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6 months ago

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

The maximum and minimum voltage of an amplitude modulated signal are 60V and 20V respectively. The percentage modulation index will be:

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alok kumar singh

Contributor-Level 10

Percentage modulation = $\frac{(V_{m a x} - V_{m i n})}{(V_{m a x} + V_{m i n})} * 100 %$

Percentage modulation = $(\frac{60 - 20}{80 + 20}) * 100$

= 50%

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6 months ago

Class 12th Ray Optics and Optical Instruments

A point source of light S, placed at a distance 60cm infront of the centre of a plane mirror of width 50cm, hangs vertically on a wall. A man walks inform of the mirror along a line parallel to the mirror at a distance 1.2m from it (see in the figure). The distance between the extreme points where he can the image of the light source in the mirror is………….cm

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Payal Gupta

Contributor-Level 10

Maximum Distance = 3 (x + x) = 6x = 3 * 2x = 3 * 50 = 150 cm

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6 months ago

Class 12th Gravitation

A body is projected vertically upward from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be:

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alok kumar singh

Contributor-Level 10

$V_{C} = \sqrt[]{\frac{2 G M}{R}}$

Using conservation of Mechanical Energy

$\Rightarrow \frac{- G M m}{R} + \frac{1}{2} * m (\frac{V e^{2}}{9}) = \frac{- G M m}{(R + h)}$

$\Rightarrow \frac{1}{R + h} = \frac{8}{9 R} \Rightarrow h = \frac{R}{8} = \frac{6400}{8} = 800 k m$

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6 months ago

Class 12th Dual Nature of Radiation and Matter

Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : y. The value of x is

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Payal Gupta

Contributor-Level 10

Case I :- $2 ? - ? = \frac{1}{2} m v_{1}^{2} . . . . . . . . (i)$

Case II :- $10 ? - ? = \frac{1}{2} m v_{2}^{2} . . . . . . . . (i i)$

$\Rightarrow \frac{1}{9} = \frac{v_{1}^{2}}{v_{2}^{2}}$

$\Rightarrow v_{1} : v_{2} = 1 : 3$

x = 1

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6 months ago

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

A monoatomic gas at pressure P and volume V is suddenly compressed to one eight of its original volume. The final pressure at constant entropy will be:

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Vishal Baghel

Contributor-Level 10

Constant Entropy means Adiabatic process

$P v^{y} = C$

$P_{1} {v_{1}}^{y} = P_{2} {v_{2}}^{y}$

$P_{2} = P_{1} {(\frac{v_{1}}{v_{2}})}^{y} = P {(\frac{v}{\frac{1}{8} v})}^{y}$

$P_{2} = P (8) \frac{5}{3}$

P₂ = 32 P

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6 months ago

Class 12th Thermodynamics

The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $V = k T^{\frac{2}{3}} .$ The work done when temperature changes by 90 K will be xR. The value of x is…….

[R = universal gas constant]

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Payal Gupta

Contributor-Level 10

$V = k T^{\frac{2}{3}}$

$T V^{- \frac{3}{2}} = K$

$\Rightarrow γ - 1 = - 3 / 2$

$\Rightarrow γ = - 1 / 2$

Work done = $\frac{n R Δ T}{- y + 1} = \frac{1 * R * 90}{3 / 2} = 6 R$

n = 60

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6 months ago

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

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R = 8.3 JK^-1 mol^-1)

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Vishal Baghel

Contributor-Level 10

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

$= 7 * \frac{3}{2} * 8.3 * 40$

$= 3486 J$

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6 months ago

Class 12th Oscillations

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{α}{β} T .$ The value of a is

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Payal Gupta

Contributor-Level 10

Total oscillation = $\frac{1}{2} + \frac{1}{8} = \frac{5}{8}$

$\Rightarrow t i m e t a k e n = \frac{T}{2} + \frac{T}{12} = \frac{7 T}{12}$

= 7

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6 months ago

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

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New answer posted

6 months ago

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

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be:

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Vishal Baghel

Contributor-Level 10

x = A sin $ω t$

$\Rightarrow v = A ω c o s ω t$

$\Rightarrow v = \pm ω \sqrt[]{A^{2} - x^{2}}$

$\frac{v^{2}}{ω^{2}} + x^{2} = A^{2}$

Elliptical

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