by Sunday | Nov 15 | Physics JAMB
A. 2.0 x 106 NC -1 B. 2. 5 x 106 NC -1 C. 1.0 x 107 NC -1 D. 4.0 x 107 NC -1 Correct Answer: Option B – 2. 5 x 106 NC -1 Explanation v = E x d, => E = v/d i.e E 107/4 = 2.5 x 106 NC -1
by Sunday | Nov 15 | Physics JAMB
A. 3 Ω B. 2 Ω C. 1 Ω D. 6 Ω Correct Answer: Option A – 3 Ω Explanation 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 =...
by Sunday | Nov 15 | Physics JAMB
A. 4.5Ω B. 1.0Ω C. 8.0Ω D. 2.0Ω Correct Answer: Option D = 2.0Ω Explanation Efficiency = power out put/power input x 100/1 = power out put/ power + that due x 100/1 to internal resistance => 75/1 = 12R x 100/1 12(R+r) i.e 75/100 = 6/6+r => 75 (6+r) = 100/6 i.e...
by Sunday | Nov 15 | Physics JAMB
A. 2.0J B. 4. 0J C. 12.0J D. 2.4J Correct Answer: Option D = 2.4J Explanation 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...
by Sunday | Nov 15 | Physics JAMB
A. 24cm B. 16cm C. 12cm D. 8cm Correct Answer: Option B – 16cm Explanation 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...