datavorous/entrance-exam-dataset huggingface dataset:
When two bodies collide elastically. The force of interaction between them is

Options:
A Conservative
B Non - conservative
C Either conservative or non - conservative
D Zero

The correct answer is:
Conservative

Solution:
When non - conservative force acts on a body. Then, mechanical energy is converted into non - mechanical energy and vice - versa.
In the case of elastic collision, kinetic energy of system remains constant. Hence, 1 is correct.

Topic: Physics, Center of Mass Momentum and Collision, NEET
When two bodies collide elastically, then

Options:
A Kinetic energy of the system alone is conserved
B Only momentum is conserved
C Both energy and momentum are conserved
D Neither energy nor momentum is conserved

The correct answer is:
Both energy and momentum are conserved

Solution:
In collisions in absence of any external force the momentum is conserved. As the collision is elastic so there will not be any deformation in the shapes of the bodies so there will not be any loss in energy so the energy will remain constant.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
When two bodies collide with each other such that their kinetic energy remains constant. Their collision is said to be

Options:
A elastic
B inelastic
C Both (a) and (b)
D None of the above

The correct answer is:
elastic

Solution:
When kinetic energy is conserved in a collision, then the collision is said to be elastic.

Topic: Physics, Center of Mass Momentum and Collision, COMEDK, COMEDK 2021
When two compounds ACl3 and DCl3 of two elements A and D are mixed together a compound ADCl6 is formed. Structural analysis showed that ADCl6 is an ionic compound. Given that DCl3 is trigonal planar and ACl3 is trigonal pyramidal. If anion has see-saw shape then shape of cation formed is

Options:
A Linear
B Bent
C Pentagonal bipyramidal
D Trigonal planar

The correct answer is:
Linear

Solution:
⇒ As DCl3 is trigonal planar, hence, 'D' belongs to boron family.
⇒ As ACl3 is trigonal pyramidal, hence, 'A' belongs to nitrogen family.
⇒ DCl3 + ACl3 → [DCl2]+ [ACl4]–
or [ACl2]+ [DCl4]–
If anionic part is [ACl4]– which has Sea - saw structure then cationic part will be [DCl2]+ where hybridization of D will be 'sp' and hence, shape will be linear.
⇒ If anionic part is DCl 4 - i.e, it is of tetrahedral shape the cation part will be ACl 2 + which will be bent with sp2 hybridisation.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
When two compounds ACl3 and DCl3 of two elements A and D are mixed together a compound ADCl6 is formed. Structural analysis showed that ADCl6 is an ionic compound. Given that DCl3 is trigonal planar and ACl3 is trigonal pyramidal, predict the shape of the anion DCl 4 - .

Options:
A See-saw
B Perfect tetrahedral
C Square planar
D None of these

The correct answer is:
Perfect tetrahedral

Solution:
â As DCl3 is trigonal planar, hence, 'D' belongs to boron family.
â As ACl3 is trigonal pyramidal, hence, 'A' belongs to nitrogen family.
â DCl3 + ACl3 â [DCl2]+ [ACl4]â
or [ACl2]+ [DCl4]â
if anionic part is [ACl4]â which has See - saw structure then cationic part will be [DCl2]+ where hybridization of D will be 'sp' and hence, shape will be linear.
â If anionic part is DCl 4 - i.e, it is of tetrahedral shape the cation part will be ACl 2 + which will be bent with sp2 hybridisation.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
When two dice are thrown n number of times, the probability of getting a doublet atleast once is greater than 80% and the least value of n is λ, then the value of λ is equal to

Options:
A 6
B 7
C 8
D 9

The correct answer is:
9

Solution:
Probability of getting a doublet =636=16=p (Let)
Let, q=1-p=56
Probability of getting a doublet atleast once >45
1-56n>45
â56n<15
ânâ¥9

Topic: Mathematics, Probability, JEE Main
When two different materials A and B having atomic number Z1 and Z2 are used as the target in Coolidge γ-ray tube at different operating voltage V1 and V2 respectively their spectrums are found as below.

