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Calculation:

9^th harmonic of closed pipe = $\frac{9 V_{1}}{4 \ell_{1}}$

4^th harmonic of open pipe = \(\frac{2 \mathrm{~V}{2}}{\ell{2}}\)

∴ \(\frac{9 \mathrm{~V}{1}}{4 \ell{1}}=\frac{2 \mathrm{~V}{2}}{\ell{2}}\)

∴ $\frac{9}{4 \ell_{1}} \sqrt{\frac{\mathrm{~B}}{\rho_{1}}}=\frac{2}{\ell_{2}} \sqrt{\frac{\mathrm{~B}}{\rho_{2}}} \Rightarrow \frac{\ell_{2}}{\ell_{1}}=\frac{8}{9} \sqrt{\frac{\rho_{1}}{\rho_{2}}}$

$\ell_{2}=\ell_{1} \times \frac{8}{9} \times \frac{1}{4}=\frac{20}{9} \mathrm{~cm}$

Calculation:

$\mathrm{I}=\frac{\mathrm{MR}^{2}}{2}-\left[\frac{\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^{2}}{2}+\frac{\mathrm{M}}{4}\left(\frac{\mathrm{R}}{2}\right)^{2}\right]$

$\mathrm{I}=\frac{13}{32} \mathrm{MR}^{2}$

Calculation: 

$\mathrm{F}_{1}=\frac{\mathrm{GMm}}{(2 \mathrm{R})^{2}}\quad\quad...(1)$

$\mathrm{F}_{2}=\frac{\mathrm{GMm}}{(2 \mathrm{R})^{2}}-\left(\frac{\mathrm{G}\left(\frac{\mathrm{M}}{27}\right) \mathrm{m}}{\left(\frac{4 \mathrm{R}}{3}\right)^{2}}\right)$

$\mathrm{F}_{2}=\frac{11}{48} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}\quad\quad...(2)$

F₁ : F₂ = 12 : 11

CONCEPT:

• Photoelectric effect: When the light of a sufficiently small wavelength is incident on the metal surface, electrons are ejected from the metal instantly. This phenomenon is called the photoelectric effect.

Photo-electric cell

• The photo-electric cell works on the principle of the photoelectric effect.
• The photo-electric cell consists of an evacuated glass tube containing two electrodes emitter (C) and Collector (A).
• The emitter is always kept at a negative potential.
• The collector is made of a metal rod and is always kept at a positive potential.
• When the light of frequency more than the threshold frequency of material of emitter is made incident on the emitter, photo-emission takes place.
• Then the photo­electrons are attracted to the collector which is positive with respect to the emitter. Thus current flows in the circuit. If the intensity of incident radiation is increased the photoelectric current increases.

Calculation:

The energy of incident light is E = $\frac{1240}{\lambda}=\frac{1240}{550} \simeq 2.25$ eV

Thus it is less than 2.5 eV (Li) and greater than 1.9 eV (Cs).

Photoelectric effect only can be possible for Cs.

Explanation:

We know that 

B = μ₀ni

$\left[\frac{\mathrm{B}}{\mu_{0}}\right]=[\mathrm{ni}]=\mathrm{L}^{-1} \mathrm{~A}^{1}$

Cakculation:

\(\frac{1}{2} \mathrm{mv}{\mathrm{A}}^{2}=\frac{1}{2} \mathrm{mv}{\mathrm{B}}^{2}+\mathrm{mgh}\)

⇒ $\frac{1}{2} m(5 g \ell)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^{2}+\mathrm{mg} \frac{\ell}{2}$

⇒ $\frac{5 \mathrm{mg} \ell}{2}-\frac{\mathrm{mg} \ell}{2}=\mathrm{KE}_{\mathrm{B}}$

⇒ $\mathrm{KE}_{\mathrm{B}}=2 \mathrm{mg} \ell$

\(\frac{1}{2} \mathrm{mv}{\mathrm{C}}^{2}=\frac{1}{2} \mathrm{mv}{\mathrm{D}}^{2}+\mathrm{mg} \frac{\ell}{2}\)

⇒ \(\mathrm{KE}{\mathrm{C}}=\frac{1}{2} \mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\mathrm{mg} \ell \Rightarrow \frac{\mathrm{KE}{\mathrm{B}}}{\mathrm{KE}_{\mathrm{C}}}=2\)

CONCEPT:

Cell:

• The cell converts chemical energy into electrical energy.
• Cells are of two types:

