Physics Ncert Solutions Class 12th

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

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

Explanation- angular resolution of the eye? = 5.8 x 10^-4

linear distance between two dots is l= 2.54/300=0.84 $*$ 10^-2cm

? = l/z

z=l/? = $\frac{0.84*{10}^{-2}}{5.8*{10}^{-4}}$ = 14.5cm

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

Explanation- As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm. The diffraction occur if the wavelength of waves is nearly equal to slit width.

As the wavelength of light waves is 7000 *10^-10 m to 4000 $*$ 10^-10m. The slit width is very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

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

Explanation- We know that the sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. Due to the large distance the radius of wavefront can be considered as large (infinity) and hence, wavefront is almost plane.

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

E=E₀sinwt

And current I=I₀sin (wt $\frac{+}{-}\varnothing$ )

Power, P= EI = E₀sinwt = E₀I₀sinwtsin (wt+ $\varnothing$ )

= $\frac{E_0{I}_0}{2}$  [cos $\varnothing -cos(2wt+\varnothing )$ ]………… (1)

Pav= $\frac{V_0}{\sqrt[]{2}}\frac{{\mathrm{I}}_0}{\sqrt[]{2}}$ cos $\varnothing$ =V_rmsI_rmscos $\varnothing$ ………. (2)

From eqn 1 ……But when cos $\varnothing$ < cos (2wt+$\mathrm{\varnothing )}$

P<0 power can be nagative of an AC source

From eqn 2……… P_av>0

cos $\varnothing$ =R/Z>0 average power can not be negative

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (c, d)

Explanation- in given figure the slope of curve A is greater than slope B. so decay in A is faster than B . also wavelength of A is grater than B.

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

Explanation- The point image I₁, due to L₁ is at the focal point. Now, due to the converging lens L₂, let final image formed is I which is point image, hence the wavefront for this image will be of spherical symmetry.

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

Explanation- When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of

wavefront is in accordance with Huygen's principle.

So, Huygen's principle is valid for longitudinal sound waves also.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b, d)

Explanation- this is only possible when rate of decay of 1 is almost double the rate of decay of other.so in this case initial decay of A must be twice of B. also  Initial rate of decay of B is same as the rate of decay of A at t = 2h and  λ_A < _B