datavorous/entrance-exam-dataset huggingface dataset:
As \( \lambda \) varies, the circles \( x^{2}+y^{2}+2 a x+2 b y+2 \lambda(a x-b y)=0 \) form a coaxial system. Then, the equation of the circle
which are orthogonal of the circles of the above system, is

Options:
A \( x^{2}+y^{2}+c\left(\frac{x}{a}+\frac{y}{b}\right)+c=0 \)
B \( x^{2}+y^{2}+\frac{c}{2}\left(\frac{x}{a}-\frac{y}{b}\right)+c=0 \)
C \( x^{2}+y^{2}+c\left(\frac{x}{a}-\frac{y}{b}\right)+c=0 \)
D \( x^{2}+y^{2}+\frac{c}{2}\left(\frac{x}{a}+\frac{y}{b}\right)+c=0 \)

The correct answer is:
\( x^{2}+y^{2}+\frac{c}{2}\left(\frac{x}{a}+\frac{y}{b}\right)+c=0 \)

Solution:
We know that, S+λL=0, λ∈R represents a family of coaxial circles having common radical axis L=0.

So, the circles

x2+y2+2ax+2by+2λax-by=0 …i,

represents a family of coaxial system of circles having common radical axis ax-by=0.

Let, x2+y2+2gx+2fy+c=0 …ii

be the circle orthogonal to the above system of coaxial circles.

Applying the condition of orthogonality of equations i & ii we get,

2ga1+λ+fb1-λ=c+0

⇒2ag+bf-c+2λag-bf=0∀λ∈R …iii

The equation iii is true for all values of λ, provided,

2ag-bf=0 and 2ag+bf-c=0

⇒ag=bf and ag+bf=c2

⇒ag=bf=c4

⇒g=c4a and f=c4b.

On substituting the values of g, f and c in equation ii, we get,

x2+y2+c2xa+yb+c=0.

This is the equation of the circle orthogonal to the given coaxial system of circles.

Topic: Mathematics, Circle, JEE Main
As per Einstein's theory, energy equivalent of $1 \mathrm{gm}$ (One gram) substance will be

Options:
A $5.6 \times 10^{26} \mathrm{MeV}$
B $9.09 \times 10^{13} \mathrm{MeV}$
C $5.6 \times 10^{29} \mathrm{MeV}$
D $5.6 \times 10^{32} \mathrm{MeV}$

The correct answer is:
$5.6 \times 10^{26} \mathrm{MeV}$

Solution:
Given, mass $\mathrm{m}=1 \mathrm{gm}=10^{-3} \mathrm{~kg}$
speed of light $\mathrm{c}=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$
Energy $E=\mathrm{mc}^2=\left(10^{-3} \mathrm{~kg}\right)\left(3 \times 10^8 \mathrm{~ms}^{-1}\right)^2=9 \times 10^{13}$ Joule
1 electron volt $=1.602 \times 10^{-19}$ Joule
So the given energy is $=6.242 \times 10^{18} \times 9 \times 10^{13} \mathrm{eV}$
Converting it in $\mathrm{MeV}$, we need to divide it by $10^{-3}$
The answer will be approx $5.6 \times 10^{26} \mathrm{Mev}$

Topic: Physics, Nuclear Physics, JEE Main
As per given figure A, B and C are the first, second and third excited energy levels of hydrogen atom respectively. If the ratio of the two wavelengthsi.e. λ1λ2 is 74n, then the value of n will be

Options:

The correct answer is:
5

Solution:
For λ1

1λ1=RZ2(122-132)=R536 ...(i)

For λ2,

1λ2=RZ2(132-142)=R7144 ...(ii)

Dividing (ii) by (i)

λ1λ2=7144536=720

So,

74n=720⇒n=5

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
As per given figure, a weightless pulley P is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if g=10 m s-2)

Options:
A 43+1 N
B 43+1 N
C 43-1 N
D 43-1 N

The correct answer is:
43+1 N

Solution:
The magnitude of acceleration of both the blocks will be same and is given by

a=4 g sin 60∘-g sin 30∘4+1

=203-55

a=43-1 m s-1

Applying Newton's second law for second block (along the direction parallel to the incline on which it is placed)

T-5=1×a

⇒T=5+43-1

=43+1 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (24 Jan Shift 1)
As per given figures, two springs of spring constants K and 2 K are connected to mass m. If the period of oscillation in figure (a) is 3 s, then the period of oscillation in figure (b) will be x s. The value of x is _____ .

