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

A small transparent slab containing material of µ = 1.5 is placed along AS₂(figure). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

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Vishal Baghel

Contributor-Level 10

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

Explanation- as the refractive index of the class $μ$ , the path difference will be calculated as $? x$ =2dsin $θ$ +( $μ - 1$ )L

For principal maxima ,(path difference is zero)

2dsin $θ$ ₀+( $μ - 1$ )L=0

Sin $θ$ ₀= - $\frac{L (μ - 1)}{2 d}$ = $\frac{- L (0.5)}{2 d}$

Sin $θ$ ₀=-1/16

OP=Dtan $θ$ ₀= Dsin $θ$ ₀=-D/16

For pat $?$ h difference $? \frac{λ}{2}$

2dsin $θ$ ₁+0.5L= $? \frac{λ}{2}$

Sin $θ$ ₁= $\frac{? \frac{λ}{2} - 0.5 L}{2 d}$ = $\frac{? \frac{λ}{2} - \frac{d}{8}}{2 d}$

= $\frac{\frac{λ}{2} - \frac{λ}{8}}{2 λ}$ = $?$ 1/4 -1/16

So two possible values $\frac{1}{4} - \frac{1}{16} = \frac{3}{16}$ and- $\frac{1}{4} - \frac{1}{16}$ = $- \frac{5}{16}$

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

Consider the LCR circuit shown in figure find the net current I and the phase of i. show that i=V/Z. find the impedance Z for this circuit.

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Payal Gupta

Contributor-Level 10

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

Total current i=i1+i2

Vmsinwt=Ri1

I1= $\frac{V_{m} s i n w t}{R}$

If q2 is charge on the capacitor at any time t then for series combination of C and L

Applying KVL in below circuit

So q2=qmsin(wt $\frac{q_{2}}{C} + \frac{L d i_{2}}{d t} - V_{m} s i n w t = 0$ )……….1

qm[ $+ \emptyset$ sin(wt+ $\frac{q_{2}}{C} + \frac{L d_{2} q_{2}}{d t^{2}} = V_{m} s i n w t$ )]= Vmsinwt

if $\frac{d q_{2}}{d t} = - q_{m} w^{2} s i n ? (w t + \emptyset)$

qm= $u s e t h i s v a l u e i n e q n 1$

from above equation i2= $\frac{1}{C} + L (- w^{2})$

i2= $\emptyset$ when $\emptyset = 0$

so total current I = i1+i2

I= $\frac{V_{m}}{(\frac{1}{c} - L w^{2})}$

I= i1+i2= Ccos $\frac{d q_{2}}{d t} = w q_{m} \cos ? (w t + \emptyset)$ +Csin $\frac{w V_{m} c o s ? (w t + \emptyset)}{\frac{1}{c} - L w^{2}}$

I= Csin(wt+ $\emptyset = 0, i_{2} = \frac{V_{m} c o s w t}{\frac{1}{w C} - L w}$ )

C= $\frac{V_{m} s i n w t}{R} + \frac{V_{m} c o s w t}{\frac{1}{w C} - L w}$

$\emptyset s i n w t$ 1/2

$\emptyset c o s w t$ = tan-1 $\emptyset$

$\sqrt[]{A^{2} + B^{2}}$ 1/2

This is the expression for impedance.2

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

Figure shown a two slit arrangement with a source which emits unpolarised light. P is a polariser with axis whose direction is not given. If I 0 �� 0 is the intensity of the principal maxima when no polariser is present, calculte in the present case, the intensity of the principal maxima as well as the first minima.

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Vishal Baghel

Contributor-Level 10

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

Explanation- when polariser is not used

A=A_perp+A

letA₁= asinwt and A₂=asin(wt+ $\emptyset$ )

now superposition principle for perpendicular polariser

A_R= asinwt+ asin(wt+ $\emptyset$ )

A_R=a(2cos $\emptyset / 2$ sin(wt+ $\emptyset$ ))

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

This eqn is also same for parallel polariser

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

And we know that intensity is directly proportional to square of amplitude

(A_R)²= (A_perp)²+(A)²

So resultant intensity is

I=4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt + 4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt

I= 8(a)²cos^{2
$\emptyset / 2$}(1/2) &nb

...more

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

Explanation- when polariser is not used

A=A_perp+A

letA₁= asinwt and A₂=asin(wt+ $\emptyset$ )

now superposition principle for perpendicular polariser

A_R= asinwt+ asin(wt+ $\emptyset$ )

A_R=a(2cos $\emptyset / 2$ sin(wt+ $\emptyset$ ))

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

This eqn is also same for parallel polariser

A_R=2acos $\emptyset / 2$ sin(wt+ $\emptyset$ )

And we know that intensity is directly proportional to square of amplitude

(A_R)²= (A_perp)²+(A)²

So resultant intensity is

I=4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt + 4(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt

I= 8(a)²cos^{2
$\emptyset / 2$}(1/2) (after integrating sin value we got 1/2)

I= 4(a)²cos^{2
$\emptyset / 2$}

So I= 4(a)²

But when there is a polariser

Perpendicular polariser is blocked

I=2(a)²cos^{2
$\emptyset / 2 \frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t + \emptyset)$}dt + (a)^{2
$\frac{1}{T} \int_{0}^{T} {s i n}^{2} (w t)$}

I=2(a)²cos^{2
$\emptyset / 2 +$}1/2(a)²

When $\emptyset = 0$

I_max=2(a)²+1/2(a)²

I_max=5/2(a)²

I_max=5/2(I/4)=5I/8

At first minima it becomes I/8

less

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a year ago

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

MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transmitted if

(i) Power is transmitted at 220V. Comment on the feasibility of doing this.

