Class 12th

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New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Pitch = (Vcosθ)T = (Vcosθ)2πm/eB
= (4*10? cos60°) (2π (1.67*10? ²? )/ (0.3*10? ¹? *1.69*10¹? )
= 4 cm

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

From Given equation
µ = 0.6
A? = µA?
(A? - A? )/2 = A? = 5
(A? + A? )/2 =?
A? - A? = 3
From equation (1)+ (2)
A? = 8
From equation (1)- (2)
A? = 2

New answer posted

4 months ago

0 Follower 21 Views

V
Vishal Baghel

Contributor-Level 10

Heat loss; ΔH = U? - U? = 1/2 (C? / (C? +C? ) (V? -V? )²
= 1/2 * (5*2.5)/ (5+2.5) (220-0)²µJ
= 5/ (2*3) * 22*22*100*10? J
= 5*11*22/3 * 10? J = 1210/3 * 10? J = 1210/3 * 10? ³ J * 4 * 10? ²
According to questions
x/100 = 4*10? ²
So, x=4
Note: But given answer by JEE Main x=36

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

hc/λ = Φ + eV
hc/3λ = Φ + eV/4
from (1) and (2)
hc/λ (1-1/3) = 3/4 eV
eV = 8/9 hc/λ
eV = 8/9 (hc/λ - Φ)
Φ = hc/9λ
λ? = 9λ ∴k=9

New answer posted

4 months ago

0 Follower 8 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

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

4 months ago

0 Follower 2 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

4 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

4 months ago

0 Follower 4 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

4 months ago

0 Follower 22 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

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