac{3 \sqrt{3}}{2} \hat{\mathbf{j}} \quad F=-\left(-\frac{\hat{\mathbf{i}}}{2}\right)=\frac{1}{2} \hat{\mathbf{i}}\end{aligned}$
Hence,
$|F|=0.5 \mathrm{~N}$

Topic: Physics, Mathematics in Physics, JEE Main
Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is

Options:
A $\sqrt{3} \mathrm{~N}$
B $0.5 \mathrm{~N}$
C $1.5 \mathrm{~N}$
D $\frac{\sqrt{3}}{4} \mathrm{~N}$

The correct answer is:
$0.5 \mathrm{~N}$

Solution:
Net force along $x$-axis zero.
Let the unknown force be $F$ along $x$-axis
$$
\begin{aligned}
& \therefore F+1 \sin 30^{\circ}+2 \sin 30^{\circ}-4 \sin 30^{\circ}=0 \\
& \therefore|F|=0.5 \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Mathematics in Physics, NEET, NEET 2008 (Mains)
Three forces F1=10 N, F2=8 N, F3=6 N are acting on a particle of mass 5 kg. The forces F2 and F3 are applied perpendicularly so that particle remains at rest. If the force F1 is removed, then the acceleration of the particle is

Options:
A 7 m s–2
B 0.5 m s–2
C 4.8 m s–2
D 2 m s–2

The correct answer is:
2 m s–2

Solution:
When the force F1 is removed, the particle starts accelerating because of the resultant force between F2 and F3.

The resultant force can be calculated as follows:

F=F22+F32=10 N

Hence, the acceleration of the particle is given by

a=10 N5 kg=2 m s–2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Three forces $\vec{P}, \vec{Q}$ and $\vec{R}$ acting along IA, IB and IC, where I is the incentre of a $\triangle A B C$, are in equilibrium. Then $\vec{P}: \vec{Q}: \vec{R}$ is

Options:
A
$\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}$

B
$\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}$

C
$\sec \frac{A}{2}: \sec \frac{B}{2}: \sec \frac{C}{2}$

D
$\operatorname{cosec} \frac{A}{2}: \operatorname{cosec} \frac{B}{2}: \operatorname{cosec} \frac{C}{2}$

The correct answer is:
$\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}$

Solution:
By Lami's theorem
$$
\begin{aligned}
& \vec{P}: \vec{Q}: \vec{R}=\sin \left(90^{\circ}+\frac{A}{2}\right): \sin \left(90^{\circ}+\frac{B}{2}\right): \sin \left(90^{\circ}+\frac{C}{2}\right) \\
& \Rightarrow \cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2} .
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
Three forces start acting simultaneously on a particle moving with velocity, $\overrightarrow{\mathrm{v}}$. These forces are represented in magnitude and direction by the three sides of a triangle $\mathrm{ABC}$. The particle will now move with velocity

Options:
A
less than $\overrightarrow{\mathrm{v}}$

B
greater than $\overrightarrow{\mathrm{v}}$

C
$|\mathrm{v}|$ in the direction of the largest force $\mathrm{BC}$

D
$\overrightarrow{\mathrm{v}}$, remaining unchanged

The correct answer is:
$\overrightarrow{\mathrm{v}}$, remaining unchanged

Solution:
According to triangle law of forces, the resultant force is zero.
In presence of zero external force, there is no change in velocity

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2003
Three harmonic waves having equal frequency v and same intensity I0 , have phase angles 0,π4 and -π4 respectively. When they are superimposed the intensity of the resultant wave is close to:

Options:
A 5.8I0
B 0.2I0
C 3I0
D I0

The correct answer is:
5.8I0

Solution:
AR=2+1A
IR=2+12I0
=3+22I0=5.8I0

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

Options:
A
$\frac{2}{9}$

B
$\frac{1}{9}$

C
$\frac{8}{9}$

D
$\frac{7}{9}$

The correct answer is:
$\frac{1}{9}$

Solution:
For a particular house being selected
Probability $=\frac{1}{3}$
Prob(all the persons apply for the same house) $=\left(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\right) 3=\frac{1}{9}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2005
Three identical balls are connected by light inextensible strings with each other as shown and placed over a smooth horizontal table. Length of each string is l .
At moment t=0 , ball B is imparted a velocity v0 perpendicular to the string and then the system is left on its own. The speed of A at the instant just before it collides with C is

