Tags Class 12th

Class 12th

Get insights from 11.9k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question

11.9k

Questions

Discussions

Active Users

Followers

New answer posted

11 months 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________

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

Please find the solution below:

0 0

New answer posted

11 months 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__________.

0 Follower 14 Views

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) = $\frac{8}{20} k Ω$

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

0 0

New answer posted

11 months 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________.

0 Follower 6 Views

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

0 0

New answer posted

11 months ago

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:

0 Follower 2 Views

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

= $2 * 1 0^{4} H_{2} = 20 * 1 0^{3} H_{2} = 20 K H_{2}$

0 0

New answer posted

11 months 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:

0 Follower 8 Views

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

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

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

0 0

View All 2 Answers

New answer posted

11 months 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?

0 Follower 64 Views

alok kumar singh

Contributor-Level 10

0 0

New answer posted

11 months 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)

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

According to question, we can write

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

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

0 0

View All 2 Answers

New answer posted

11 months ago

Class 12th 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

0 Follower 3 Views

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 = $λ = \frac{1}{t_{a v}}$

0 0

New answer posted

11 months ago

Class 12th Physics

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:

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

According to question, we can write

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

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

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

0 0

New answer posted

11 months ago

Class 12th Dual Nature of Radiation and Matter

Let K₁and K₂ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $λ_{1} a n d λ_{2},$ respectively are incident on a metallic surface. If $λ_{1} = 3 λ_{2}$ then:

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

E Energy with which photons are incident on a metallic surface.

$λ_{1} = 3 λ_{2} \Rightarrow λ_{1} > λ_{2}$

E $α \frac{1}{λ},$ E₁ < E₂

We know E₁ = $?_{0} + k_{1}$

$\frac{h c}{λ_{1}} = ? + k_{1}$

k₁ = $\frac{h c}{λ_{1}} - ?_{0} - - - - (i)$

& E₂ = $?_{0} + k_{2}$

$\frac{h c}{λ_{2}} = ?_{0} + k_{2}$

$\frac{3 h c}{λ_{1}} = ? + k_{2}$

k₂ = $\frac{3 h c}{λ_{1}} - ?_{0}$ (i)

from (1) & (2)

$k_{z} = 3 [k_{1} + ?_{0}] - ?_{0} = 3 k_{1} + 2 ?_{0}$

k₂ > $3 k_{1} \Rightarrow \frac{k_{2}}{3} > k_{1}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
703k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience