Ask & Answer: India's Largest Education Community

This is a short answer type question as classified in NCERT Exemplar

L= 0.01H

R= 1ohm

Voltage =200V

Frequency = 50Hz

Z= $\sqrt[]{R^2+X_L^2}$ = $\sqrt[]{1^2+({2\times 3.14\times 50\times 0.01)}^2}$ = $\sqrt[]{10.86}$ =3.3ohm

tan $\varnothing =\frac{wL}{R}$ = $\frac{2\pi fL}{R}=\frac{2\times 3.14\times 50\times 0.01}{1}$ = 3.14

$\varnothing$ = tan-13.14=72

$\varnothing$ =72 $\frac{\pi }{180}rad$

$?t$ = $\varnothing$ /w= $\frac{72\pi }{180\times 2\pi \times 50}$ = $\frac{1}{250}$ s

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a

Explanation- due to refraction from the glass medium there is a phase change of $\pi$

$?t$ = $\frac{o{p}^{\text{'}\text{'}}}{v}$ = $\frac{\frac{d}{cosr}}{\frac{c}{n}}=\frac{nd}{ccosr}$

According to snells law n= sin $i$ /sinr

Cosr = $\sqrt[]{1-{sin}^{2}r}=\sqrt[]{1-\frac{{sin}^2\theta }{n^2}}$

$?t$ = $\frac{nd}{c({\frac{{1-sin}^2\theta }{n^2})}^{1/2}}=\frac{n^{2}d}{c}$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

Phase difference = $?\varnothing$ = $\frac{2\pi }{T}\times ?t$ = $\frac{2\pi nd}{\lambda }$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

So net difference = $?\varnothing +\pi$ = $\frac{4\pi d}{\lambda }$ ( $1-\frac{1}{n^2}{sin}^2\theta$ )-1/2+ $\pi$

Praxis Tech School Kolkata concluded the placements for the Class of 2023 with the highest package of INR 23 LPA. The salary packages offered during Praxis Tech School placements 2023 are tabulated below:

This is a short answer type question as classified in NCERT Exemplar

For a Direct Current (DC),

1ampere= 1coulomb/sec

An AC current changes direction with the source frequency and the attractive force would average to zero. Thus, the AC ampere must be defined in terms of some property that is independent of the direction of current.

Joule’s heating effect is such property and hence it is used to define rms value of AC.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- ( a)

Explanation- given width of slit is 10⁴10^-10m= 10^-6

Wavelength of sunlight varies from 4000A⁰ to 8000A⁰

As the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear i.e., mixing of colours form white patch at the centre.

This is a short answer type question as classified in NCERT Exemplar

(a) We know that P= VI

that is curve of power will be having maximum amplitude, equals to multiplication of amplitudes of voltage (V)and current (I) curve. So, the curve will be represented by A.

(b) As shown by shaded area in the diagram, the full cycle of the graph consists of one positive and one negative symmetrical area. Hence average power over a cycle is zero.

(c) As the average power is zero, hence the device may be inductor (L) or capacitor (C) or the series combination of L and C.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (c)

Explanation-Consider the diagram the light beam incident from air to the glass slab at Brewster's angle (ip ). The incident ray is unpolarised and is represented by dot (.).

The reflected light is plane polarised represented by arrows.

As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

The basic eligibility criteria for BTech course admissions is Class 12 with a minimum of 55% aggreagte. The accepted entrance exams are VITEEE,  JEE Main,  AP EAMCET,  MHT CET,  TS EAMCET,  CUET,  TS3 India, and others. All the candidates who are eligible for admissions must visit the official website and apply for timely admissions. Admissions to various courses are based on the merit and scores secured by students in entrance tests. 

Following are some tips to improve the writing and presentation skills for the MBSE Class 12th board exams.

• Students must write the answers neatly. The hand writing should be legible.
• Write the answers with proper heading and subheading.
• Highlight your important points.
• Cross out the errors neatly with single line.
• Divide the time for each question to avoid rushing.  
• Practice writing the previous year question papers.

