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Physics Ncert Solutions Class 12th

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Study the circuits (a) and (b) shown in figure and answer the following questions.

(a) Under which conditions would the rms currents in the two circuits be the same?

(b) Can the rms current in circuit (b) be larger than that in (a)?

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Payal Gupta

Contributor-Level 10

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

(I_rms)a= $\frac{V_{r m s}}{R}$ = $\frac{V}{R}$

(I_rms)b= $\frac{V_{r m s}}{Z}$ = $\frac{V}{\sqrt[]{R^{2} + (X_{L} - X_{C}) 2}}$

When $(I r m s) a$ = (I_rms)b

Then R= $\sqrt[]{R^{2} + (X_{L} - X_{C}) 2}$

XL=Xc resonance condition

As Z>R $\frac{(I r m s) a}{(I r m s) b}$ = $\frac{\sqrt[]{R^{2} + (X_{L} - X_{C}) 2}}{R}$ = $\frac{Z}{R}$ >1

(I_rms)a > (I_rms)b

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Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact
(a) Nuclear forces have short range
(b) Nuclei are positively charged
(c) The original nuclei must be completely ionized before fusion can take place
(d) The original nuclei must first break up before combining with each other

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, b)

Explanation- fusion process are almost impossible at ordinary temperature because nuclei are positive charge and forces between them is of short range.

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Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used have light nuclei. Heavy nuclei will not serve the purpose, because
(a) They will break up
(b) Elastic collision of neutrons with heavy nuclei will not slow them down
(c) The net weight of the reactor would be unbearably high
(d) Substances with heavy nuclei do not occur in liquid or gaseous state at room temperature

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation- the moderator used have light nuclei . because when we use heavy nuclei then there will be collision of proton which does not slow down the process.

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance.

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Payal Gupta

Contributor-Level 10

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

Inductive reactance XL= 2 $π ϑ$ L

And capacitive reactance Xc= $\frac{1}{2 π ϑ C}$

For very high frequencies XL= infinity and Xc=0

When reactance of the circuit is infinite it will be considered as open circuit . when reactance of a circuit is zero it will be considered as short circuited.

So C1, C2 = shorted and L1, L2=opened

So R_eff= R₁+R₂

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Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Heavy stable nuclei have more neutrons than protons. This is because of the fact that .

(a) Neutrons are heavier than protons
(b) Electrostatic force between protons are repulsive
(c) Neutrons decay into protons through beta decay
(d) Nuclear forces between neutrons are weaker than that between protons

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation – a nucleus contain more number of neutron than protons because the repulsive force is more if it contain more proton and become more unstable.

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Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

To ensure almost 100% transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is MgF2 (n=138). What should the thickness of the film be so that at the centre of the visible spectrum (5500 Å) there is maximum transmission.

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Vishal Baghel

Contributor-Level 10

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

Explanation- refractive index = 1.38 refractive index = 1.5

^{$λ = 5500 A$
0}

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass interface and a part transmitted. A part of the

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

Explanation- refractive index = 1.38 refractive index = 1.5

^{$λ = 5500 A$
0}

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass interface and a part transmitted. A part of the

reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r 1. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence, rays r 1 and r2 shall dominate the behaviour. If incident light is to be transmitted through the lens, r 1 and r2 should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence, there is no phase change on reflection. The optical path difference between r₂ and r₁ is

n(AD+CD)-AB

if d is the thickness of the film , then AD=CD=d/cosr

AB=AC sini

AC/2= d tanr

AC= 2dtanr

Hence AB= 2d tanr sini

Thus the optical path difference = $\frac{2 n d}{c o s r}$ -2dtanrsini

= 2. $\frac{s i n i d}{s i n r c o s r}$ - 2d $\frac{s i n r}{c o s r} s i n i$

= 2dsin{ $\frac{1 - {s i n}^{2} r}{s i n r c o s r}$ }

= 2ndcosr

For path difference $\frac{λ}{2}$

2ndcosr= $\frac{λ}{2}$

ndcosr= $\frac{λ}{4}$

for photographic lenses, the sources are normally in vertical plane

i=r=0⁰

ndcos0= $\frac{λ}{4}$

d= $\frac{λ}{4 n}$ =5500/4 $* 1.38$ = 1000A⁰

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Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Tritium is an isotope of hydrogen whose nucleus triton contains 2 neutrons and 1 proton. Free neutrons decay into p + e + n . If one of the neutrons in Triton decays, it would transform into He³ nucleus. This does not happen. This is because
(a) Triton energy is less than that of a He³nucleus
(b) The electron created in the beta decay process cannot remain in the nucleus
(c) Both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a He³ nucleus.
(d) Free neutrons decay due to external perturbations which is absent in triton nucleus

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation – tritium contain 1 proton and 2 neutron but when it disintegrates and remove 1 beta particle it will now contain 2 proton and 1 neutron so it will change into ₂He³ but its energy will be less from it.

