Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question
5.6k

Questions

0

Discussions

28

Active Users

0

Followers

New answer posted

4 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

New answer posted

4 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

The excess pressure inside a soap bubble is given by ΔP = 4T/R, where T is the surface tension and R is the radius.
The pressures are given as P? = 1.01 atm and P? = 1.02 atm. Let the atmospheric pressure be P? = 1 atm.
ΔP? = P? - P? = 1.01 - 1 = 0.01 atm = 4T/R?
ΔP? = P? - P? = 1.02 - 1 = 0.02 atm = 4T/R?
Dividing the two equations: (ΔP? /ΔP? ) = (R? /R? )
0.01 / 0.02 = R? /R? ⇒ R? /R? = 2
The ratio of their volumes is V? /V? = ( (4/3)πR? ³ ) / ( (4/3)πR? ³ ) = (R? /R? )³ = 2³ = 8.
The ratio is 8:1.

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 6 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Angular width of a fringe is YDSE

 

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 5 Views

A
alok kumar singh

Contributor-Level 10

The law of radioactive decay is N = N? e? λt, where N is the amount remaining at time t.
Given that at time t, N/N? = 9/16.
So, 9/16 = e? λt
At time t/2, the fraction remaining will be N'/N?
N' = N? e? λ ( t/2 ) = N? (e? λt)¹/²
Substituting the value of e? λt:
N' = N? (9/16)¹/² = N? (3/4)
The fraction remaining is N'/N? = 3/4.

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

For a rotating loop, the induced emf is ε = ε? sin (ωt), where the peak emf ε? = NABω.
Here, Area A = πab. The number of turns N = 1.
So, ε? = B (πab)ω
The average power loss due to Joule heating in a resistor R is given by:
P_avg = (ε_rms)² / R
Where ε_rms = ε? /√2
P_avg = (ε? ²/2) / R = (B (πab)ω)² / (2R) = π²a²b²B²ω² / (2R)

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 681k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.