datavorous/entrance-exam-dataset huggingface dataset:
In a compound microscope the focal length of objective lens is $1.2 \mathrm{~cm}$ and focal length of eye piece is $3.0 \mathrm{~cm}$. When object is kept at $1.25 \mathrm{~cm}$ in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:

Options:
A
200

B
100

C
400

D
150

The correct answer is:
200

Solution:
Given : $f_0=1.2 \mathrm{~cm} ; f_e=3.0 \mathrm{~cm}$ $u_0=1.25 \mathrm{~cm} ; M_{\infty}=$ ?
From $\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}$
$$
\Rightarrow \quad \frac{1}{1.2}=\frac{1}{v_0}-\frac{1}{(-1.25)}
$$
$$
\begin{aligned}
&\Rightarrow \quad \frac{1}{v_0}=\frac{1}{1.2}-\frac{1}{1.25} \\
&\Rightarrow v_0=30 \mathrm{~cm}
\end{aligned}
$$
Magnification at infinity,
$$
\begin{aligned}
&M_{\infty}=-\frac{v_0}{u_0} \times \frac{D}{f_e} \\
&=\frac{30}{1.25} \times \frac{25}{3} \quad(\because D=25 \mathrm{~cm} \text { least distance }
\end{aligned}
$$
of distinct vision)
$$
=200
$$
Hence the magnifying power of the
compound microscope is 200

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (11 Apr Online)
In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is ______

Options:

The correct answer is:
6.25

Solution:
M=v0u01+Dfe

M=Lf01+Dfe

100=2011+25fe ; 5=1+25fe

4=25fe ; fe=254=6.25 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (04 Sep Shift 1)
In a compound microscope, the objective lens and eyepiece have a focal length of 0.95 cm and 5 cm respectively and are kept at a distance of 20 cm. If the last image is formed at a distance of 25 cm from the eyepiece, calculate magnifying power.

Options:
A 94
B 84
C 75
D 88

The correct answer is:
94

Solution:
f0=0.95 m=V0u01+Dfe
fe=5 cm ue=256
For eyepiece
1+Dfe=VeUe u=Ve6
V0=20-256=956m0=956-0.950.95≈946
mnet=946×6=94

Topic: Physics, Ray Optics, JEE Main
In a compound oxide ion have ccp arrangement, cations A are present in 18 of the tetrahedral voids and cations B occupy 12 of the octahedral voids. What are the formula of the compound?

Options:
A A2BO4
B AB2O4
C ABO2
D A2BO2

The correct answer is:
AB2O4

Solution:
Oxide ions are forming the lattice
Effective number of oxide ions =4ccp
∴ Tetrahedral voids =2×4=8
And effective number of A atom =18×8=1
Number of octahedral voids =4
Effective number of B atom =12×4=2
∴ Molecular formulae =AB2O4

Topic: Chemistry, Solid State, JEE Main
In a conductor, if the number of conduction electrons per unit volume is 8.5×1028  m-3 and mean free time is 25 fs (femto second), it’s approximate resistivity is: me=9.1×10-31 kg

Options:
A 10-7 Ω m
B 10-6 Ω m
C 10-5 Ω m
D 10-8 Ω m

The correct answer is:
10-8 Ω m

Solution:
We know i=neAvd
⇒i=VR=VAρl=EAρ
ρ=EAi=EAneAvd
Drift velocity can be written as vd=eEmeτ
ρ=EAmeneAeEτ=mene2τ
=9.1×10-318.5×1028×1.6×10-192×25×10-15
≈1.65×10-8 Ω m

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Apr Shift 2)
In a conference of 8 persons, if each person shake hand with the other one only, then the total number of shake hands shall be

Options:
A $64$
B $56$
C $49$
D $28$

The correct answer is:
$28$

Solution:
Total number of shake hands when each person shake hands with the other once only $=8 C_2 = 28$ ways.

