Class 11th
Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th
Follow Ask QuestionQuestions
Discussions
Active Users
Followers
New answer posted
7 months agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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 agoContributor-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)
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 66k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers


