2007 - JAMB Physics Past Questions and Answers - page 4

The fundamental frequency is related to tension as

F ∝ √T

Since the cells are connected in parallel the net e.m.f is that of one cell, E.

However the effective internal resistance should be added as parallel resistance.

=> ¹/_v + ¹/₂₅ + ¹/₂₅ + ¹/₂₅ = 1.2

∴ v = 0.83Ω

∴ Effective e.m.f and internal resistance are respectively = 1.5V and 0.83Ω

Period is the time to complete one cycle of 360^o

=> Periodic Time T ≡ 1 cycle of 360^o

∴ ^T/₄ = ³⁶⁰/₄ = 90^o

∴ Instantaneous e.m.f E = E^o sin ωt

= E^o sin 90

But sin 90 = 1: => E = E^o x 1

= E^o

The decay process illustrated above shows that the lost β -particle to decay to the X by increase its atomic number ( from 92 to 93) and leaving the mass no. unaffected. There after, x decayed by -α emission to give Y, by decreasing the mass no. by 2 ( from 93 to 91).

Thus , the emission is β and α particles respectively

The inductive reactance is given as

Xl = 2πfl

= 2 x ⁵⁰/_π x 0.1

= 10Ω

Again from ohms law,

R = ^V/_I = ⁷⁵/_1.5 = 50Ω

∴ Thus the inductive reactance and the resistance are respectively 10Ω, and 50Ω

From faradays law, M = itz

But mass, M = volume x density

and volume = cross sectional area x thickness

M = 118.8cm² x 10^-6 x 9 x 10³

∴ 118.8cm² x 10^-6 x 9 x 10³ = 10 x t x 3.3 x 10^-7

∴ (118.8 x 9) x 10^-7 = 10 x t x 3.3 x 10^-7

∴ t = (\frac{118.8 \times 9}{ 3.3})

t = 324 seconds

Convert to minutes by dividing by 60

t = (\frac{324}{60})

t = 5.4 sec

Efficiency = ( \frac{W(120N)INPUT}{WORK OUTPUT} \times \frac{100}{1} )

And for incline plane

V.R =( \frac{1}{Sin\theta} \

= V.R = \frac{1}{Sin30} =\frac{1}{0.5} = 2 \\

\text{therefore } \frac{60}{120} = \frac{M.A}{2}<br />

 =M.A = \frac{120}{100} = 1.2\\

\text{but M.A } = \frac{load}{effort} = \frac{120}{E}\

 \text{therefore} \frac{120}{E} = 1.2 \\

 = E =\frac{120}{1.2 } = 100N \ \)<br />

Therefore Effort up the plane = 100N

from hooke's law F=Ke

(

\frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore}

\frac{4}{0.02} = \frac{15}{e_2} )

e₁ = ( \frac{0.02 x 15}{4} = 0.075M )

The new length of the string is 4 + 0.075

= 4.075

= 4.08

Mathematically Boyle's law = PV = K

Where P = Pressure

and V = Volume

By definition Pressure = ( \frac{F}{A} = \frac{F}{M \times M} )

Again Workdone = force x displacement

therefore ( PV = \frac{F}{A} \times V = \frac{F}{M^2} \times \frac{M^3}{1} \

PV = \frac{F}{x\times x} \times \frac{x \times x \times x}{1} = F \times x )

= force x displacement