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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 4 Views

R
Raj Pandey

Contributor-Level 9

C? = dQ/ndT = (dU + pdV)/ndT
= C? + (pdV/ndT)
C? - C? = pdV/ndT = R - For ideal Gas [Box: PV = nRT, pdV = nRdT]
C? - C? = 1.1R - For Non – Idea gas (for Real gas)
And Real gas behaves as ideal gas at high temperature & low pressure.
∴ T_B > T_A

New answer posted

10 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

v = ω√ (A² - x²)
K.E. = ½mv² = ½mω² (A² - x²)
Total Energy T.E = ½mω²A²
At x = A/2
K.E. = ½mω² (A² - (A/2)²) = ½mω²A² (3/4)
K.E./T.E. = (½mω²A² (3/4) / (½mω²A²) = 3/4

New answer posted

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

For adiabatic process
T? V? ¹ = T? V? ¹
⇒ T? (Al? )? ¹ = T? (Al? )? ¹
T? /T? = (l? /l? )? ¹ = (l? /l? )? /³? ¹ = (l? /l? )?

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

New answer posted

10 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

By Einstein's equation of photoelectric effect,
hc/λ = hc/λ? + eVs
Now hc/λ = hc/λ? + e*4.8 . (i)
hc/ (2λ) = hc/λ? + e*1.6 . (ii)
Multiply (ii) by 2:
hc/λ = 2hc/λ? + e*3.2 . (iii)
Equating (i) and (iii):
hc/λ? + 4.8e = 2hc/λ? + 3.2e
1.6e = hc/λ?
Substitute this into (i)
hc/λ = 1.6e + 4.8e = 6.4e
Substitute this into (ii)
(6.4e)/2 = hc/λ? + 1.6e
3.2e = hc/λ? + 1.6e
1.6e = hc/λ?
From (i): hc/λ = hc/λ? + 3 (hc/λ? ) = 4hc/λ?
λ? = 4λ

New answer posted

10 months ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

Dimension of B = [L? ¹]
Dimension of D = [T? ¹]
Dimension of A = [MLT? ²]
The dimensional formula of AD/B
= [MLT? ²] [T? ¹]/ [L? ¹] = [ML²T? ³]

New answer posted

10 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Maximum value of emf is measured when pointer is at B with I? = 0.
So, I_AB = 6 / (20 + 0.1 * 1000) = 6/120 = 1/20 A
V_AB = E = I_AB * R_AB
= (1/20) * 100
= 5V

New answer posted

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

v =? t +? t²
ds/dt =? t +? t²
? this =? (? t +? t²)dt from 1 to 2
s = [? t²/2 +? t³/3] from 1 to 2
s = [? (2)²/2 +? (2)³/3] - [? (1)²/2 +? (1)³/3]
s = [2? + 8? /3] - [? /2 +? /3]
s = 3? /2 + 7? /3

New answer posted

10 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

η = 1 - T? /T?
1/6 = 1 - T? /T? (i)
2η = 1/3 = 1 - (T? -62)/T? (ii)
By (i) and (ii)
(T? /T? ) = 5/6
1/3 = 1 - (T? /T? ) + 62/T? = 1 - 5/6 + 62/T?
1/3 = 1/6 + 62/T?
1/6 = 62/T?
T? = 62 * 6 = 372K = 99°C

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