Class 12th

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New answer posted

8 months ago

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Payal Gupta

Contributor-Level 10

l=π2π2dx (1+ex) (sin6x+cos6x) ……. (i)

=20dt4+t2=2 (tan1 (t2))0=π

New answer posted

8 months ago

0 Follower 51 Views

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Payal Gupta

Contributor-Level 10

f (x)= {sin (x+2)x+2, x (2, 1)0, x (1, 0]2x, x (0, 1)1, otherwise

LHD=Lth0f (0h)f (0)h=0

RHD=Lth0f (0+h)f (0)h=2

Hence f (x) is not differentiable at x = 1, 0, 1

m=2, n=3

New answer posted

8 months ago

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Payal Gupta

Contributor-Level 10

x + y + az = 2………… (i)

3x + y + z = 4 ………… (ii)

x + 2z = 1 ……………. (iii)

x = 1, y = 1, z = 0

(for unique solution a 3  

( α , 1 ) , ( 1 , α ) a n d ( 1 , 1 ) are collinear.

α 1 1 α = 1 α 1 1 1 α 2 = 0 α = ± 1                

Sum of absolute value = 2

New answer posted

8 months ago

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Payal Gupta

Contributor-Level 10

e2x4=0or6e2x5ex+1=0

x=ln2x=ln (13), ln (12)

ln3, ln2

New answer posted

8 months ago

0 Follower 9 Views

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Payal Gupta

Contributor-Level 10

given N = 1000 turns

A = 1m2

ω=1rev/sec

=1*2πrad/sec

V=d? Bdt=ddt (NBAcosωt)

140*227

= 20 * 22

= 440 volts

New answer posted

8 months ago

0 Follower 8 Views

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Payal Gupta

Contributor-Level 10

d=43 cm (Lateral shift)

By Snell's law

μairsin60°=μgsinθ

1*32=3sinθ

sinθ=12

= 30°

tan30°=dt=43t

13=43t

t = 12 cm

New answer posted

8 months ago

0 Follower 5 Views

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Payal Gupta

Contributor-Level 10

forA, t12=4sec

=mA0e0.6934*16

mA=mA0e4*0.693 - (1)

for B,

for B,  t12=8sec

mB=mB0e2*0.693 - (2)

mAmB=mAOmBOe4*0.693e2*0.693

mAmB=25100=x100x=25

New answer posted

8 months ago

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Payal Gupta

Contributor-Level 10

IL=VRL=51kΩ

lL=5*103A=5mA

New answer posted

8 months ago

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Payal Gupta

Contributor-Level 10

AB = 10 m

RAB=20Ω

LAC=250cm

= 2.5 m

l=2520+30

E=x10=2510

New answer posted

8 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

v=Cμrεr=3*1086.25*1=3*1082.5=1.25*108m/sec

Asfλ=f (5*103*4)=1.25*108f=6.25GHz

So,  f6.

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