datavorous/entrance-exam-dataset huggingface dataset:
In a uniform magnetic field of induction B , a wire, in the form of a semicircle of radius r , rotates about the diameter of the circle with angular frequency ω . The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R , the mean power generated per period of rotation is

Options:
A B π r 2 ω 2 R
B B Ï r 2 Ï 2 8 R
C B π r ω 2 2 R
D B π r ω 2 2 8 R

The correct answer is:
B π r 2 ω 2 8 R

Solution:
Magnetic flux linked

= BA cos Ï t = B Ï r 2 cos Ï t 2
â´ Induced emf e = - d Ï dt = - 1 2 B Ï r 2 Ï sin Ï t
â´ Power = e 2 R = B 2 Ï 2 r 4 Ï 2 sin 2 Ï t 4 R
= B Ï r 2 Ï 2 4 R sin 2 Ï t
âµ < sin 2 Ï t > = 1 / 2
â´ Mean power generated
= B Ï r 2 Ï 2 4 R Ã 1 2 = B Ï r 2 Ï 2 8 R .

Topic: Physics, Electromagnetic Induction, JEE Main
In a uniform magnetic field of induction B, a wire in the form of a semicircle of radius r rotates about the diameter of the circle with angular frequency Ï. If the total resistance of the circuit is R, the mean power generated per period of rotation is

Options:
A BÏr2Ï2R
B BÏr2Ï25 Rt
C BÏrÏ22Â R
D BÏrÏ228 R

The correct answer is:
BÏrÏ228 R

Solution:
The flux associated with coil of area A and magnetic induction B is
Ï=BAcosâ¡Î¸
=12BÏr2cosâ¡Ït âµA=12Ïr2
â´einduced=-dÏdt
= -ddt12BÏr2cosâ¡Ït
=12BÏr2Ïsinâ¡Ït
â´power p=einduced2R
= B2Ï2r4Ï2sin2â¡Ït4R
Hence,Pmean=<p>
=B2Ï2r4Ï24R.12 âµ<sinâ¡Ït>=12
=BÏr2Ï28R

Topic: Physics, Electromagnetic Induction, JEE Main
In a uniform magnetic field of induction $B$ a wire in the form of semicircle of radius $r$ rotates about the diameter of the circle with angular frequency $\omega$. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R$ the mean power generated per period of rotation is

Options:
A
$\frac{B \pi r^2 \omega}{2 R}$

B
$\frac{\left(B \pi r^2 \omega\right)^2}{2 R}$

C
$\frac{(\mathrm{B} \pi \mathrm{r} \omega)^2}{2 \mathrm{R}}$

D
$\frac{\left(B \pi r \omega^2\right)^2}{8 R}$

The correct answer is:
$\frac{\left(B \pi r^2 \omega\right)^2}{2 R}$

Solution:
Magnetic flux $=B A \cos \theta=B \cdot \frac{\pi r^2}{2} \cos \omega t$
$$
\begin{aligned}
& \therefore \varepsilon_{\text {ind }}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{1}{2} \mathrm{~B} \pi \mathrm{r}^2 \omega \sin \omega \mathrm{t} \\
& \therefore \mathrm{P}=\frac{\varepsilon_{\text {ind }}^2}{\mathrm{R}}=\frac{\mathrm{B}^2 \pi^2 \mathrm{r}^4 \omega^2 \sin ^2 \omega \mathrm{t}}{4 \mathrm{R}}
\end{aligned}
$$
Now, $ < \sin ^2 \omega t>=1 / 2$ (mean value)
$$
\therefore\langle\mathrm{P}\rangle=\frac{\left(\mathrm{B} \pi \mathrm{r}^2 \omega\right)^2}{8 \mathrm{R}} \text {. }
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2004
In a uniform magnetic field of strength 0.15 T, a short bar magnet of magnetic moment m=0.32 J T-1 is placed. The potential energy of the magnet in its unstable equilibrium position is

Options:
A 4.8×10-2 J
B 9.6×10-2 J
C 2.8×10-2 J
D 1.2×10-2 J

The correct answer is:
4.8×10-2 J

Solution:
For the unstable equilibrium, the angle between the magnetic moment and magnetic field is 180o. ( âµ In this position it will be in a direction perpendicular to magnetic field thus maximum torque will act on it.)
Î¸ = 1 8 0 o
Potential energy of the magnet
U = - mB cos 180o
= - 0.32 x 0.15 (- 1) = 4.8 x 10-2 J
Thus, for the unstable equilibrium the potential energy is 4.8 x 10-2 J.

