To prepare a print,the time taken is 5, sec d | vedclass

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(C) To develop a print,a fixed amount of energy is required. The total light energy incident on the photo print is proportional to the intensity of li

Vedclass · Accepted Answer

$12.8$

Vedclass · Answer

(C) To develop a print,a fixed amount of energy is required. The total light energy incident on the photo print is proportional to the intensity of light multiplied by time.Intensity $I \propto \frac{1}{r^2}$,where $r$ is the distance from the source.Thus,the energy $E = I \cdot A \cdot t = k \cdot \frac{1}{r^2} \cdot t$,where $k$ is a constant.Since the energy required for the print is constant,we have $\frac{t_1}{r_1^2} = \frac{t_2}{r_2^2}$.Given: $t_1 = 5\, sec$,$r_1 = 0.25\, m$,$r_2 = 40\, cm = 0.40\, m$.Substituting the values: $t_2 = t_1 	imes \left( \frac{r_2}{r_1} ight)^2$.$t_2 = 5 	imes \left( \frac{0.40}{0.25} ight)^2 = 5 	imes (1.6)^2$.$t_2 = 5 	imes 2.56 = 12.8\, sec$.

To prepare a print,the time taken is $5\, sec$ due to a lamp of $60\, W$ at a $0.25\, m$ distance. If the distance is increased to $40\, cm$,what is the time taken to prepare the similar print in seconds?

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