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New answer posted

4 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

U = mV (r) = -Cm/r

F = -dU/dr = -Cm/r² ⇒ The force which provides required centripetal force

⇒ mv²/r = Cm/r² ⇒ r = C/v²

New answer posted

4 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

I = (V/R) (1 - e^ (-t/τ) where τ = L/R

E = (1/2)LI² = (1/2)L (V²/R²) (1 - e^ (-t/τ)² ⇒ Energy stored in inductor

According to Question, we can write
(1/4) * (1/2) (LV²/R²) = (1/2) (LV²/R²) (1 - e^ (-t/τ)²
⇒ 1/4 = (1 - e^ (-t/τ)² ⇒ 1/2 = 1 - e^ (-t/τ) ⇒ e^ (-t/τ) = 1/2

⇒ t/τ = ln (2) ⇒ t = τln (2) = (L/R)ln (2)

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

R = mv / (qB) = √ (2mK) / (qB) ⇒ K = (R²q²B²) / (2m)

⇒ K? /Kα = (R? ²q? ²B² / 2m? ) * (2mα / (Rα²qα²B²) = (mα/m? ) * (R? /Rα)² * (q? /qα)²

⇒ K? /Kα = (4) * (2)² * (1/2)² = 4:1

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

R = (2μ sinθ) / (1.22λ) . (1)

According to de-Broglie's hypothesis, we can write
λ = h / (mv) . (2)

With the help of equations (1) and (2), we can write
R = (2μmv sinθ) / (1.22h) ⇒ R? /R? = m? /m? = 1837 ⇒ R? = 1837R?

  • Concept involved: Resolving Power of Microscope
  • Topic: Optics
  • Difficulty level: Moderate
  • Point of Error: Formula

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

  • Concept involved: Electric and magnetic field lines
  • Topic: Magnetics and Magnetic Material
  • Difficulty level: Moderate
  • Point of Error: Fact

New answer posted

4 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

(v/v? ) + (x/x? ) = 1 ⇒ v = - (v? /x? )x + v?

⇒ a = dv/dt = - (v? /x? ) (v) = - (v? /x? ) [- (v? /x? )x + v? ] ⇒ a = (v? ²/x? ²)x - v? ²/x?

  • Concept involved: Graph of kinematics
  • Topic: Kinematics
  • Difficulty level: Moderate
  • Note: IIT-Jee-2005
  • Point of Error: Writing Equation of straight line and differentiation

New answer posted

4 months ago

0 Follower 2 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

When white light passes through a cool gas, atoms absorb specific wavelengths. This produces a continuous background with dark lines at those absorbed wavelengths. This is atomic absorption spectra. But when atoms are excited, they release photons at specific wavelengths, producing a dark background with bright lines. That is emission spectra. 

New answer posted

4 months ago

0 Follower 1 View

S
Syed Aquib Ur Rahman

Contributor-Level 10

Every element has a unique set of spectral lines as its electrons occupy specific energy levels. What scientists know for certain is that these unique patterns act like fingerprints. It helps in identifying elements in stars, flames, or unknown samples. For instance, Helium was discovered in the Sun's spectrum before it was found on Earth.

New answer posted

4 months ago

0 Follower 3 Views

S
Syed Aquib Ur Rahman

Contributor-Level 10

We see a discrete emission spectrum when electrons inside excited atoms or ions in gases fall back from higher energy levels to lower ones. Every transition releases a photon of a specific wavelength. Usually the spectrum appears as sharp, bright lines. Dark gaps separate them. These lines are unique to each element.

On the other hand, a continuous emission spectrum is produced when hot solids, liquids, or dense gases emit radiation. This happens because of the collective motion of their atoms and electrons. We don't see sharp lines. Instead the spectrum shows a smooth and unbroken spread of all wavelengths. 

New answer posted

4 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

While the particle moves from mean position to displacement, half of its amplitude, its phase changes by π/6 rad. So,
Time taken, t = (π/6)/ω = T/12 = (2/12)s = (1/6)s
a = 6

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