Options:
A v03
B 2v03
C 2v05
D 2v09

The correct answer is:
2v03

Solution:
Using conservation of linear momentum along the initial direction of motion of B we get,

mv0=3mv

v=v03

The total kinetic energy of the system will remain the same
12mv02=12mv2+212mv2+v12

where v1 is the component of velocity of A and C perpendicular to the initial direction of motion of B

v02=v032+2v032+2v12

v12=v023

Velocity of A=v2+v12=23v0

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three identical bar magnets are rivetted together at centre in the same plane as shown in figure. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the figure. Determine the poles of the remaining two.

Options:

None

Solution:
The system will be in stable equilibrium if the net force on the system is zero and net torque on the system is also zero. That is possible only when the poles of the remaining two magnets are as given figure.
The north pole of magnet $(1)$ is equally attracted by south poles of (2) and (3) magnets placed at equal distance.

Topic: Physics, Magnetic Properties of Matter
Three identical bar magnets each of magnetic moment M are arranged in the form of an equilateral triangle such that at two vertices like poles are in contact. The resultant magnetic moment will be

Options:
A Zero
B 2M
C 2 M
D M3

The correct answer is:
2M

Solution:
Magnetic dipole moment is a vector quantity.
Mres=3M2+M2
Mres=2M

Topic: Physics, Magnetic Properties of Matter, JEE Main
Three identical bars A, B and C are made of different magnetic materials. When kept in a uniform magnetic field, the field lines around them look as follows:
Make the correspondence of these bars with their material being diamagnetic ( $\mathrm{D})$, ferromagnetic (F) and paramagnetic $(\mathrm{P})$ :

Options:
A
$\mathrm{A} \leftrightarrow \mathrm{D}, \mathrm{B} \leftrightarrow \mathrm{P}, \mathrm{C} \leftrightarrow \mathrm{F}$

B
$\mathrm{A} \leftrightarrow \mathrm{F}, \mathrm{B} \leftrightarrow \mathrm{D}, \mathrm{C} \leftrightarrow \mathrm{P}$

C
$\mathrm{A} \leftrightarrow \mathrm{P}, \mathrm{B} \leftrightarrow \mathrm{F}, \mathrm{C} \leftrightarrow \mathrm{D}$

D
$\mathrm{A} \leftrightarrow \mathrm{F}, \mathrm{B} \leftrightarrow \mathrm{P}, \mathrm{C} \leftrightarrow \mathrm{D}$

The correct answer is:
$\mathrm{A} \leftrightarrow \mathrm{F}, \mathrm{B} \leftrightarrow \mathrm{D}, \mathrm{C} \leftrightarrow \mathrm{P}$

Solution:
Diamagnetic materials are repelled in an external magnetic field.
Bar $B$ represents diamagnetic materials.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2014 (11 Apr Online)
Three identical blocks A, B and C are placed on a horizontal frictionless surface. The blocks B and C are at rest but A is approaching towards B with a speed 10 m s-1. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after the collision is

