A) 1 cm, 9.3 cm
B) 1 cm, 10.5 cm
C) 2 cm, 12.8 cm
D) 2 cm, 15.2 cm
Correct Answer: A
Solution :
\[{{h}_{0}}=\sqrt{{{h}_{1}}{{h}_{2}}}=\sqrt{4.1\times 0.48}\] | |
\[=\sqrt{0.7\times 3\times 0.16\times 3}=3\times 0.4\times \sqrt{0.7}\]\[=1.2\times \sqrt{0.7}\]\[=1\operatorname{cm}\] | |
For case 1 | |
\[a\frac{{{h}_{i}}}{{{h}_{0}}}=\frac{2.1}{1}=2.1\operatorname{cm}=\frac{y}{x}\] | |
\[y=2.1x\] | ? (i) |
\[y-x=15\operatorname{cm}\] |
\[2.1x-x=15\] |
\[1.1x=15\] |
\[x=\frac{150}{11}\operatorname{cm}=13.6cm\] |
\[\frac{1}{2.1x}-\frac{1}{-x}=\frac{1}{f}\]\[\frac{2.1+1}{2.1x}=\frac{1}{f}\Rightarrow f=\frac{2.1x}{3.1}=\frac{21}{31}\times \frac{150}{11}=9.3\operatorname{cm}\] |
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