The correct relation is

Options:
A V1>V2 and Z1>Z2
B V1<V2 and Z1<Z2
C V1<V2 and Z1>Z2
D V1>V2 and Z1<Z2

The correct answer is:
V1>V2 and Z1<Z2

Solution:
λ min=hceV
⇒ λ∝1V
âµ λ2> λ1 (see graph)
⇒ V1>V2
f=aZ-b Moseley's law
f ∝Z-12
⇒ λ ∝1Z-12 (âµf∝1λ)
λ1> λ2 (see graph for characteristic lines)
⇒ Z2> Z1

Topic: Physics, Electromagnetic Waves, JEE Main
When two ice cubes are pressed over each other, they unit to form one cube. Which of the following force is responsible for holding them together

Options:
A Vander Waal's forces
B Hydrogen bond formation
C Covalent attraction
D Dipole-dipole attraction

The correct answer is:
Hydrogen bond formation

Solution:
In ice cube all molecules are held by inter molecular hydrogen bond.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor $R$, the rate of heat produced in $R$ is $J_1$. When the same batteries are connected in parallel across $R$, the rate is $J_2$. If $J_1=2.25 J_2$ then the value of $R$ in $\Omega$ is

Options:

The correct answer is:
4

Solution:
In series, $i=\frac{2 E}{2+R}$
$$
\therefore J_1=i^2 R=\left(\frac{2 E}{2+R}\right)^2 \cdot R
$$
In parallel, $i=\frac{E}{0.5+R}$
$$
\begin{aligned}
\therefore \quad J_2 & =i^2 R=\left(\frac{E}{0.5+R}\right)^2 \cdot R \\
\frac{J_1}{J_2} & =2.25=\frac{4(0.5+R)^2}{(2+R)^2}
\end{aligned}
$$
or $1.5=\frac{2(0.5+R)}{(2+R)}$
Solving we get, $R=4 \Omega$
$\therefore$ The answer is 4 .

Topic: Physics, Current Electricity, JEE Main
When two resistance R1 and R2 connected in series and introduced into the left gap of a meter bridge and a resistance of 10 Ω is introduced into the right gap, a null point is found at 60 cm from left side. When R1 and R2 are connected in parallel and introduced into the left gap, a resistance of 3 Ω is introduced into the right-gap to get null point at 40 cm from left end. The product of R1 R2 is ______ Ω2

Options:

The correct answer is:
30

Solution:
A meter bridge consists of a wire of length 1 m and of uniform cross-sectional area stretched taut and clamped between two thick metallic strips bent at right angles with two gaps across which resistors are to be connected. It works on the principle of the Wheatstone bridge. Therefore, we can write

R1+R210=6040⇒R1+R2=15 Ω

Then for the second case, we can write

R1R2R1+R2×3=4060⇒R1R2=30 Ω2

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
When two soap bubbles of radii a and b(b>a) coalesce, the radius of curvature of common surface is:

Options:
A abb-a
B a+bab
C b-aab
D aba+b

The correct answer is:
abb-a

Solution:
Excess pressure at common surface is given by

Pex=4T1a-1b=4Tr

∴1r=1a-1 b

r=abb-a

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (17 Mar Shift 1)
When two soap bubbles of radius $r_1$ and $r_2\left(r_2 \gt r_1\right)$ coalesce, the radius of curvature of common surface is

Options:
A $r_2-r_1$
B $\frac{r_2-r_1}{r_1 r_2}$
C $\frac{r_1 r_2}{r_2-r_1}$
D $r_2+r_1$