Topic: Mathematics, Permutation Combination, JEE Main
In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K-1 , the numerical value for the enthalpy of combustion of the gas in kJ mol-1 is

Options:

The correct answer is:
9

Solution:
Energy released by combustion of 3.5 g gas
=2.5×(298.45-298)kJ
Energy released by 1 mole of gas =2.5×0.453.528=9 kJmol-1

Topic: Chemistry, Thermodynamics (C), JEE Main
In a constant volume calorimeter, \( 3.5 \mathrm{~g} \) of a gas with molecular weight \( 28 \) was burnt in excess oxygen at \( 298.0 \mathrm{~K} \). The temperature of the calorimeter was found to increase from \( 298.0 \mathrm{~K} \) to \( 298.45 \mathrm{~K} \) due to the combustion process. Given that the heat capacity of the calorimeter is \( 2.5 \mathrm{~kJ} \mathrm{~K}^{-1} \), the numerical value for the enthalpy of combustion of the gas in \( \mathrm{kJ} \) \( \mathrm{mol}^{-1} \) is

Options:

The correct answer is:
9

Solution:
Energy released by combustion of 3.5 g gas
=2.5×(298.45-298)kJ
Energy released by 1 mole of gas =2.5×0.453.528=9 kJmol-1

Topic: Chemistry, Thermodynamics (C), JEE Main
In a continuous printing process, paper roll is wrapped on a cylindrical core which is free to rotate about a fixed horizontal axis. The paper is drawn into the process at a constant speed v, as shown in the figure. If r is the radius of paper on the roll at any given time and b is the thickness of the paper, then the angular acceleration of the roll at this instant is

Options:
A v2b2π2r3
B

zero because v is constant

C v2b2Ïr3
D cannot be calculated from given information

The correct answer is:
v2b2πr3

Solution:
In a continuous printing process suppose the total length of paper is l.

If r1 is the inner radius of roll then, Ïr2-Ïr12=lÃb ...(i)
And at this moment, v=rÏ ...(ii)
Differentiating equation (i) w.r.t. time,
âÏÃ2rdrdt=dldtÂ·b
âdrdt=v.b2Ïr ...(iii)
Differentiating equation (ii) w.r.t time
âdvdt=rdÏdt+Ïdrdt
â0=rÎ±+Ïdrdt âµdrdtis-ve
âdrdt=rÎ±Ï
âÎ±=Ïrdrdt ...(iv)
Using (iii) and (iv), vb.Ï2Ïr2=v2b2Ïr3=Î±

Topic: Physics, Rotational Motion, JEE Main
In a Coolidge tube, the potential difference used to accelerate the electron is increased from 24.8 kV to 49.6 kV. As a result, the difference between the wavelength of KÎ±-line and minimum wavelength becomes two times. Then the wavelength of KÎ±-line is
hce=12.4 kV Aâ

Options:
A 0.75 A∘
B 1.50 A∘
C 0.25 A∘
D 0.50 A∘

The correct answer is:
0.75 A∘

Solution:
Let âE be the energy of kα-line then hcλkα=ΔEkα
⇒ λkα=hcΔEkα
Now λ1 min=hce×24.8 kV=12.424.8 A∘=12 A∘ and λ2 min=hce×49.6 kV=12.449.6 A∘=14 A∘
⇒ 2λkα-12=λkα-14
⇒ 2λkα-1=λkα-14
⇒ λkα=1-14
⇒ λkα=0.75 A∘

Topic: Physics, Atomic Physics, JEE Main
In a Coolidge tube, the potential difference used to accelerate the electrons is increased from 24.8 kV to 49.6 kV. As a result, the difference between the wavelength of KÎ±-line and minimum wavelength becomes two times. The initial wavelength of the KÎ±-line is [Take hce=12.4 kV AÂ°]

Options:
A 32A°
B 34A°
C 52A°
D 54A°

The correct answer is:
34A°

Solution:
Let âE be the energy of kα-line then

hcλKα=ΔEKα
⇒ λKα=hcΔEKα
Now the cutoff wavelengths in the two cases are

λ1min=hce×24.8 kV=12.424.8A°=12A°

λ2min=hce×49.6 kV=12.449.6A°=14A°
⇒ 2λkα-12=λkα-14
⇒ 2λkα-1=λkα-14
⇒ λkα=1-14
⇒ λkα=34Å

Topic: Physics, Atomic Physics, JEE Main
In a crystal, the atoms are located at the position of

Options:
A Maximum P.E.
B Minimum P.E.
C Zero P.E
D Infinite P.E.

The correct answer is:
Minimum P.E.

Solution:
Lowest potential energy level provides stable arrangement.