Topic: Physics, Magnetic Properties of Matter, JEE Main
In a uniformly charged sphere of total charge $Q$ and radius $R$, the electric field $E$ is plotted as a function of distance from the centre. The graph which would correspond to the above will be

Options:
A

B

C

D

The correct answer is:

Solution:
$\overrightarrow{\mathrm{E}}_{\text {inside }}=\left(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{R}^3}\right) \overrightarrow{\mathrm{r}}$
$\overrightarrow{\mathrm{E}}_{\text {outside }}\left(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}^3}\right) \overrightarrow{\mathrm{r}}$
$\therefore$

Topic: Physics, Electrostatics, JEE Main
In a uniformly charged sphere of total charge $Q$ and radius $R$, the electric field $E$ is plotted as a function of distance from the centre. The graph which would correspond to the above will be

Options:
A

B

C

D

The correct answer is:

Solution:
$\overrightarrow{\mathrm{E}}_{\text {inside }}=\left(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{R}^3}\right) \overrightarrow{\mathrm{r}}$
$\overrightarrow{\mathrm{E}}_{\text {outside }}\left(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}^3}\right) \overrightarrow{\mathrm{r}}$
$\therefore$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2012 (Offline)
In a u-tube as shown in a figure, water and oil are in the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are $15 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. The density of the oil is $\left[\right.$ take $\left.\rho_{\text {water }}=1000 \mathrm{~kg} / \mathrm{m}^3\right]$

Options:
A $1200 \mathrm{~kg} / \mathrm{m}^3$
B $750 \mathrm{~kg} / \mathrm{m}^3$
C $1000 \mathrm{~kg} / \mathrm{m}^3$
D $1333 \mathrm{~kg} / \mathrm{m}^3$

The correct answer is:
$750 \mathrm{~kg} / \mathrm{m}^3$

Solution:
According to Pascal's law "Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel."
In the given situation as shown in the figure below
Pressure due to water column of height $15 \mathrm{~cm}=$ Pressure due to oil column of height $20 \mathrm{~cm}$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{h}_{\mathrm{w}} \rho_{\mathrm{w}} \mathrm{g} & =\mathrm{h}_0 \rho_0 \mathrm{~g} \\
15 \rho_w & =20 \rho_0 \Rightarrow \rho_0=\frac{15}{20} \rho_\omega \\
\rho_0 & =\frac{15}{20} \times 1000\left(\because \text { given, } \rho_w=1000 \mathrm{~kg} \mathrm{~m}^{-3}\right) \\
& =750 \mathrm{kgm}^{-3}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
In a vernier calliper N divisions of vernier scale coincides with (Nâ1) divisions of the main scale (in which length of one division is 1 mm). The least count of the instrument should be (in cm):

Options:
A 1 Nâ1
B 1 10N
C N−1
D N

The correct answer is:
1 10N

Solution:
Least count 1 MSD−1VSD
=1MSD−( N−1 N )MSD
∵NVSD=( N−1 )MSD
∴1VSD=( N−1 N )MSD
∴ Least count = 1 N MSD=( 1 N )×( 1 10 )cm=( 1 10 N )

Topic: Physics, Units and Dimensions, NEET
In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its $4^{\text {th }}$ division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is $0.04 \mathrm{~mm}$ then how many main scale divisions are there in $1 \mathrm{~cm}$ ?

Options:
A 10
B 5
C 20
D 40

The correct answer is:
20

Solution:
$\begin{aligned}
& 50 \mathrm{VSD}=49 \mathrm{MSD} \\
& 1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD} \\
& \mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD} \\
& \mathrm{LC}=1 \mathrm{MSD} \cdot \frac{49}{50} \mathrm{MSD}=\frac{\mathrm{MSD}}{50} \ldots . . . . . .(1) \\
& \text {Now, } | \text { zero error }|=| 0-4(\mathrm{~L} . \mathrm{C}) | \\
& 0.04=|-4(\mathrm{~L} . \mathrm{C})| \Rightarrow \mathrm{L} . \mathrm{C}=0.01 \mathrm{~mm} \\
& \mathrm{0 . 0 4}=|-\mathrm{4}(\mathrm{L} . \mathrm{C})| \Rightarrow \mathrm{L} . \mathrm{C} \ldots . . .(2) \\
& =\mathrm{0 . 0 1} \mathrm{mm}
\end{aligned}$
Using (1) and (2)
$\begin{aligned}
& \frac{M S D}{50}=0.01 \\
& M S D=0.5 \mathrm{~mm} \\
& M S D=\frac{1 \mathrm{~cm}}{\mathrm{~N}} \\
& \mathrm{~N}=\frac{1 \mathrm{~cm}}{0.5 \mathrm{~mm}}=\frac{100 \mathrm{~mm}}{5 \mathrm{~mm}} \\
& \mathrm{~N}=20
\end{aligned}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2024 (06 Apr Shift 2)
In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be _____ ×10-2 mm

