Ask & Answer: India's Largest Education Community

1000+ExpertsQuick ResponsesReliable Answers

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post




Ok

All Questions

New Question

9 months ago

0 Follower 2 Views

New Question

9 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Structure of Bithinol is;

New Question

9 months ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

Total 3 streo- isomers

New Question

9 months ago

0 Follower 18 Views

P
Payal Gupta

Contributor-Level 10

Mole of polyhydric alcohol = 1.84×10392=2.0×105mole.

Mole of H2 gas produced = 1.34422400=6.0×105mole.

No of OH gp present = 6.0×1052.0×105=3

New Question

9 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

Only H2S2O8 has per-oxo bond.

New Question

9 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Oxidation state of Co = 3

And co-ordination No = 6

Sum = 3 + 6 = 9

New Question

9 months ago

0 Follower 14 Views

P
Payal Gupta

Contributor-Level 10

λt=lnNoNt

0.693t1/2×t=ln1Nt

Or 0.693×10030=ln1Nt

1Nt=10

Nt=110=0.1

Or Nt = 1 × 101 μg

New Question

9 months ago

0 Follower 3 Views

New Question

9 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

ΔTf=Kf×m

(156155.1)=2×1.8MW×1000 (62.5×0.8)

0.9 = 2×1.8×1000MW×50

MW=80gm/mol

New Question

9 months ago

0 Follower 15 Views

P
Payal Gupta

Contributor-Level 10

Following have the sp3d2 hybridisation

New Question

9 months ago

0 Follower 3 Views

A
abhishek gaurav

Contributor-Level 9

Candidates need to take entrance for these courses,

New Question

9 months ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

Po2 = PTotalPN2

= 25 – 20 = 5bar

Po2nO2nO2+nNe×PTotal

5= (x/32x/32+20020)×25x=80gm

New Question

9 months ago

0 Follower 2 Views

A
abhishek gaurav

Contributor-Level 9

New Question

9 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

C = 500 μF,  V = 100 v, L = 50 mH

In this LC – oscillation

q = q0 cos ωt

i=dqdt=q0ωsinωtω=12c=150×103×5×104

10005=200

So, imaxq0ω=500×106×100×200

= 10A

New Question

9 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

λA=25λ,  λB=16λ

At t = 0 NA = NB = N0

after t = 1aλ:NBNA=N0e16λtN0e25λt=e (25λ16λ)t

e = e (9λ1aλ)

9a=1a=9

New Question

9 months ago

0 Follower 2 Views

A
abhishek gaurav

Contributor-Level 9

At the time of admission in Biyani Girls B.Ed. College the candidates are required to upload their caste certificate (in case of SC/ST), copies of marksheet of Class 10, 12, graduation and a formal passport size photograph with a white background.

New Question

9 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

 i (zenermax)=25mA

20 – imax R – 8 = 0

imax R = 12

At minimum zener current  (μA):

20iminRiminRL=0

RRL=128=32

lminR=12

iminRL=8

At maxm zener current –

20imaxR8=0

iL=O {asizmaxm=25mA}

imaxR = 12v

25 × 103 R = 12

R=12×10325=12×40=480Ω

New Question

9 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

geff = g – (ρwρb)g

T=2πlg

T'=2πlgeff

T'=54T

54×10

55sec

So, x = 5

New Question

9 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

m = 10g

l=50cm

A = 2mm2

Y = 1.2 × 1011N/m2

Δx=x×105m

As,  TA=YΔxl

Δx=TlAY

Tl=V2m

V2mAY

=3600×10×1032×106×1.2×1011=1800×103×1061.2

=1812×104=32×104=15×105m

So, x = 15

New Question

9 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Modulating signal → 2sin (6.28 × 106)t

Carrier signal     → 4 sin (12.56 × 109)t

Register to get relevant
Questions & Discussions on your feed

Login or Register

Ask & Answer
Panel of Experts

View Experts Panel

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.