1. Primary cell: This type of cell cannot be recharged.
2. Secondary cell: This type of cell can be recharged.

• For a cell of emf E and internal resistance r,

⇒ E - V = Ir

Where I = current, and V = potential difference across external resistance

Cells in series:

• If the number of cells are connected end to end that the positive terminal of one cell is connected to the negative terminal of the succeeding cell then it is called a series arrangement of cells.
• The equivalent emf of cells in series arrangement is given as,

⇒ Eeq = E1 + E2 +...+ En

• The equivalent internal resistance of cells in a series arrangement is given as,

⇒ req = r1 + r2 +...+ rn

Cells in parallel:

• If the number of cells is connected such that the positive terminals are connected together at one point and the negative terminals of these cells are connected together at another point then it is called a parallel arrangement of cells.
• The equivalent emf of cells in a parallel arrangement is given as,

$⇒ E_{eq}=\frac{E_1/r_1+E_2/r_2+E_3/r_3+...+E_n/r_n}{1/r_1+1/r_2+1/r_3+...+1/r_n}$

• The equivalent internal resistance of cells in a parallel arrangement is given as,

$⇒ \frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+...+\frac{1}{r_n}$

Calculation:

= $\rm \frac{\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{r_{2}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}}=\varepsilon$

Explanation:

Forward Biased:

• The diode is said to be forward-biased when the P region is connected to a more positive terminal and N region is connected to the negative terminal of the battery.
• A forward-biased diode offers very little resistance to the flow of current and hence a large current flow through the diode.
• When forward biased, the diode acts as an ON switch.

Reversed Biased:

• A diode is said to be reversed biased when the P terminal is connected to lower potential than the N terminal
• A reverse-biased diode offers very large resistance to the flow of current and hence a small minority of current flows through the diode

How to identify forward and reversed biased diode:

1. If the input voltage is greater than the other end voltage, then it is forward-biased and vice-versa.
2. If the voltage at the p-section is greater than the n-section, then is forward biased and vice-versa.

CONCEPT:

Capacitor:

• The capacitor is a device in which electrical energy can be stored.
• In a capacitor two conducting plates are connected parallel to each other and carrying charges of equal magnitudes and opposite sign and separated by an insulating medium.
• The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​

​Parallel plate capacitor:

• A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance.
• The space between the two plates can either be a vacuum or an electric insulator such as glass, paper, air, or semi-conductor called a dielectric.​
• The electric field intensity at the outer region of the parallel plate capacitor is always zero whatever be the charge on the plate.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor remains the same at every point.
• The electric field intensity in the inner region between the plates of a parallel plate capacitor is given as,

$\Rightarrow E=\frac{σ}{\epsilon_o}=\frac{Q}{A\epsilon_o}$

• The potential difference between the plates is given as,

$\Rightarrow V=\frac{Qd}{A\epsilon_o}$

• The capacitance C of the parallel plate capacitor is given as,

$\Rightarrow C=\frac{Q}{V}=\frac{A\epsilon_o}{d}$

Where A = area of the plates, d = distance between the plates, Q = charge on the plates, and σ = surface charge density

Calculation:

Δq = (KC - C)V

= 40 × 10^-6 × 100

= 4000 × 10^-3 = 4mC

$\Delta \mathrm{U}=\frac{1}{2} \mathrm{C}^{\prime} \mathrm{V}^{2}-\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2}(\mathrm{~K}-1) \mathrm{CV}^{2}$

= $\frac{1}{2} \mathrm{CV}^{2}(2-1)$

= $\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \times 40 \times 10^{-6} \times 10000$

= 0.2 J

Calculation:

P_eq = 2P_ℓ + P_m

\(-\frac{1}{\mathrm{f}{\mathrm{Q}}}=\frac{2}{\mathrm{f}{\ell}}-\frac{1}{\mathrm{f}_{\mathrm{m}}}\)

= $\frac{4(\mu-1)}{R}-\frac{2}{-R}=\frac{1}{R}(4 \mu-4+2)$

$-\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}(4 \mu-2)$

⇒ $\frac{1}{f_{e q}}=\frac{-1}{R}(4 \mu-2)$

$\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{R}}{2}$

R = $2 f_{\mathrm{eq}}=-2\left(\frac{\mathrm{R}}{4 \mu-2}\right)=\frac{-\mathrm{R}}{(2 \mu-1)}$