Options:

The correct answer is:
5

Solution:
Vernier constant is defined as the difference between the value of one main scale division and one vernier scale division. It is also known as the least count of vernier calliper.

The main scale division of the given vernier calliper is MSD=1 cm20=0.5 mm.

LC=MSDn=0.5 mm10=0.05 mm=5×10-2 mm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2022 (26 Jun Shift 1)
In a vernier callipers if each centimetre of the main scale is divided into q equal parts and p vernier scale divisions coincide with p-1 main scale divisions, then what will be the least count?

Options:
A pqp-1
B qp
C q-1pq
D 1pq

The correct answer is:
1pq

Solution:
1 MSD=1q

p×VSD=p-1×MSD

⇒1 VSD=p-1pq

L.C.=1 MSD-1VSD

⇒1q-p-1pq=1pq

Topic: Physics, Experimental Physics, JEE Main
In a vernier callipers, $(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If 1 MSD represents 0.1 mm , the vernier constant (in cm ) is:

Options:
A $\frac{1}{100(N+1)}$
B 100 N
C $10(N+1)$
D $\frac{1}{10 N}$

The correct answer is:
$\frac{1}{100(N+1)}$

Solution:
$V . C=M S D-V S D$....(1)
given : $(N+1) V S D=N$ MSD
$\operatorname{VSD}=\left(\frac{N}{N+1}\right) \mathrm{MSD}$....(2)
From (1) and (2)
$\begin{aligned}
& V \cdot C=(\mathrm{MSD})-\frac{N}{N+1}(\mathrm{MSD}) \\
& =\operatorname{MSD}\left(1-\frac{N}{N+1}\right)=\frac{\mathrm{MSD}}{N+1} \\
& =\frac{0.01}{N+1}=\frac{1}{100(N+1)}
\end{aligned}$

Topic: Physics, Experimental Physics, NEET, NEET 2024
In a vessel containing $\mathrm{SO}_3, \mathrm{SO}_2$ and $\mathrm{O}_2$ at equilibrium, some helium gas is introduced so that the total pressure increases while temperature and volume remain constant. According to Le-Chatelier.principle the dissociation of $\mathrm{SO}_3$

Options:
A Increases
B Decreases
C Remains unaltered
D Changes unpredictably

The correct answer is:
Remains unaltered

Solution:
The volume is kept constant. Hence, the concentration of the gases does not change. The partial pressure of the gases in the reaction does not change, even though we have increased the total pressure by adding helium. Therefore, there is no change in equilibrium.

Topic: Chemistry, Chemical Equilibrium, JEE Main
In a wave, the path difference corresponding to a phase difference of $\varphi$ is

Options:
A $\frac{\pi}{2 \lambda} \phi$
B $\frac{\pi}{\lambda} \phi$
C $\frac{\lambda}{2 \pi} \phi$
D $\frac{\lambda}{\pi} \phi$

The correct answer is:
$\frac{\lambda}{2 \pi} \phi$

Solution:
For $2 \pi$ phase difference $\rightarrow$ Path difference is $\lambda$
$\therefore$ For $\varphi$ phase difference $\rightarrow$ Path difference is $\frac{\lambda}{2 \pi} \times \phi$

Topic: Physics, Wave Optics, JEE Main
In a Wheatstone bridge (see fig.), Resistances $P$ and $Q$ are approximatelyequal. When $R=400 \Omega$, the bridge is balanced. On interchanging $\mathrm{P}$ and $\mathrm{Q}$, the value of $\mathrm{R},$ for balance, is $405 \Omega$. The value of $Y$ is close to

Options:
A $401.5 \mathrm{ohm}$
B $404.5 \mathrm{ohm}$
C $403.5 \mathrm{ohm}$
D $402.5 \mathrm{ohm}$

The correct answer is:
$402.5 \mathrm{ohm}$

Solution:
$402.5 \mathrm{ohm}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (11 Jan Shift 1)
In a Wheatstone bridge, three resistances \( P, Q \) and \( R \) are connected in the three arms and the fourth arm is formed by two resistances \( S_{1} \) and \( S_{2} \) connected in parallel. The condition for the bridge to be balanced will be

