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Class 12th Physics

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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Class 12th Physics

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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a year ago

Class 12th Physics Moving Charges and Magnetism

In a Young's double slit experiment, an angular width of the fringe is 0.53° on a screen place at 2m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is $\frac{1}{α} .$ The value of a is__________.

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alok kumar singh

Contributor-Level 10

we know, = $\frac{λ}{d}$ $μ = \frac{C}{V} = \frac{C}{υ λ}$

$λ_{1} = \frac{C}{μ_{1} υ_{1}}$ [with change in medium frequency refnain constaint]

$λ_{2} = \frac{C}{μ_{2} υ_{2}}$

$\frac{λ_{2}}{λ_{1}} = \frac{μ_{1}}{μ_{2}} = \frac{5}{7}$ $λ_{2} = \frac{5}{7} λ_{1}$

$θ_{1} = \frac{λ_{1}}{d_{1}}$

2 = $\frac{λ_{2}}{d_{2}}$

$\frac{θ_{2}}{θ_{1}} = \frac{λ_{2}}{λ_{1}} [? d_{1} = d_{2}]$

2 $= \frac{5}{7} * θ_{1}$

$θ_{2} = \frac{5}{7} * 0.35$

$θ_{2} = \frac{1}{4} = \frac{1}{α}$

= 4

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a year ago

Class 12th Physics

In the following unclear reaction,

D $\overset{α}{\to} D_{1} \overset{β^{-}}{\to} D_{2} \overset{α}{\to} D_{3} \overset{γ}{\to} D_{4}$

Mass number of D is 182 and atomic number is 74. Mass number and atomic number of D₄ respectively will be________

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alok kumar singh

Contributor-Level 10

Please find the solution below:

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a year ago

Class 12th Current Electricity Analysis

A zener of breakdown voltage V_Z = 8V and maximum zener current, I_ZM = 10mA is subjected to an input voltage V_i = 10V with series resistance R = 100 $Ω$ In the given circuit R_L represents the variable load resistance. The ratio maximum and minimum value of R_L is__________.

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alok kumar singh

Contributor-Level 10

Across zener diode & RL

8 = I_LR_L

8 = $(20 - 10) R_{L (m a x)}$

$R_{L m a x} = \frac{8}{10} k Ω$

At max current in loop (1)

10 = i_max R + 8

i_max= $\frac{2}{R} = \frac{2}{100} = 0.02 A$

= 20 mA

R_L (max) = $\frac{8}{10} k Ω$

At minimum curre3nt through zener

i_max R_L (minimum) = 8

R_L (minimum) = $\frac{8}{20} k Ω$

$\frac{R_{L m a x}}{R_{L m i n}} = 2$

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a year ago

Class 12th Physics

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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Class 12th Physics

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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New answer posted

a year ago

Class 12th Physics

Two point charge Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is:

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alok kumar singh

Contributor-Level 10

According to question, we can write

$F_{C A} = F_{C B} = \frac{K q Q}{x^{2} + {(\frac{d}{2})}^{2}} \Rightarrow F = 2 F_{C A} c o s θ = \frac{2 K q Q x}{{[x^{2} + {(\frac{d}{2})}^{2}]}^{\frac{3}{2}}}$

For maxima of force

$\frac{d F}{d x} = 0, s o$

$x = \frac{d}{2 \sqrt[]{2}}$

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New answer posted

a year ago

Class 12th Physics

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?

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alok kumar singh

Contributor-Level 10

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New answer posted

a year ago

Class 12th Physics Moving Charges and Magnetism

At what temperature a gold ring of diameter 6.230cm be heated so that it can be fitted on a wooden bangle of diameter. 6.241cm? Both the diameters have been measured at room temperature (27°C).

(Given : coefficient of linear thermal expansion of gold a_L = 1.4 * 10^-⁵ K^-¹)

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alok kumar singh

Contributor-Level 10

According to question, we can write

$d = d_{0} (1 + α Δ T) \Rightarrow 6.241 = 6.230 (1 + 1.4 * 1 0^{- 5} * Δ T)$

$\Rightarrow T = 126.18 + 27 = 153.18 ° C$

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