question_answer

Two candles of equal heights, 15 cm, each are placed in between vertical screens at a distance of 10 cm from each other and also from the nearer screen.

Candle A burns completely in \[1.5\text{ }h\]and candle B burns completely in 1 h. Shadows of A and B moves with speeds

A) \[\frac{1}{4}cm\,{{s}^{-\,1}},\frac{1}{8}cm\,{{s}^{-1}}\]       

B) \[\frac{1}{12}cm\,{{s}^{-\,1}},\frac{1}{12}cm\,{{s}^{-\,1}}\]

C) \[\frac{1}{3}cm\,{{s}^{-\,1}},\frac{1}{12}cm\,{{s}^{-\,1}}\]   

D) \[\frac{1}{3}cm\,{{s}^{-\,1}},\frac{1}{4}cm\,{{s}^{-\,1}}\]

Correct Answer: C

Solution :

Let candles burn down by \[{{l}_{1}}\]and \[{{l}_{2}}\] and the shadows shorten by \[{{x}_{1}}\]and \[{{x}_{2}}.\]

From similar triangles \[\Delta ABG,\]\[\Delta DCG\]and \[\Delta FEG,\]we have

\[\frac{{{x}_{1}}-{{x}_{2}}}{3a}=\frac{{{l}_{1}}-{{x}_{2}}}{2a}=\frac{{{l}_{2}}-{{x}_{2}}}{a}\]

\[\Rightarrow \]\[2\,({{x}_{1}}-{{x}_{2}})=3\,({{l}_{1}}-{{x}_{2}})=6\,({{l}_{2}}-{{x}_{2}})\]

where, \[{{l}_{1}}=\frac{h}{{{t}_{1}}}=\frac{15}{1.5\times 60}=\frac{1}{6}cm{{s}^{-\,1}}\]

and \[{{l}_{2}}=\frac{h}{{{t}_{2}}}=\frac{15}{1\times 60}=\frac{1}{4}cm{{s}^{-\,1}}\]

Let \[{{x}_{1}}={{v}_{A}}\]and \[{{x}_{2}}={{v}_{B}}.\]Then,

\[2\,({{v}_{A}}-{{v}_{B}})=3\left( \frac{1}{6}-{{v}_{B}} \right)=6\left( \frac{1}{4}-{{v}_{B}} \right)\]

\[\Rightarrow \]\[{{v}_{A}}=\frac{1}{3}cm{{s}^{-\,1}}\]

\[{{v}_{B}}=\frac{1}{12}cm{{s}^{-\,1}}\]