Answer:
Option a.
Total area of the cells to generate 6.7 kW, A = 61.4
![m^(2)](https://img.qammunity.org/2020/formulas/physics/high-school/a8bdskptcop3l9g8p0t7hyfjtlohmn36.png)
Given:
The efficiency of the cell,
= 9.1%
Power generated by the cells, P = 6.7 kW
Intensity of light, I = 1200 W/
![m^(2)](https://img.qammunity.org/2020/formulas/physics/high-school/a8bdskptcop3l9g8p0t7hyfjtlohmn36.png)
Solution:
Using the formula for intensity of light, I =
![(P)/(\eta _(cell)A)](https://img.qammunity.org/2020/formulas/physics/college/sz3689t3rplfb9eq415txrdo69ghr8go5h.png)
rearranging the above formula for Area, A:
Area, A =
![(P)/(I\eta _(cell))](https://img.qammunity.org/2020/formulas/physics/college/d3zhzlw9mqw8bwmbgjrikem0p0du4g12hb.png)
Now, putting values in the above formula:
Area, A =
![(6.7* 10^(3))/(1200* 0.091)}](https://img.qammunity.org/2020/formulas/physics/college/7j99rqb5113qj41p646hvxyrcwh0z2zvxo.png)
Area, A = 61.355 ≈ 61.4
![m^(2)](https://img.qammunity.org/2020/formulas/physics/high-school/a8bdskptcop3l9g8p0t7hyfjtlohmn36.png)