Question:
A man \(1.6\mathrm{m}\) high walks at the rate of \(30\mathrm{m/min}\) away from a lamp which is \(4\mathrm{m}\) above ground. How fast is the man's shadow lengthening?
A man \(1.6\mathrm{m}\) high walks at the rate of \(30\mathrm{m/min}\) away from a lamp which is \(4\mathrm{m}\) above ground. How fast is the man's shadow lengthening?
Show Hint
Use similar triangles to relate shadow length to distance from lamp.
Updated On: Apr 23, 2026
- \(22\mathrm{m/min}\)
- \(20\mathrm{m/min}\)
- \(15\mathrm{m/min}\)
- \(25\mathrm{m/min}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Formula / Definition}
\[ \frac{4}{1.6} = \frac{x + l}{l} \]
Step 2: Calculation / Simplification}
\(\frac{4}{1.6} = 2.5 = \frac{x+l}{l} \Rightarrow 2.5l = x + l \Rightarrow 1.5l = x\)
\(l = \frac{2}{3}x\)
\(\frac{dl}{dt} = \frac{2}{3} \frac{dx}{dt} = \frac{2}{3} \times 30 = 20 \mathrm{m/min}\)
Step 3: Final Answer
\[ 20\mathrm{m/min} \]
\[ \frac{4}{1.6} = \frac{x + l}{l} \]
Step 2: Calculation / Simplification}
\(\frac{4}{1.6} = 2.5 = \frac{x+l}{l} \Rightarrow 2.5l = x + l \Rightarrow 1.5l = x\)
\(l = \frac{2}{3}x\)
\(\frac{dl}{dt} = \frac{2}{3} \frac{dx}{dt} = \frac{2}{3} \times 30 = 20 \mathrm{m/min}\)
Step 3: Final Answer
\[ 20\mathrm{m/min} \]
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