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Concept:

Vernier Caliper

• Vernier Caliper is a precision instrument that can be used to measure internal and external distances accurately.
• A vernier caliper consists of two main parts:
• the main scale engraved on a solid L-shaped frame and the vernier scale that can slide along the main scale.
• It works on the principle of vernier and can measure the dimensions to an accuracy of 0.02 mm.

Parts of a Vernier caliper:

• Outside jaws: Used to measure the external diameter or width of an object
• Inside jaws: Used to measure the internal diameter of an object
• Locking screw: to lock the jaws
• Adjusting screw: to take an accurate measurement of the workpiece.

Explanation:

In general one vernier scale division is smaller than one main scale division but in some modified cases it may be not correct.

Also least count is given by one main scale division / number of vernier scale division for normal vernier calliper.

Calculation:

Total charge inside the cube

= $\frac{\lambda \frac{\mathrm{a}}{2}}{4}=\frac{\lambda \mathrm{a}}{8}$

∴ \(\phi=\frac{\mathrm{q}{\mathrm{in}}}{\varepsilon{0}}=\frac{\lambda \mathrm{a}}{8 \varepsilon_{0}}\)

Calculation:

The circuit can be considered as 

∴ R_eq = 0.5 + $\frac{0.5 \times 2}{2+0.5}=\left(\frac{5}{10}+\frac{10}{25}\right) \Omega$

= $\frac{45}{50}=\frac{9}{10}=0.9$

∴ i = $\frac{0.9}{0.9}=1 \mathrm{~A}$

CONCEPT:

Young's double-slit experiment

• Young’s double-slit experiment helped in understanding the wave nature of light.
• The original Young’s double-slit experiment used diffracted light from a single monochromatic source of light.
• The light that comes from the monochromatic source is passed into two slits to be used as two coherent sources.
• At any point on the screen at a distance ‘y’ from the center, the waves travel distances l1 and l2 to create a path difference of Δl at that point.
• If there is a constructive interference on the point then the bright fringe occurs.
• If there is a destructive interference on the point then the dark fringe occurs.
• The distance of the nth bright fringe from the central fringe is given as,

$⇒ y=\frac{nλ D}{d}$

Where d = distance between slits, D = distance between slits and screen, and λ = wavelength

​

Explanation:

β (fringe width) = $\frac{λ \mathrm{D}}{\mathrm{~d}}$

In denser medium, λ ↓ ⇒ β ↓

⇒ fringe come closer 

Also, µ = $\frac{c}{V}$ ⇒ V = $\frac{c}{μ}$

Frequency remains same, 

⇒ μ = $\frac{\lambda_{\text {vac. }} \mathrm{f}}{\lambda_{\text {med }} \mathrm{f}}$ ⇒ $\lambda_{\text {med }}$ = $\frac{\lambda_{\mathrm{vac} .}}{\mu}$

Calculation:

We know that

$\frac{\mathrm{d} \theta}{\mathrm{dt}}=\sigma \mathrm{eAT}^{4} \Rightarrow \mathrm{P} \propto \mathrm{AT}^{4}$

\(\frac{\mathrm{P}{\text {smaller }}}{\mathrm{P}{\text {larger }}}=\frac{(0.2)^{2} \times 800^{4}}{(0.8)^{2} \times 400^{4}}\)

$\frac{1}{16} \times 16=1$

∴ P_larger = P_smaller 

Calculation:

$\rm\Delta Q_{1}=m \times S_{1} \times \Delta T=10^{-3} \times 2100 \times 10=21 \mathrm{~J}$

\(\rm\Delta \mathrm{Q}{2}=\mathrm{m} \times \mathrm{L}{\mathrm{f}}=10^{-3} \times 3.35 \times 10^{5}=335 \mathrm{~J}\)

$\rm\Delta Q_{3}=m \times S_{w} \times \Delta T=10^{-3} \times 4180 \times 100=418 \mathrm{~J}$

\(\rm \Delta \mathrm{Q}{4}=\mathrm{m} \times \mathrm{L}{\mathrm{v}}=10^{-3} \times 2.25 \times 10^{6}=2250 \mathrm{~J}\)

$\rm\Delta Q_{5}=\mathrm{m} \times \mathrm{S}_{\mathrm{v}} \times \Delta \mathrm{T}=10^{-3} \times 1920 \times 10=19.2 \mathrm{~J}$

$\rm \Delta Q_{\mathrm{net}}=3043.2 \mathrm{~J}$

Concept:

de Broglie wavelength:

• Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.
• The wavelength of material waves is also known as the de Broglie wavelength.
• de Broglie wavelength can be calculated from Planks constant h divided by the momentum of the particle.

$\lambda = \frac {h}{mv}$

where λ is de Broglie wavelength, h is Plank's constant, and mv is the momentum.

• If different particles have the same velocity, then the wavelength is inversely proportional to the mass of that particle.

Calculation:

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$

$\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$

$\mathrm{mv}=\frac{\mathrm{nh}}{2 \pi \mathrm{r}}$

$\lambda=\frac{2 \pi \mathrm{rh}}{\mathrm{nh}}$

$\lambda \propto \frac{\mathrm{r}}{\mathrm{n}}$

$\frac{\lambda_{1}}{\lambda_{4}}=\frac{r_{1} n_{4}}{n_{1} r_{4}}=\frac{5.3 \times 10^{-11} \times 4}{1 \times 84.8 \times 10^{-11}}$

$\frac{\lambda_{1}}{\lambda_{4}}=\frac{1}{4}$

Calculation:

⇒ p_eq = p₁ + p₂ + p₃  

⇒ p1 = $\left(\frac{1}{3\left|\mathrm{R}_{1}\right|}\right)$

⇒ p2 = $\left(\frac{1}{2}\right)\left(\frac{1}{-\left|R_{1}\right|}-\frac{1}{-\left|R_{2}\right|}\right)$

⇒ p2 = \(\frac{1}{2}\left(\frac{1}{\left|\mathrm{R}{2}\right|}-\frac{1}{\left|\mathrm{R}{1}\right|}\right)\)

⇒ p3 = \(\left(\frac{1}{3}\right)\left(\frac{1}{-\left|\mathrm{R}{2}\right|}-\frac{1}{\infty}\right)=-\frac{1}{3\left|\mathrm{R}{2}\right|}\)

⇒ peq = \(\frac{1}{3}\left(\frac{1}{\left|\mathrm{R}{1}\right|}-\frac{1}{\left|\mathrm{R}{2}\right|}\right)-\frac{1}{2}\left(\frac{1}{\left|\mathrm{R}{1}\right|}-\frac{1}{\left|\mathrm{R}{2}\right|}\right)\)

=  \(-\frac{1}{6}\left(\frac{1}{\left|\mathrm{R}{1}\right|}-\frac{1}{\left|\mathrm{R}{2}\right|}\right)\)

Calculation:

⇒ t = $\frac{\ell}{V_{x}}=\frac{10 × 10^{-2}}{10^{6}}=10^{-7}$

⇒ V_y = u_y + a_yt

⇒ Vy = 0 + $\frac{\mathrm{eE}}{\mathrm{~m}} × 10^{-7}$

⇒ Vy = $\frac{1.6 × 10^{-19}}{9.1 × 10^{-31}} × 9.1 × 10^{-2} × 10^{-7}$

⇒ Vy =  16 × 10⁴

Explanation:

Resistivity is independent of temperature for wire-bound resistors