Clini Launch Research Institute offers PG Diploma in Clinical Research, Bioinformatics, AI and ML in Healthcare, Biostatistics, Healthcare Management and others. The institute offers the respective course for the durations varying from 6 months to 12 months. Furthermore, the eligibility criteria of the respective course includes a minimum age limit of 18 years. Apart from this, candidates must secure a minimum aggregate of 60% in graduation. 

This is a short answer type question as classified in NCERT Exemplar

Praxis Tech School is a well known institute in Kolkata for offering placement opportunities to its PGP Data Science with GEN AI and ML students. The key highlights of Praxis Tech School Kolkata placements for the Class of 2023 are tabulated below:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- T₂P= T₂O+OP= D+x

T₁P= T₁O-OP= D-x

S₁P $\frac{\mathit{\lambda }}{2}$ = $\sqrt[]{{{(S}_1{T}_1)}^2+{\left(P{T}_1\right)}^2}=\sqrt[]{D^2+{(d-x)}^2}$

S₂P= $\sqrt[]{{{(S}_2{T}_2)}^2+{\left(P{T}_2\right)}^2}=\sqrt[]{D^2+{(d+x)}^2}$

The minima will occur when S₂P- S₁P= (2n-1) $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{D^2+{(d+x)}^2}$ - $\sqrt[]{D^2+{(d-x)}^2}$ = $\frac{\mathit{\lambda }}{2}$

$x=D$

[D²+4D²]^1/2-[D²+0]^1/2= $\frac{\mathit{\lambda }}{2}$

[5D²]^1/2- [D²]^1/2= $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{5}$ D-D= $\frac{\mathit{\lambda }}{2}$

$\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}$ we get D= 0.404 $\mathit{\lambda }$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- resolving power =1/d = $\frac{2sin\beta }{1.22\lambda }$  = d_min= $\frac{1.22\lambda }{2sin\beta }$  where $\lambda$  is the wavelength of light

So d_min= $\frac{1.22\times 5500\times {10}^{-10}}{2sin\beta }$

$\lambda$ = $\frac{12.27}{\sqrt[]{V}}$ = 0.12 $\times$ 10^-9m

$\frac{d\text{'}min}{dmin}$ = $\frac{0.12\times {10}^{-9}}{5500\times {10}^{-10}}$ = 0.2 $\times$ 10^-3

This is a short answer type question as classified in NCERT Exemplar

I_rms= $\sqrt[]{\frac{1^2+2^2}{2}}$ = $\sqrt[]{\frac{5}{2}}$ = 1.58A = 1.6 approx

This value is indicating in graph

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented By arrows.

Polarisation by reflection occurs when the angle of incidence Is the Brewster’s angle

So tani_b = $\frac{{\mu }_1}{{\mu }_2}$  where $\mu$ ₂< $\mu$ ₁

When the light rays travels in such a medium, the critical angle is

Sini_c= $\frac{{\mu }_2}{{\mu }_1}$

Where $\mu$ ₂< $\mu$ ₁

As tani_b > sini_c for large angles i_bc

Thus the polarisation by reflection occurs definitely.

Praxis Tech School, Kolkata has a good record of placements for its PGP-Data Science program. According to the report, the the highest and average packages offered to PGP Data Science students during Praxis Tech School placement 2023 stood at INR 23 LPA and INR 13.50 LPA, respectively. Check out the key highlights of of Praxis Tech School Kolkata placements for the Class of 2023 in the table below:

This is a short answer type question as classified in NCERT Exemplar

In the figure

Band width = w₂-w₁ where these two corresponds to frequencies at which magnitude of current is 1/ $\sqrt[]{2}$ times of maximum value

I_rms= $\frac{I_{max}}{\sqrt[]{2}}$ = 1/ $\sqrt[]{2}$ = 0.7A

From the diagram the corresponding frequencies are 0.8 and 1.2rad/s