Class 11th

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New answer posted

7 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

7 months ago

0 Follower 4 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

7 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

7 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

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Pressure outside is 0.
Here, Pin = 4T/r
By, P? V? + P? V? = PV (isothermal process, Boyle's law applied to the mass of gas inside)
(4T/r? ) (4/3 πr? ³) + (4T/r? ) (4/3 πr? ³) = (4T/r) (4/3 πr³)
r? ² + r? ² = r²
r = √r? ² + r? ²

New answer posted

7 months ago

0 Follower 26 Views

R
Raj Pandey

Contributor-Level 9

dQA = mACA dTA
dQB = mBCB dTB
dQA/dt = dQB/dt
⇒ mACA (dTA/dt) = mBCB (dTB/dt)
⇒ (dTA/dt)/ (dTB/dt) = (mBCB)/ (mACA)
Assuming mA = mB
(dTA/dt)/ (dTB/dt) = CB/CA
CA/CB = (dTB/dt)/ (dTA/dt) = (90/6)/ (120/3) = 15/40 = 3/8

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

a = F/m = (8î + 2? )
r = ut + ½ at²
r = 0 + ½ (8î + 2? ) 10²
r = 400î + 100?

New answer posted

7 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

g = (4/3)πRρG . (i) (ρ = density)
Now ρ = M/V = M/ (4/3)πR³)
R³ = M/ (4/3)πρ) => R ∝ (M/ρ)¹/³
From equation (i)
g ∝ Rρ ∝ (M/ρ)¹/³ρ = M¹/³ρ²/³
For planet, M' = 2M, ρ' = ρ
g'/g = (M'/M)¹/³ (ρ'/ρ)²/³ = (2)¹/³ (1)²/³ = 2¹/³
W' = mg' = m (2¹/³g) = 2¹/³ (mg) = 2¹/³W

New answer posted

7 months ago

0 Follower 20 Views

R
Raj Pandey

Contributor-Level 9

For ball (a): I? = Δp? = 2mu
For ball (b): I? = Δp? = 2mu cos 45°
I? /I? = 2mu / (2mu cos 45°) = √2

New answer posted

7 months ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

By t = mx² + nx
dt/dx = 2mx + n
V = dx/dt = 1/ (2mx + n)
a = v (dv/dx) = V (d/dx (1/ (2mx+n) = V [-2m/ (2mx+n)²] = -2mV³
So, Retardation will be (2mv³)

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