2001 - JAMB Physics Past Questions and Answers - page 2

A driving mirror is always a convex mirror which is always of negative focal length and always forms a virtual image

Thus if r = 1m = 100cm

f = 1/2 = -50

∴ if U = 4cm = 400cm

f = -50cm

∴1/u + 1/v = 1/f

1/400 + 1/v = -1/50

∴ -1/v = 1/400 + 1/50

-1/v = (1 + 8)/400 = 9/400

∴ v = -400/9

v = 4/9

P1 = 300Nm^-2; P2 = 120Nm^-2

T1 = 127 + 273; T2 = -73 + 273

= 400K; = 200K

Let the initial volume V1; and the final volume = V2

∴ P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

∴ V2/V1 = P1T2/P2T1

∴ V2/V1 = (300 x 200)/(120 x 400) = 5/4

∴ V2 : V1 = 5 : 4

If F = 85.5Hz, V = 342ms^-1,

Then wavelength = V/F = 342/85.5 = 4m

Then if wavelength = 4m

=> wavelength/2 = 4/2 = 2m = NN,

Thus NA = wavelength/4 = 4/4 = 1m.

For a string fixed at both ends and plucked at the middle, the fundamental frequency is given

by f₀ = v/λ, where v = velocity of sound, λ = wave length.

At its fundamental, note, the the length of the string is given l = λ/2, => λ₀ = 2l, where λ₀ = wave length fundamental note. for the first over tone, λ₁ = l.

For the second over tone, λ₂ = 21/3 = (2 x 24)/3 = 16cm.

i.e. λ₂ = 16cm.

P = F/A => F = PA = PV/L

W = FD = (PV/L) x L = PV

= 2 x 10⁵

(2 x 10^-6 x 6)

= 2 x 10⁵ x 1.2 x 10^-5

= 2.4 J

Efficiency = power out put/power input * 100/1

= power out put/ power + that due * 100/1

to internal resistance

=> 75/1 = 1²R *100/1

        1<sup>2</sup>(R+r)<br />

i.e 75/100 = 6/6+r => 75 (6+r) = 100/*6

i.e 450 + 75r = 600

75r =600 - 450

i.e r = 150/75 = 2Ω

i.e r = 2Ω

1 = V/R = E/R+r , if r = 2Ω

V = 3/5 E

=> 3E/5R = E/R+2

i.e 3E(R+2) = 5RE

3RE + 6E = 5RE

i.e 3R + 6 = 5R

6 = 5R - 3R

=> 2r = 6 i.e R = 6/2 = 3Ω

v = E*d, => E = v/d

i.e E 10⁷/4 = 2.5 * 10⁶ NC^-1

Efficiency = power output/power Input

=> 80/100 = power output/1.5 * 250

i.e power output = 1.5*250 *80/100 =300w