datavorous/entrance-exam-dataset huggingface dataset:
A thin circular ring of Mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be:

Options:
A $\frac{M \omega}{4 m}$
B $\frac{M \omega}{M+4 m}$
C $\frac{(M+4 m) \omega}{M}$
D $\frac{(M-4 m) \omega}{M+4 m}$

The correct answer is:
$\frac{M \omega}{M+4 m}$

Solution:
According to law of conservation of angular momentum
$$
\begin{aligned}
& M r^2 \omega=\left(M r^2+4 m r^2\right) \omega^{\prime} \\
& \Rightarrow \omega^{\prime}=\frac{M \omega}{M+4 m}
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
A thin circular ring of mass $\mathrm{m}$ and radius $\mathrm{R}$ is rotating about its axis with a constant angular velocity $\omega$. Two objects each of mass $M$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity $\omega^{\prime}=$

Options:
A
$\frac{\omega \mathrm{m}}{(\mathrm{m}+2 \mathrm{M})}$

B
$\frac{\omega(\mathrm{m}+2 \mathrm{M})}{\mathrm{m}}$

C
$\frac{\omega(m-2 M)}{(m+2 M)}$

D
$\frac{\omega m}{(m+M)}$

The correct answer is:
$\frac{\omega \mathrm{m}}{(\mathrm{m}+2 \mathrm{M})}$

Solution:
$L_i=L_f$
$m R R^2 \omega=\left(m R^2+2 M R^2\right) \omega^{\prime}$
$\omega^{\prime}=\left(\frac{m \omega}{m+2 M}\right)$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2006
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become:

Options:
A ωMM+m
B ωM+2 mM
C ωMM+2 m
D ωM-2 mM+2 m

The correct answer is:
ωMM+2 m

Solution:
Using conservation of angular momentum
⇒Mr2ω=Mr2+2mr2ω'

⇒ω'=MωM+2 m.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with constant angular velocity $\omega$. Two objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by

Options:
A $\frac{(M+2 m) \omega}{2 m}$
B $\frac{2 \mathrm{M} \omega}{\mathrm{M}+2 \mathrm{~m}}$
C $\frac{(M+2 m) \omega}{M}$
D $\frac{M \omega}{M+2 m}$

The correct answer is:
$\frac{M \omega}{M+2 m}$

Solution:
In the absence of external torque, angular momentum remain constant
$$
\begin{aligned}
\mathrm{L} & =\mathrm{I} \omega=\mathrm{I}^{\prime} \omega^{\prime} \\
\therefore \quad \mathrm{MR}^2 \omega & =(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^2 \omega^{\prime} \\
\omega^{\prime} & =\frac{\mathrm{M} \omega}{(\mathrm{M}+2 \mathrm{~m})}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with constant angular velocity $\omega$. Two objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by

Options:
A $\frac{(M+2 m) \omega}{2 m}$
B $\frac{2 \mathrm{M} \omega}{\mathrm{M}+2 \mathrm{~m}}$
C $\frac{(M+2 m) \omega}{M}$
D $\frac{M \omega}{M+2 m}$

The correct answer is:
$\frac{M \omega}{M+2 m}$

Solution:
In the absence of external torque, angular momentum remain constant
$$
\begin{aligned}
\mathrm{L} & =\mathrm{I} \omega=\mathrm{I}^{\prime} \omega^{\prime} \\
\therefore \quad \mathrm{MR}^2 \omega & =(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^2 \omega^{\prime} \\
\omega^{\prime} & =\frac{\mathrm{M} \omega}{(\mathrm{M}+2 \mathrm{~m})}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2010 (Mains)
A thin circular ring of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$. If two objects each of mass $m$ be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

Options:
A $\frac{\omega(M-2 m)}{M+2 m}$
B $\frac{\omega M}{M+2 m}$
C $\frac{\omega(M+2 m)}{M}$
D $\frac{\omega M}{M+m}$

