JAMB Physics Past Questions & Answers - Page 11

A pyrometer is a type of remote-sensing thermometer used to measure the temperature of a surface. It does this by measuring the thermal radiation or infrared energy being emitted from the object. Therefore, a pyrometer thermometer measures temperature from the thermal radiation emitted by objects.

Quantity of heat required to raise the temperature of the ice cube from $-10^oC$ to $0^oC$
⇒H=mc►ø = 0.02 × 2100 × 0 -(-10)

⇒ H = 0.02 × 2100 × (0 + 10) = 0.02 × 2100 × 10

⇒ H = 420 J

quantity of heat required to melt ice at $0^oC$

⇒ H = mL = 0.02 × 3.34 × $10^5$

⇒H=6680 j

Quantity of heat required to raise the temperature of the melted ice cube (water) from $0^oC$ to $-10^oC$

⇒H=mc►ø = 0.02 × 4200 × (10-0)

⇒H = 0.02 × 4200 × 10

⇒H = 840j

;420 + 6680 + 840 = 7940 j

 

A= $0.8m^2$    d= 20mm =$\frac{20}{1000}$ = 0.02m

v =120v;  $ε_oA$= $8.85 × 10^-12 fm^-1$

C = $\frac{ε_oA}{d}$

C =$\frac{8.85 × 10^-12×0.8}{0.02}$

C= $3.54 × 10^-10 F$

Q= CV

⇒  $3.54 × 10^-10$ × 120 = $4.25×10^-8c$

     Q=$42.5 × 10^-9c$  = 42.5nC

      

$r_1$ = 4.5cm , $P_1$ =is the total pressure on the bubble at a depth of 12m from the surface.

$P_1$ = 12 + 10.34 =22.34m

 $V_1$ = $\frac{4}{3}$π× $r^3_1$

= $\frac{4}{3}× π×{4.5^3cm^3}$

$P_2$ = 10.34m

$V_2$  = $\frac{4}{3} {π}{r^3_2}$

from boyles law:

$P_1V_1$  = $P_2V_2$ 

⇒ 22.34× $\frac{4}{3}× π×{4.5^3}$ = 10.34 × $\frac{4}{3}×π×{r^3_2}$

⇒ 22.34 × $4.5^3$ = 10.34 × $r^3_2$

⇒ $r^3_2 = \sqrt[3]{196.88}$

⇒ $r_2$ = 5.82cm

ε= 80%, L=200× 10 = 2000N, $d_L$=10m
 

$d_E$= 110-10 = 100m, E=?

ε = $\frac{work done on the load}{work done by the effort}$  × 100%

⇒  work done on load = 2000 ×10 =20,000 j


⇒ work done by effort = E × 100  = 100E

⇒ 80 =   $\frac{20,000}{100E}$ × 100%

⇒ 80 =   $\frac{20,000}{E}$ 

⇒ E = $\frac{20,000}{80}$

              ⇒  E = 250N