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

a year ago

Class 12th Physics

Draw the output signal Y in the given combination of gates.

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

$y = \bar{\bar{A} + B} = A . \bar{B}$

0 0

New answer posted

a year ago

Class 12th Electric Flux

Given below are two statements:

Statement I : An electric dipole is placed at the centre of a hollow sphere. The flux of electric field through the sphere is zero but the electric field is not zero anywhere in the sphere.

Statement II : If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (

In the light of the above statements, choose the correct answer from the option given below:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

All the charge given to a conducting sphere resides on outer surface.

0 0

New answer posted

a year ago

Class 12th Maths

If $e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + . . . \infty) l o g_{e} 2}$ satisfies the equation t² – 9t + 8 = 0, then the value of $\frac{2 s i n x}{s i n x + \sqrt[]{3} c o s x} (0 < x < \frac{π}{2}) i s$

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

$e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + . . . . . \infty)} l o g_{e} 2$

$= e^{\frac{c o s^{2} x}{1 - c o s^{2} x} l o g_{e} 2} = e^{c o t^{2} x l o g_{e} 2} = 2^{c o t^{2} x}$

t² – 9t + 8 = 0

(t – 8) (t – 1) = 0

$t = 2^{c o t^{2} x} = 8 = 2^{3}$

$\Rightarrow c o t^{2} x = 3 = c o t^{2} \frac{π}{6}$

$\frac{2 s i n x}{s i n x + \sqrt[]{3} c o s x} = \frac{2 * \frac{1}{2}}{\frac{1}{2} + \sqrt[]{3} * \frac{\sqrt[]{3}}{2}} = \frac{1}{2}$

0 0

New answer posted

a year ago

Class 12th Atoms

The recoil speed of a hydrogen atom after it emits a photon in going from n = 5 state to n = 1 state will be:

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{5^{2}}) = 13.6 * \frac{24}{25} e V$

$\Rightarrow \frac{h c}{λ} = 13.6 * \frac{24}{25} e V . . . . . . . . . . (1)$

With the help of conservation of linear momentum, we can write

$\frac{h}{λ} = m_{H} v_{H} \Rightarrow \frac{h c}{λ} = c m_{H} v_{H} \Rightarrow v_{H} = \frac{\frac{h c}{λ}}{c m_{H}} = \frac{13.6 * \frac{24}{25} * 1.6 * 1 0^{- 19}}{3 * 1 0^{8} * 1.67 * 1 0^{- 27}} = 4.17 m / s$

0 0

New answer posted

a year ago

Class 12th Physics

A wire of $1 Ω$ has a length of 1m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is:

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

$R_{i} = \frac{ρ l}{A}$

$R_{f} = \frac{ρ (1.25 l)}{(A / 1.25)} = {(1.25)}^{2} * \frac{ρ l}{A}$

$\Rightarrow R_{f} = 1.5625 * R_{i}$

$\Rightarrow \frac{R_{f} - R_{l}}{R_{l}} * 100 = 56.25 %$

0 0

New question posted

a year ago

Class 12th Physics Ray Optics and Optical Instruments

Given below are two statements : one is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image.

Reason R : Magnification is achieved as the small object can be kept much closer to the eye that 25cm and hence it subtends a large angle.

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

0 Follower 3 Views

New answer posted

a year ago

Class 12th Physics Electromagnetic Waves

A radiation is emitted by 1000W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2m.Th efficiency of the bulb is 1.25%. The value of peak electric field at P is x * 10^-1 V/m. Value of x is----------. (Rounded – off to the nearest integer)

[Take $ε_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} m^{- 2}, c = 3 * 1 0^{8} m s^{- 1}]$

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

$l = \frac{1}{2} c ε_{0} E_{0}^{2} = η * (\frac{P}{4 π r^{2}})$ Here $η$ is the efficiency

$E_{0} = \sqrt[]{\frac{η P}{2 π r^{2} c ε_{0}}} = \sqrt[]{\frac{1.25}{100} * \frac{1000}{2 * 3.14 * 4 * 3 * 1 0^{8} * 8.85 * 1 0^{- 12}}}$

$\Rightarrow E_{0} = \sqrt[]{\frac{12.5}{8 * 3.14 * 3 * 8.85 * 1 0^{- 4}}} = 13.69 V m = 136.9 * 1 0^{- 1} V / m$

0 0

New answer posted

a year ago

Class 12th Maths

If f : R ->R is a function defined by f(x) = $[x - 1] c o s (\frac{2 x - 1}{2}) π,$ where[.] denotes the greatest integer function, then f is:

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

$f (x) = [x - 1] c o s (\frac{2 x -}{2}) π$

$I f x = k, k \in I$

then f(x) = 0 as $c o s (\frac{2 k - 1}{2}) π = 0, \forall k \in I$

LHL = $\underset{h \to 0}{L t} [k - h - 1] c o s (\frac{2 k - 2 h - 1}{2}) π = \underset{h \to 0}{L t} (k - 2) c o s (\frac{2 k - 1}{2}) π \to 0$

$R H L = \underset{h \to 0}{L t} [k + h - 1] c o s (\frac{2 k + 2 h - 1}{2}) π = \underset{h \to 0}{L t} (k - 1) c o s (\frac{2 k - 1}{2}) π \to 0$

$∴ f (x)$ is continuous $\forall x \in R$

0 0

New answer posted

a year ago

Class 12th Electromagnetic Induction

An aeroplane, with its wings spread 10m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 * 10^-⁴Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be………….

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Bv = B sin 60°

$B_{v} = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} * v * l = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2} * 180 * \frac{5}{18} * 1 = 108.25 * 1 0^{- 3} v o l t s$

0 0

New answer posted

a year ago

Class 12th Physics Alternating Current

In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be--------.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

As we know that Q = $\frac{ω L}{R}$

$\Rightarrow \frac{Q'}{Q} = \frac{L'}{R'} * \frac{R}{L} = (\frac{L'}{L}) * (\frac{R}{R'}) = 2 * 2 \Rightarrow Q' = 4 Q = 4 * 100 = 400$

0 0

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

On Shiksha, get access to

67k Colleges
1.2k Exams
717k 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