The correct answer is:
$\frac{\omega M}{M+2 m}$

Solution:
Key Idea Apply law of conservation of angular momentum.
$$
\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2
$$
In the given case
$$
\begin{aligned}
\mathrm{I}_1 & =\mathrm{MR}^2 \\
\mathrm{I}_2 & =\mathrm{MR}^2+2 \mathrm{mR}^2 \\
\omega_1 & =\omega
\end{aligned}
$$
Then
$$
\omega_2=\frac{I_1}{I_2} \omega=\frac{M}{M+2 m} \omega
$$

Topic: Physics, Rotational Motion, JEE Main
A thin circular ring of mass $\mathrm{M}$ and radius $\mathrm{R}$ is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity $\omega$. If two objects each mass $m$ be attached gently to the opposite ends of a diameter of the ring, the ring, will then rotate with an angular velocity:

Options:
A $\frac{\omega \mathrm{M}}{\mathrm{M}+\mathrm{m}}$
B $\frac{\omega(M-2 m)}{M+2 m}$
C $\frac{\omega \mathrm{M}}{\mathrm{M}+2 \mathrm{~m}}$
D $\frac{\omega(M+2 m)}{M}$

The correct answer is:
$\frac{\omega \mathrm{M}}{\mathrm{M}+2 \mathrm{~m}}$

Solution:
Apply conservation of angular momentum.
$$
\begin{aligned}
\mathrm{L}_{\mathrm{i}} & =\mathrm{L}_{\mathrm{f}} \\
\mathrm{MR}^2 \omega & =(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^2 \omega^{\prime} \\
\omega^{\prime} & =\frac{\mathrm{M} \omega}{\mathrm{M}+2 \mathrm{~m}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2009 (Mains)
A thin conducting ring of radius $\mathrm{R}$ is given a charge $+\mathrm{Q}$. The electric field at the centre $\mathrm{O}$ of the ring due to the charge on the part $\mathrm{AKB}$ of the ring is $\mathrm{E}$. The electric field at the centre due to the charge on the part $\mathrm{ACDB}$ of the ring is

Options:
A 3 E along $\mathrm{OK}$
B 3 E along $\mathrm{KO}$
C E along $\mathrm{OK}$
D E along KO

The correct answer is:
E along $\mathrm{OK}$

Solution:
$$
\begin{aligned}
\vec{E}_0 & =0 \\
\Rightarrow \quad \vec{E}_{A K B}+\vec{E}_{A C D B} & =0 \\
\vec{E}_{A C D B} & ={ }^{(-)} \vec{E}_{A K B} \\
& =-E(\text { along } K O) \\
& =E(\text { along } O K)
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2008 (Mains)
A thin converging lens of focal length f=25 cm forms the image of an object on a screen placed at a distance of 75 cm from the lens. Now the screen is moved closer to the lens by a distance of 25 cm. The distance through which the object has to be shifted so that its image on the screen is sharp again is

Options:
A 37.5 cm
B 16.25 cm
C 12.5 cm
D 13.5 cm

The correct answer is:
12.5 cm

Solution:
According to the first condition
f=25 cm, V=75 cm
1f=1v-1u
125=175-1u
1u=175-125
1u=1-375
u= -752= -37.5 cm
According to the second condition
v1=50 cm, f=25 cm
1f=1v1-1u1
125=150-1u1
1u1=150-125
â 1u1=1-250
â u1= -50 cm
So, the image is sharp again for,
Îu=u1-u
=50-37.5
=12.5 cm

Topic: Physics, Ray Optics, JEE Main
A thin converging lens of focal length $f=25 \mathrm{~cm}$ forms the image of an object on a screen placed at a distance of $75 \mathrm{~cm}$ from the lens. The screen is moved closer to the lens by a distance of $25 \mathrm{~cm}$. The distance through which the object has to be shifted, so that its image on the screen in sharp again is

Options:
A $37.5 \mathrm{~cm}$
B $16.25 \mathrm{~cm}$
C $12.5 \mathrm{~cm}$
D $13.5 \mathrm{~cm}$

