Solution

Applying Kirchoff's loop rule to closed loop ADBA
-I₁R₁ + 0 + I₂R₂ = 0 (Iᵍ= 0) .........(i)
For loop CBDC
-I₂R₄ + 0 + I₁R₃ = 0 ..........(ii)
⇒ From equation (i) I₁/I₂ = R₁/R₂
From equation (ii)
I₁/I₂ = R₄/R₃
Therefore, R₁/R₂ = R₄/R₃

Solution

(a) Radar systems used in aeroplane navigation can benefit from microwaves.
Special vacuum tubes, such as klystrons, magnetrons, and Gunn diodes, create these rays.
(b) Muscle strain is treated using infrared waves.
Hot bodies and molecules generate these rays.
(c) X-rays are utilised in medicine as a diagnostic tool. When high-energy electrons collide with a metal with a high atomic number, these rays are created.

Solution

(i) Angular magnification
m = fₒ/fₑ
= 15/10⁻² = 1500
Angular size of the moon = (3.48 × 10⁶/3.8 × 10⁸) = 3.48/3.8 × 10⁻² radian
Therefore, Angular size of the image = (3.48/3.8 × 10⁻² × 1500) radian
(ii) Diameter of the image = 3.48/3.8 × 15 × focal length of eye piece
= 3.48/3.8 × 15 × 1 cm
= 13.7 cm

Solution

Einsteins Photoelectric equation is given by
hν = 𝜙ₒ + kₘₐₓ
or hν = hνₒ + 1/2 mv²ₘₐₓ
Important features are given by:
(i) kₘₐₓ depends linearly on frequency ν
(ii) Existence of threshold frequency for the metal suraface.
hν = 𝜙ₒ + kₘₐₓ
hc/λ₁ = hc/λₒ + kₘₐₓ .........(i)
hc/λ₂ = hc/λₒ + 2 kₘₐₓ ...........(ii)
From (i) and (ii)
2hc/λ₁ - hc/λ₂ = hc/λₒ
1/λₒ = (2/λ₁ - 1/λ₂)
λₒ = λ₁λ₂/ 2λ₂ -λ₁
Work function 𝜙ₒ = hc/λₒ = hc(2λ₂ -λ₁)/λ₁λ₂/

Solution

Only a minor percentage of the incident α-particle rebound was recovered. This demonstrate that the atom's mass is concentrated in a compact volume in the shape of the nucleus and provides an estimate of the nucleus' size.
Radius of the nucleus
R = Rₒ A^(1/3)
Density = mass/ volume where, m: mass of one nucleon, A: Mass number
= mA/4/3 πR³
= mA/4/3π(Rₒ A^(1/3))³
= 3m/4πRₒ³
It means that the nuclear matter density is independent of A.

Solution

Nuclear fission is the splitting of a heavy nucleus into smaller fragments, while fusion is the merging of light nuclei to generate a heavy nucleus.
In both processes, the daught nuclei's binding energy per nucleon is greater than that of the parent nuclei. Energy is released as a result of the difference in binding energy. Some mass is turned into energy in both processes.
Energy released
Q = [m(c²₁H ) + m(₁³H) - m(₂⁴He) - m(n)] × 931.5 MeV
= [2.014102+3.016049- 4.002603 - 1.008665] × 931.5 MeV
= 0.018883 × 931.5 MeV
= 17.59 MeV

Solution

Solution

Solution

We know that, energy stored in a capcitor is
E = 1/2 CV²
In series combination
0.045 = 1/2 c₁c₂/c₁+c₂ (100)²
⇒ c₁c₂/c₁+c₂ = 0.09 × 10⁻⁴ .......(i)
In parallel combination
0.25 = 1/2 (C₁ + C₂) (100)²
⇒ C₁ + C₂ = 0.5 × 10⁻⁴ .......(ii)
On simplifying (i) and (ii)
C₁C₂ = 0.045×10⁻⁸
(C₁-C₂)² = (C₁+C₂)² - 4 C₁C₂
= (0.5 × 10⁻⁴)² - 4 × 0.045×10⁻⁸
= 0.25 ×10⁻⁸ - 0.180 ×10⁻⁸
(C₁-C₂)² = 0.07 ×10⁻⁸
(C₁-C₂)² = 2.6 ×10⁻⁵ = 0.26 × 10⁻⁴ .........(iii)
From (ii) and (iii)
⇒ C₁ = 0.38× 10⁻⁴ F and C₂ = 0.12 × 10⁻⁴ F
Chrages on capacitor C₁ and C₂ in parallel combination
Q₁ = C₁V = (0.38 × 10⁻⁴ × 100) = 0.38 × 10⁻² C
Q₂ = C₂V = ( 0.12 × 10⁻⁴ × 100) = 0.12 ×10⁻² C

Solution

When a current carrying coil is put in a magentic feld, it experiences torque, which causes the coil to rotate and produce an angular deflection.