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New answer posted
5 months agoContributor-Level 10
Least count = Pitch / (Total divisions on circular scale)
In 5 revolution, distance travelled = 5mm
? In 1 revolution, distance travelled = 1mm
So least count = 1/50 = 0.02mm = 0.002 cm
So, Assertion is not correct but reason is correct
New answer posted
5 months agoContributor-Level 9
r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? * v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? * v? = 0
⇒ (10αt²î + 5β (t-5)? ) * (20αtî + 5β? ) = 0
⇒ 50αβt² (î*? ) + 100αβt (t-5) (? *î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second
New answer posted
5 months agoContributor-Level 10
A→B & D→C (isothermal process)
So, TA = TB & TD = TC. Now B→C & D→A (adiabatic process)
|WBC| = nR/ (γ-1) (TB - TC)
|WAD| = nR/ (γ-1) (TA - TD) = nR/ (γ-1) (TB - TC)
∴ |WBC| = |WAD|
New answer posted
5 months agoContributor-Level 10
1 litre, T = 300K, P = 2 atm, KE = 2*10? J/molecule, no of molecule =?
No. of molecules = (no of moles) * NA = nNA
Also, n = PV/RT = PV/ (NAkT)
KE = (3/2)kT = 2*10? J [Given]
kT = (4/3)*10?
P = 2 atm = 2 * 1.013 * 10? N/m²
vol = 1 lit = 10? ³ m³
No. of molecules = PV/kT = (2*1.013*10? * 10? ³)/ (4/3)*10? ) ≈ 1.5 * 10¹¹
New answer posted
5 months agoContributor-Level 10
Polygon law is applicable in both the situation given but the equation given in the reason is not useful in explaining the assertion.
New answer posted
5 months agoContributor-Level 10
v = 1/√με = 1/√ (µ? µ? ε? ε? ) = c/√ (µ? ε? )
v = 3*10? / √ (1*81)
v = 3*10? / 9 m/s
= 0.33 * 10? m/s
= 3.33 * 10? m/s
New answer posted
5 months agoContributor-Level 10
v = √2gh velocity of efflux.
F = v ( dm/dt ) = v (aρv) = aρv² = 2aρgh
fr = µR = µAhρg
For just sliding, for = F
µAhρg = 2aρgh
or µ = 2a/A
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