1999 - JAMB Physics Past Questions and Answers - page 5

From Ohms Law, E = I(R +X)

=> E = I₁(R₁ + X)

again, E = I₂(R₂ + X)

=> I₁(R₁ + X) = I₂(R₂ + X)

∴ 5(8 + X) = 7(5 + X)

40 + 5X = 35 + 7X

7X - 5X = 40 - 35

2X = 5

X = ⁵/₂

= 2.5Ω

Energy stored in a capacitor is given by

E = ¹/₂CV²

But effective capacitance for the parallel arrangement is given by

C = C₁ + C₂

= 2 + 3

= 5μF = 5 x 10^-6 x (100)²

∴E = ¹/₂CV²

= ¹/₂ x 5 x 10^-6 x 10⁴

2.9 x 10^-2J

Power = IV = I²R = ^V2/_R

In terms of e.m.f E, of a cell we have

IE = I²R = E²/R

∴P₁ = ^E2/_R; P₂ = ^E2/_R2/^E2/_R2

= ^E2/_R1 x ^R₂/_E² = ^R₂/_R1

∴^P₁/_P2 = ^R₂/_R1

The effective V =

√ V_R² + V_L² 

=

√ 6² + 8² 

=

√ 36 + 64 

=

√ 100 

10V

Let the forces be F₁, and F₂.

F₁ = m₁ a₁ = 3 x 2 = 6N

F₂ = m₂ a₂ = 1 x 9 = 9N

F₁F₂ = 69 = 23

Thus ratio of forces is 2: 3

F = (MV²)/r = Mω²r

F = (10)² x (100) x 0.2

F =20 N

From principle of conservation of linear momentum,

(0.05 x 200) - (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).

Thus V = 10m/s.

K.E = 1/2 (0.05 + 0.95) x 10²

K.E = 1/2 (1 x 100) = 50 J

Height h = 1/2 (U²/g) = 10²/20 = 5m

Energy before hitting ground is = P.E at the height of 15m.

P.E = mgh = 0.1 x 10 x 15 = 15 J