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

Draw the effective equivalent circuit of the circuit shown in figure, at very high frequencies and find the effective impedance

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Payal Gupta

Contributor-Level 10

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

If we consider a L-C circuit analogous to a harmonically oscillating spring block system. The electrostatic energy $\frac{1}{2}$ CV2 is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\frac{1}{2}$ LI2 is analogous to kinetic energy.

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4 months ago

Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Ten

The optical properties of a medium are governed by the relative permittivity (εr) and relative permeability (μ_r). The refractive index is defined as $\sqrt[]{μ_{r} ε_{r}}$ = n.For ordinary material ε_r > 0 and μ_r > 0 and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and r < 0. Since then such 'metamaterials' have been produced in the laboratories and their optical properties studied. For such materials n = $\sqrt[]{μ_{r} ε_{r}}$ . As light enters a medium of such refractive index the phases travel away from the direction of propagation. (i) According to the description above show that if rays of light enter such a medium from air (refractive index =1) at an angle θ in 2nd quadrant, them the refracted b

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The optical properties of a medium are governed by the relative permittivity (εr) and relative permeability (μ_r). The refractive index is defined as $\sqrt[]{μ_{r} ε_{r}}$ = n.For ordinary material ε_r > 0 and μ_r > 0 and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and μr < 0. Since then such 'metamaterials' have been produced in the laboratories and their optical properties studied. For such materials n = – $\sqrt[]{μ_{r} ε_{r}}$ . As light enters a medium of such refractive index the phases travel away from the direction of propagation. (i) According to the description above show that if rays of light enter such a medium from air (refractive index =1) at an angle θ in 2nd quadrant, them the refracted beam is in the 3rd quadrant. (ii) Prove that Snell's law holds for such a medium.

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Vishal Baghel

Contributor-Level 10

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

Explanation-All points with the same optical path length must have the same phase.

So – $\sqrt[]{μ_{r} ε_{r}} A E$ =BC-– $\sqrt[]{μ_{r} ε_{r}} C D$

BC= $\sqrt[]{μ_{r} ε_{r}}$ (CD-AE)

BC>0, si must be greater than AD

But in other figure

– $\sqrt[]{μ_{r} ε_{r}} A E = B C - \sqrt[]{μ_{r} ε_{r}} C D$

So BC= – $\sqrt[]{μ_{r} ε_{r}} (C D - A E)$

But clearly here BE is less than zero

To proving snells law we know that

BC=ACsin $θ$ and CD-AE=ACsin $θ$

So n= sini/sinr

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Physics Ncert Solutions Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

M_e and M_s denote the atomic masses of the parent and he daughter nuclei respectively in radioactive decay. The Q value for a beta^-1 decay is Q₁ and that for beta+ decay is Q₂.if m_e denotes the mass of an electron, then which of the following statements is correct?

(a) Q₁=(M_x-M_y)c² and Q₂=[M_x-M_y-2m_e]c²

(b) Q₁=(M_x-M_y)c² and Q₂=[M_x-M_y]c²

(c) Q₁=(M_x-M_y-2m_e)c² and Q₂=[M_x-M_y-2C_e]c²

(d) Q₁=(M_x-M_y+2m_e)c² and Q₂=[M_x-M_y+2m_e]c²

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer –a

Explanation- $β$ ⁺ decay represents

_zX^{A
$\to$}_z-1Y^A+₁e^o+v+Q₂

Q₂= [m_n (_zX^A)-m_n (_z-1Y^A)-m_e]c²

= (M_x-M_y-2m_e)c²

^{$β$
-}decay represents

_zX^{A
$\to$}_z+1Y^A+-₁e^o+v+ $\propto$ ₁

_{$\propto$
1}= [m_n (_zX^A)-m_n (_z+1Y^A)-m_e]c²

= (M_x-M_y)c²

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