Class 12th

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New answer posted

10 months 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

10 months ago

0 Follower 2 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

10 months ago

0 Follower 4 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

10 months ago

0 Follower 19 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

10 months 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.

New answer posted

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

Bandwidth
= (ω? + ω? ) - (ω? - ω? )
= 2ω?
= 2 (2πf? )
= 4πf?
= 2π (2f? )
= 2π (2*10? ) rad/s
= 2 * 10? Hz
= 200 kHz

New answer posted

10 months ago

0 Follower 15 Views

V
Vishal Baghel

Contributor-Level 10

T sin θ = (1/4πε? ) * q²/ (2lsinθ)²
T cos θ = mg
∴ tan θ = q² / (4πε? mg * 4l²sin²θ)
[tan θ ≈ θ, for small angle]
So, θ³ = q² / (16πε? mgl²)
θ = ( q² / (16πε? mgl²) )¹/³
Also separation = 2l sin θ ≈ 2lθ
= 2l ( q² / (16πε? mgl²) )¹/³
= ( 8q²l³ / (16πε? mgl²) )¹/³
= ( q²l / (2πε? mg) )¹/³

New answer posted

10 months ago

0 Follower 92 Views

A
alok kumar singh

Contributor-Level 10

Each side of the square has a resistance of 16/4 = 4Ω.
The side AC has the 9V, 1Ω source.
The other three sides (AB, BD, DC) form a path with resistance 4+4+4 = 12Ω.
This is in parallel with the side AC (4Ω).
Total resistance of the loop part: (12*4)/ (12+4) = 48/16 = 3Ω.
Total resistance of the circuit: R_total = 3Ω + 1Ω (internal) = 4Ω.
Total current from source I = V/R_total = 9/4 A.
This current splits.
Current through path ABDC, I? = I * (R_AC / (R_ABDC + R_AC) = (9/4) * (4/16) = 9/16 A.
Potential at B: V_B = V_A - I? R_AB = 9 - (9/16)*4 = 9 - 9/4 = 27/4 V.
Potential at D: V_D = V_C + I? R_CD = 0 + (9/16)*4 = 9/4 V.
Potential drop a

...more

New answer posted

10 months ago

0 Follower 22 Views

A
alok kumar singh

Contributor-Level 10

Binding Energy = (Δm)c²
= [Zmp + (A-Z)mn - MAl]c²
= [ (13*1.00726 + 14*1.00866) - 27.18846] u
= [ (13.09438 + 14.12124) - 27.18846] u
= [27.21562 - 27.18846] u
= 0.02716 u
= 0.02716 x = 27.16x * 10? ³

New answer posted

10 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Applying nodal analysis at point X with potential xV:
(x-20)/5 + (x-0)/2 + (x-20)/5 = 0
2 (x-20) + 5x + 2 (x-20) = 0
9x - 80 = 0 => x = 80/9 V
Potential drop across 2Ω resistor = x = 80/9 V.
(Note: There seems to be a discrepancy in the provided solution and standard circuit analysis. The provided solution calculates the current junction, not a single point.)
The solution provided in the image calculates as:
(x-0)/5 + (x-20)/5 + (x-20)/2 = 0
2x + 2 (x-20) + 5 (x-20) = 0
9x - 140 = 0
=> x = 140/9 V
Potential drop across 2Ω = 20 - x = 20 - 140/9 = 40/9 V

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