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6 months ago

Physics Class 12th

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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6 months ago

Physics 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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New answer posted

6 months ago

Physics 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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New answer posted

6 months ago

Physics 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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New answer posted

6 months ago

Physics Units and Measurement

The speed of light in media 'A' and 'B' are 2.0 * 10¹⁰ cm/s and 1.5 * 10¹⁰ cm/s respectively. A ray of light enters from the medium B to an incident angle 'q'. If the ray suffers total internal reflection, then

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alok kumar singh

Contributor-Level 10

$\frac{s i n θ_{C}}{V_{B}} = \frac{s i n 90 °}{V_{A}} \Rightarrow s i n θ_{C} = \frac{V_{B}}{V_{A}} = \frac{1.5 * 1 0^{10}}{2.0 * 1 0^{10}} = \frac{3}{4}$

According to question, we can write

$θ > θ = s i n^{- 1} (\frac{3}{4})$

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New answer posted

6 months ago

Physics Class 12th

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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6 months ago

Physics 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?

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alok kumar singh

Contributor-Level 10

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New answer posted

6 months ago

Physics 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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6 months ago

Physics Physics Nuclei

Following statements related to radioactivity are given below:

(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.

(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.

(C) Slope of the graph of log_e (no. of undecayed nuclei) Vs, time represents the reciprocal of mean life time $(τ) .$

(D) Product of decay constant $(λ)$ and half-life time $(T_{1 / 2})$

...more

Following statements related to radioactivity are given below:

(A) Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.

(B) The number of un-decayed nuclei in the radioactive sample decays exponentially with time.

(C) Slope of the graph of log_e (no. of undecayed nuclei) Vs, time represents the reciprocal of mean life time $(τ) .$

(D) Product of decay constant $(λ)$ and half-life time $(T_{1 / 2})$ is not constant.

less

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alok kumar singh

Contributor-Level 10

We know, N = $N_{0} e^{- λ t}$

Where, N Number of un-decayed Nuclei

No Initial No. Of Nuclei

N $e^{- λ t}$

$N = N_{0} {\bar{e}}^{λ t}$

Taking log both side log N = log N0 + loge $e^{- λ t}$

log N = log No $λ t$

Slope = $λ = \frac{1}{t_{a v}}$

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New answer posted

6 months ago

Physics Class 12th

A small toy starts moving from the position of rest under a constant acceleration. If it travels a distance of 10m in t s, the distance travelled by the toy in the next t s will be:

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alok kumar singh

Contributor-Level 10

According to question, we can write

10 = $\frac{1}{2} a t^{2} . . . . . . . . . . . . . . . (1) a n d$

$10 + x = \frac{1}{2} a {(2 t)}^{2} . . . . . . . . . . . . . . . . (2)$

$\Rightarrow X = \frac{1}{2} A [{(2 t)}^{2} - t^{2}] = 3 (\frac{1}{2} a t^{2}) = 30 m$

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