Options:
A 5.6 m s-1
B 6 m s-1
C 8 m s-1
D 10 m s-1

The correct answer is:
5.6 m s-1

Solution:
For collision between blocks A and B,
e=vB-vAuA-uB=vB-vA10-0=vB-vA10
∴vB-vA=10e=10×0.5=5 ….(i)
From principle of momentum conservation,
mAuA+mBuB=mAvA+mBvB
m×10+0=mvA+mvB
∴ vA+vB=10 ….(ii)
Adding eqs. (i) and (ii), we get
vB=7.5 m s-1 …(iii)
Similarly for collision between B and C,
vC-vB=7.5e=7.5×0.5=3.75
∴ vC-vB=3.75 m s-1 ...(iv)
Adding Eqs. (iii) and (iv) we get
2vC=11.25
∴ vC=11.252=5.6 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three identical blocks of masses $\mathrm{m}=2 \mathrm{~kg}$ are drawn by a force $\mathrm{F}=10.2 \mathrm{~N}$ with an acceleration of $0.6 \mathrm{~ms}^{-2}$ on a frictions surface, then what is the tension (in $\mathrm{N}$ ) in the string between the blocks $B$ and $C$?

Options:
A
9.2

B
7.8

C
4

D
9.8

The correct answer is:
7.8

Solution:
Apply Newton's second law
$$
\mathrm{F}-\mathrm{T}_{\mathrm{ab}}=m a ; \mathrm{T}_{\mathrm{ab}}-\mathrm{T}_{\mathrm{bc}}=\mathrm{ma} \quad \therefore \mathrm{T}_{\mathrm{bc}}=7.8 \mathrm{~N}
$$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2002
Three identical capacitors are combined differently. For the same voltage to each combination, the one that stores the greatest energy is

Options:
A Two in parallel and the third in series with it
B Three in series
C Three in parallel
D Two in series and third in parallel with it

The correct answer is:
Three in parallel

Solution:
$U=\frac{1}{2} C V^2$
Now if $V$ is constant, then $U$ is greatest when ' $C_{e q}$ ' is maximum. This is when all the three are in parallel.

Topic: Physics, Capacitance, JEE Main
Three identical charged balls each of charge 2C are suspended from a common point P by silk threads of 2 m each (as shown in figure). They form an equilateral triangle of side 1 m.

The ratio of net force on a charged ball to the force between any two charged balls will be

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

The correct answer is:
3:1

Solution:
We have force between any two charge F1=F2=q24πε0l2=F

Resultant electrostatics force FR=F12+F22+2F1F2cos60°

=F2+F2+2F2×12=3F2

So, the ratio of net force on a charged ball to the force between any charged balls is

⇒FRF=3

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Three identical particle A,B and C of mass 100 kg each are placed in a straight line with AB=BC=13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately

Options:
A 21G
B 100G
C 59G
D 42G

The correct answer is:
100G

Solution:
Given here, m=100 kg

Gravitational force, FAP=Gm21322,

FBP=Gm2132 and FCP=Gm21322

Net gravitational force,

Fnet=FBP+FAPcos45°+FCPcos45°

=G×1041321+12

Fâ100G

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Three identical resistors with resistance R=12 Ω and two identical inductors with sell inductance L=5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be________A.

Options:

The correct answer is:
3

Solution:
We know that after long time an inductor offers zero resistance i.e., in steady state condition.

Here all three identical resistors are in parallel, so equivalent resistance is, Req=123=4 Ω

Now, current through the battery is I=124=3 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (24 Jan Shift 2)
Three identical rods $A, B$ and $C$ are placed end to end. A temperature difference is maintained between the free ends of $A$ and $C$. The thermal conductivity of $B$ is thrice that of $C$ and half of that of $A$. The effective thermal conductivity of the system will be $\left(K_{A}\right.$ is the thermal conductivity of $\left.\operatorname{rod} A\right)$

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

The correct answer is:
$\frac{1}{3} K_{A}$

Solution:
$\begin{array}{ll}\text { Given, } & K_{B}=K_{A} / 2, \\ \text { and } & K_{B}=3 K_{C} \\ \therefore & K_{C}=K_{A} / 6\end{array}$
Rods are in series form so
$$
\begin{gathered}
\frac{L}{K}=\frac{l_{1}}{K_{A}}+\frac{l_{2}}{K_{B}}+\frac{l_{3}}{K_{C}} \\
\frac{3 l}{K}=\frac{l}{K_{A}}+\frac{l}{K_{A} / 2}+\frac{l}{K_{A} / 6} \\
\frac{3 l}{K}=\frac{9 l}{K_{A}} \\
K=\frac{K_{A}}{3}
\end{gathered}
$$

