A) 64.4%
B) 34.4%
C) 53%
D) 55.6%
Correct Answer: D
Solution :
Let I be the luminous intensity of bulb the initial illuminance is: \[E=\frac{I}{{{(2)}^{2}}}\] ?..(1) Final illuminance is given by \[E'=\frac{I}{{{(3)}^{2}}}\] ..?.(2) Percentage change in illuminance is given by \[\frac{\Delta E}{E}\times 100=\frac{E'-E}{E}\times 100%\] \[=\left( \frac{E'}{E}-1 \right)\times 100%\] \[=\left[ \frac{{{\left( \frac{I}{3} \right)}^{2}}}{{{\left( \frac{I}{2} \right)}^{2}}}-1 \right]\times 100%\] \[=\left( \frac{4}{9}-1 \right)100%=-\frac{5}{9}\times 100%\] \[=-55.6%=55.6%\] decreaseYou need to login to perform this action.
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