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

0

Discussions

49

Active Users

0

Followers

New answer posted

10 months ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

Flux as a function of time Φ = B.A = ABcos (ωt) emf induced,
e = -dΦ/dt = ABωsin (ωt)
Maximum value of emf = ABω = πR²Bω
= 3.14*0.1*0.1*3*10? * (0/0.2) = 15

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

ρ? = 98 * 10?
ρ? = 2.65 * 10?
ρc = 1.724 * 10?
ρT = 5.65 * 10?

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

ΔP = dsinθ = dθ
dy/D = (10? ³ * 1.270mm)/1m = 1.27µm

New answer posted

10 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

∫? ² |x-1|-x|dx
Let f (x)=|x-1|-x|
= {|1-2x|, x≤1; 1, x≥1}
A = 1/2+1=3/2
∫? ¹/² (1-2x)dx+∫? /? ¹ (2x-1)+∫? ²1dx
= [x-x²]? ¹/²+ [x]? ² = 3/21dx
= [x-x²]? ¹/²+ [x]? ² = 3/2

New answer posted

10 months ago

0 Follower 50 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

10 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

|a|=|b|=|c|=1
|a-b|²+|a-c|²=8
⇒|a|²+|b|²-2a.b+|a|²+|c|²-2a.c=8
⇒4-2 (a.b+a.c)=8
a.b+a.c=-2
|a+2b|²+|a+2c
=|a|²+4|b|²+4a.b+|a|²+4|c|²+4a.c
=10+4 (a.b+a.c)
=10-8=2

New answer posted

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

With weak field ligands Mn (II) will be of high spin and with strong field ligands it will be of low spin.
Ni (II) tetrahedral complexes will be generally of high spin due to sp³ hybridisation. Mn (II) is of light pink color in aqueous solution.

New answer posted

10 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Let B? be the event where Box-I is selected and B? →where box-II selected
P (B? )=P (B? )=1/2
Let E be the event where selected card is non prime.
For B? : Prime numbers: {2,3,5,7,11,13,17,19,23,29}
For B? : Prime numbers: {31,37,41,43,47}
P (E)=P (B? )*P (E/B? )+P (B? )P (E/B? )
= 1/2*20/30+1/2*15/20
Required probability:
P (B? /E) = (P (E/B? )P (B? )/P (E) = (1/2*20/30)/ (1/2*20/30+1/2*15/20) = (2/3)/ (2/3+3/4) = 8/17

New answer posted

10 months ago

0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

x > d path is straight line
-y = 1/2 at²
x-d = V?t ⇒ t=(x-d)/V?
-y = 1/2 a((x-d)/V?)²
-y/at = (1/2at)(a²t²)
at = V?
(-y/at) = (d/2V?)
(x-d)/V? = d/V?
(-y - 1/2 a t²)/(at) = (x-d - V?t)/V?
y = (-qEd/mV?)(x/V? - d/2V?) ; y = (qEd/mV?²)(d/2 - x)

 

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

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 702k Reviews
  • 1850k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.