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Physics Ncert Solutions Class 12th

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Physics Ncert Solutions Class 12th Class 12th

A baseband signal of 3.5 MHz frequency is modulated with a carrier signal of 3.5 GHz frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal?

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Vishal Baghel

Contributor-Level 10

Baseband Signal frequency 3.5 MHz

& Carrier signal frequency = 3.5 GHz

$υ_{C} = 3.5 G H z = 3.5 * 1 0^{9}$

$λ = \frac{C}{υ_{c}} = \frac{3 * 1 0^{8}}{3.5 * 1 0^{9}} = \frac{3}{3.5} * 10$

$= \frac{30}{35} * 10$

= $\frac{60}{7}$ $\frac{}{}$

Size of antenna = $\frac{λ}{4} = \frac{60}{7 * 4}$ $\frac{}{} \frac{}{}$

= 21.4 mm

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Physics Ncert Solutions Class 12th Class 12th

A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 * 10^-2 Am^-1 at a point, what will be the approximate magnitude of electric field intensity at that point ?

(Given : Permeability of free space µ₀ = 4p * 10^-⁷ NA^-², speed of light in vacuum c = 3 * 10⁸ ms^-¹)

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Vishal Baghel

Contributor-Level 10

Velocity of wave in a medium = $\frac{1}{\sqrt[]{μ_{m} \in_{m}}}$ $\frac{}{_{}_{}}$

$\begin{array}{l} v = \frac{1}{\sqrt[]{μ_{0} μ_{r} \in_{r} \in_{0}}} \\ v = \frac{3 * 1 0^{8}}{\sqrt[]{1.61 * 6.44}} = 0.931 * 1 0^{8} m / s e c \end{array}$

$\begin{array}{l} B = μ_{m} H \\ = μ_{r} μ_{0} H \\ = 1.61 * 4 π * 1 0^{- 7} * 4.5 * 1 0^{- 2} \\ = 1.61 * 4 π * 4.5 * 1 0^{- 9} \end{array}$

$\frac{E}{B} = v$

E = vB = 0.931 * 10⁸ * 1.61 * 4p * 4.5 * 10^-⁹

= [0.931 * 1.61 * 4p * 4.5] * 10^-¹

= 8.476

» 8.476 v m^-¹

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Physics Ncert Solutions Class 12th Electrostatics

A vertical electric field of magnitude 4.9 * 10⁵ N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will

(Given g = 9.8 m/s²)

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Vishal Baghel

Contributor-Level 10

qE = mg

$\Rightarrow q = \frac{m g}{E} = \frac{0.1}{1000} * \frac{9.8}{4.9 * 1 0^{5}}$

q = 2 * 10^-⁹ C

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Physics Ncert Solutions Class 12th Class 12th

The displacement current of 4.425 $μ A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10⁶ Vs^-¹. The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is x * 10^-³. The value of x is,

(Permittivity of free space, $E_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2})__________$

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Payal Gupta

Contributor-Level 10

According to definition of displacement current, we can write

$I_{d} = ε_{0} \frac{d ?}{d t} = ε_{0} \frac{d (E S)}{d t} = ε_{0} \frac{d (\frac{V}{l} S)}{d t} = \frac{ε_{0} S}{l} (\frac{d V}{d t})$

$\Rightarrow l = \frac{8.85 * 1 0^{- 12} * 40 * 1 0^{- 4} * 1 0^{6}}{4.425 * 1 0^{- 6}} 8 * 1 0^{- 3} m$

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Physics Ncert Solutions Class 12th Class 12th

A potential barrier of 0.4V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 * 10⁵. The speed with which electron enters the p side will be $\frac{x}{3} * 1 0^{5} m s^{- 1}$ the value of x is__________.

(Given mass of electron = 9 * 10^-³¹ kg, charge on electron = 1.6 * 10^-¹⁹ C.)