Topic: Chemistry, Solid State, JEE Main
In a cubic container of inner side length $10 \mathrm{~cm}$, nitrogen gas of $100 \mathrm{kPa}$ pressure is maintained at $300 \mathrm{~K}$. If the pressure inside the gas is increased to $300 \mathrm{kPa}$ by adding oxygen gas, the ratio of number of $\mathrm{N}_2$ to $\mathrm{O}_2$ molecules in the container is

Options:
A 0.5
B 3.0
C 1.5
D 0.33

The correct answer is:
0.5

Solution:
As $V$ and $T$ are same for both gases, we can write
$$
\left.\begin{array}{rl}
p_1 V & =n_1 R T \\
p_2 V & =n_2 R T
\end{array}\right\} \text { where } p_1 \text { and } p_2 \text { are partial }
$$
pressures of the gases.
$\left.\Rightarrow \frac{p_1}{p_2}=\frac{n_1}{n_2}\right\}$ where, 1 - for oxygen and 2- for nitrogen.
$$
\Rightarrow \frac{100}{200}=\frac{n_1}{n_2} \Rightarrow \frac{n_1}{n_2}=\frac{1}{2}=0.5
$$

Topic: Physics, Kinetic Theory of Gases, TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids one occupied by X ions while one half of the octahedral voids are occupied by Y ions. Then formula of the oxide will be

Options:
A X5Y4O10
B X4Y5O10
C XY2O4
D X2YO4

The correct answer is:
X4Y5O10

Solution:
Number of O2- ions = n
Number of tetrahedral voids = 2n
Number of octahedral voids = n
Number of X ions =2n5
Number of Y ions =n2
X : Y : O 2n 5  :  n 2  : n 4 : 5 : 10
Simplest formula =X4Y5O10 .

Topic: Chemistry, Solid State, JEE Main
In a cuboid of dimension 2L×2L×L , a charge q is placed at the centre of the surface S having area of 4L2 . The flux through the opposite surface to S is given by

Options:
A q12∈0
B q3∈0
C q2∈0
D q6∈0

The correct answer is:
q6∈0

Solution:
Assume a similar cuboid on top of the given cuboid to make symmetry about charge. Thus, it becomes a cube of length 2L.

Flux passing through whole cube is ϕ=q∈0(Gauss's Law).

Since, total faces are 6, so flux passing through shaded face =q6∈0.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (29 Jan Shift 1)
In a Cu-voltameter, mass deposited in 30s is m gm. If the time-current graph is shown in the following figure

What is the electrochemical equivalent of Cu?

Options:
A m/2
B m/3
C m/4
D m63.5

The correct answer is:
m/3

Solution:
Total charge passed in 30 sec.
=12×100×10+100×10+100×10+12×100×10
=3×100×10mA×s
=3000×10-3amp s=3C
W=ZQ
So Z=WQ=m3

Topic: Chemistry, Electrochemistry, JEE Main
In a cyclotron, a charged particle

Options:
A
undergoes acceleration all the time

B
speeds up between the dees because of the magnetic field

C
speeds up in a dees

D
slows down within a dee and speeds up between dees

The correct answer is:
undergoes acceleration all the time

Solution:
There is crossed electric and magnetic field between dees so the charged particle accelerates by electric field between dees towards other dees.
So, the charged particle undergoes acceleration as
(i) speeds up between the dees because of the oscillating electric field.
(ii) speed remain the same inside the dees because of the magnetic field but direction undergoes change continuously.
Hence, the charge particle accelerates inside and between Dees always.

Topic: Physics, Magnetic Effects of Current
In a cylinder-piston arrangement, the air is under a pressure P1. A soap bubble of radius r lies inside the cylinder. Soap bubble has surface tension T. The radius of soap bubble is to be reduced to half. Find the pressure P2 to which air should be compressed isothermally.

Options:
A P 1 + 4 T r
B 4 P 1 + 1 2 T r
C 8 P 1 + 2 4 T r
D P 1 + 2 T r

The correct answer is:
8 P 1 + 2 4 T r

Solution:
Using PiVi = Pf Vf (for the air bubble)
we have
P 1 + 4 T r 4 3 π r 3 = P 2 + 4 T r / 2 4 3 π r 2 3
Solving this, we get
P 2 = 8 P 1 + 2 4 T r

Topic: Physics, Mechanical Properties of Fluids, JEE Main
In a cylindrical water tank, there are two small holes $A$ and $B$ on the wall at a depth of $h_1$, from the surface of water and at a height of $h_2$ from the bottom of water tank. Surface of water is at height of $h_2$ from the bottom of water tank. Surface of water is at heigh $H$ from the bottom of water tank. Water coming out from both holes strikes the ground at the same point $S$. Find the ratio of $h_1$ and $h_2$