The correct answer is:
$\frac{r_1 r_2}{r_2-r_1}$

Solution:
The correct option is $\mathbf{3}$
$\frac{r_1 r_2}{r_2-r_1}$
If $r_1$ and $r_2$ are the radii of smaller and larger bubble and $P_0$ is the atmospheric pressure, then the pressure inside them will be
$P_1=P_0+\frac{4 T}{r_1}$ and $P_2=P_0+\frac{4 T}{r_2}$.
Now as $r_1 \lt r_2 \therefore P_1 \gt P_2$
So for interface
$\Delta P=P_1-P_2=4 T\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$ \ldots (i)
As excess pressure acts from concave to convex side, the interface will be concave towards the smaller bubble and convex towards larger bubble and if $r$ is the radius of interface.
$\Delta P=\frac{4 T}{r}$ \ldots (ii)
From (i) and (ii) $\frac{1}{r}=\frac{1}{r_1}-\frac{1}{r_2}$
$\therefore$ Radius of the interface $r=\frac{r_1 r_2}{r_2-r_1}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
When two sound waves are superimposed, beats are produced when they have

Options:
A Different amplitudes and phases
B Different velocities
C Different phases
D Different frequencies

The correct answer is:
Different frequencies

Solution:
For producing beats, their must be small difference in frequency.

Topic: Physics, Waves and Sound, JEE Main
When two sound waves travel in the same direction in a medium, the displacements of a particle located at ' $x$ ' at time ' $t$ ' is given by :
$\begin{aligned}
& y_1=0.05 \cos (0.50 \pi x-100 \pi t) \\
& y_2=0.05 \cos (0.46 \pi x-92 \pi t)
\end{aligned}$
where $y_1, y_2$ and $x$ are in meters and $t$ in seconds.
The speed of sound in the medium is :

Options:
A
$92 \mathrm{~m} / \mathrm{s}$

B
$200 \mathrm{~m} / \mathrm{s}$

C
$100 \mathrm{~m} / \mathrm{s}$

D
$332 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$200 \mathrm{~m} / \mathrm{s}$

Solution:
Standard equation
$y(x, t)=A \cos \left(\frac{\omega}{V} x-\omega t\right)$
From any of the displacement equation
Say $\mathrm{y}_1$
$\frac{\omega}{\mathrm{V}}=0.50 \pi$ and $\omega=100 \pi$
$\therefore \frac{100 \pi}{\mathrm{V}}=0.5 \pi$
$\therefore \mathrm{V}=\frac{100 \pi}{0.5 \pi}=200 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2013 (09 Apr Online)
When two surfaces are coated with a lubricant, then they

Options:
A Stick to each other
B Slide upon each other
C Roll upon each other
D None of these

The correct answer is:
Slide upon each other

Solution:
When two surfaces are coated with a lubricant:
1. Two surfaces will slide upon each other after they are coated with a lubricant.
2. Coating lubricant on the surfaces will make them smoother, so the friction will be reduced, and due to the reduction of friction, they will slide upon each other, when one is pushed on the other.

Topic: Physics, Laws of Motion, JEE Main
When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2 . When the tuning forks are sounded again, 6 beats per seconds are heard. If the frequency of fork 1 is $200 \mathrm{~Hz}$, then what was the original frequency of fork 2 ?

Options:
A
$200 \mathrm{~Hz}$

B
$202 \mathrm{~Hz}$

C
$196 \mathrm{~Hz}$

D
$204 \mathrm{~Hz}$

The correct answer is:
$196 \mathrm{~Hz}$

Solution:
$$
\left|f_1-f_2\right|=4
$$
Since mass of second tuning fork increases so $f_2$ decrease and beats increase so
$$
\begin{aligned}
& f_1>f_2 \\
& \Rightarrow f_2=f_1-4=196
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2005
When two waves of almost equal frequency $\mathrm{n}_{1}$ and $\mathrm{n}_{2}$ are produced simultaneously, then the time interval between succesive maxima is

Options:
A $\frac{1}{n_{1}+n_{2}}$
B $\frac{1}{n_{1}}+\frac{1}{n_{2}}$
C $\frac{1}{n_{1}}-\frac{1}{n_{2}}$
D $\frac{1}{n_{1}-n_{2}}$