Options:

The correct answer is:
2

Solution:
In figure (a), the springs are connected in series. Therefore,

Keq=K1K2K1+K2=K×2KK+2K=2K3

Therefore, the time period

Ta=2πmKeq=2πm2K3=2π3m2K

In figure (b), the springs are connected in parallel. Therefore,

Keq=K1+K2=3K

Therefore, the time period

Tb=2πmKeq=2πm3K

Now,

TaTb=92⇒Tb=23Ta=2 s

Hence, x=2

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (26 Jul Shift 2)
As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:

Options:
A K3FeCN6<K2CrO4<KBr=KNO3=AlCl3
B K3FeCN6<K2CrO4<AlCl3<KBr<KNO3
C AlCl3>K3FeCN6>K2CrO4>KBr=KNO3
D K3FeCN6>AlCl3>K2CrO4>KBr>KNO3

The correct answer is:
K3FeCN6<K2CrO4<KBr=KNO3=AlCl3

Solution:
According to hardy-schultz rule,
Coagulation value or flocculation value ∝1Coagulation power

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (08 Jan Shift 1)
As per the given figure, two blocks each of mass 250 g are connected to a spring of spring constant 2 N m-1. If both are given velocity v in opposite directions, then maximum elongation of the spring is

Options:
A v22
B v2
C v4
D v2

The correct answer is:
v2

Solution:
Let the maximum elongation of the spring be x.

Using energy conservation

Loss in kinetic energy of both blocks=Gain in spring energy

12mv2×2=12kx2

⇒0.25v2=12×2×x2

⇒14v2=12×2×x2

∴x=v2

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (26 Jul Shift 1)
As per the given figure, two plates A and B of thermal conductivity K and 2K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm2 for each plate. The equivalent thermal conductivity of the compound plate is 1+5αK, then the value of α will be _____ .

Options:

The correct answer is:
21

Solution:
The thermal resistance of rod 1 will be R1=l1KA=4KA and rod 2 will be R2=l22KA=2.52KA.

Since they are connected in series, Req=R1+R2=10.52KA.

If equivalent thermal conductivity is Keq, then Req=l1+l2KeqA=6.5KeqA. Therefore,

10.52KA=6.5KeqA⇒Keq=2621K=1+521K

Hence, α=21.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (29 Jun Shift 1)
As per the given graph, choose the correct representation for curve A and curve B
{Where XC= Reactance of pure capacitive circuit connected with A.C. source
XL= Reactance of pure inductive circuit connected with A.C. source
R= Impedance of pure resistive circuit connected with A.C. source
Z= Impedance of the LCR series circuit }

Options:
A A=XC,B=R
B A=XL,B=R
C A=XL,B=Z
D A=XC,B=XL

The correct answer is:
A=XC,B=XL

Solution:
The formula to calculate the inductive reactance is

XL=ωL ...1

The formula to calculate capacitive reactance is

XC=1ωC ...2

As can be seen from equation (1), the inductive reactance varies directly with the frequency and equation (2) indicates that capacitive reactance varies inversely with the frequency.

Hence, curve A represents the capacitive reactance and curve B represents the inductive reactance.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2023 (11 Apr Shift 1)
As per the postulates of Werner's theory for coordination compounds, which of the following is correct?

Options:
A Primary valencies are ionisable.
B Secondary valencies are ionisable.
C Only primary valencies are non-ionisable.
D Primary and secondary valencies are non-ionisable.

The correct answer is:
Primary valencies are ionisable.

Solution:
As per the postulates of Werner's theory for coordination compounds, metal atoms exhibit two types of valencies, i.e., primary valency and secondary valency. The primary valency is ionisable, whereas the secondary valency is non-ionisable.

Topic: Chemistry, Coordination Compounds, JEE Main
As secondary growth proceeds, in a dicot stem, the thickness of

Options:
A sapwood increases
B heartwood increases
C both sapwood and heartwood increases
D both sapwood and heartwood remains the same.