Options:
A \( \frac{P}{Q}=\frac{2 R}{S_{1}+S_{2}} \)
B \( \frac{P}{Q}=\frac{R\left(S_{1}+S_{2}\right)}{S_{1} S_{2}} \)
C \( \frac{P}{Q}=\frac{R\left(S_{1}+S_{2}\right)}{2 S_{1} S_{2}} \)
D \( \frac{P}{Q}=\frac{R}{S_{1}+S_{2}} \)

The correct answer is:
\( \frac{P}{Q}=\frac{R\left(S_{1}+S_{2}\right)}{S_{1} S_{2}} \)

Solution:
As we can see, S1 & S2 are parallel, so equivalent resistance,

1S=1S1+1S2

⇒1S=S2+S1S1S2

⇒S=S1S2S1+S2...eq 1

For wheatstone stone bridge, ratio of resistance along a arm remains equal to ratio of resistance in another arm.
So, the balance condition will be:

PQ=RS....eq 2

Putting eq 1 in eq 2

PQ=R(S1+S2)S1S2

Topic: Physics, Current Electricity, JEE Main
In a Wheatstone's bridge all the four arms have equal resistance $R$. If the resistance of the galvanometer arm is also $R$, the equivalent resistance of the combination as seen by the battery is:

Options:
A $\frac{R}{4}$
B $\frac{R}{2}$
C $R$
D $2 R$

The correct answer is:
$R$

Solution:
Resistance seen by the battery $=$ Equivalent resistance between $A$ and $B$ is $R$.

Topic: Physics, Current Electricity, JEE Main
In a Wheatstone's bridge, there resistances $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ connected in the three arms and the fourth arm is formed by two resistances $\mathrm{S}_1$ and $\mathrm{S}_2$ connected in parallel. The condition for bridge to be balanced will be

Options:
A
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}}{\mathrm{S}_1+\mathrm{S}_2}$

B
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{2 \mathrm{R}}{\mathrm{S}_1+\mathrm{S}_2}$

C
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$

D
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{2 \mathrm{~S}_1 \mathrm{~S}_2}$

The correct answer is:
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$

Solution:
$\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{R}\left(\mathrm{S}_1+\mathrm{S}_2\right)}{\mathrm{S}_1 \mathrm{~S}_2}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2006
In a Wheatstone's network, P=2 Î©, Q=2 Î©, R=2 Î© and S=3 Î©. Find the resistance (in Î© ) with which S is to be shunted, in order that the bridge gets balanced.

Options:

The correct answer is:
6

Solution:
Let a resistance r Î© be shunted with resistance S, so that the bridge is balanced.
Let S' be the resultant resistance of S and r, then
In balanced position
PQ=RSâ²
22=2Sâ²
â´ Sâ²=2 Î©
Now,
1Sâ²=1S+1r
1r=1Sâ²-1S=12-13=3-26
1r=16
r=6 Î©

Topic: Physics, Current Electricity, JEE Main
In a woody dicotyledonous tree, which of the following parts will mainly consist of primary tissues?

Options:
A Stem and root
B All parts
C Shoot tips and root tips
D Flowers, fruits and leaves

The correct answer is:
Shoot tips and root tips

Solution:
Primary tissues are those meristematic tissues, which are derived directly from embryonal tissues.
e.g., shoot apex and root apex:

Topic: Biology, Anatomy of Flowering Plants, JIPMER, JIPMER 2007
In a woody dicotyledonous tree, which of the following parts will mainly consist of primary tissues?

Options:
A all parts
B stem and root
C flowers, fruits and leaves
D shoot tips and root tips.

The correct answer is:
flowers, fruits and leaves

Solution:
The tissues which develop from apical meristem and procambium are known as primary tissues and these constitute the primary structures of the plant body. The development of the primary tissues causes the stem to grow in length and to some extent in thickness. Stem and root in dicotyledons show secondary growth as they have development of secondary vascular tissues (i.e., secondary xylem and secondary phloem) in them. Flowers, fruits and leaves are primary tissues.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2005
In a workshop, there are five machines and the probability of any one of them to be out of service on a day is 14 . If the probability that at most two machines will be out of service on the same day is 343k, then k is equal to

Options:
A 178
B 174
C 172
D 4

The correct answer is:
178

Solution:
Given $n=5, p=\frac{1}{4}$ and $q=1-p=\frac{3}{4}$

Required probability = when no machine has fault + when one machine has fault + when two machines have fault.