The correct answer is:
$\frac{1}{n_{1}-n_{2}}$

Solution:
Time interval between two suecessive maxima $=$ time interval between two successive beats $=\frac{1}{n_{1}-n_{2}}$

Topic: Physics, Waves and Sound, JEE Main
When unpolarised light is passed through crossed polaroids, then light passing
through first polaroid

Options:
A also passes through second polaroid.
B partially passes through second polaroid.
C is blocked by second polaroid.
D passes with greater intensity.

The correct answer is:
is blocked by second polaroid.

Solution:
Intensity passed by the second polaroid is given by $I=I_{0} \cos ^{2} \theta$ when $\theta=90^{\circ}, I=0$

Topic: Physics, Wave Optics, JEE Main
When unpolarized light is incident at an angle of 60° on a transparent medium from air. The reflected ray is completely polarized. The angle of refraction in the medium is

Options:
A 30°
B 60°
C 90°
D 45°

The correct answer is:
30°

Solution:
By Brewster’s law at complete polarization, refracted ray and reflected ray are perpendicular to each other.

As angle of reflection will be =60°. Therefore, angle of refraction will be =180°-60°-90°=30°.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (31 Jan Shift 2)
When Uranium is bombarded with neutrons, it undergoes fission. The fission reaction can be written as :
${ }_{92} \mathrm{U}^{235}+{ }_0 n^1 \rightarrow{ }_{56} \mathrm{Ba}^{141}+{ }_{36} \mathrm{Kr}^{92}+3 x+\mathrm{Q}($ energy)
where three particles named $x$ are produced and energy $\mathrm{Q}$ is released. What is the name of the particle $x$ ?

Options:
A
electron

B
$\alpha$-particle

C
neutron

D
neutrino

The correct answer is:
neutron

Solution:
Nuclear fission equation
$\begin{aligned}
{ }_{92} \mathrm{U}^{135}+{ }_0 \mathrm{n}^1 \longrightarrow & { }_{56} \mathrm{Ba}^{141}+{ }_{36} \mathrm{Kr}^{92} \\
& +3{ }_0 \mathrm{n}^1+\mathrm{Q}(\text { energy })
\end{aligned}$
Hence particle $\mathrm{x}$ is neutron.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2013 (09 Apr Online)
When UV light of wavelength $300 \mathrm{~nm}$ is incident on the metal surface having work function $2.13 \mathrm{eV}$, electron emission takes place. The stopping potential is:
(Given hc $=1240 \mathrm{eV} \mathrm{nm}$ )

Options:
A $1.5 \mathrm{~V}$
B $4.1 \mathrm{~V}$
C $2 \mathrm{~V}$
D $4 \mathrm{~V}$

The correct answer is:
$2 \mathrm{~V}$

Solution:
$\begin{aligned} & \frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{e} . \mathrm{V}_{\mathrm{s}} \\ & \Rightarrow \frac{1240}{300} \mathrm{eV}-2.13 \mathrm{eV}=\mathrm{eVs} \\ & \Rightarrow 4.13 \mathrm{eV}-2.13 \mathrm{eV}=\mathrm{eVs} . \\ & \Rightarrow \text { So, } \quad \mathrm{V}_{\mathrm{s}}=2 \mathrm{volt}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (06 Apr Shift 2)
When vapour condenses into liquid

Options:
A It absorbs heat
B It liberates heat
C Its temperature increases
D Its temperature decreases

The correct answer is:
It liberates heat

Solution:
In vapor to liquid phase transition, heat liberates.

