Physics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b, d)

Explanation-Consider the pattern of the intensity shown in the figure

(i) As intensities of all successive minima is zero, hence we can say that two sources S₁ and S₂ are having same intensities.

(ii) As width of the successive maxima (pulses) increases in continuous manner, we can say that the path difference (x) or phase difference varies in continuous manner.

(iii) We are using monochromatic light in YDSE to avoid overlapping and to have very clear pattern on the screen.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- d

Explanation- According to question, there is a hole at point P₂. From Huygen's principle, wave will propagates from the sources S₁ and S₂. Each point on the screen will acts as secondary sources of wavelets. Now, there is a hole at point P₂ (minima). The hole will act as a source of fresh light for the slits S₃ and S₄.

Therefore, there will be a regular two slit pattern on the second screen

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

P_s=60W, I_s= 0.54 A

P=VI, V= 60/0.54= 110V

Voltage in the secondary is less than voltage in primary so transformer is step down transformer

As transformation ratio is, k = E_s/E_p= I_p/I_s

110/220=I_p/0.54A

I_p= 0.27A

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer – (c)

Explanation- For the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength. In a Young's double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filteration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen. Hence, in this case there shall be no interference fringes.

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

L= 0.01H

R= 1ohm

Voltage =200V

Frequency = 50Hz

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

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

$\varnothing$ = tan-13.14=72

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

$?t$ = $\varnothing$ /w= $\frac{72\pi }{180*2\pi *50}$ = $\frac{1}{250}$ 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 = $?\varnothing$ = $\frac{2\pi }{T}*?t$ = $\frac{2\pi nd}{\lambda }$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

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

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 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.