Options:
A
Depends on $H$

B
$1: 1$

C
$2: 2$

D
$1: 2$

The correct answer is:
Depends on $H$

Solution:
Range is same for both holes
$$
\therefore 2 \sqrt{\left(\mathrm{H}-h_1\right) h_1}=2 \sqrt{\left(\mathrm{H}-h_2\right) h_2}
$$
Squaring both sides,
$$
\begin{aligned}
& 4\left(H-h_1\right) \mathrm{h}_1=4\left(H-h_2\right) h_2 \\
& H h_1-h_1^2=H h_2-h_2^2
\end{aligned}
$$
On solving we get,
$$
H=h_1+h_2
$$
Hence, the ratio of $\frac{h_1}{h_2}$ depends on $H$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2012 (26 May Online)
In a cymose inflorescence the main axis:

Options:
A has unlimited growth
B bears a solitary flower
C has unlimited growth but lateral branches end in flowers
D terminates in a flower.

The correct answer is:
terminates in a flower.

Solution:
In the cymose type of inflorescence, the main axis does not grow continuously. A flower is present terminally on the main axis. The flowers are borne in a basipetal order. The main axis has limited growth. The growth of the main axis is limited and the lateral axis also terminates into flower and that limits the growth of the axis. Flowers are present in the basipetal succession, i.e. older flower is present terminally and the younger ones at the base. This type of arrangement is known as centrifugal with the older flowers present in the middle and the younger ones at the periphery. In racemose types of inflorescence, the main axis grows continuously and flowers are present laterally on the floral axis. Flowers are present in an acropetal manner.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2013 (Karnataka)
In a detritus food chain, decomposition of detritus depends upon all, except

Options:
A oxygen non-availability
B chemical composition of detritus
C temperature and soil moisture
D soil microbes

The correct answer is:
oxygen non-availability

Solution:
Food chain starting with dead organic matter is Detritus food chain. Detritus is the dead organic material which includes the parts or bodies of dead organisms and their faecal matter. These organic contents are broken down to inorganic substances by the process called decomposition. It is brought about by decomposers. Decomposition is a mainly an oxygen dependent process and its rate is dependent on chemical composition and climatic factors. Temperature and moisture play an important role in rate of decomposition. Enough supply of oxygen and proper moisture accelerates the decomposition process.

Topic: Biology, Ecosystem, NEET
In a developing embryo, analyse the consequences if cell divisions are not followed by cell differentiation.

Options:

None

Solution:
Cell division and cell differentiation occurs through embryogenesis. Cell division occurs through increase in the number of cells in the developing embryo, while cell differentiation takes part in the modifications of group of cells to form specialised tissues and organs to form an organism.
If cell differentiation does not occur in embryogenesis, the embryo cannot develop into a new organism. It will only remain as a mass of cells.

Topic: Biology, Reproduction in Organisms
In a diffraction pattern due to a single slit of width a the first minimum is observed at an angle 30o when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of:

Options:
A sin-1⁡14
B sin-1⁡23
C sin-1⁡12
D sin-1⁡34

The correct answer is:
sin-1⁡34

Solution:
asin⁡30=λ
asin⁡θ=3λ2
sin⁡θsin⁡30=32
sin⁡θ=32×12
sin⁡θ=34
θ=sin⁡34

Topic: Physics, Wave Optics, JEE Main
In a diode, the diffusion current is \( i_{i f f u s i o n} \) and drift current is \( i_{i d i f i t} \), then match the column.
\(\begin{array}{|l|l|} \hline \text{Column-I} & \text{Column-II} \\ \hline \text{(1) Diode with no biasing} & (p) \mathrm{i}_{\text {diffusion }}>>\mathrm{i}_{\text {drift }} \\ \text{(2) Diode in forward bias} & \text{(q) } \mathrm{i}_{\text {drift }}> >\mathrm{i}_{\text {diffusion }} \\ \text{(3) Diode in reverse bias without} & (r) \dot{\mathrm{I}}_{\mathrm{drift}} >\mathrm{i}_{\text {diffusion }} \\ \text{breakdown} & \\ \text{(4) Diode in reverse bias with} & (s) \dot{\mathrm{I}}_{\mathrm{drift}}=\mathrm{i}_{\text {diffusion }}\\ \text{breakdown} & \\ \hline \end{array}\)

Options:
A \( A \rightarrow s, B \rightarrow q, C \rightarrow r, D \rightarrow p \)
B \( A \rightarrow s, B \rightarrow p, C \rightarrow r, D \rightarrow q \)
C \( A \rightarrow s, B \rightarrow q, C \rightarrow p, D \rightarrow r \)
D \( A \rightarrow q, B \rightarrow p, C \rightarrow r, D \rightarrow s \)

