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