Topic: Physics, Thermal Properties of Matter, JEE Main
When vapours of an alcohol are passed over hot copper, alcohol is converted into alkene quickly, the alcohol is

Options:
A Primary
B Secondary
C Tertiary
D None of these

The correct answer is:
Tertiary

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
When vector A→=2i^+3j^+2k^ is subtracted from vector B→, it gives a vector equal to 2j^. Then the magnitude of vector B→ will be:

Options:
A 5
B 3
C 6
D 33

The correct answer is:
33

Solution:
It is given that A→=2i^+3j^+2k^.

By the given condition,

B→-(2i^+3j^+2k^)=2j^

⇒B→=2i^+5j^+2k^

The magnitude is given by the formula,B→=B2x+B2y+B2z
⇒|B→|=4+25+4=33

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
When water is converted into ice, its entropy

Options:
A Increases
B Decreases
C Remains unchanged
D First decreases and then increases

The correct answer is:
Decreases

Solution:
Entropy is a measure of disorder. When water is converted into ice, disorder decreases, hence entropy decreases.

Topic: Physics, Thermodynamics, NEET
When X amperes of current is passed through molten AlCl3 for 96.5 s. 0.09 g of aluminium is deposited. What is the value of X?

Options:
A 10 A
B 20 A
C 30 A
D 40 A

The correct answer is:
10 A

Solution:
Current i =?
Equivalent weight of Al =273 =9
w=Z.i.t or i=wZ.t = 0.09 Ã965009 Ã96.5
= 10 A

Topic: Chemistry, Electrochemistry, JEE Main
When $X$ amperes of current is passed through molten $\mathrm{AlCl}_3$ for $96.5 \mathrm{~s}, 0.09 \mathrm{~g}$ of aluminium is deposited. What is the value of $X$ ?

Options:
A 10 ampere
B 20 ampere
C 30 ampere
D 40 ampere

The correct answer is:
10 ampere

Solution:
Current $i=$ ?
Equivalent weight of $\mathrm{Al}=\frac{27}{3}=9$
$w=z i t$
$i=\frac{w}{z t}=\frac{0.09 \times 96500}{9 \times 96.5}=10$ ampere

Topic: Chemistry, Electrochemistry, JEE Main
When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3; a dark green product is formed which disproportionate in acidic solution to afford a dark purple solution. X is:

Options:
A Mn
B Cr
C V
D Ti

The correct answer is:
Mn

Solution:
When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3; a dark green product is formed which disproportionate in acidic solution to afford a dark purple solution. X is Mn

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2018 (16 Apr Online)
When ' \(x\) ' \(\times 10^{-2} \mathrm{~mL}\) methanol (molar mass \(=32 \mathrm{~g}\); density \(=0.792 \mathrm{~g} / \mathrm{cm}^3\)) is added to \(100 \mathrm{~mL}\) water (density \(=1 \mathrm{~g} / \mathrm{cm}^3\)), the following diagram is obtained.
\(x=\) ______ (nearest integer).
[Given : Molal freezing point depression constant of water at \(273.15 \mathrm{~K}\) is \(1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)]

Options:

The correct answer is:
543

Solution:
$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{f}}=273.15-270.65=2.5 \mathrm{~K} \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \Rightarrow 2.5=1.86 \times \frac{\mathrm{n}}{0.1} \\ & \Rightarrow \mathrm{n}=0.1344 \text { moles } \\ & \Rightarrow \mathrm{w}=0.1344 \times 32=4.3 \mathrm{~g} \\ & \text { Volume }=\frac{4.3}{0.792}=5.43 \mathrm{ml}=543 \times 10^{-2} \mathrm{ml}\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (06 Apr Shift 2)
When xenon hexafluoride is treated with AsF 5 to form an ionic compound, the hybridisation of Xe and As will be respectively

Options:
A sp 3 d 2 , sp 3 d
B sp 3 d, sp 3 d 2
C sp 3 d 3 , sp 3 d 2
D sp 3 d 2 , sp 3 d 2

The correct answer is:
sp 3 d 2 , sp 3 d 2

Solution:
XeF 6  +  AsF 5 → [XeF 5 ] +   [AsF 6 ] −

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
When you make ice cubes, the entropy of water

Options:
A Does not change
B Increases
C Decreases
D May either increase or decrease depending on the process used

The correct answer is:
Decreases

Solution:
During formation of ice cubes orderness increases, ie, disorderness decreases, hence entropy decreases.

