2001 - JAMB Physics Past Questions and Answers - page 3

Total power in watt = 605 + 4100

= 300 + 400

= 700w

Total power in kilo watt = 700/100

= 0.7 killo w

Unit of electricity in killo watt-hour

= 0.7 * 20hrs

= 14 okwh

i.e If I kwh costs # 10.00

14kwh = 10.00 * 14 = #140

In the Daniel cell,

i. Depolarize = CuSO₄ solution

ii. Positive electrode = Copper container

iii Negative electrode = zinc rod/plate

Electrostatic force = F_E = e²/4λEod²

Gravitational force = F_G = GM²/d²

let the internal resistance of the cell = r,

therefore E = I (R+r)

(i) E = 0.4(4+r), (ii) E = 0.2 (10+r)

therefore 0.4 (4+r)= 0.2 (10+r)

1.6 + 0.4r = 2.0 * 0.2r

0.4r - 0.2r = 2.0 - 1.6

therefore 0.2r = 0.4 => r = 0.4/0.2 = 2Ω

therefore r = 2Ω

³⁹₁₉K ³⁹₁₉K.

Since neither the mass number (39), nor the atomic number (19) is affected by the emission,the particles emitted can only be gamma ray, which is not a charge particle, but simply a radiation.

Electricity conduction through gases is usually done under low pressure (I) and high voltage (III). Thus /B/

The given a.c. current is $I(t) = 2 \sin(\omega t)$. The d.c. equivalent of an a.c. current is the root-mean-square (rms) value of the current. The rms value of a sinusoidal current is given by:

$I_{rms} = \frac{I_{peak}}{\sqrt{2}}$

In this case, the peak current, $I_{peak} = 2A$. Therefore,

$I_{rms} = \frac{2}{\sqrt{2}} = \sqrt{2}A$

For the transformer, V ∝ ∩

=> V _s ∝ ∩_s; V_p ∝ ʌ_p

=> V_p/V_v = ∩_p/∩_s, therefore 120/1200 = ∩_p/1000

therefore ∩_p * 1200 = 120 * 1000

∩_p = 120*1000/1200 = 100

The ability of a material to shield against ionizing radiation is related to its density and atomic number. Materials with higher density and higher atomic number are more effective at absorbing and attenuating radiation.

• Lead (Pb) has a high atomic number and high density.
• Iron (Fe) has a lower atomic number and density than lead.
• Manganese (Mn) has a lower atomic number and density than iron.
• Aluminum (Al) has a significantly lower atomic number and density than iron.

Therefore, lead provides the greatest shield against ionizing radiation.