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

The kinetic energy of emitted electron is E when the right incident on the metal has wavelength $λ$ . To double the kinetic energy, the incident light must have wavelength:

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Vishal Baghel

Contributor-Level 10

$\frac{h c}{λ} - ? = E$ -(i)

$\frac{h c}{λ'} - ? = 2 E$ -(ii)

$h c (\frac{1}{λ'} - \frac{1}{λ}) = E$

$λ' = \frac{h c λ}{E λ + h c}$

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

Match List – I with List – II:

List – I List – II

(A) UV rays (i) Diagnostic tool in medicine

(B) X-rays(ii) Water purification

(C) Microwave(iii) Communication, Radar

(D) Infrared wave(iv) Improving visibility in foggy daya

Choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

Based an theoretical data.

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

A coil of inductance 1H and resistance 100 $Ω$ is connected to a battery of 6.V. Determine approximately:

(A)The time elapsed before the current acquires half of its steady – state value.

(B)The energy stored in the magnetic field associated with the coil at instant 15ms after the circuit is switched on. (Given In 2 = 0.693, e3/2 = 0.25)

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Vishal Baghel

Contributor-Level 10

L = 1H, R = 100 $Ω$

$A s, i = i_{0} e^{- t / τ}$

$F o r i = \frac{i_{0}}{2} \Rightarrow \frac{i_{0}}{2} = i_{0} e^{- t / τ}$

$\Rightarrow - l n 2 = - t / τ$

$i = \frac{6}{100} e^{- \frac{15}{1 / 100}} = . 06 e^{- 1500 * 1 0^{- 3}} = 0.06 e^{- 1.5} = 0.06 * 0.25 = . 0.015 - A$

So, $U = \frac{1}{2} L i^{2} = \frac{1}{2} * 1 * {(15 * 1 0^{- 3})}^{2} = \frac{1}{2} (225) * 1 0^{- 6} = 112.5 * 1 0^{- 6} J = 0.1125 m J$

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

Orange light of wavelength $6000 * 10^{- 10} m$ illuminates a single slit of width $0.6 * 10^{- 4} m$ . the maximum possible number of diffraction minima produced on both sides of the central maximum is

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alok kumar singh

Contributor-Level 10

Sol. Angular momentum conservation:

$\begin{array}{r} \Rightarrow I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{f} \\ \Rightarrow \frac{{M R}^{2}}{2} ω_{o} = (\frac{{M R}^{2}}{2} + \frac{{M R}^{2}}{8}) ω_{f} \\ \Rightarrow ω_{f} = \frac{4}{5} ω_{o} \\ \Rightarrow {K E}_{final} = \frac{1}{2} (I_{1} + I_{2}) ω_{f}^{2} = \frac{{M R}^{2} ω^{2}}{5} \\ \Rightarrow K E_{initial} = \frac{1}{2} I, ω_{0}^{2} = \frac{{M R}^{2} ω^{2}}{4} \\ \Rightarrow % loss \Rightarrow 20 % . \end{array}$

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

Four resistance $40 Ω, 60 Ω, 90 Ω$ , and $110 Ω$ make the arms of a quadrilateral ABCD. Across $A C$ is a battery of emf $40 V$ and internal resistance negligible. The potential difference across $B D$ in $V$ is

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alok kumar singh

Contributor-Level 10

10553.33

Sol. $\frac{1}{λ_{min}} = R [\frac{1}{1} - \frac{1}{\infty}] = R [n = \infty \to n = 1]$

$\frac{1}{λ_{m a x}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} [n = 2 \to n = 1]$

$\Rightarrow Δ λ \Rightarrow \frac{4}{3 R} - \frac{1}{R} \Rightarrow \frac{1}{3 R} = 340$

For Paschan $\Rightarrow \frac{1}{λ_{min}} = R [\frac{1}{9}] [n = \infty \to h = 3]$

$\Rightarrow \frac{1}{λ_{m a x}} = R [\frac{1}{9} - \frac{1}{16}] = \frac{7 R}{144}$

$\Rightarrow Δ λ = \frac{81}{7 R}$

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

An alternating emf E = 440 sin 100t is applied to a circuit containing an inductance of $\frac{\sqrt[]{2}}{π} H .$ If an a.c. ammeter is connected in the circuit, its reading will be:

