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6 months ago

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

The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 k Ω$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 k Ω .$ The voltage gain of the transistor is________.

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alok kumar singh

Contributor-Level 10

Voltage gain = $\frac{Δ l_{C}}{Δ l_{B}} * \frac{R_{C}}{R_{B}}$

$\frac{5 * 1 0^{- 3}}{100 * 1 0^{- 6}} * \frac{2}{0.5}$

= 200

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6 months ago

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

In the circuit shown below, maximum zener diode current will be _________mA.

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Vishal Baghel

Contributor-Level 10

$l = \frac{120 - 60}{4000} = 0.015 A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA

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6 months ago

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

A uniformly heavy rod of mass 20kg, cross sectional area 0.4 m² an length 20m is hanging from a fixe support. Neglecting the lateral contraction. The elongation in the rod due to its own weight is x * 10^-9m. The value of x i________.

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alok kumar singh

Contributor-Level 10

$\int_{0}^{Δ l} d y = \int_{0}^{l} (\frac{m g}{l}) * \frac{d x}{A Y}$

$\Rightarrow Δ l = \frac{m g l}{2 A Y} = 25 * 1 0^{- 9} m$

$\Rightarrow x = 25.00$

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6 months ago

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

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: ${}_{1}^{2}X +_{1}^{2} X = \frac{4}{2} Y$

The binding energies per nucleon for are 1.1 Me V and 7.6

Me V respectively. The energy released in this process is________ Me V.

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alok kumar singh

Contributor-Level 10

Energy released = Binding energy of product – Binding energy of reacts :

$Δ E = 7.6 * 4 - (1.1 * 4)$

= 26 MeV

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6 months ago

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

In the hydrogen spectrum, $λ$ be the wavelength of first transition line of Lyman series. The wavelength difference will be $" a λ "$ between the wavelength of 3rd transition line of Paschen series and that 2nd transition line B Balmer series where a = __________.

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Vishal Baghel

Contributor-Level 10

For first line of Lyman

$\frac{1}{λ} = R (1 - \frac{1}{4}) = R \frac{3}{4}$

$λ = \frac{4}{3 R}$

3rd line (Paschen)

$\frac{1}{λ_{3}} = R (\frac{1}{3^{2}} - \frac{1}{6^{2}}) = \frac{R}{9} * \frac{3}{4}$

2nd line (B almer)

${\frac{1}{λ}}_{2} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{R}{4} * \frac{3}{4}$

$a (\frac{4}{3 R}) = \frac{20}{3 R} \Rightarrow a = 5$

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6 months ago

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

The graph between $\frac{1}{u} a n d \frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is 1.5 and its both the surfaces have same radius of curvature R. The value of R will be _________cm. (where u = object distance, v = image distance)

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Vishal Baghel

Contributor-Level 10

For point B, $\frac{1}{u} = - 0.10 c m^{- 1}, \frac{1}{v} = 0$

Thus u = 10 cm, v = $\infty$

i.e., f = 10 cm

$\frac{1}{10} = (1.5 - 1) (\frac{2}{R}) \Rightarrow \frac{1}{R} = \frac{1}{10}$

R = 10 cm

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6 months ago

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

The effective current I in the given circuit at very high frequencies will be _________A.

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Vishal Baghel

Contributor-Level 10

At very high frequencies

$X_{C} = \frac{1}{ω C} \to 0$

$X_{L} = ω L \to \infty$

Thus equivalent circuit

$Z = 1 + 2 + 2 = 5 Ω$

$l = \frac{220}{5} = 44 A$

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6 months ago

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

In the given figure, the face AC of the equilateral prism is immersed in a liquid of refractive index 'n'. For incident angle $60 °$ at the side AC, the refracteel light beam just grazes along face AC. The refractive index of the liquid $n = \frac{\sqrt[]{x}}{4} .$

The value of x is_______.

(Given refractive index of glass = 1.5)

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alok kumar singh

Contributor-Level 10

1.5 * sin 60 = R * sin 90

$\Rightarrow 1.5 * \frac{\sqrt[]{3}}{2} = n = \frac{\sqrt[]{x}}{4}$

x = 27

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6 months ago

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

The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the centre will be________m.

Given, the length of the rod is $10 \sqrt[]{3} m .$

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alok kumar singh

Contributor-Level 10

$k = \sqrt[]{\frac{l}{m}} = \sqrt[]{\frac{m l^{2} / 12}{m}} = \frac{l}{2 \sqrt[]{3}}$

$k = \frac{10 \sqrt[]{3}}{2 \sqrt[]{3}} = 5$

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6 months ago

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

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{2}$ is 40cm. Now as unknown resistance x is connected in series with P and new balancing length is found to be 80cm measured from the same end. Then the value of x will be _________ $Ω.$

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Vishal Baghel

Contributor-Level 10

Initially, $\frac{P}{Q} = \frac{40}{60} = \frac{2}{3}$ - (1)

Finally, $\frac{P + x}{Q} = \frac{80}{20} = \frac{4}{1}$

(2) (1)

$\frac{P + x}{P} = \frac{4 * 3}{2} = 6$

$\Rightarrow 1 + \frac{x}{P} = 6$

$\frac{x}{P} = 5$

x = 5 P = 5 x 4 = 20 $Ω$

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