(ii) A step-up transformer is used to boost the voltage to 11000V, power transmitted, then a step-down transformer is used to bring voltage to 220 V. (ρ_cu= 1.7*10^-8 SI unit)

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Payal Gupta

Contributor-Level 10

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

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New answer posted

a year ago

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

An electrical device draws 2KW power from AC mains (voltage 223V(rms)=) $\sqrt[]{50000} V$ . The current differs (lags) in phase by $\emptyset$ (tan $\emptyset = - 3 / 4$ ) as compared to voltage . find

(a) R

(b) X_L-X_C(c) I_m. Another device has twice the values for R,X_candX_L. how the answer affected?

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Payal Gupta

Contributor-Level 10

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

Power drawn, p= 2KW= 2000W

tan $\emptyset$ =-3/4

P=V2/Z

Z=V2/P= $\frac{223 * 223}{2 * 10^{3}}$ =25

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

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

625 = $R^{2} + ({X_{L} - X_{C})}^{2}$ ………… (1)

tan $\emptyset$ = $\frac{X_{L} - X_{C}}{R}$ = ¾

XL-XC= 3R/4

Use this in eqn 1

625= R2+ (3R/4)2 = R2+9R2/16

625 = 25R2/16, R= 20ohm

XL-XC=15ohm

Im= $\sqrt[]{2}$ I= $\sqrt[]{2} V / Z$ = $\frac{223 * \sqrt[]{2}}{25}$ = 12.6A

If R, XL and Xc are all doubled, tan $\emptyset$ does not change . Z is doubled, current is halved. So power is also halved.

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a year ago

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

The gravitational force between a H atom and a another particle of mass m will be given newton's law

F=GMm/r² where r in km

(a) M=m_proton+m_electron

(b) M=m_proton+m_electron-

(c) M is not relate to the mass of the hydrogen atom

(d) M=m_proton+m_electron-(|V|)= magnitude of the potential energy of electron in H atom.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b

Explanation- F=GMm/r²

M=effective mass of hydrogen atom=mass of electron +mass of proton -B²/c

B.E of hydrogen atom = 13.6eV

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

Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 year. After 1 year,
(a) All the containers will have 5000 atoms of the material
(b) All the containers will contain the same number of atoms of the material but that number will only be approximately 5000
(c) The containers will in general have different numbers of the atoms of the material but their average will be close to 5000
(d) None of the containers can have more than 5000 atoms

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer (c)

Explanation- it is a process in which metal decays spontaneously . so during a span of 1 year it will decay almost half of its original value so it will come close to 5000.

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

A nuclide 1 is said to be the mirror isobar of nuclide 2 if Z₁=N₂ and Z₂ = N₁ (a) What nuclide is a mirror isobar of ₁₁²³Na ? (b) Which nuclide out of the two mirror isobars have greater binding energy and why?

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alok kumar singh

Contributor-Level 10

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

Explanation- a nuclide 1 is said to be mirror isobar of nuclide 2, if Z₁=N₂ and Z₂= N₁

Now ₁₁Na²³, ₁₂Mg²³ for which Z₂=12=N₁ and N₂ =23-12=11=Z₁

(b) As ₁₂Mg²³ contain even number of protons against ₁₁Na²³ which has odd number of protons (11), therefore Mg has greater binding energy than sodium.

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

Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10^-15

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alok kumar singh

Contributor-Level 10

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

Explanation- Each particle (neutron and proton) present inside the nucleus is called nucleon .

Let $λ = 10^{- 15} m$

$λ = \frac{h}{p}$

and KE= PE

$E = h c / λ$ = $\frac{6.6 * 10^{- 34} * 3 * 10^{8}}{10^{- 15} * 1.6 * 10^{- 19}}$ eV

K.E= 10⁹eV

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a year ago

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

A piece of wood from the ruins of an ancient building was found to have a ¹⁴C activity of 12 disintegrations per minute per gram of its carbon content. The ¹⁴C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of ¹⁴C is 5760 yr.

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alok kumar singh

Contributor-Level 10

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

Explanation- R= 12dis/min per gm and R₀= 16dis/min per gm T_1/2= 5760yr

R= $e^{- λ t}$ $e^{λ t}$

By taking log

$λ t l o g e = l o g \frac{R_{0}}{R}$

$λ t$ = log₁₀16/12 $* 2.303$

$t = \frac{2.303 (l o g 4 - l o g 3)}{λ}$

= 2.303 (0.6020-4.771)5760/0.6931

= 2391.20yr

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