The correct answer is:
$12.5 \mathrm{~cm}$

Solution:
According to the first condition,
$$
\begin{aligned}
f & =25 \mathrm{~cm}, v=75 \mathrm{~cm} \\
u & =? \\
\frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\
\frac{1}{25} & =\frac{1}{75}-\frac{1}{u} \\
\frac{1}{u} & =\frac{1}{75}-\frac{1}{25} \\
\frac{1}{u} & =\frac{1-3}{75} \\
u & =-\frac{75}{2}=-37.5 \mathrm{~cm}
\end{aligned}
$$
According to the second condition
$$
\begin{aligned}
v_1=50 \mathrm{~cm}, f & =25 \mathrm{~cm}, u_1=? \\
\frac{1}{f} & =\frac{1}{v_1}-\frac{1}{u_1} \\
\frac{1}{25} & =\frac{1}{50}-\frac{1}{u_1} \\
\frac{1}{u_1} & =\frac{1}{50}-\frac{1}{25} \\
\Rightarrow \quad \frac{1}{u_1} & =\frac{1-2}{50} \\
\Rightarrow \quad u_1 & =-50 \mathrm{~cm}
\end{aligned}
$$
So, the screen is sharp again is
$$
\begin{aligned}
\Delta u & =u_1-u \\
& =50-37.5 \\
& =12.5 \mathrm{~cm}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
A thin convex lens L (refractive index =1.5 ) is placed on a plane mirror M. When a pin is placed at A, such that OA=18 cm, its real inverted image is formed at A itself, as shown in figure. When liquid of refractive index Î¼l is put between the lens and the mirror, the pin has to be moved to A', such that OA'=27âcm, to get its inverted real image at A' itself. The value of Î¼l will be

Options:
A 43
B 3
C 32
D 2

The correct answer is:
43

Solution:
1f1=12×218
1f2=μ1-1-18
When μ1 is filled between lens and mirror,
P=218-218μ1-1=2-2μ1+218
⇒Fm=-182-μ1
2=6-3μ1
3μ1=4
μ1=43

Topic: Physics, Ray Optics, JEE Main
A thin convex lens made from crown glass μ = 3 2 has focal length f⁡ . When it is measured in two different liquids having refractive indices 4 3 and 5 3 , it has the focal lengths f⁡ 1 and f⁡ 2 respectively. The correct relation between the focal lengths is :

Options:
A fâ¡ 1 = fâ¡ 2 < fâ¡
B fâ¡ 1 > fâ¡ and fâ¡ 2 becomes negative
C fâ¡ 2 > fâ¡ and fâ¡ 1 becomes negative
D fâ¡ 1 and fâ¡ 2 both become negative

The correct answer is:
fâ¡ 1 > fâ¡ and fâ¡ 2 becomes negative

Solution:
Let [ 1 R 1 - 1 R 2 ] = x
&therefore; 1 f = [ 3 2 1 - 1 ] [ 1 R 1 - 1 R 2 ] ⇒ f = 2 x
When in liquid 1
1 f 1 = [ 3 2 4 3 - 1 ] [ 1 R 1 - 1 R 2 ] ⇒ f 1 = 8 x
When in liquid 2
1 f 2 = [ 3 2 5 3 - 1 ] [ 1 R 1 - 1 R 2 ] ⇒ f 2 = - 10 x

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (06 Apr)
A thin convex lens of focal length $10 \mathrm{~cm}$ is placed in contact with a concave lens of same material and of same focal length. The focal length of combination will be

Options:
A Zero
B Infinity
C 10 cm
D 20 cm

The correct answer is:
Infinity

Solution:
$f=\frac{f_1 f_2}{f_1+f_2}=\frac{10(-10)}{10+(-10)}=\frac{-100}{10-10}={ }_{\infty}$

Topic: Physics, Ray Optics, JEE Main
A thin convex lens of focal length f is put on a plane mirror as shown in the figure. When an object is kept at a distance a from the lens-mirror combination, its image is formed at a distance a3 in front of the combination. The value of a is:

