JAMB Physics Past Questions & Answers - Page 68

Water 1 => M = 150g = 0.15kg,

  θ$_2$ = 60$^o$C; θ$_1$ = θ

  Q$_1$ = MC (θ$_2$ – θ$_1$)

  = 0.15C (60 - θ)

  Water 2 => M = 300g = 0.3kg

  θ1 = 20$^o$C, θ$_2$ = θ

  Q2 = MC (θ$_2$ – θ$_1$)

  = 0.3C (θ – 20)

  Combining Q1 and Q2

  0.15C (60 - θ) = 0.3C (θ – 20)

  9 – 0.15θ = 0.3θ – 6

  0.3θ + 0.15θ = 9 + 6

  0.45 θ = 15

  θ = 33.33$^o$C

In case of small drops of mercury, the gravitational potential energy is negligible in comparison to the potential energy due to surface tension.Consequently, to keep the drop in equilibrium, the mercury drop’s surface tends to contract so that its surface area will be the least for a sphere and the drops will be spherical.

  But in the case of bigger drops of mercury, the potential energy due to gravity is predominant over the potential energy due to surface tension.Consequently, to keep equilibrium , the mercury drop tends to assume minimum potential energy as possible, the drop becomes oval in shape and lower center of gravity.

P = 40W

  R = 90W

  V = ?

  P = $\frac{V^2}{R}$

  V$^2$ = PR = 40 x 90

  V$^2$ = 3000

  V = √3600

  V = 60N

Fuse wires have a low melting point so that if a huge amount of current is passed through it to any appliance than that wire would melt and stop the appliance from getting damaged by not letting the current pass any further.

Recall V = F × $\lambda$

= 260 × 1.29

= 335.4ms$^{-1}$

Echo time = $\frac{2}{2}$ = 1 sec

Distance = v x t = 335.4 x 1

= 335.4m