=C05345+C1514344+C25142343

=2431024+4051024+2701024=9181024=459512=27×1764×8

⇒343×k=343×178

∴k=178

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (07 Jan Shift 2)
In a workshop, there are five machines and the probability of any one of them to be out of service on a day is 14 . If the probability that at most two machines will be out of service on the same day is 343k, then k is equal to

Options:
A 178
B 174
C 172
D 4

The correct answer is:
178

Solution:
Required probability = when no machine has fault + when one machine has fault + when two machines have fault.
=C05345+C1514344+C25142343
=2431024+4051024+2701024=9181024=459512=27×1764×8
⇒343×k=343×178
∴k=178

Topic: Mathematics, Probability, JEE Main
In a workshop, there are five machines and the probability of any one of them to be out of service on a day is \( \frac{1}{4} \). If the
probability that at most two machines will be out of service on the same day is \( \left(\frac{3}{4}\right)^{3} k \), then \( k \) is equal to

Options:
A \( \frac{17}{8} \)
B \( \frac{17}{4} \)
C \( \frac{\sqrt{\frac{1}{2}}}{2} \)
D \( 4 \)

The correct answer is:
\( \frac{17}{8} \)

Solution:
Required probability = when no machine has fault + when one machine has fault + when two machines have fault.

=C05345+C1514344+C25142343

=2431024+4051024+2701024=9181024=459512=27×1764×8

⇒343×k=343×178

∴k=178

Topic: Mathematics, Probability, JEE Main
In a YDSE bi-chromatic light of wavelengths, 400 nm and 560 nm are used. The distance between the plane of the slits is 0.1 mm and the distance between the slits and the screen is 1 m. The minimum distance between two successive regions of complete darkness is

Options:
A 4 mm

B 5.6 mm
C 14 mm

D 28 mm

The correct answer is:
28 mm

Solution:
Let n th minimum of 400nm coincides with m th minimum of 560 nm, then
For nth minima Î´=2n-1Î»2
2 n - 1 4 0 0 2 = 2 m - 1 5 6 0 2
2 n - 1 2 m - 1 = 7 5 = 1 4 1 0 = 2 1 1 5
i.e., 4 th minimum of 400 nm coindes with 3rd minimum of 560 nm.
Distance of nth minima from center =2n-1 Î»D2d
Location of this minimum is,
Y 1 = 2 Ã 4 - 1 1 0 0 0 4 0 0 Ã 1 0 - 6 2 Ã 0 Â· 1 = 1 4 mm
Next 11th minimum of 400 nm will coincide with 8th
minimum of 560 nm.
Location of this minimum is,
Y 2 = 2 Ã 1 1 - 1 1 0 0 0 4 0 0 Ã 1 0 - 6 2 Ã 0 Â· 1 = 4 2 mm
â´ Required distance = Y 2 - Y 1 = 2 8 mm
Hence, the correct option is (28 mm)

Topic: Physics, Wave Optics, JEE Main
In a Young's double slit experiment 15 fringes are observed on a small portion of the screen when light of wavelength 500 nm is used. Ten fringes are observed on the same section of the screen when another light source of wavelength λ is used. Then the value of λ is (in nm ) __________.

Options:

The correct answer is:
750

Solution:
15×500×Dd=10×λ2×Dd
λ2=15×50 nm
λ2=750 nm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (09 Jan Shift 2)
In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is _____ nm.

Options:

The correct answer is:
630

Solution:
Given here: λ1=560×10-9 m, β1=7.2×10-3 m and β2=8.1×10-3 m.

Let d be the width and D be the distance between slit and screen then fringe width, β=λDd.

Thus, from above, β∝λ

So, the wavelength of second light is λ2=β2β1λ1=98λ1

∴ λ2=98×560=630 nm.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (28 Jul Shift 1)
In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:

Options:
A 8
B 9
C 12
D 6

The correct answer is:
12

Solution:
Wavelength of light used is inversely proportional to number of fringes observed.

Hence, we can write:

n1λ1=n2λ2⇒n2=n1λ1λ2=8×600×10-9400×10-9=12

Topic: Physics, Wave Optics, JEE Main
In a Young's double slit experiment, if there is no initial phase-difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has path difference.

