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Physics Ncert Solutions Class 12th

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

In the LCR circuit shown in figure the AC driving voltage isV=V_msinwt

Write the equation of motion for q(t) describe subsequent motion of charges.

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

Four identical monochromatic sources A, B, C, D as shown in the (figure)

produce waves of the same wavelength λ and are coherent. Two receiver

R1 and R2 are at great but equal distances from B.

(i) Which of the two receivers picks up the larger signal?

(ii) Which of the two receivers picks up the larger signal when B is turned off?

(iii) Which of the two receivers picks up the larger signal when D is turned off?

(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?

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Vishal Baghel

Contributor-Level 10

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

Explanation-consider the disturbance at the receiver R₁ which is at a distance d from B

Y_A= acos(wt) and path difference is $\frac{λ}{2}$ hence phase difference is $π$ .

Thus the wave R₁ because of B

Y_B= acos(wt- $π$ )= - acoswt here path difference is $λ$ and hence phase difference is $2 π$

Thus R₁ because of C

Y_c= acos(wt-2 $π$ )= acoswt

(i)let the signal picked up at R₂ from B be Y_B= a₁cos(wt)

The path difference between signal at D and that B is $\frac{λ}{2}$

Y_D= -a₁cos(wt)

The path difference between signal at A and that atB is

$\sqrt[]{d^{2} + {(\frac{λ}{2})}^{2}}$ -d = d( ${1 + \frac{λ^{2}}{4 D^{2}})}^{1 / 2}$ -d = $\frac{λ^{2}}{8 d^{2}}$

$a s d ? λ$ therefore path difference os 0

$p h a s e d i f f e r e n c e = \frac{2 π}{γ} \frac{λ^{2}}{8 d^{2}}$

_{$Y$
A}=a₁co

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

Explanation-consider the disturbance at the receiver R₁ which is at a distance d from B

Y_A= acos(wt) and path difference is $\frac{λ}{2}$ hence phase difference is $π$ .

Thus the wave R₁ because of B

Y_B= acos(wt- $π$ )= - acoswt here path difference is $λ$ and hence phase difference is $2 π$

Thus R₁ because of C

Y_c= acos(wt-2 $π$ )= acoswt

(i)let the signal picked up at R₂ from B be Y_B= a₁cos(wt)

The path difference between signal at D and that B is $\frac{λ}{2}$

Y_D= -a₁cos(wt)

The path difference between signal at A and that atB is

$\sqrt[]{d^{2} + {(\frac{λ}{2})}^{2}}$ -d = d( ${1 + \frac{λ^{2}}{4 D^{2}})}^{1 / 2}$ -d = $\frac{λ^{2}}{8 d^{2}}$

$a s d ? λ$ therefore path difference os 0

$p h a s e d i f f e r e n c e = \frac{2 π}{γ} \frac{λ^{2}}{8 d^{2}}$

_{$Y$
A}=a₁cos(wt- $\emptyset$ )

_{$Y$
C}=a₁cos(wt- $\emptyset$ )

Therefore signal picked up by R₂ is

Y_A+Y_b+Y_c+Y_d= y = 2a₁cos(wt- $\emptyset$ )

|y|² = 4a₁²cos²(wt- $\emptyset$ )

=2a₁². Thus R₁ picks up the larger signal.

(ii) if B is swiched off

R₁ picks up y = acoswt

^{IR1=
$\frac{1}{2}$}^a2

R₂ picks up y = a coswt

I_R2= $\frac{1}{2}$ a²

(iii) thus R₁ and R₂ pick up the same signal

If D is switched off

R₁ picks up by y = acoswt

I_R1= $\frac{1}{2}$ a²

Y= 3acoswt

I_R2=9a²/2

R₂ picks up larger signal compared to R₁

(iv) thus a signal at R₁ indicates B has been switched off and an enhanced signal at R₂indicates D has been switched off.

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

For a LCR circuit at frequency w the equation reads

Multiply the equation by i and simplify where is possible

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

When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom
(a) Do not change for any type of radioactivity
(b) Change for α and β-radioactivity but not for γ-radioactivity
(c) Change for α -radioactivity but not for others
(d) Change for β-radioactivity but not for others

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (b)

Explanation – because alpha and beta particle have some charge and mass so their energy level will change but in case of gamma particle it does not contain any charge.

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Physics Ncert Solutions 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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Physics Ncert Solutions 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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Physics Ncert Solutions 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

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Physics Ncert Solutions 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

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Physics Ncert Solutions 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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Physics Ncert Solutions 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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