A) 5s
B) 10s
C) 20s
D) 25s
Correct Answer: C
Solution :
Amount of light needed for photographic print \[E\times t=\frac{I}{{{r}^{2}}}\times t\] As die quality of photographic print is same hence \[\frac{{{I}_{1}}}{{{r}_{1}}^{2}}\times {{t}_{1}}=\frac{{{I}_{2}}}{{{r}_{2}}^{2}}\times {{t}_{2}}\] Given, \[{{I}_{1}}=60\,cd\], \[{{r}_{1}}=2\,m\], \[{{t}_{1}}=10\,\,s\], \[{{I}_{2}}=120\,cd\], \[{{r}_{2}}=4\,m\] \[\therefore \] \[\frac{60}{{{2}^{2}}}\times 10=\frac{120}{{{4}^{2}}}\times {{t}_{2}}\] \[\Rightarrow \] \[150=\frac{15}{2}\times {{t}_{2}}\] \[\Rightarrow \] \[{{t}_{2}}=20\,s\]
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