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Class 12th Chemistry NCERT Exemplar Solutions Class 12th Chapter Twelve

Find the ratio of energies of photos product due to transition of an electron of hydrogen atom from its (i) second permitted energy level to the first level. and (ii) the highest permitted energy level to the first permitted level.

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Vishal Baghel

Contributor-Level 10

$Δ E_{1} = - \frac{E_{0}}{4} + \frac{E_{0}}{1} = \frac{3}{4} E_{0}$

$Δ E_{2} = 0 - (- E_{0}) = E_{0}$

$\frac{Δ E_{1}}{Δ E_{2}} = \frac{3}{4}$

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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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9 months ago

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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9 months ago

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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9 months ago

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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9 months ago

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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