The correct answer is:
\( A \rightarrow s, B \rightarrow p, C \rightarrow r, D \rightarrow q \)

Solution:
Diode with no biasing ⇒ Idrift =idiffusion

Diode in forward bias ⇒İdiffusion >> idrift

Diode in reverse bias without breakdown

⇒ İdrift > idiffusion

Diode in reverse bias with breakdown

⇒ idrift >> idiffusion

Topic: Physics, Semiconductors, JEE Main
In a discrete data $\frac{\text { lth }}{4}$ of the observations are equal to $a$, another $\frac{\text { th }}{4}$ of the observations are equal to $-a$. Out of the remaining, half of them are equal to $b$ and the rest are equal to $-b$. If the variance of all the observations is $(a b)$, then

Options:
A $a^2=4 b^2$
B $a=-2 b$
C $a=b$
D $a=-3 b$

The correct answer is:
$a=b$

Solution:
$\begin{aligned} & \text { Given, } \frac{1}{4} \text { th observation }=a \\ & \text { another } \frac{1}{4} \text { th observation }=-a \\ & \quad \frac{1}{4} \text { th observation }=b \Rightarrow \frac{1}{4} \text { th observation }=-b \\ & \therefore \quad \bar{x}=\frac{a-a+b-b}{n}=0 \\ & \text { Variance }=a b \\ & \quad a b=\frac{\Sigma x_i^2}{n}-(\bar{x})^2=\frac{\Sigma x_i^2}{n} \\ & \Rightarrow a b=\frac{\frac{n}{4}\left(a^2+a^2\right)+\frac{n}{4}\left(b^2+b^2\right)}{n} \Rightarrow a b=\frac{a^2+b^2}{2} \\ & \Rightarrow \quad a^2+b^2-2 a b=0 \Rightarrow(a-b)^2=0 \Rightarrow a=b\end{aligned}$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 2)
In a double – slit experiment, at a certain point on the screen the path difference between the two interfering waves is 18th of a wavelength. The ratio of the intensity of light at that point to that at the center of a bright fringe is:

Options:
A 0.853
B 0.672
C 0.568
D 0.760

The correct answer is:
0.853

Solution:
I=I0cos2∆ϕ2
II0=cos22πλ×∆X2=cos2π8;II0=0.853

Topic: Physics, Wave Optics, JEE Main, JEE Main 2020 (08 Jan Shift 2)
In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D=0.2 m, the minimum distance between the slits S1 and S2 is α mm. Write the value of 10α to the nearest integer.

Options:

The correct answer is:
2

Solution:
According to the given diagram, the path difference for minima at P is given by

∆x=2D2+d2-2D ...1

And, from the condition of minima, it follows that

∆x=λ2 ...2

Equations (1) and (2) imply that

2D2+d2-2D=λ2

⇒D2+d2-D=λ4

⇒D2+d2=λ4+D

⇒D2+d2=D2+λ216+Dλ2

⇒d2=Dλ2+λ216

⇒d2=0.2×400×10-92+16×10-1416

⇒d2≈400×10-10

∴d=20×10-5 m=0.20 mm

Therefore, 10α=2.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (29 Jan Shift 1)
In a double slit experiment, the distance between slits is increased 10 times whereas their distance from screen is halved, then what is the fringe width?

Options:
A It remains same
B Becomes $1 / 10$
C Becomes $1 / 20$
D Becomes $1 / 90$

The correct answer is:
Becomes $1 / 20$

Solution:
Let $\lambda$ be wavelength of monochromatic light, $d$ the distance between coherent sources, and $D$ the distance between screen and source, then fringe width is
$\quad \beta=\frac{D \lambda}{d}$
Given, $d_{1}=d, D_{1}=D, d_{2}=10 d, D_{2}=\frac{D}{2}$
$\therefore \quad \quad \beta_{2}=\frac{\frac{D}{2} \lambda}{10 d}=\frac{D \lambda}{20 d}$
$\Rightarrow \quad \beta_{2}=\frac{W}{20}$

Topic: Physics, Magnetic Properties of Matter, MHT CET, MHT CET 2007
In a double slit experiment, when a thin film of thickness t having refractive index μ is introuduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is (λ is the wavelength of the light used):

Options:
A λ2(μ-1)
B λ(2μ-1)
C 2λ(μ-1)
D λ(μ-1)

The correct answer is:
λ(μ-1)