Topic: Physics, Thermodynamics, JEE Main
When you walk through a metal detector carrying a metal object in your pocket, it raises an alarm. This phenomenon works on

Options:
A Electromagnetic induction
B Resonance in ac circuits
C Mutual induction in ac circuits
D interference of electromagnetic waves

The correct answer is:
Resonance in ac circuits

Solution:
Metal detector works on the principle of transmitting an electromagnetic signal and analyses a return signal from the target. So it works on the principle of resonance in AC circuit.

When the LC circuit detects any resonating frequency from any metal which is near to it, electric field will be created which will lead to induces current in the coil and changes in the signal flow through the coil.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (25 Jul Shift 2)
When Young's double slit experiment is performed in air, the Y-coordinates of central maxima and $10^{\text {th }}$ maxima are $2 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. If the apparatus is immersed in a liquid of refractive index 1.5 , the corresponding Y-coordinates will be

Options:
A $2 \mathrm{~cm}, 7.5 \mathrm{~cm}$
B $3 \mathrm{~cm}, 6 \mathrm{~cm}$
C $2 \mathrm{~cm}, 4 \mathrm{~cm}$
D $\frac{4}{3} \mathrm{~cm}, \frac{10}{3} \mathrm{~cm}$

The correct answer is:
$2 \mathrm{~cm}, 4 \mathrm{~cm}$

Solution:
For YDSE apparatus in air
For central maxima
$\begin{aligned}
& \frac{y_0 d}{D}-d \sin \theta=0 \\
& \Rightarrow \sin \theta=\frac{y_0}{D}=\frac{2}{D} ... (i)
\end{aligned}$
For $10^{\text {th }}$ maxima
$\begin{aligned}
& \frac{\mathrm{y}_{10} \mathrm{~d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda \\
& \Rightarrow \frac{5 \mathrm{~d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda
\end{aligned}$
From equation (i), we have
$\begin{aligned}
& \frac{5 d}{D}-\frac{2 d}{D}=10 \lambda \\
& \Rightarrow 10 \lambda=\frac{3 d}{D} \Rightarrow \lambda=\frac{3 d}{10 D}
\end{aligned}$
Now, when the YDSE apparatus is immersed in a liquid of refractive index 1.5 , then it's wave length $\lambda^{\prime}=\frac{\lambda}{1.5}$
As the y coordinate of central maximum doesn't depend upon wavelength, therefore it will remain same as in air, $\mathrm{y}_0^{\prime}=\mathrm{y}_0=2 \mathrm{~cm}$
For $10^{\text {th }}$ maxina
$\begin{aligned}
& \frac{\mathrm{y}_{10}^{\prime} \mathrm{d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda^{\prime} \\
& \Rightarrow \frac{\mathrm{y}_{10}^{\prime} \mathrm{d}}{\mathrm{D}}-\frac{\mathrm{d} \times 2}{\mathrm{D}}=\frac{10 \times 3 \mathrm{~d}}{1.510 \mathrm{D}} \\
& \Rightarrow \mathrm{y}_{10}^{\prime}=4 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Wave Optics, JEE Main
When Young's double slit experiment is performed in air, the Y-coordinates of central maxima and $10^{\text {th }}$ maxima are $2 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. If the apparatus is immersed in a liquid of refractive index 1.5 , the corresponding Y-coordinates will be

Options:
A $2 \mathrm{~cm}, 7.5 \mathrm{~cm}$
B $3 \mathrm{~cm}, 6 \mathrm{~cm}$
C $2 \mathrm{~cm}, 4 \mathrm{~cm}$
D $\frac{4}{3} \mathrm{~cm}, \frac{10}{3} \mathrm{~cm}$

