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