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Vishal Baghel

Contributor-Level 10

E = 440 sin 100 t $ω = 100 π$

$L = \frac{\sqrt[]{2}}{π} H .$ $X_{L} = ω L = 100 π . \frac{\sqrt[]{2}}{π} = 100 \sqrt[]{2} Ω .$

$= \frac{220}{100} = 2.2 A$

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

Two metallic wires of identical dimensions are connected in series. If $σ_{1} a n d σ_{2}$ are the conductivities of these wires respectively, the effective conductivity of the combination is:

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Vishal Baghel

Contributor-Level 10

R = R1 + R2

$\Rightarrow \frac{2 l}{σ A} = \frac{l}{σ_{1} A} + \frac{l}{σ_{2} A} \Rightarrow \frac{2}{σ} = \frac{1}{σ_{1}} + \frac{1}{σ_{2}} \Rightarrow σ = \frac{2 σ_{1} σ_{2}}{σ_{1} + σ_{2}}$

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

The change in the magnitude of the volume of an ideal gas when a small additional pressure $Δ P$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $Δ T$ at constant pressure. The initial temperature and pressure of the gas were $300 K$ and 2 atm. respectively. If $| Δ T | = C | Δ P |$ then value of $C$ in (K/atm) is .

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alok kumar singh

Contributor-Level 10

Sol.

$\begin{array}{l} A H \to \frac{H}{2} & H = \frac{\sqrt[]{3} a}{2} \\ A G \to \frac{H}{\sqrt[]{3}} & A l = \frac{H}{2 \sqrt[]{3}} \\ M_{A B C} = M & M_{A D E} = \frac{M}{4} \end{array}$

$\Rightarrow I_{G} = \frac{M a^{2}}{12} - [\frac{M}{4} \frac{{[\frac{a}{2}]}^{2}}{12} + \frac{M}{4} {[\frac{a}{2 \sqrt[]{3}}]}^{2}] = (\frac{11}{16}) \frac{{M a}^{2}}{12}$

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Class 12th Sequences and Series

If a₁, a₂, a₃…. and b₁, b₂, b₃…. are A.P., and a₁ = 2, a₁₀ = 3, a₁b₁ = 1 – a₁₀b₁₀, then a₄b₄ is equal to…….

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Payal Gupta

Contributor-Level 10

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

...more

$a_{1}, a_{2}, a_{3} . . . . . .$ are in A.P. (Let common difference is d₁)

$b_{1}, b_{2}, b_{3} . . . .$ are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$? a_{1} b_{1} = 1$

$∴ b_{1} = \frac{1}{2}$

$a_{10} b_{10} = 1$

$∴ b_{10} = \frac{1}{3}$

Now,

$a_{10} = a_{1} + 9 d_{1} \Rightarrow d_{1} = \frac{1}{9}$

$b_{10} = b_{1} + 9 d_{2} \Rightarrow d_{2} = \frac{1}{9} [\frac{1}{3} - \frac{1}{2}] = - \frac{1}{54}$

Now

$a_{4} = 2 + \frac{3}{9} = \frac{7}{3}$

$b_{4} = \frac{1}{2} - \frac{3}{54} = \frac{4}{9}$

$∴ a_{4} b_{4} = \frac{28}{27}$

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Class 12th Sets

Let S = $2 + \frac{6}{7} + \frac{12}{7^{2}} + \frac{20}{7^{3}} + \frac{30}{7^{4}} + . . . .$ Then 4S is equal to

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Payal Gupta

Contributor-Level 10

$S = 2 + \frac{6}{7} + \frac{12}{7^{2}} + \frac{20}{7^{3}} + \frac{30}{7^{4}} + . . . . . . . . (i)$

$\frac{1}{7} S = \frac{2}{7} + \frac{6}{7^{2}} + \frac{12}{7^{3}} + \frac{20}{7^{4}} + . . . . . . . . (i i)$

(i) - (ii)

$\frac{6}{7} S = 2 + \frac{4}{7} = \frac{6}{7^{2}} + \frac{8}{7^{3}} + . . . . . . . . . . . (i i i)$

$\frac{6}{7^{2}} S = \frac{2}{7} + \frac{4}{7^{2}} + \frac{6}{7^{3}} + . . . . . . . . (i v)$

(iii) – (iv)

$= 2 + [\frac{1}{1 - \frac{1}{7}}] = 2 (\frac{7}{6})$

$∴ 4 S = 8 {(\frac{7}{6})}^{3} = {(\frac{7}{3})}^{3}$

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