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New answer posted

a year ago

0 Follower 35 Views

V
Vishal Baghel

Contributor-Level 10

Polygon law is applicable in both the situation given but the equation given in the reason is not useful in explaining the assertion.

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

v = 1/√με = 1/√ (µ? µ? ε? ε? ) = c/√ (µ? ε? )
v = 3*10? / √ (1*81)
v = 3*10? / 9 m/s
= 0.33 * 10? m/s
= 3.33 * 10? m/s

New answer posted

a year ago

0 Follower 14 Views

V
Vishal Baghel

Contributor-Level 10

v = √2gh velocity of efflux.
F = v ( dm/dt ) = v (aρv) = aρv² = 2aρgh
fr = µR = µAhρg
For just sliding, for = F
µAhρg = 2aρgh
or µ = 2a/A

New answer posted

a year ago

0 Follower 13 Views

R
Raj Pandey

Contributor-Level 9

MV? = 3MV? = p
λ = h/mv = h/p
λ? /λ? = 1

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

1/Ceq = 1/C? + 1/C? + 1/C?
1/Ceq = 1/ (K Aε? /d) + 1/ (3K Aε? /2d) + 1/ (5K Aε? /3d)
1/Ceq = d/ (K Aε? ) + 2d/ (3K Aε? ) + 3d/ (5K Aε? )
1/Ceq = (d/K Aε? ) * (1 + 2/3 + 3/5)
1/Ceq = (d/K Aε? ) * (15+10+9)/15) = 34d / (15K Aε? )
Ceq = 15K Aε? / 34d

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

Let vel of A and B just after collision be VA & VB respectively.
m * 9 + 0 = m * VA + 2mVB
9 = VA + 2VB
again e = (VB - VA)/ (9-0) = 1
9 = VB - VA
From (i) & (ii)
VB = 6 m/s & VA = -3 m/s
Now, for B and C collision (completely inelastic):
2m * 6 + 0 = (2m + 2m)Vc
12m = 4mVc
Vc = 3 m/s

New answer posted

a year ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

t? /? = 3 days = 72 hours
dN/dt = λN = (ln2/t? /? ) N
= (0.693 * 6.02*10²³ * 2*10? ³) / (72 * 3600 * 198)
= 1.618 * 10¹³
= 16.18 * 10¹² disintegration/second

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

V = E (1 - e? /τ)
Where τ = RC = 100 * 10? = 10? sec
50V = 100V (1 - e? /10? )
1/2 = 1 - e? ¹?
1/2 = e? ¹?
-ln2 = -10? t
t = ln2/10?
t = 0.693 * 10? sec

 

New answer posted

a year ago

0 Follower 20 Views

R
Raj Pandey

Contributor-Level 9

dC? = (ε? + kx)A / dx [For 0 < x < d/2]
1/C? = ∫ dx / (ε? + kx)A) from 0 to d/2
= (1/Ak) [ln (ε? + kx)] from 0 to d/2
= (1/kA) ln (1 + kd/ (2ε? )
C? = kA / ln (1 + kd/ (2ε? )

Similarly dC? = (ε? + k (d-x)A / dx [For d/2 ≤ x ≤ d]
C? = kA / ln (1 + kd/ (2ε? )
Clearly, C? = C? = C
For series combination:
C_eq = C? / (C? + C? ) = C/2 = kA / (2ln (2ε? + kd)/2ε? )

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

i = constant
v ∝ E & E ∝ 1/r² So, E increases with decrease in the radius.
also v ∝ E
So, drift speed increases.

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