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The illuminance of a surface varies inversely with the square of its distance from the light source. If the illuminance of a surface is 120 lumens per square meter when its distance from a certain light source is 6 meters, by how many meters should the distance of the surface from the source be increased to reduce its illuminance to 30 lumens per square meter?(a) 3(b) 6(c) 12(d) 15(e) 18

User Eironeia
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Answer: b) 6

Explanation:

Given : The illuminance of a surface varies inversely with the square of its distance from the light source.

i.e. for d distance and l luminance , we have


d^2* l=k , where k is constant. (1)

If the illuminance of a surface is 120 lumens per square meter when its distance from a certain light source is 6 meters.

From (1), we have


\Rightarrow (6)^2* 120=k\\\\\Rightarrow\ k=4320 (2)

For the distance (d) corresponds to the illuminance to 30 lumens per square meter , we have


d^2* 30=k

Put value of k , we get


d^2* 30=4320\\\\\Rightarrow\ d^2=(4320)/(30)=144\\\\\Rightarrow\ d^2=144\\\\\Rightarrow\ d= 12

Then , the number of meters should the distance of the surface from the source be increased= 12 meters- 6 meters = 6 meters.

User Paulchen
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