Physics Questions and Answers

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

let the masses of the two objects be M1 and M2
and their distance apart = r

Therefore the force acting F1 = \( \frac{GM1M2}{r^2} \)

for a new distance of r/2, we have a force of

F2 = \( \frac{GM2M2}{r/2} \\

\text{Thus } F1 \times r^2 = GM1M1 \text{and} \\
F2 \times (r/2)^2 = GM1M1 \\

\text{Since } GM1M1 = GM1M1 \\

\implies F1 \times r^2 = F2 \times (r/2)^2 \\

\text{Therefore } 200 \times r^2 = F2 \times r^2/4 \\

F2 = \frac{4 \times 200 \times r^2}{r^2} = 800N \)