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Physics NCERT Exemplar Solutions Class 12th Chapter Five

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics Thermal Properties of Matter

As per the given figure, two plates A and B of thermal conductivity K and 2K are joined together of form a compound plate. The thickness of plates are 4.0cm and 2.5cm respectively and the area the area of cross-section is 120 cm² for each plate. The equivalent thermal is 120 cm² for each plate. The equivalent thermal conductivity of the compound plat is $(1 + \frac{5}{α}) K,$ then the value of a will be________.

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alok kumar singh

Contributor-Level 10

k_{eq=
$\frac{L_{1} + L_{2}}{\frac{L_{1}}{k_{1}} + \frac{L_{2}}{k_{2}}} = \frac{4 + 2.5}{\frac{4}{k} + \frac{2.5}{2 k}}$}

_{keq=
$\frac{6.5}{\frac{8 + 2.5}{2 k}} = \frac{6.5}{10.5} * 2 k = \frac{13 * 2}{21} k$}

_{=
$(1 + \frac{5}{51}) k$}

_{=
$(1 + \frac{5}{α})$
=21}

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Electromagnetic Waves

The intensity of the light from a bulb incident on a surface is 0.22 W/m². The amplitude of the magnetic field in this light – wave is__________ * 10^-9 T.

(Given : Permittivity of vacuum $\in_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} - m^{- 2},$ speed of light in vacuum c = 3 * 10⁸ ms^-1)

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alok kumar singh

Contributor-Level 10

$\frac{B_{0}^{2}}{2 μ_{0}} C$

$B_{0}^{2} = \frac{I 2 μ_{0}}{C}$

$B_{0}^{2} = \frac{0.22 * 2 * 4 * 1 0^{- 7}}{9 * 108}$

B₀ = 42.9 * 10^-9

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2 months ago

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

Two coil require 20 minutes and 60 minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be_________ min.

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alok kumar singh

Contributor-Level 10

$H_{1} = \frac{v^{2}}{R_{1}} t_{1} = \frac{v^{2}}{R_{1}} * 20$

H2 $= \frac{v^{2}}{R_{2}} * t_{2} = \frac{v^{2}}{R_{2}} * 60$

from question,

H1 = H2

$\frac{v^{2}}{R_{1}} * 20 = \frac{v^{2}}{R_{2}} * 60$

$\frac{R_{2}}{R_{1}} = 3$

Now connected is parallel

Req = $\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

H = $(\frac{v^{2}}{R_{1}} + \frac{v^{2}}{R_{2}}) t'$

H = $v^{2} [\frac{R_{1} + R_{2}}{R_{1} R_{2}}] t'$

Now, According to question

H = H1 + H2

$v^{2} (\frac{R_{1} + R_{2}}{R_{1} R_{2}}) t' = \frac{v^{2}}{R_{1}} * 20$

$t' [1 + \frac{\frac{R_{2}}{R_{1}}}{R_{2}}] = \frac{20}{R_{1}}$

$t' [1 + 3] = 20 * 3$

$t' = \frac{60}{4} = 15 m i n$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Communication Systems

Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an signal requires a bandwith of 8 kHz, how many channels can be accommodated for transmission:

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alok kumar singh

Contributor-Level 10

$v = f λ$

$f = \frac{v}{λ} = \frac{3 * 1 0^{8}}{1 0^{3} * 1 0^{- 9}} = 3 * 1 0^{14} H_{z}$

Channels = $\frac{2 % o f 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{2 % o f 3 * 1 0^{14} H_{z}}{8 * 1 0^{3}}$

$\frac{= \frac{2 x}{100} 3 * 1 0^{14}}{8 * 1 0^{3}} = 75 * 1 0^{7}$

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2 months ago

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

Using Young's double slit experiment, a monochromatic light of wavelength 5000 $\overset{o}{A}$ produces fringes of fringe width 0.5mm. If another monochromatic light of wavelength 6000 $\overset{o}{A}$ is used and the separation between the slits is doubled, then the new fringe width will be:

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alok kumar singh

Contributor-Level 10

$β_{1} = \frac{λ_{1} D_{1}}{d_{1}}$

$β_{2} = \frac{λ_{2} D_{2}}{d_{2}}$

$\frac{β_{1}}{β_{2}} = \frac{λ_{1}}{d_{1}} * \frac{d_{2}}{λ_{2}}$

$\frac{0.5}{β_{2}} = \frac{5000 * d_{1}}{d_{1} * 600}$

$β_{2} = \frac{0.5 * 6}{10}$

2 = 0.3mm

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics Electrostatic Potential and Capacitance

A positive charge particle of 100mg is thrown in opposite direction to a uniform electric field of strength 1 * 10⁵ NC^-1, If the charge on the particle is $40 μ C$ and the initially velocity is 200 ms^-1, how much distance it will travel before coming to the rest momentarily:

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alok kumar singh

Contributor-Level 10

a = $\frac{q E}{m} = \frac{40 * 1 0^{? 6} * 1 0^{5}}{100 * 1 0^{? 3} * 1 0^{? 3}}$

= 40 * 10³ m/sec²

v² = u² + 2a s

$? 0 = {(200)}^{2} ? 2 * 40 * 1 0^{3} s$

$s = \frac{200 * 200}{2 * 40 * 1 0^{3}} = 0.5 m$

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Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics Electrostatic Potential and Capacitance

A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will:

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alok kumar singh

Contributor-Level 10

UI= $\frac{1}{2} c v^{2}$

$U_{i} = \frac{1}{2} \frac{\in_{0} A}{d} * k v^{2}$

$U_{i} = \frac{1}{2} \frac{\in_{0} A}{d} * 10 v^{2}$

$U_{f} = \frac{1}{2} \frac{\in_{0} A}{d} * 15 v^{2}$

$\frac{U_{f}}{U_{i}} = \frac{3}{2}$

$\frac{U_{f} - U_{i}}{U_{i}} * 100 = (\frac{3}{4} - 1) * 100$

$\frac{1}{2} * 100$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Waves

A longitudinal wave is represented by x = 10 sin 2p $(n t - \frac{x}{λ}) c m .$ The maximum particle velocity will be four times the wave velocity if the determined value of wavelength is equal to:

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alok kumar singh

Contributor-Level 10

x = 10 $2 ? (n t ? \frac{x}{?}) c m$

V (wave velocity) = $\frac{? ?}{2 ?}$

v_max = 10 * 2p n

10 * 2pn = 4 * $\frac{? ?}{2 ?}$

10 * 2pn = $4 \frac{2 ? ? n ?}{2 ?}$

x = 5?

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Magnetism and Matter

At a certain place the angle of dip is 30° and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at the certain place, is:

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alok kumar singh

Contributor-Level 10

$B c o s δ = B_{H}$

B cos 30° = 0.5

$B = \frac{0 . * 2}{\sqrt[]{3}} = \frac{1}{\sqrt[]{3}}$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Five Laws of Motion

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-¹ and 40 ms-¹ respectively. The velocity of the third piece will be:

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alok kumar singh

Contributor-Level 10

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

&nb

...more

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

M * 0 = M₂v_y + M₃v'_y

v_y = $\frac{- M_{2}}{M_{3}} v_{y}$

v_y = $\frac{- 1}{2} * 40$

v_y = -20m/sec

^{$\vec{v} = - 15 \hat{i} - 20 \hat{j}$}

v = $\sqrt[]{1 5^{2} + 2 0^{2}} = \sqrt[]{225 + 400} = 25 m / s e c$

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