The correct answer is:
heartwood increases

Solution:
In a woody stem, due to secondary growth, secondary xylem constitutes the bulk of the stem and is commonly called wood. The secondary xylem consists of vessels, tracheids, wood parenchyma and wood fibres. Width of secondary xylem increases continuously with the age of the plant. In older stems, the wood (or secondary xylem) gets differentiated into two zones - the outer light coloured younger sapwood and the inner darker older heartwood. The xylem vessels and tracheids of heartwood become non-functional because they get filled with resins, tannins, gums etc. produced by adjacent living parenchymatous cells and due to tyloses. Changes in the nature of sapwood occur with the death of parenchyma cells and with the completion of these changes, heartwood is formed. Thus, the conversion of sapwood into heartwood is a progressive phenomenon which continues throughout the life of the plant. Initially, both sapwood and heartwood increase in size but as a tree reaches a certain age, amount of heartwood increases continuously and the amount of sapwood remains almost constant.

Topic: Biology, Anatomy of Flowering Plants, AIIMS, AIIMS 2016
As show in the figure, in steady state, the charge stored in the capacitor is _____×10-6C.

Options:

The correct answer is:
10

Solution:
At the steady state, there will be no current through the capacitor. Therefore, potential difference across capacitor VC=10R+rR=10×100100+10=10011 V

Charge on capacitor is q=CVC

=1.1×10-610011

⇒q=10 μC=10×10-6 C.

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (27 Jul Shift 2)
As shown in diagram, AB is a rod of length 30 cm and area of cross section 1.0 cm2 and thermal conductivity 336 SI units. The ends A and B are maintained at temperatures 20 Â°Cand 40 Â°C, respectively. A point C of this rod is connected to a box D, containing ice at 0 Â°C, through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is (assume latent heat of fusion for ice Lf=80 cal g-1)

Options:
A 84 mg s-1
B 84 g s-1
C 20 mg s-1
D 40 mg s-1

The correct answer is:
40 mg s-1

Solution:
Thermal resistance of AC = L KA = 0.1 3 3 6 Ã 1 0 - 4 = 1 0 3 3 3 6 = R let
Thermal resistance of BC = 0.2 3 3 6 Ã 1 0 - 4 = 2 R
Heat flow rates are
H 1 = 2 0 R ; H 2 = 4 0 2 R = 2 0 R
undefined
= 1 3 4 4 0 1 0 3 = 13.44 W
Rate of melting of ice
=HLf=13.44/4.280 g s-1=40 mg s-1

Topic: Physics, Thermal Properties of Matter, JEE Main
As shown in fig. when a spherical cavity (centred at $O$ ) of radius 1 is cut out of a uniform sphere of radius $R$ (centred at $C$ ), the centre of mass of remaining (shaded) part of sphere is at $G$, i.e on the surface of the cavity. $R$ can be determined by the equation :

Options:
A $\left(R^2+R-1\right)(2-R)=1$
B $\left(R^2-R+1\right)(2-R)=1$
C $\left(R^2+R+1\right)(2-R)=1$
D None of these

The correct answer is:
$\left(R^2+R+1\right)(2-R)=1$

Solution:
Option $3:\left(R^2+R+1\right)(2-R)=1$
CONCEPT:
Centre of mass (COM):
- Centre of mass of a body is a physical point that can be imagined to hold all of the mass of the body.
- In other words, the point along which if any external force is applied there will be only be transitional motion of the body and no rotational motion.
Mathematically, $x_{c m}=\sum m_i x_i / \sum m_i$ and $Y_{c m}=\sum m_i y_i / \sum m_i$
where, $\mathrm{m}_{\mathrm{i}}=$ mass of the ith particle, $\mathrm{x}_{\mathrm{i}}$ and $\mathrm{y}_{\mathrm{i}}$ are the $\mathrm{x}$ and $\mathrm{y}$ coordinates of the body.
EXPLANATION:
The center of the sphere is at $\mathrm{C}$ and the center of the cavity is at $\mathrm{O}$.
We use the superposition principle to solve the problem.
The sphere with a spherical cavity can be imagined as the sum of the sphere with a positive mass and a smaller sphere with a negative mass.
Let's take the mass density of the sphere as $+\rho$
and take the mass density of the cavity as - $\rho$
So, the mass of the sphere $m_1=$ volume $\times$ density $=(4 / 3) \pi R^3 \rho$ and the mass of the cavity $m_2=$ volume $\times$ density $=(4 / 3) \pi(1)^3(-\rho)=-(4 / 3) \pi \rho$
(where $\rho$ is the density of the material)
We take $C$ as origin $(x, y=0,0)$
The center of mass of the sphere is at $C(0,0)$.
From the figure $C O=C P-O P=R-1$
The center of mass of the cavity is at $O(R-1,0)$
Centre of mass of the whole system (sphere + cavity)
$\mathrm{X}_{\mathrm{cm}}=\frac{0 \times m_1+(R-1) \times m_2}{m_1+m_2}=\frac{(R-1) \times \frac{4}{3} \pi(1)^3(-\rho)}{\frac{4}{3} \pi R^3 \rho+\frac{4}{3} \pi(1)^3(-\rho)}=-\frac{(R-1)}{R^3-1}$
Again, $G C=G O \cdot C O=1 \cdot(R-1)=2 \cdot R$
according to the question centre of mass is at $G$ i.e. at ( $2-R, 0)$
$\Rightarrow X_{c m}=2 \cdot R$
$\begin{aligned} & \Rightarrow \frac{(R-1)}{R^3-1}=2-R \\ & \Rightarrow \frac{(R-1)}{(R-1)\left(R^2+R+1\right)}=2-R \\ & \Rightarrow \frac{1}{\left(R^2+R+1\right)}=2-R \\ & \Rightarrow\left(R^2+R+1\right)(2-R)=1\end{aligned}$
$R$ can be determined by the above equation.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
As shown in figure, a 70 kg garden roller is pushed with a force of F→=200 N at an angle of 30° with horizontal. The normal reaction on the roller is (Given g=10 m s-2)