Topic: Physics, Thermal Properties of Matter, KCET, KCET 2011
Three identical rods AB,CD and PQ are joined as shown P and Q are mid points of AB and CD respectively. Ends A, B, C and D are maintained at 0Â°C,100Â°C, 30Â°C and 60Â°C respectively. The direction of heat flow in PQ is

Options:
A From P to Q
B From Q to P
C Heat does not flow in PQ
D Data not sufficient

The correct answer is:
From P to Q

Solution:
assume that there is no 'PQ' Rod
Then TP=1002=50 âC
TQ=45 âC
TP>TQâ Heat flow from P to Q

Topic: Physics, Thermal Properties of Matter, NEET
Three identical rods AB, CD and PQ are joined as shown P and Q are midpoints of AB and CD respectively. Ends A, B, C and D are maintained at 0Â°C,100Â°C, 30Â°C and 60Â°C respectively. The direction of heat flow in PQ is

Options:
A From P to Q
B From Q to P
C Heat does not flow in PQ
D Data not sufficient

The correct answer is:
From P to Q

Solution:
assume that there is no 'PQ' Rod
Then TP=1002=50 âC
TQ=45 âC
TP>TQâ Heat flow from P to Q

Topic: Physics, Thermal Properties of Matter, NEET
Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be x m. The value of x is

Options:

The correct answer is:
2

Solution:
The diagram represents the locations of the masses as mentioned in the question.

For point 1, r1→=0i^+0j^ for point 2, r2→=3i^+0j^ and for point 3, r3→=0i^+3j^.

The formula for centre of mass is

r→com=m1r1→+m2r2→+m3r3→m1+m2+m3

⇒r→com=M0i^+0j^+M3i^+M3j^3M

⇒r→com=i^+j^

r→com=12+12=2=x

⇒x=2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Three identical spherical shells, each of mass m and radius r are placed as shown in the figure. Consider an axis XXâ² which is touching the two shells and passing through diameter of the third shell. Moment of inertia of the system consisting of these three spherical shells about XXâ² axis is:

Options:
A 115 mr2
B 3 mr2
C 165mr2
D 4 mr2

The correct answer is:
4 mr2

Solution:
Total MI of the system,
I=I1+I2+I3
I2=I3=23mr2+mr2=5mr23
I1=23mr2
â´I=2Ã5mr23+23mr2
=12mr23=4 mr2

Topic: Physics, Rotational Motion, JEE Main
Three identical square plates rotate about the axes shown in the figure in such a way that their kinetic energies are equal. Each of the rotation axes passes through the centre of the square. Then the ratio of angular speeds $\omega_{1}: \omega_{2}: \omega_{3}$ is

Options:
A 01:01:01
B $\sqrt{2}: \sqrt{2}: 1$
C $1: \sqrt{2}: 1$
D $1: 2: \sqrt{2}$

The correct answer is:
$\sqrt{2}: \sqrt{2}: 1$

Solution:
We know that
$$
K=\frac{1}{2} \mathrm{l} \omega^{2}
$$
where $K=$ kinetic energy
$$
\begin{array}{l}
I=\text { moment of inertia } \\
\omega=\text { angular speed }
\end{array}
$$
So, $\quad \omega \propto \frac{1}{\sqrt{I}}$
$$
\begin{aligned}
\omega_{1}: \omega_{2}: \omega_{3} &=\frac{1}{\sqrt{I_{1}}}: \frac{1}{\sqrt{I_{2}}}: \frac{1}{\sqrt{I_{3}}} \\
&=\frac{1}{\sqrt{1}}: \frac{1}{\sqrt{1}}: \frac{1}{\sqrt{2}}=\sqrt{2}: \sqrt{2}: 1
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
Three identical thin rods, each of length $L$ land mass $\mathrm{m}$, are welded perpendicular to one another as shown in figure. The assembly is rotated about an axis that passes through the end of one rod and is parallel to another. The moment of inertia of this structure about this axis is

