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Physics Ray Optics and Optical Instruments

A parallel beam of light is allowed to afall on a transparent spherical globe of diameter 30cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is_________ mm.

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alok kumar singh

Contributor-Level 10

When object is at infinity

u = $\infty, μ_{1} = 1, R = 15 c m$

v = ? $μ_{2} = 1.5$

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1}{\infty} + \frac{1.5}{v} =$ $\frac{1.5 - 1}{15}$

$\frac{1.5}{v} =$ $\frac{0.5}{15}$

v =

45cm

Now for Refraction through C2

u = + 15cm

v = ?

$μ_{1} = 1.5$

$μ_{2} = 1$

R = 15

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1.5}{15} + \frac{1}{v} = \frac{1 - 1.5}{- 15}$

$\frac{- 1}{10} + \frac{1}{v} = \frac{1}{30}$

$\frac{1}{v} = \frac{1}{30} + \frac{1}{10} = \frac{4}{10}$

$v = \frac{30}{4} = + 7.5 c m$

from centre = 15 + 7.5 = 22.5 cm = 225mm

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Physics Semiconductor Electronics

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100 $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$ and input resistance is 1 $k Ω,$ the value of power gain is x * 10⁴. The value of x is_____________.

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alok kumar singh

Contributor-Level 10

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} * \frac{R_{0}}{R_{i}}$

= ${(\frac{10 * 1 0^{- 3}}{100 * 1 0^{- 6}})}^{2} * \frac{2}{1}$

= 2 * 104 = x * 104

= 2

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Physics Physics Nuclei

$\sqrt[]{d_{1}} a n d \sqrt[]{d_{2}}$ are the impact parameters corresponding to scattering angles 60° and 90° respectively, when an a particle is approaching a gold nucleus. For d₁ = x d₂, the value of x will be________.

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alok kumar singh

Contributor-Level 10

$\sqrt[]{d}$ $c o t \frac{θ}{2}$

cot² 30° = x cot² 45°

x = 3

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Physics Thermodynamics

300 cal. Of heat is given to a heat engine and it rejects 225 cal. of heat. If source temperature is 227°C, then the temperature of sink will be__________°C.

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alok kumar singh

Contributor-Level 10

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

...more

T₁ = 227°C

= 500k

T₂ =?

$\frac{Q_{1}}{T_{1}} = \frac{Q_{2}}{T_{2}}$

$\frac{300}{500} = \frac{225}{T_{2}}$

$T_{2} = \frac{500 * 225}{300}$

= 5 * 75

T₂ = 375k

T₂ = 102°C

less

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

The variation of applied potential and flowing through a given wire is shown in figure. The length of wire is 31.4cm. The diameter of wire is measured as 2.4cm. The resistivity of the given wire is measured as x * 10^-3 $Ω c m .$ The value of x is_________.

[Take p = 3.14]

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alok kumar singh

Contributor-Level 10

R = tanq = tan 45° = $\frac{ρ L}{A}$

$? 1 = ρ * \frac{31.4}{π {(1.2)}^{2}}$

$ρ = \frac{3.14 * 1.2 * 1.2}{31.4}$

$ρ = 12 * 12 * 1 0^{- 3}$

$ρ = 144 * 1 0^{- 3} = x * 1 0^{- 3} Ω - c m$

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Physics Physics Oscillations

A body is performing simple harmonic with an amplitude of 10cm. The velocity if the body was tripled by air Jet when it is at 5cm from its mean position. The new amplitude vibration is $\sqrt[]{x} c m .$ The value of x is__________.

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alok kumar singh

Contributor-Level 10

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

at x = 5, A = 10

$v' = 3 v = 3 ω \sqrt[]{A^{2} - 5^{2}} = ω \sqrt[]{A'^{2} - 5^{2}}$

$= 3 \sqrt[]{A^{2} - 5^{2}} = \sqrt[]{A'^{2} - 5^{2}}$

$\sqrt[]{1 0^{2} - 5^{2}} = \sqrt[]{A'^{2} - 25}$

$A'^{2} = 25 + 9 * 75$ $A'^{2} = 700$

$A' = \sqrt[]{700} c m$

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Physics 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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6 months ago

Physics 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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Physics 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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Physics 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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