2020 - JAMB Physics Past Questions and Answers - page 4

Heat energy = Electrical Energy 

Hence,

MC\(\theta\) = power x time 

5 x 1 x 10\(^{-2}\) x C = 100 x 20 

C = \(\frac{200}{5 \times 10^2}\) 

= 4 x 10\(^4\) J/kg/\(^oC\) 

Pressure at H = 40m

Pressure at G, P\(_1\) = 20m

\(\frac{P_2}{P_1} = \frac{40}{20}\)

= 2

V.E = 6, load = 120N 

Efficiency = M.A x \(\frac{1}{V.R}\) x 100%

40 = \(\frac{120}{effort}\) x \(\frac{1}{6}\) x \(\frac{100}{1}\) 

Effort = 50N

R\(_T\) = \(\frac{V}{1} = \frac{12}{2}\) = 6\(\Omega\) 

R\(_T\) = 3 + \(\frac{R \times 6}{R + 6}\) 

R = 6\(\Omega\) 

To determine which instrument consumes the highest current, we need to consider the power rating of each instrument. The power (P) consumed by an electrical appliance is related to the voltage (V) and current (I) by the formula: P = V * I Therefore, the current (I) can be calculated as: I = P / V Let's analyze each option: * **I. Electric iron:** The power rating is not explicitly given, but we have no information about the electric iron in the question, therefore, we cannot determine its power rating. * **II. Television set:** The power rating is 110W, and the voltage is 220V. Therefore, the current drawn would be: I = 110W / 220V = 0.5A * **III. Torch light:** The power rating is 30W, and the voltage is 6V. Therefore, the current drawn would be: I = 30W / 6V = 5A * **IV. Immersion heater:** The power rating is 500W, and the voltage is 250V. Therefore, the current drawn would be: I = 500W / 250V = 2A Comparing the currents, the immersion heater draws the highest current. Therefore, the correct answer is IV.

Let N = count rate at zero time

N \(\frac{3 min}{\to}\)  \(\frac{N}{2}\) \(\frac{3 min}{\to}\) \(\frac{N}{4}\)

\(\frac{N}{4}\) 

= 400

N = \(\frac{400}{N/4}\) x N 

N = 1600