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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

KIndly go through the solution

 

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]

New answer posted

4 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

Binding energy = (50M? + 70M? – Msn)C²
= (50.3915 + 70.6069 – 119.902199)U²
= (1.0962U)C²
= 931 * 1.0962MeV

Binding energy per neucleon
= (931 x 1.0962)/120 MeV
= 8.5MeV

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

I = 0.8 kgM², |μ? | = 20Am²
U? = -μ? ⋅ B? = 0
Uf = -μBcos (30°) = -20 x 4 x √3/2
U? – Uf = 40√3 = ½ I ω² = 0.4ω²
ω² = 100√3; ω = 10 (3¹/? )

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

By conservation of energy
(K.E. + P.E.)? = (K.E. + P.E.)?
0 + (1/ (4πε? a³) * 2P * P = 2 * (1/2)mv² + 0
v = √ (p² / (2πε? ma³) = (P/a) * √ (1 / (2πε? ma)

New answer posted

4 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

I? ∝ R? ² & I? ∝ R? ²
I? /I? = (R? /R? ) ² = 1/α² = 1/16
α = 4

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Sol. eVslop = hc/λ - φ

Slope of curve

tan θ = hc/e = constant
as intensity of incident radiation is increased, there will be no effect on graph

New answer posted

4 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

In steady state rate of flow of heat in all three rods are same.
dQ/dt = (k? A (100 – 70)/? = (k? A (70 – 20)/? = (k? A (20 – 0)/?
30k? = 50k? = 20k?
∴ k? : k? = 2:3 & k? : k? = 2:5

New answer posted

4 months ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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