Physics

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

(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

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. 

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.

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. 

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

Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 * 10 * (10 - 5) = 10m/s

Potential difference across resistor at time t = V = 30/3 = 10V
Current, I = 10 / (5 * 10? ) = 2? A

For Wire-A
F / (πr_A²) = Y (l_A / 2) . (1)
For Wire-B
F / (πr_B²) = Y (l_B / 4) . (2)
From equations (1) and (2), we can write
(l_A / (2r_A²) = (l_B / (4r_B²) ⇒ l_B/l_A = 2 (r_B/r_A)² ⇒ x = l_B = 32

Angle from direction of flow = 90° + θ = 120°
⇒ θ = 30°
sin 30° = u/v ⇒ u/10 = 1/2 ⇒ u = 5 m/s