Options:
A $5 \frac{\lambda}{2}$
B $10 \frac{\lambda}{2}$
C $9 \frac{\lambda}{2}$
D $11 \frac{\lambda}{2}$

The correct answer is:
$9 \frac{\lambda}{2}$

Solution:
In a YDSE, the path difference for $\mathrm{n}$ th minima is given by
$$
\Delta y=(2 n-1) \frac{\lambda}{2}
$$
For 5 th minima, $\mathrm{n}=5$
$$
\therefore \quad \Delta y=[2(5)-1] \frac{\lambda}{2}=\frac{9 \lambda}{2}
$$

Topic: Physics, Wave Optics, JEE Main
In a Young's double slit experiment, light of 500nm is used to produce and interference pattern. When the distance between the slits is 0.05mm, the angular width (in degree) of the fringes formed on the distance screen is close to :

Options:
A 0.17°
B 0.57°
C 1.7°
D 0.07°

The correct answer is:
0.57°

Solution:
β0=λd=500×10−95×10−5=10−2 rad

∴β0=0.57°

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (03 Sep Shift 1)
In a Young's double slit experiment slit separation 0.1 mm, one observes a bright fringe at angle 140 rad by using light of wavelength λ1. When the light of wavelength λ2 is used a bright fringe is seen at the same angle in the same set up. Given that λ1 and λ2 are in visible range 380 nm to 740 nm, their values are:

Options:
A 400 nm, 500 nm
B 380 nm, 525 nm
C 625 nm, 500 nm
D 380 nm, 500 nm

The correct answer is:
625 nm, 500 nm

Solution:
d sinθ=nλ
∴λ=dsinθn=10-4×140×1n
∴λ=2.5×10-6n=2500×10-9n=2500nnm
Putting n=1, λ1=2500 nm
n=2, λ2=1250 nm
n=3, λ3=833 nm
n=4, λ4=625 nm
n=5, λ5=500 nm
n=6, λ6=357 nm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (10 Jan Shift 1)
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

Options:
A 15.6 mm
B 1.56 mm
C 7.8 mm
D 9.75 mm

The correct answer is:
7.8 mm

Solution:
Given,
Wavelength of first light is, λ1=650 nm
Wavelength of second light is, λ2=520 nm
Screen distance is, D=150 cm
Slit width is, d=0.5 mm
For first wavelength, location of fringe from central line,
y1=n1λ1Dd for bright fringes
For second wavelength, location of fringe from central line,
y2=n2λ2Dd for bright fringes
Condition of coincide of fringe is,
y1=y2
⇒n1λ1=n2λ2
⇒n1×650=n2×520
⇒n1n2=45
For the minimum value of n1 and n2
n1=4 and n2=5
y1min=4×650 nm×1.50 m0.5×10-3 m
=7800×10-6 m
=7.8 mm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2017 (02 Apr)
In a Young's double slit experiment, the central point on the screen is

Options:
A Bright
B Dark
C First bright and then dark
D First dark and then bright

The correct answer is:
Bright

Solution:
In the normal adjustment of young's double slit experiment, path difference between the waves at central location is always zero, so maxima is obtained at central position.

Topic: Physics, Wave Optics, JEE Main
In a Young's double slit experiment, the fringe width is \( \beta \). If the entire arrangement is now placed inside a liquid of refractive index \( \mu \), the fringe width will become,

Options:
A \( \mu \beta \)
B \( \frac{\beta}{\mu} \)
C \( \frac{\beta}{\mu+1} \)
D \( \frac{\beta}{\mu-1} \)

The correct answer is:
\( \frac{\beta}{\mu} \)

Solution:
The fringe width,

β=λDd⇒β∝λ.

When the entire arrangement is placed inside a liquid, the wavelength of light changes. The new wavelength becomes,

λ'=λμ

ββ'=λλ' ⇒ ββ'=λλμ

β'=βμ.

Topic: Physics, Wave Optics, JEE Main
In a Young’s double slit experiment, the intensities at two points, for the path difference λ4 and λ3 (λ being the wavelength of light used) are I1 and I2 respectively. If I0 denotes the intensity produced by each one of the individual slits, then I1+I2I0=______.

Options:

The correct answer is:
3

Solution:
As we know, resultant intensity is given by,

I=4I0 cos2Δϕ2 and phase difference can be written as ∆ϕ=2πλ∆x.

For first case:

I1=4I0 cos22πλ×λ4=4I0 cos2π4=2I0

For second case:

I2=4I0 cos22πλ×λ3=4I0 cos22π3=I0

Therefore, required value ofI1+I2I0=3.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2023 (30 Jan Shift 2)