Solution:
Shift =μ-1 tDd
Fringe width (β)=λDd
(μ-1)  tDd=λDd
t=λμ-1

Topic: Physics, Wave Optics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
In a double slit experiment, when the distance between slits is increased 10 times, while their distance from the screen is halved, then the fringe width

Options:
A becomes $\left(\frac{1}{20}\right)$ times the original
B becomes $\left(\frac{1}{90}\right)$ times the original
C remains the same
D becomes $\left(\frac{1}{10}\right)$ times the original

The correct answer is:
becomes $\left(\frac{1}{20}\right)$ times the original

Solution:
Case 1 Fringe width, $\beta_1=\frac{\lambda D_1}{d_1}$
Case 2 Fringe width, $\beta_2=\frac{\lambda D_2}{d_2}$
$\begin{array}{ll}\text { Given, } d_2= & 10 d_1, D_2=\frac{D_1}{2} \\ \Rightarrow & \frac{\beta_1}{\beta_2}=\frac{\lambda D_1}{d_1} \times \frac{d_2}{\lambda D_2} \\ \Rightarrow & \frac{\beta_1}{\beta_2}=\frac{D_1}{D_2} \times \frac{d_2}{d_1}=2 \times 10 \\ \Rightarrow & \frac{\beta_1}{\beta_2}=20 \\ \Rightarrow & \beta_2=\frac{1}{20} \beta_1\end{array}$

Topic: Physics, Wave Optics, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5×10-2 m towards the slits, the change in fringe width is 3×10-3 cm. If the distance between the slits is 1 mm, then the wavelength of the light will be____nm.

Options:

The correct answer is:
600

Solution:
We know that the fringe width in YDSE is β=λDd

⇒Δβ=λdΔD

⇒λ=ΔβdΔD=3×10-3×10-2×10-35×10-2=6×10-7 m=600 nm

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
In a double-slit experiment, the distance between slits is increased 10 times, whereas their distance from screen is halved, then the fringe width

Options:
A Becomes 120th
B Becomes 190th
C Remains same
D Becomes 110th

The correct answer is:
Becomes 120th

Solution:
We know that, fring width
β=Dλd ....(i)
When distance slits is increased 10 times and distance between between the source and screen is halved, then new fringe width,
β′=D′λd′
= Dλ 2×10×d
=120⋅Dλd
=β20 [from Equation (i)]

Topic: Physics, Wave Optics, NEET
In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double-slit within the central maxima of a single-slit pattern?

Options:
A 0.2 mm

B 0.1 mm

C 0.5 mm

D 0.02 mm

The correct answer is:
0.2 mm

Solution:
d =1 mm=1×10-3 m
D=1 m., λ=500×10-9 m
Width of the central maximum = width of 10 maxima in YDSE
2λDa=10λDd
a=0.2d=0.2 mm

Topic: Physics, Wave Optics, JEE Main
In a ellipse, its foci and the ends of its major axis are equally spaced, if the length its semi-minor axis is $2 \sqrt{2}$, then the length of its semi-major axis is :

Options:
A 4
B $2 \sqrt{3}$
C $\sqrt{10}$
D 3

The correct answer is:
3

Solution:
$A S^{\prime}=S S^{\prime}=S B$ $\Rightarrow e=1 / 3$ $\Rightarrow a=3$

Topic: Mathematics, Ellipse, JEE Main
In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atom B is missing from one of the face centered points, the formula of the ionic compound is :

Options:
A A2B3
B AB2
C A5B2
D A2B5

The correct answer is:
A2B5

Solution:
A atoms at corners = 8 × 1 8 = A (one atom)
B atom is missing at One face centre then will present at 5 faces
At face centre atom contribute 1/2 part
Number of B atom at Face centre = 5× 1 2 (B atom)
So formula of compound = A2B5

Topic: Chemistry, Solid State, JEE Main, JEE Main 2014 (09 Apr Online)
In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of $B$ is missing from one of the face centred points, the formula of the compound is

Options:
A
$\mathrm{AB}_2$

B
$A_2 B_3$

C
$\mathrm{A}_2 \mathrm{~B}_5$

D
$A_2 B$

The correct answer is:
$\mathrm{A}_2 \mathrm{~B}_5$

Solution:
Effective no.of $A$ atoms $=\frac{1}{8} \times 8=1$
Effective no.of $B$ atoms $=\frac{1}{2} \times 5$ (One is missing) $=\frac{5}{2}$
Therefore formula is $A_1 B_{\frac{5}{2}}=A_2 B_5$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2011