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Physics NCERT Exemplar Solutions Class 12th Chapter Four

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Physics Alternating Current

An inductor of 0.5 mH, a capacitor of 200 $μ F$ and a resistor of 2 $Ω$ are connected in series with a 220V ac source. If the current is in phase with the emf, the frequency of ac source will be__________ * 10² Hz.

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Payal Gupta

Contributor-Level 10

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2 π \sqrt[]{L C}} = \frac{1}{2 π \sqrt[]{(0.5 * 1 0^{- 3}) * (200 * 1 0^{- 6})}}$

$\Rightarrow f = \frac{1 0^{4}}{2 π \sqrt[]{10}} \approx 5 * 1 0^{2} H z$ $[T a k i n g π = \sqrt[]{10}]$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Physics Wave Optics

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by 5 * 10^-²m towards the slits, the change in fringe width is 3 * 10^-³cm. If the distance between the slits is 1mm, then the wavelength of the light will be_____________ nm.

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Payal Gupta

Contributor-Level 10

According to Young's double slit experiment, we can write

$β = \frac{λ D}{d} \Rightarrow Δ β = β_{2} - β_{1} = \frac{λ}{d} Δ D$

$\Rightarrow λ = \frac{d Δ β}{Δ D} = \frac{1 * 1 0^{- 3} * 3 * 1 0^{- 5}}{5 * 1 0^{- 2}} = 60 * 1 0^{- 8} m = 600 n m$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Physics Nuclei

The half life of a radioactive substance is 5 years. After x years a given sample of the radioactive substance gets reduced to 6/25% of its initial value. The value of x is___________.

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Payal Gupta

Contributor-Level 10

According to Nuclear activity, we can write

$N_{0} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{2} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{4} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{8} \overset{t_{\frac{1}{2}}}{\to} \frac{N_{0}}{16} = (0.0625) N_{0}$

Time required = 4 * $t_{\frac{1}{2}} = 20 y r s$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Class 11th

The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I₁. The same rod is into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ then the value of x will be__________.

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Payal Gupta

Contributor-Level 10

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} * \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four 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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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four 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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3 months ago

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

Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $Ω$ is used to measure the potential difference across 500 $Ω$ resister, the reading of the voltmeter will be_________ V.

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Payal Gupta

Contributor-Level 10

According to Kirchhoff's Law, we can write

20 + 2000I + 600 * 5I = 0 $\Rightarrow I = \frac{20}{5000} A$

Reading of voltmeter = 2000I = 2000 * $\frac{20}{5000} = 8 v o l t$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Physics Waves

In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning form of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms^-¹. The third resonance is observed when the air column has a length of ____________cm.

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Payal Gupta

Contributor-Level 10

According to Concept of resonance tube, we can write

$\frac{λ}{4} + e = l_{1} a n d \frac{3 λ}{4} + e = l_{2} \Rightarrow \frac{λ}{2} = l_{2} - l_{1} \Rightarrow \frac{V}{2 ν} = l_{2} - l_{1}$

$\Rightarrow l_{2} = \frac{v}{2 ν} + l_{1} = \frac{336}{400} + 0.20 = 0.84 + 0.20 = 1.04 m = 104 c m$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Physics Mechanical Properties of Fluids

A small spherical ball of radius 0.1mm and density 10⁴kgm^-³ falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity inside water, then the value of h will be_________ m.

(Given g = 10 ms^-², viscosity of water = 1.0 * 10^-⁵ N-sm^-²).

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Payal Gupta

Contributor-Level 10

mg = F_B + F_v $\Rightarrow \frac{4 π}{3} r^{3} ρ g = \frac{4 π}{3} r^{3} σ g + 6 π η r v$

$\Rightarrow v = \frac{2 r^{2} (ρ - σ) g}{9 η} = \frac{2 * 0.1 * 0.1 * 1 0^{- 6} * (1 0^{4} - 1 0^{3}) * 10}{9 * 1.0 * 1 0^{- 5}}$

$\Rightarrow h = \frac{400}{2 g} = 20 m$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Four Units and Measurement

The Vernier constant of Vernier calipers is 0.1 mm and it has zero error of (-0.05) cm. White measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The correct diameter will be__________ * 10^-2 cm.

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Payal Gupta

Contributor-Level 10

Least count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 102 cm.

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