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Payal Gupta

Contributor-Level 10

According to Work energy theorem, we can write

$K_{f} - K_{i} = W_{E l e c t r i c F o r c e} \Rightarrow \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = - e V \Rightarrow v^{2} = v_{0}^{2} - \frac{2 e V}{m}$

$\Rightarrow v^{2} = {(6.0 * 1 0^{5})}^{2} - \frac{2 * 1.6 * 1 0^{- 19}}{9 * 1 0^{- 31}} = \frac{324 - 128}{9} * 1 0^{10} = \frac{196}{9} * 1 0^{10} \Rightarrow V = \frac{14}{3} * 1 0^{5} m / s$

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Physics Ncert Solutions Class 12th Class 12th

All resistances in figure are $1 Ω$ each. The value of current 'I' is $\frac{a}{5} A .$ The value of a is___________.

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alok kumar singh

Contributor-Level 10

I = $\frac{3 * 8}{15 R} = \frac{8}{5 R} = \frac{a}{5}$

$\Rightarrow \frac{8}{5 * 1} = \frac{a}{5}$

a = 8

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Physics Ncert Solutions Class 12th Class 12th

In the given circuit, the magnitude of V_L and V_C are twice that of V_R. Given that f = 50 Hz. the inductance of the coil is $\frac{1}{K π} m H .$ The value of K is_________.

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alok kumar singh

Contributor-Level 10

Given

V_L = V_C= 2V_R, f = 50Hz

$L = \frac{1}{k π} m H$

since, V_L = V_C.

then

$V_{n e t} = \sqrt[]{{(V_{L} - V_{C})}^{2} + {(V_{R})}^{2}}$

V_R = V_Net= 220V

I = $\frac{V_{N e t}}{Z} = \frac{220}{\sqrt[]{{(X_{L} - X_{C})}^{2} + 1 2^{2}}}$

$I = 44 A$

$V_{L} = I x_{L} = 2 v_{R}$

= I _xL = 440

x_L = 10

WL = 10

2πfL = 10

$L = \frac{1}{\frac{1}{100} π} k = \frac{1}{100} = 0.01 \approx 0$

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Physics Ncert Solutions Class 12th Class 12th

A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22mm, 1.23mm, 1.19mm and 1.20mm. The percentage error is $\frac{x}{121} % .$ The value of x is________.

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alok kumar singh

Contributor-Level 10

Given x₁ = 1.22mm

x₂ = 1.23mm

x₃ = 1.19mm

& x₄ = 1.20mm

$x_{m e a n} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = \frac{1.22 + 1.19 + 1.20}{4} = 1.21$

$| Δ x_{1} | = | x_{m e a n} - x_{1} | = | 1.21 - 1.22 | = 0.01$

$| Δ x_{2} | = | x_{m e a n} - x_{2} | = | 1.2 | - | 1.23 | = 0.02$

$| Δ x_{3} | = | x_{m e a n} - x_{3} | = | 1.21 - 1.19 | = 0.02$

$| Δ_{4} | = | x_{m e a n} - x_{4} | = | 1.21 - 1.20 | = 0.01$

${(Δ x)}_{m e a n} = \frac{0.01 + 0.02 + 0.02 + 0.01}{4}$

$= \frac{0.06}{4}$

$% e r r o r = \frac{Δ x_{m e a n}}{x_{m e a n}} = \frac{0.06}{4 * 1.21} * 100$

$= \frac{600}{4 * 121}$

$= \frac{150}{121} %$

x = 150

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Physics Ncert Solutions Class 12th Class 12th

Amplitude modulated wave is represented by V_AM= $10 [1 + 0.4 c o s (2 π * 1 0^{4} t)]$ $c o s (2 π * 1 0^{7} t) .$ Total bandwidth of the amplitude modulated wave is:

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alok kumar singh

Contributor-Level 10

$V_{A M} = 10 [1 + 0.4 c o s (2 π * 1 0^{4} t)] c o s (2 π * 1 0^{7} t)$

Main wave frequency (carrier frequency)

Bandwidth = 2 fm

= $2 * 1 0^{4} H_{2} = 20 * 1 0^{3} H_{2} = 20 K H_{2}$

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Physics Ncert Solutions Class 12th Class 12th

In the given circuit the input voltage V_in is shown in figure. The cut-in voltage of p-n junction diode (D₁ or D₂) is 0.6 V. Which of the following output voltage (V₀) waveform across the diode is correct?