Options:
A 2f
B 32f
C 3f
D f

The correct answer is:
2f

Solution:
Equivalent power of combination is
Peff=2Plens+Pmirror ...(i)
∵Pmirror =0 for plane mirror and Plens=1f
Also combination behaves as a mirror. So. Peff=1fmirror
In equation (i)
-1fmirror =2f⇒fmirror=-f2
By mirror formula
1fmirror =1u+1v
Here u= -a, v=-a3, fmirror= -f2
⇒-2f=1-a+3-a=-4a
⇒-2f=-4a⇒a=2f

Topic: Physics, Ray Optics, JEE Main, JEE Main 2015 (11 Apr Online)
A thin convex lens of refractive index 1.5 has 20 cm focal length in air. If the lens is completely immersed in a liquid of refractive index 1.6, its focal length will be

Options:
A -160 cm
B -100Â cm
C +160 cm
D +100Â cm

The correct answer is:
-160 cm

Solution:
From lens formula
1f=Î¼ga -11R1-1R2
=1.5-11R1-1R2 .....(i)
Also, Î¼gl =Î¼gÎ¼l=1.51.6
â´ 1fâ²=Î¼g-1l 1R1-1R2
1fâ²=1516-11R1-1R2 ...(ii)
Dividing Equation (i) by Equation (ii), we get
ffâ²=1516-11.5-1=-18
f'=â8f
=â160âcm

Topic: Physics, Ray Optics, JEE Main
A thin cylindrical rod PQ of length L and density d1 is pivoted at its lowest point P, inside a stationary homogeneous and non-viscous liquid of density d2. The rod is always fully submerged inside the liquid and is free to rotate in a vertical plane about a horizontal axis passing through P. If d1<d2, then the time period of small angular oscillations of the rod about its vertical equilibrium position will be

Options:
A T=2π2L3gd1d2-d1
B T=2πL3gd2d2-d1
C T=2π2Lgd1d2-d1
D T=2π2L3gd2-d1d1

The correct answer is:
T=2π2L3gd1d2-d1

Solution:
Consider the diagram in a displaced position.

The weight and upthrust FB, both pass through the centre of gravity G

W=πr2L d1g

FB=πr2Ld2g

The restoring torque about the pivot is

τ=-πr2Lgd2-d1L2θ

α=-πr2Lgd2-d1L2θπr2Ld1 L23

α=-3g2Ld2-d1d1θ

T=2π2L3gd1d2-d1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A thin disc of mass M and radius R has mass per unit area σr=kr2 where r is the distance from its centre. Its moment inertia about an axis going through its centre of mass and perpendicular to its plane is:

Options:
A MR23
B MR22
C MR26
D 2MR23

The correct answer is:
2MR23

Solution:
dm=σ⋅dA=σ 2πr dr
=2πk r3 dr
M=∫dm=2πk∫0Rr3dr
=2πkR44=πkR42
dI=dm r2=2πkr5 dr
I=2πk∫0Rr5dr=2πkR66=2πkR63×2

since, M=πkR42

so,
I=2MR23

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A thin disc of radius b=2a has a concentric hole of radius a in it (see figure). It carries uniform surface charge σ on it. If the electric field on its axis at a height hh<<a from its centre is given as Ch then the value of C is

Options:
A Ï4 aÏµ0
B ÏaÏµ0
C Ï5aÏµ0
D Ï2aÏµ0

The correct answer is:
Ï4 aÏµ0

Solution:
∵ At the axial point of a uniformly charged disc electric field is given by,
E=σ2ϵ01-cosθ
By superposition principle, when inner disc is removed , then, electric field due to remaining disc is,
E=σ2ϵ0 1-cosθ2-1-cosθ1
=σ2ϵ0cosθ1-cosθ2
=σ2ϵ0hh2+a2 -hh2+b2
=σ2ϵ0ha1+h2a2-hb1+h2b2
∵h ≪a and b.
∴E=σ2ϵ0ha-hb
=σ2ϵ0ha-h2a=σh4ϵ0a
⇒C=σ4aϵ0.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2015 (10 Apr Online)
A thin film of soap solution Î¼s=1.4 lies on the top of a glass plate Î¼g=1.5. When visible light is incident almost normal to the plate, two wavelengths 420 nm and 630 nm are strongly reflected. What is the minimum thickness (in nm) of the soap film?

