2001 - JAMB Physics Past Questions and Answers - page 5

For parallel of 5Ω and 10Ω
¹/_R = ¹/₁₀ + ¹/₅ = ¹⁺²/₁₀ = ³/₁₀

∴R = ¹⁰/₃

Total R = 3¹/₃ + 3 x 1 = 7¹/₂

∴I = 6/7¹/₃ = ⁶/₁ x ³/₂₂ = ¹⁸/₂₂ = ⁹/₁₁

∴I = ⁹/₁₁

Energy = ¹/₂CV²

∴Energy of P = ¹/₂ x 2 x V² = V²

Energy of Q = ¹/₂ x 4 x V² = 2V²

∴ ^P/_Q = ^V2/_2V² = ¹/₂ => P:Q = 1:2

C = ^Q/_V => Q = CV : ∴V = ^Q/_C

∴V₁ = ^Q/_C1; V₂ = ^Q/_C;

V₃ = ^Q/_C3

for series
¹/_C = ¹/_C1 + ¹/_C2 + ¹/_C3

= ¹/₂ + ¹/₆ + ¹/₃

= ⁶/₆ = 1

∴ C = 1μF = 1.0 x 10^-6F

∴ Q = CV = 1.0 x 10^-6 x 12

= 12.0 x 10^-6C

∴V₁ = ^Q/_C1

= 6V

V₂ = ^Q/_C2

= 2V

V₃ = ^Q/_C3

∴ 6V, 2V and 4V

0.0 E since one of the electric fields points in the opposite direction (left)

for a parallel R and X, when R = 5Ω,

and when R = 2Ω

Thus since E = I(R+r), and if r = 0 then E = IR

∴(i) E = 3(5X/(5+x)); (ii) E = 6(2X(2+x))

∴3(5X/(5+X)) = 6(2X/(2+X))

=> (5X/(5+X)) = 2(2x/(2+X))

∴ (5X/(5+X)) = (4X/(2+X))

5x(2+x) = 4x(5+x)

10x + 5x² = 20x + 4x²

5x² - 4x² = 20x - 10x

=>x² = 10x

∴x² - 10x = 0

x(x - 10) = 0

∴ x = 0 or 10

∴ x = 10Ω

I_r.m.s = ^V_r.m.s/_Z
But Z =

√ R² + X_L² 

=

√ 4² + 3² 

=

√ 16 + 9 

=

√ 25 

∴I_r.m.s = ²⁴⁰/₅

= 48.0A

in vibrations in string, the fundamental frequency happens to be the first harmonic, and is given as

f₀ = V/2l.

Where V = Velocity of the sound wave, I = Length of the string.

The second harmonic is the first overtone given by

f₁ = V/l = 2f₀ ;

and the third harmonic happens to be the second overtone, given by

f₂ = 3V/2l = 2f₀.

Thus if V = 120m/s and l = 1.5m

Thus f₁ = 2f₀ = 2 x 40 = 80Hz

Thus f₂ = 3f₀ = 3 x 40 = 120Hz

Thus we have in order, 40Hz, 80Hz, 120Hz

Electrical Energy = IVt

= 4 x 240 x (1 x 60)

= 57600J

= 5.76 x 10⁴J

The angle of dip dip is the angle which the resultant earth's magnetic field makes with the horizontal.

At the magnetic equator, which is entirely along the horizontal this angle is zero