JEE Advanced · Physics · 21. EMI

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Modern trains are based on Maglev technology in which trains are magnetically leviated, which runs its EDS Maglev system.
There are coils on both sides of wheels. Due to motion of train, current induces in the coil of track which levitate it. This is in accordance with Lenz's law. If trains lower down then due to Lenz's law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force disc to gravity. The advantage of Maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds.
Disadvantage of Maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it leviated and as it moves forward according to Lenz law, there is an electromagnetic drag force.
Question:
What is the advantage of this system?

A No friction hence no power consumption
B No electric power is used
C Gravitation force is zero
D Electrostatic force draws the train

Verified Solution

Answer & Solution

Correct Answer

(A) No friction hence no power consumption

Step-by-step Solution

Detailed explanation

No solution is required.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From JEE Advanced

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The reactions of \(\mathrm{Cl}_{2}\) gas with cold-dilute and hot-concentrated \(\mathrm{NaOH}\) in water give sodium salts of two (different) oxoacids of chlorine, \(\boldsymbol{P}\) and \(\boldsymbol{Q}\), respectively. The \(\mathrm{Cl}_{2}\) gas reacts with \(\mathrm{SO}_{2}\) gas, in presence of charcoal, to give a product \(\boldsymbol{R}\). \(\boldsymbol{R}\) reacts with white phosphorus to give a compound \(\boldsymbol{S}\). On hydrolysis, \(\boldsymbol{S}\) gives an oxoacid of phosphorus, \(\boldsymbol{T}\).
Question:
\(P\) and \(Q\), respectively, are the sodium salts of

JEE Advanced 2013 Medium

The wave function,

ψ_{n, l, m_{l}}

is a mathematical function whose value depends upon spherical polar coordinates

(r, θ, ϕ)

of the electron and characterized by the quantum numbers

n, l

and

m_{l}

. Here

r

is the distance from the nucleus,

θ

is colatitude and

ϕ

is azimuth. In the mathematical functions given in the Table,

Z

is the atomic number and

a_{o}

is Bohr radius

Column 1	Column 2	Column 3
(I) 1s orbital	\(\begin{array}{l}\text { (i) } \psi_{ n , 1, m_1} \\ \propto\left(\frac{ Z }{ a _{ o }}\right)^{\frac{3}{2}} e ^{-\left(\frac{ Zr }{ e _{ o }}\right)}\end{array}\)	(P)
(II) 2s orbital	(ii) One radial node	(Q) Probability density at the nucleus \(\propto \frac{1}{ a _{ o }^3}\).
(III) \(2 p _{ z }\) orbital	\(\begin{array}{l}\text { (iii) } \psi_{ n }, 1, m_{ l } \\ \propto\left(\frac{ Z }{ a _{ o }}\right)^{\frac{5}{2}} re ^{-\left(\frac{ Zr }{2 a _{ o }}\right)} \cos \theta\end{array}\)	(R) Probability density is maximum at the nucleus.
(IV) \(3 d_{ z }^2\) orbital	(iv) \(x y\)-plane is a nodal plane	(S) Energy needed to excite an electron from \(n =2\) state to \(n =4\) state is \(\frac{27}{32}\) times the energy needed to excite are electron from \(n =2\) state to \(n =6\) state.

For the given orbital in column 1, the only correct combination for any hydrogen-like species is

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