Options:

The correct answer is:
450

Solution:
For reflection at the air - soap solution interface, the phase difference is π
For reflection at the interface of soap solution to glass also, there will be a phase difference of π
∴ The condition for the maximum intensity
= 2μt=nλ
For n, nλ1=n-1 λ2
n×420=n-1630
∴ n630-420=630
⇒ n210=630
⇒ n=630210
⇒ n=3
This is the maximum order, where they coincide
2×1.4×t=3×420
⇒ t=3×4202×1.40=450 nm

Topic: Physics, Wave Optics, JEE Main
A thin flexible wire of length L is connected to a battery (negligible in size as compared to the wire) to form a closed loop. The battery drives a current I in the wire. When the system is put in a uniform magnetic field of strength B perpendicular to the plane in which the wire is placed, the wire takes the shape of a circle. The tension in the wire is

Options:
A IBL
B IBLÏ
C IBL2Ï
D IBL4π

The correct answer is:
IBL2π

Solution:
L=2ÏR
R=L2Ï
2TsindÎ¸=Fm
For small angles, sin(dÎ¸)≈dÎ¸
2TdÎ¸=IdLBsin90Â°
=I R .dÎ¸.B
T=IRB=ILB2Ï

Topic: Physics, Magnetic Effects of Current, JEE Main
A thin glass plate of thickness is $\frac{2500}{3} \lambda$ ( $\lambda$ is wavelength of light used) and refractive index $\mu=1.5$ is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :

Options:
A
$2: 1$

B
$1: 4$

C
$4: 1$

D
$4: 3$

The correct answer is:
$4: 1$

Solution:
No Solution is given.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2013 (25 Apr Online)
A thin glass prism of angle $9^{\circ}$ with refractive index 1.4 is combined with another glass prism of refractive index 1.6 as shown in the figure. The combination of the prism provides dispersion without deviation. Determine the angle $(A)$ of the second prism.

Options:
A $9^{\circ}$
B $12^{\circ}$
C $6^{\circ}$
D $4^{\circ}$

The correct answer is:
$6^{\circ}$

Solution:
For dispersion without deviation through a prism combination,
$\delta_{m_1}=\delta_{m_2} \Rightarrow A_1\left(\mu_1-1\right)=A_2\left(\mu_2-1\right)$
Here, $A_1=9^{\circ}, \mu_1=1.4$ and $\mu_2=1.6$
$\therefore \quad A=A_2=9 \times \frac{(1.4-1)}{(1.6-1)}=9 \times \frac{0.4}{0.6}=6^{\circ}$

Topic: Physics, Ray Optics, JEE Main
A thin glass (refractive index 1.5) lens has an optical power of -5 D in the air. Its optical power in a liquid medium with refractive index 1.6 will be

Options:
A 1 D

B -1 D

C 25 D

D -25 D

The correct answer is:
1 D

Solution:
1fa=(μ-1)(1R1-1R2)
=1.5-11R1-1R2 …………..i
and1fm=μg-μmμm(1R1-1R2)
1fm=1.51.6-11R1-1R2 ………..ii
Thus,
fmfa=1.5-11.51.6-1=-8
fm= -8×fₐ
=-8×-15 ∵fa=1p=-15m
=1.6 m
∴Pm=μfm
=1.61.6=1 D

Topic: Physics, Ray Optics, JEE Main
A thin glass (refractive index 1.5) lens has optical power of $-5 D$ in air. Its optical power in a liquid medium with refractive index $1.6$ will be

