JAMB Physics Past Questions & Answers - Page 146

Original length = 20m
first new length = 20.01m

first increase in length = 20.01m - 20m = 0.01m

Second new length = 20.05m
second increase in length = 20.05 - 20m = 0.05m

From hooke's law = $\frac{F_1}{e_1} = \frac{F_2}{e_2} \implies \frac{50}{0.01} = \frac{F_2}{0.05} \\ \implies F_2 = \frac{50 \times 0.05}{0.01} = 250N$

E = V + v

where E = emf of cell.
V = terminal p.d
v = loss voltage of the cell.

But from ohm's law V=IR ; v = Iv

$\implies$ lost voltage = V = Ir
= 4 x 0.5
= 2.0v

From the relation refractive index = Sin $\left(\frac{A+Dm}{2}\right)$

Where A = refractive angle = 60⁰, Sin$\left(\frac{A}{2}\right)$

$\implies 1.4 = \frac{\text{Sin}\left(\frac{60^0 + Dm}{2}\right)}{\text{Sin}\left(\frac{60}{2}\right)} \\ = \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{\text{Sin}30^0} \\ = \frac{\text{Sin}\left(\frac{60 + Dm}{2}\right)}{0.5}\\ \text{Therefore} 1.4 \times 0.5 = \text{Sin}\left(\frac{60 + Dm}{2}\right) \\ \text{Therefore} \left(\frac{60 + D_m}{2}\right)= \text{Sin}^{-1} 0.7000 \\ = 44^0 12^1 \\ 60^0 + D_m = 88^0 24^1 \\ D_m = 88^0 24^1 - 60^0 \\ = 28^0 24 \\ 29^0$

For Hooke's law F = Ke

$\implies F/e = K \\ f1/e1=f2/e2 \\ \text{Let the original length} = t_0 \\ \text{therefore} e1 = (36 - t_0)cm ; e2 = (46 - t_0)cm \\ \text{if f1} = 40N \text{f2} = 60N \\ \text{Then } \frac{40}{36 - t_0} = \frac{60}{45 - t_0} \\ \implies 40(45 - t_0) = 60(36 - t_0) \\ \text{therefore } 1800 - 40t_0 \\ 2160 - 60t_0 \\ 60t_0 - 40t_0 = 2160 - 1800 \\ 20t_0 = 360 \\ t_0 = 18cm$

l =l1 + l2 = 3 + 6 = 9H