Options:
A 8002 N
B 600 N
C 800 N
D 2003 N

The correct answer is:
800 N

Solution:
Component of the force perpendicular to the ground,

Fsin30°.

Now for equilibrium along vertical direction, we get

N=mg+Fsin30°=70×10+200×12=800 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (31 Jan Shift 1)
As shown in figure, a cuboid lies in a region with electric field E=2x2i^-4yj^+6k^ N C-1. The magnitude of charge within the cuboid is nε0 C.

The value of n is ______ (if dimension of cuboid is 1×2×3 m3)

Options:

The correct answer is:
12

Solution:
Given:

E→=2x2i^-4yj^+6k^

For top and bottom faces, as value of x=0 & y=0, electric field becomes constant and therefore net flux through these two faces will be zero(ϕz=0).

Now, Ex=2x2, Ey=-4y and Ez=6.

Change in net flux through face at x=1 and at x=0 parallel to y-z plane can written as, ϕx=2×12×2×3-0=12.

Similarly, net flux through face at y=2 and y=0 parallel to x-z plane can be written as, ϕy=-4×2×1×3-0=-24

ϕnet=12-24=-12

Using Gauss's law,

ϕnet=qε0

⇒q=12ε0

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (30 Jan Shift 2)
As shown in figure the tension in the horizontal cord is $30 \mathrm{~N}$. The weight $W$ and tension in the string $O A$ in Newton are

Options:
A $30 \sqrt{3}, 30$
B $30 \sqrt{3}, 60$
C $60 \sqrt{3}, 30$
D None of these

The correct answer is:
$30 \sqrt{3}, 60$

Solution:
From the figure $T \sin 30^{\circ}=30$...(i)
$T \cos 30^{\circ}=W$...(ii)
By solving equation (i) and (ii) we get
$W=30 \sqrt{3} N$ and $T=60 N$

Topic: Physics, Mathematics in Physics, JEE Main
As shown in figure, the tension in the horizontal cord is 30 N. The weight w and tension in the string OA in newton are

Options:
A 303, 30
B 303,60
C 603, 30
D None of the above

The correct answer is:
303,60

Solution:
Horizontal and vertical components of tension are shown in the figure.
From the figure, Tsinâ¡30o=30 .....(i)
Tcosâ¡30o=w .....(ii)
By solving Equations (i) and (ii), we get
w=303 N
and T=60 N

Topic: Physics, Laws of Motion, JEE Main
As shown in figure, the two parallel conducting rails, in a horizontal plane, are connected at one end by an inductor of inductance L. A slider (metallic) of mass m, is imparted a velocity v0, upon the rails, as shown in figure. The period of oscillation of the conducting rod is

Options:
A π2mLBl
B 2πmL3Bl
C 2πmLBl
D 3πmL2Bl

The correct answer is:
2πmLBl

Solution:
From energy conservation (there are no dissipative forces)-

E=12mv2+12LI2

⇒m2×2vdvdt+L2×2IdIdt=0

mvdvdt=-ILdIdt

The emf induced in an inductor is

E=LdIdt=Blv

⇒LdIdt=Bldxdt⇒LdI=Bldx⇒I=BlLx

The force acting on a current-carrying conductor is

mdvdt=-IBl

⇒dvdt=-IBlm

Substituting the value of I we get

dvdt=-BlLxBlm⇒a=-B2l2mLx

We got acceleration in the form of

a=-ω2x

this implies that the motion is SHM, with angular frequency

ω=B2l2mL

The time period of motion is
⇒ T=2πmLBl

Topic: Physics, Electromagnetic Induction, JEE Main
As shown in figures, three identical polaroids P1, P2 and P3 are placed one after another. The pass axis of P2 and P3 are inclined at angle of 60º and 90º with respect to axis of P1. The source S has an intensity of 256 W m-2. The intensity of light at point O is _____W m-2.

Options:

The correct answer is:
24

Solution:
As given here, the pass axis of P2 and P3 are inclined at angle of 60º and 90º with respect to axis of P1.

Therefore, intensity of light after falling on first polaroid P1, I1=I02

Intensity of light after falling on second polaroid P2, I2=I1cos260°=I08

Intensity of light after falling on third polaroid P3, I3=I2cos230°=I0834=3I032

The intensity of light at O is 3I032=3×25632=24 W m-2

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (29 Jan Shift 1)
As shown in the figure, a battery of emf ∈ is connected to an inductor L and resistance R in series. The switch is closed at t=0. The total charge that flows from the battery, between t=0 and t=tc ( tc is the time constant of the circuit) is:

Options:
A ∈LeR2
B ∈LR21-1e
C ∈LR2
D ∈ReL2

The correct answer is:
∈LeR2

Solution:
q=∫0Tcidt
=∈Rt-e-tTc-1Tc0Tc=∈RTc+Tce-1-Tc
=∈R×1e×LR

q=∈LR2e

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (08 Jan Shift 2)
As shown in the figure, a block of mass 3 kg is kept on a horizontal rough surface of coefficient of friction 133. The critical force to be applied on the vertical surface as shown at an angle 60° with horizontal such that it does not move, will be 3x. The value of x will

g=10 m s-2; sin60°=32; cos60°=12

Options:

The correct answer is:
3.33

Solution:
Fcos60°=μN or F2=133 N …1

& N=sin60°+3g …2

From equation 1 & 2

F2=133F32+3g

⇒F=g=10 Newton =3x

So x=103=3.33

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (26 Feb Shift 1)
As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30°, with horizontal. For μs=0.25, the block will just start to move for the value of F: [Given g = 10 m·s–2]

Options:
A 33.3 N
B 25.2 N
C 20 N
D 35.7 N

The correct answer is:
25.2 N

Solution:
The free body diagram for the given scenario is shown below-

From the equilibrium of the vertical component of forces, it can be written that

N=mg-F sin 30°=mg-F2=100-F2=200-F2........................(1)

From the equilibrium of horizontal component of forces, it can be written that,

F cos 30°= μN......................(2)

Substitute the expression for the normal reaction force from equation (1) into equation (2) and solve to calculate the value of the applied force.

3F2=0.25×200-F243F=200-FF=20043+1≈25.2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (01 Feb Shift 2)
As shown in the figure, a bob of mass m is tied to a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:

Options:
A 1r4gh3
B r32gh
C 1r2gh3
D r34gh

The correct answer is:
1r4gh3

Solution:
Using work-energy theorem

mgh=12mv2+12IÏ2

mgh=12mÏ2r2+12mr22Ï2 ( since I=mr22 and v=rÏ )

mgh=34mÏ2r2

Hence, Ï=4gh3r2=1r4gh3

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (07 Jan Shift 1)
As shown in the figure, a bob of mass m is tied to a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:

Options:
A 1r4gh3
B r32gh
C 1r2gh3
D r34gh

The correct answer is:
1r4gh3

Solution:
Using work-energy theorem

mgh=12mv2+12IÏ2

mgh=12mÏ2r2+12mr22Ï2 ( since I=mr22 and v=rÏ )

mgh=34mÏ2r2

Hence, Ï=4gh3r2=1r4gh3

Topic: Physics, Laws of Motion, JEE Main
As shown in the figure a body of mass m moving vertically with speed 3 m/s hits a smooth fixed inclined plane and rebounds with a velocity vf in the horizontal direction. If angle of incline is 30Â°, the velocity vf will be