Options:
A
$1 \mathrm{D}$

B
$-1 \mathrm{D}$

C
$25 \mathrm{D}$

D
None of these

The correct answer is:
None of these

Solution:
$\frac{\mathrm{P}_{\mathrm{m}}}{\mathrm{P}_{\mathrm{air}}}=\frac{\left(\frac{\mu_{\ell}}{\mu_{\mathrm{a}}}-1\right)}{\left(\frac{\mu_{\ell}}{\mu_{\mathrm{m}}}-1\right)}$
$\mathrm{P}_{\mathrm{m}}=5 / 8 \mathrm{D}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2005
A thin hollow sphere of mass $m$ is completely filled with a liquid of mass $m$. When the sphere rolls with a velocity $v$, kinetic energy of the system is (neglect friction)

Options:
A $\frac{1}{2} m v^2$
B $m v^2$
C $\frac{4}{3} m v^2$
D $\frac{4}{5} m v^2$

The correct answer is:
$\frac{4}{3} m v^2$

Solution:
$\begin{aligned} \text { Total energy } & =\mathrm{KE}+\text { rotational } \mathrm{KE} \\ & =\frac{1}{2}(2 m) v^2+\frac{1}{3} m v^2 \\ & =\frac{4}{3} m v^2\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc

Options:
A Continuously decreases
B Continuously increases
C first increases and then decreases
D Remains unchanged

The correct answer is:
first increases and then decreases

Solution:
Moment of inertia =12 MR2+mx2
Where m=mass of insect,
and x=distance of insect from centre.
Clearly, as the insect moves along the diameter of the disc, MI first decreases, then increases.
By conservation of angular momentum, angular speed first increases, then decreases.

Topic: Physics, Rotational Motion, NEET
A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc:

Options:
A
continuously decreases

B
continuously increases

C
first increases and then decreases

D
remains unchanged

The correct answer is:
first increases and then decreases

Solution:
$$
\tau=0
$$
Angular momentum is conserve
$$
\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2 \Rightarrow \omega_2=\frac{\mathrm{I}_1 \omega_1}{\mathrm{I}_2}
$$
$\mathrm{I}_2$ first decreases and then increases
$\therefore \omega$ first increases and then decreases.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2011
A thin infinite sheet charge and an infinite line charge of respective charge densities +σ and +λ are placed parallel at 5 m distance from each other. Points P and Q are at 3π m and 4π m perpendicular distances from line charge towards sheet charge, respectively. Ep and Eq are the magnitudes of resultant electric field intensities at point P and Q respectively. If EPEQ=4a for 2|σ|=|λ|, then the value of a is _____.

Options:

The correct answer is:
6

Solution:
Let's consider the following diagram:

The formula to calculate the electric field due to an infinitely long charged sheet is given by

Es=σ2ε0 ...1

The formula to calculate the electric field due to an infinitely long line charge is given by

El=λ2πε0r ...2

Using equations (1) and (2), the electric field at P can be calculated as follows:

EP=σ2ε0-14πε02λ3π=σ2ε0-λ6ε0=σ6ε0 ...3

Similarly, the electric field at Q is given by

EQ=σ2ε0-14πε02λ4π=σ2ε0-λ8ε0=σ4ε0 ...4

Divide equation (3) by equation (4) to obtain the required ratio.

EPEQ=σ6ε0σ4ε0=46

Hence, a=6.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A thin lens focal length $f_1$ and its aperture has diameter $d$. It forms an image of intensity $I$. Now the central part of the aperture upto diameter $\frac{d}{2}$ is blocked by an opaque paper. The focal length and image intensity will change to

Options:
A $\frac{f}{2}$ and $\frac{I}{2}$
B $f$ and $\frac{I}{4}$
C $\frac{3 f}{4}$ and $\frac{I}{2}$
D $f$ and $\frac{3 I}{4}$

The correct answer is:
$f$ and $\frac{3 I}{4}$

Solution:
$I \propto A^2 \Rightarrow \frac{I_2}{I_1}=\left(\frac{A_2}{A_1}\right)^2=\frac{\pi r^2-\frac{\pi r^2}{4}}{\pi r^2}=\frac{3}{4}$
$\Rightarrow I_2=\frac{3}{4} I_1$
and focal length remains unchanged.

