Tags Class 12th

Class 12th

Get insights from 12k 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

12k

Questions

Discussions

Active Users

Followers

New answer posted

3 months ago

Class 12th Chemistry

The ${(\frac{\partial E}{\partial T})}_{P}$ of different types of half cells are as follows:

A	B	C	D
*1 10^-⁴**	*2 10^-⁴**	*0.1 10^-⁴**	*0.2 10^-⁴**

(Where E is the electromotive force )

Which of the above half cells would be preferred to be used as reference electrode?

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

$\frac{d E}{d T}$ is the temperature Co-efficient of cell. The cell having less variation of EMF, with respect to temperature have high efficiency.

0 0

New answer posted

3 months ago

Class 12th Electromagnetic Waves

The displacement current of 4.425 $μ A$ is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10⁶ Vs^-1. The area of each plate of the capacitor is 40 cm². The distance between each plate of the capacitor is x * 10^-3. The value of x is,

(Permittivity of free space, $E_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2})__________$

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

According to definition of displacement current, we can write

$I_{d} = ε_{0} \frac{d ?}{d t} = ε_{0} \frac{d (E S)}{d t} = ε_{0} \frac{d (\frac{V}{l} S)}{d t} = \frac{ε_{0} S}{l} (\frac{d V}{d t})$

$\Rightarrow l = \frac{8.85 * 1 0^{- 12} * 40 * 1 0^{- 4} * 1 0^{6}}{4.425 * 1 0^{- 6}} 8 * 1 0^{- 3} m$

0 0

New answer posted

3 months ago

Class 12th Physics

A potential barrier of 0.4V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 * 10⁵. The speed with which electron enters the p side will be $\frac{x}{3} * 1 0^{5} m s^{- 1}$ the value of x is__________.

(Given mass of electron = 9 * 10^-31 kg, charge on electron = 1.6 * 10^-19 C.)

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

According to Work energy theorem, we can write

$K_{f} - K_{i} = W_{E l e c t r i c F o r c e} \Rightarrow \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = - e V \Rightarrow v^{2} = v_{0}^{2} - \frac{2 e V}{m}$

$\Rightarrow v^{2} = {(6.0 * 1 0^{5})}^{2} - \frac{2 * 1.6 * 1 0^{- 19}}{9 * 1 0^{- 31}} = \frac{324 - 128}{9} * 1 0^{10} = \frac{196}{9} * 1 0^{10} \Rightarrow V = \frac{14}{3} * 1 0^{5} m / s$

0 0

New answer posted

3 months ago

Class 12th Physics

Two resistors are connected in series across a battery as shown in figure. If a voltmeter of resistance 2000 $Ω$ is used to measure the potential difference across 500 $Ω$ resister, the reading of the voltmeter will be_________ V.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

According to Kirchhoff's Law, we can write

-20 + 2000I + 600 * 5I = 0 $\Rightarrow I = \frac{20}{5000} A$

Reading of voltmeter = 2000I = 2000 * $\frac{20}{5000} = 8 v o l t$

0 0

New answer posted

3 months ago

Class 12th Waves

In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning form of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms^-1. The third resonance is observed when the air column has a length of ____________cm.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

According to Concept of resonance tube, we can write

$\frac{λ}{4} + e = l_{1} a n d \frac{3 λ}{4} + e = l_{2} \Rightarrow \frac{λ}{2} = l_{2} - l_{1} \Rightarrow \frac{V}{2 ν} = l_{2} - l_{1}$

$\Rightarrow l_{2} = \frac{v}{2 ν} + l_{1} = \frac{336}{400} + 0.20 = 0.84 + 0.20 = 1.04 m = 104 c m$

0 0

New answer posted

3 months ago

Class 12th Chemistry Solutions

A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$M o l a l i t y = \frac{m o l e s o f s o l u t e * 1000}{w t o f s o l v e n t (g m)}$

$M o l a l i t y = \frac{35 * 1000 * 1.46}{36.5 * 100}$ = 14.0 M

0 0

New answer posted

3 months ago

Class 12th Physics

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2 Ω .$ The value of internal resistance of each cell is

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

According to question, we can write

$\frac{2 V}{2 + 2 r} = \frac{V}{2 + \frac{r}{2}}$

$\Rightarrow 2 + 2 r = 4 + r \Rightarrow r = 2 Ω$

0 0

New answer posted

3 months ago

Class 12th Physics

If the electric potential at any point (x, y, z)m in space is given by V = 3x² volt. The electric field at the point (1, 0, 3)m will be:

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

According to relation between field and potential, we can write

^{$\vec{E} = - \frac{d V}{d x} (\hat{i}) = - \frac{d (3 x^{2})}{d x} (\hat{i}) = - 6 x (\hat{i})$}

^{${\vec{E}}_{(1, 0, 3)} = - 6 (\hat{i}) N / C$}

0 0

New answer posted

3 months ago

Class 12th Physics Electric Charge and Field

Given below two statements : One is labeled as Assertion (A) and other is labeled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negativne charge distribution .

In the light of above statements, choose the most appropriate answer from the options given below:

0 Follower 1 View

Raj Pandey

Contributor-Level 9

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

0 0

New answer posted

3 months ago

Class 12th Physics

Two capacitors having C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is not connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is:

0 Follower 3 Views

Raj Pandey

Contributor-Level 9

After switch 'S' is closed

$Q_{1} + Q_{2} = C_{1} V$ - (1)

Using KVL

$\frac{Q_{1}}{C_{1}} - \frac{Q_{2}}{C_{2}} = 0$

$Q_{1} = Q_{2} \frac{C_{1}}{C_{2}}$ - (2)

from (1) & (2)

$\Rightarrow Q_{2} [\frac{C_{1} + C_{2}}{C_{2}}] = C_{1} V \Rightarrow Q_{2} = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V$

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
681k Reviews
1800k 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