Options:
A 3 m/s
B 3 m/s
C 1 / 3 m/s
D This is not possible

The correct answer is:
3 m/s

Solution:
Velocity along surface remain same
3 sin 3 0 o = v f sin 6 0 o
3 × 1 2 = v f × 3 2
v f = 3 m/s

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
As shown in the figure, a chain of mass m is placed on a smooth quarter circular portion of radius R. End A is tied with a wedge while the remaining chain is free, then the minimum work required to be done by the external agent to make the chain horizontal keeping point A fixed, is

Options:
A mgR
B mg2RÏ
C mg22Rπ
D None

The correct answer is:
mg2Rπ

Solution:
Center of mass of the chain is at 2R Ï from horizontal
As shown in figure .......
Height gain by center of mass is 2 R Ï

Wg=-mg2RÏ
Wext.=-Wg=mg2RÏ
Alternate solution

U i = â« - dm gy
=gâ«R cosÎ¸mÏ2dÎ¸
=-2RgmÏâ«0Ï2cosÎ¸dÎ¸
=-2RgmÏ[sinÎ¸]0Ï/2
=-2mgRÏ
Uf=0âÎU=wext=2mgRÏ

Topic: Physics, Work Power Energy, JEE Main
As shown in the figure, a combination of a thin plano-concave lens and a thin plano-convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is 30 cm and refraction index of the material for both the lenses is 1.75. Both the lenses are placed at distance of 40 cm from each other. Due to the combination, the image of the object is formed at distance x= ______ cm, from concave lens.

Options:

The correct answer is:
120

Solution:
Using Lens Makers formula for the plano-concave part, we can write

1f1=(1.75-1)-130

⇒ f1=-40 cm

Using Lens Makers formula for the plano-convex part, we can write

1f2=1.75-1130⇒ f2=40 cm

Image from first lens will be virtual and on the left of first lens at focal length 40 cm. So the object for second lens will be 80 cm from second lens which is 2f2. Final image is formed at 80 cm from second lens on the right.
So, x=120.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (24 Jan Shift 1)
As shown in the figure, a current of 2 A flowing in an equilateral triangle of side 43 cm . The magnetic field at the centroid O of the triangle is :

(Neglect the effect of earth’s magnetic field)

Options:
A 43×10-4 T
B 43×10-5 T
C 3×10-4 T
D 33×10-5 T

The correct answer is:
33×10-5 T

Solution:
Perpendicular distance of centroid from any of the sides will be, dtan60°=23

⇒d=2 cm

Each side will subtend an angle of 120° at the centre. Also magnetic field due to all sides will be out of plane of paper, hence total magnetic field can be written as

B=3×μ0i4πdsin60°+sin60°

=3×10-7×22×10-2×3

=33×10-5 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (30 Jan Shift 2)
As shown in the figure, a long straight conductor with semicircular arc of radius π10 m is carrying current I = 3A. The magnitude of the magnetic field. at the center O of the arc is: (The permeability of the vacuum = 4π×10-7NA-2 )

Options:
A 6 μT
B 1 μT
C 4 μT
D 3 μT

The correct answer is:
3 μT

Solution:
The formula to calculate the magnetic field BC produced at the centre of a circular current carrying conductor is given by

BC= μoI4πrθ.......................(1)

where, μ0 is the permeability of free space, I is the current through the conductor, r is the radius of the circular conductor and θ is the angle subtended by the arc at the centre.

From the given figure, it is clear that the arc of the conductor makes an angle π at its centre.

Substitute the known values of the parameters into equation (1) to calculate the required magnetic field.

BC=4π×10-7 N A-2×3 A4×π×π10 m×π= 3×10-6 T×106 μT1 T= 3 μT

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (01 Feb Shift 2)
As shown in the figure, a metallic rod of linear density 0.45 kg m-1 is lying horizontally on a smooth incline plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be

(Use g=10 m s-2)

Options:
A 30 A
B 15 A
C 10 A
D 3 A

The correct answer is:
30 A

Solution:
For the wire to remain stationary, the net force acting on it should be zero.

Balancing the forces along the incline,

mgsin45°=ILBcos45°

⇒I=mLgB

⇒I=0.45100.15=30 A

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (28 Jul Shift 1)