Topic: Physics, Ray Optics, JEE Main
A thin lens has a focal length f1 and its aperture has a diameter d. If forms an image of intensity I. Now, the central part of the aperture up to diameter d2 is blocked by an opaque paper. The focal length and image intensity will change to

Options:
A f2 and I2
B f and I4
C 3f4 and I2
D f and 3I4

The correct answer is:
f and 3I4

Solution:
The focal length of the lens remains unchanged by changing its aperture.
The intensity of the image is proportional to the uncovered area.
IfIi=AfAi=π4d2-d24π4d2=34
If=34Ii=34I

Topic: Physics, Ray Optics, NEET
A thin lens has a focal length f and its aperture has a diameter d. It forms an image of intensity I. Now the central part of the aperture up to diameter d2 is blocked by an opaque paper. The focal length and image intensity would change to

Options:
A f2, I2
B f, I4
C 3f4, I2
D f, 3I4

The correct answer is:
f, 3I4

Solution:
The focal length remains f, as any small portion of the lens forms a complete image of the object and has the same focal length.
While the intensity of the image depends on the area as,
Intensity∝Area2
I2I1=A2A1
I2I=πd24-πd224πd24
I2I=34
I2=34I

Topic: Physics, Ray Optics, NEET
A thin lens made of glass of refractive index \( \mu=1.5 \) has a focal length equals is \( 12 \mathrm{~cm} \) in air. It is now immersed in water
\( \left(\mu=\frac{4}{3}\right) . \) Its new focal length is

Options:
A \( 48 \mathrm{~cm} \)
B \( 36 \mathrm{~cm} \)
C \( 24 \mathrm{~cm} \)
D \( 12 \mathrm{~cm} \)

The correct answer is:
\( 48 \mathrm{~cm} \)

Solution:
Focal length in air is given by
1fa= aÎ¼g-11R1-1R2
The focal length of lens immersed in water is given by
1f1= lng-11R1-1R2
When, R1,R2 are radii of curvatures of the two surfaces of lens and lng is refractive index of glass with respect to liquid.
Also, lÎ¼g = ang anl
Given, ang=1.5, fa=12 cm, anl=43
â´flfa= ang-1 lng-1
âf1 = 4Ã12=48âcm

Topic: Physics, Ray Optics, JEE Main
A thin lens made of glass (refractive index =1.5 ) of focal length f=16cm is immersed in a liquid of refractive index 1.42 . If its focal length in liquid is fl , then the ratio fl/f is closest to the integer:

Options:
A 1
B 9
C 5
D 17

The correct answer is:
9

Solution:
1fa=Î¼gÎ¼a-11R1-1R2
1fm=Î¼gÎ¼m-11R1-1R2
⇒fafm=Î¼gÎ¼m-1Î¼gÎ¼a-1=1.501.42-11.501-1=0.081.920.5
fafm=1.420.050.08=8.875â9

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A thin lens of focal length \( \mathrm{f} \) and its aperture has a diameter \( \mathrm{d} \). It forms an image of intensity I. Now the central part of the aperture upto diameter \( (\mathrm{d} 2) \) is blocked by an opaque paper. The focal length and image intensity would change to

Options:
A \( \frac{f}{2}, \frac{I}{2} \)
B f, \( \frac{I}{4} \)
C \( \frac{3 f}{4}, \frac{1}{2} \)
D f, \( \frac{3 \mathrm{I}}{4} \)

The correct answer is:
f, \( \frac{3 \mathrm{I}}{4} \)

Solution:
I ∝ aperture 2
I 1 ∝ π d 2 4 ⇒ I 2 ∝ π d 2 4 - π d 4 2
I 2 ∝ π d 2 4 1 - 1 4 ⇒ I 2 ∝ = 3 4 &CenterDot; π d 2 4
I 2 I 1 = 3 4 ⇒ I 2 = 3 4 I 1

Topic: Physics, Ray Optics, JEE Main