The correct answer is:
$2 \mathrm{~cm}, 4 \mathrm{~cm}$

Solution:
For YDSE apparatus in air
For central maxima
$\begin{aligned}
& \frac{y_0 d}{D}-d \sin \theta=0 \\
& \Rightarrow \sin \theta=\frac{y_0}{D}=\frac{2}{D} ... (i)
\end{aligned}$
For $10^{\text {th }}$ maxima
$\begin{aligned}
& \frac{\mathrm{y}_{10} \mathrm{~d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda \\
& \Rightarrow \frac{5 \mathrm{~d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda
\end{aligned}$
From equation (i), we have
$\begin{aligned}
& \frac{5 d}{D}-\frac{2 d}{D}=10 \lambda \\
& \Rightarrow 10 \lambda=\frac{3 d}{D} \Rightarrow \lambda=\frac{3 d}{10 D}
\end{aligned}$
Now, when the YDSE apparatus is immersed in a liquid of refractive index 1.5 , then it's wave length $\lambda^{\prime}=\frac{\lambda}{1.5}$
As the y coordinate of central maximum doesn't depend upon wavelength, therefore it will remain same as in air, $\mathrm{y}_0^{\prime}=\mathrm{y}_0=2 \mathrm{~cm}$
For $10^{\text {th }}$ maxina
$\begin{aligned}
& \frac{\mathrm{y}_{10}^{\prime} \mathrm{d}}{\mathrm{D}}-\mathrm{d} \sin \theta=10 \lambda^{\prime} \\
& \Rightarrow \frac{\mathrm{y}_{10}^{\prime} \mathrm{d}}{\mathrm{D}}-\frac{\mathrm{d} \times 2}{\mathrm{D}}=\frac{10 \times 3 \mathrm{~d}}{1.510 \mathrm{D}} \\
& \Rightarrow \mathrm{y}_{10}^{\prime}=4 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Wave Optics, TS EAMCET, TS EAMCET 2023 (12 May Shift 1)
When zeolite, which is hydrated sodium aluminium silicate, is treated with hard water the sodium ions are exchanged with
1. H+ ions
2. Ca2+ ions
3. SO4-2 ions
4. Mg2+ ions

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

The correct answer is:
2, 4

Solution:
Na2Al2Si2O8.xH2O (zeolite) is written as Na2Z. In it Na+ ions exchange Ca+ and Mg2+ ions of hard water to remove permanent hardness of water as follows
Na2Z+Mg2+→2Na++MgZ
Na2Z+Ca2+→2Na++CaZ

Topic: Chemistry, s Block Elements, JEE Main
When

Options:
A Propene is the major product
B Ethane and C3H7NCH32 are the only product
C Ethene and propene are obtained while ethene as the major product
D Equimolar amounts of ethane and propene are obtained

The correct answer is:
Ethene and propene are obtained while ethene as the major product

Solution:
In Hofmann elimination reaction, less substituted alkene is the main product.

Topic: Chemistry, Amines, NEET
Where are the following hormones synthesised in plants?
(a) IAA
(b) Gibberellins
(c) Cytokinins

Options:

None

Solution:
(a) IAA i.e., Indole acetic acid. It is synthesised at the growing apices of the plant, e.g., shoot tip, leaf primordia and developing seeds.
(b) Gibberellins It is synthesised in the apical shoot buds, young leaves, root tips and developing seeds.
(c) Cytokinins are synthesised mainly in routs, but synthes also occurs in the endosperm of seeds, growing embryo etc.

Topic: Biology, Plant - Growth and Development
Where can a geostationary satellite be installed

Options:
A Over any city on the equator
B Over the north or south pole
C At height R above earth
D At the surface of earth

The correct answer is:
Over any city on the equator

Solution:
Over any city on the equator

Topic: Physics, Gravitation, JEE Main