by Sunday | Nov 11 | Physics JAMB
A. 4.05m B. 4.08m C. 5.05m D. 6.08m Correct Answer: Option B = 4.08m Explanation From hooke’s law F=Ke F1/e1 = F2/e2 therefore 4/0.02 = 15/e2 e1 = 0.02×15/4 = 0.075M The new length of the string is 4 + 0.075 =...
by Sunday | Nov 11 | Physics JAMB
A. 5.4 min B. 10.8 min C. 15.0 min D. 20.0 min Correct Answer: Option A – 5.4 min Explanation 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...
by Sunday | Nov 11 | Physics JAMB
A. 0.00 B. εo/4 C. εo/2 D. εo Correct Answer: Option D = εo Explanation Period is the time to complete one cycle of 360o => Periodic Time T ≡ 1 cycle of 360o ∴ T/4 = 360/4 = 90o ∴ Instantaneous e.m.f E = Eo sin ωt = Eo sin 90 But sin 90 = 1: => E = Eo x 1 =...
by Sunday | Nov 11 | Physics JAMB
A. 1600N B. 400N C. 160N D. 40N Correct Answer: Option A = 1600N
by Sunday | Nov 11 | Physics JAMB
A. 2.0 x 10-6K-1 B. 2.1 x 10-6K-1 C. 4.2 x 10-6K-1 D. 6.3 x 10-6K-1 Correct Answer: Option C = 4.2 x 10-6K-1 Explanation If the lineal expansivity = ∝ => area expansivity β= 2∝. And cubic expansivity δ = 3∝. => 6.3 x 10-6 = 3∝ ∴ ∝ = ∴ ∝ = 6.3 x 10-6 = 2.1x 10-6...
by Sunday | Nov 11 | Physics JAMB
A. 3 J B. 4 J C. 5 J D. 6 J Correct Answer: Option A – 3J Explanation Effective component of the force along the horizontal = F cos 60o Work done by this force : 60 F cos 60 x X = 2 x 0.5 x 3 =...