Options:
A $\frac{7}{12} m L^2$
B $\frac{11}{14} m L^2$
C $\frac{5}{15} m L^2$
D $\frac{11}{12} m L^2$

The correct answer is:
$\frac{11}{12} m L^2$

Solution:
The moment of the rod on the y axis about the y axis itself is essentially zero (axis through axis centre', parallel to rod) because the rod is thin.
The moments of the rods on the $x$ and $z$ axes are each $I=\frac{1}{12} M L^2$ (Axis through centre, perpendicular to rod) from the table in the chapter.
The total moment of the three rods about the axis (and about the CM) is $I_{C M}=I_{\text {on } x \text { axis }}+I_{\text {on y axis }}+I_{\text {on } z \text { axis }}$.
$\left.=\frac{1}{12} M L^2+0+\frac{1}{12} M L^2=\frac{1}{2} m L^2 \right\rvert\,$
For the moment of the rod-combination about the axis of rotation, the parallel-axis theorem gives
$I=I_{C M}+3 m\left(\frac{L}{2}\right)^2=\left[\frac{1}{6}+\frac{3}{4}\right] m L^2\left|=\left[\frac{2}{12}+\frac{9}{12}\right] m L^2=\frac{11}{12} m L^2\right|$

Topic: Physics, Rotational Motion, JEE Main
Three identical thin rods, each of mass $m$ and length $\ell$, are joined to form an equilateral triangular frame. The moment of inertia of the frame about an axis parallel to its one side and passing through the opposite vertex is

Options:
A $\frac{5}{2} m \ell^2$
B $\frac{5}{4} m \ell^2$
C $\frac{3}{2} m \ell^2$
D $\frac{5}{3} m \ell^2$

The correct answer is:
$\frac{5}{4} m \ell^2$

Solution:
Moment of Inertia of each of rod $\mathrm{AC}$ and $\mathrm{BC}$ about the given axis $00^{\prime}$ is
$\left.I_{A C}=I_{B C}=\frac{m \ell^2}{3} \sin ^2 60^{\circ}=\frac{m \ell^2}{4} \right\rvert\,$ and $\mathrm{M}$.I. of rod $\mathrm{AB}$ about the given axis $0 0^{\prime}$ is
$I_{A B}=m\left(\frac{\ell \sqrt{3}}{2}\right)^2=\frac{3}{4} m \ell^2$
Hence, $\left.I=I_{A C}+I_{B C}+I_{A B}=\frac{m \ell^2}{4}+\frac{m \ell^2}{4}+\frac{3}{4} m \ell^2=\frac{5}{4} m \ell^2 \right\rvert\,$\n

Topic: Physics, Rotational Motion, JEE Main
Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is

Options:
A 2 σ ε 0 k &Hat;
B - 2 Ï Îµ 0 k ^
C 4 σ ε 0 k &Hat;
D - 4 σ ε 0 k &Hat;

The correct answer is:
- 2 σ ε 0 k ^

Solution:
E → P = σ 2 ε 0 - k ^ + - 2 σ 2 ε 0 k ^ + - σ 2 ε 0 k ^
= - 2 σ ε 0 k ^

Topic: Physics, Electrostatics, JEE Main
Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point $P$ is $\frac{x \sigma}{\epsilon_o}$. The value of $x$ is _______ (all quantities are measured in SI units).

Options:

The correct answer is:
2

Solution:
$\begin{aligned} \overrightarrow{\mathrm{E}}_{\mathrm{p}} & =\left(\frac{\sigma}{2 \varepsilon_0}+\frac{2 \sigma}{2 \varepsilon_0}+\frac{\sigma}{2 \varepsilon_0}\right)(-\hat{\mathrm{i}}) \\ & =-\frac{2 \sigma}{\varepsilon_0} \hat{\mathrm{i}}\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (06 Apr Shift 1)