question_answer

A man 1.6 m high walks at the rate of\[30\text{m}/\min \] away from a lamp which is\[4m\]above ground. How fast is the mans shadow lengthening?

A) \[22\text{m/min}\]                        

B) \[20\text{m/min}\]

C) \[15\text{m/min}\]        

D)        \[25\text{m/min}\]

Correct Answer: B

Solution :

Let\[PQ=4m\]be the height of pole and, \[AB=1.6m\] be height of man, Let the end of shadow and it is at a distance of\[l\]from\[A\]when the man is at a distance\[x\]from \[PQ\] at some instant. Since, \[\Delta PQR\]and\[\Delta ABR\]are similar, we have,\[\frac{PQ}{AB}=\frac{PR}{AR}\] \[\Rightarrow \]               \[\frac{4}{1.6}=\frac{x+l}{l}\] \[\Rightarrow \]               \[2x=3l\] \[\Rightarrow \]               \[\frac{2\,\,dx}{dt}=3\cdot \frac{dl}{td}\] \[\left\{ \text{given}\,\,\frac{dx}{dt}=30\,\,\text{m}/\min  \right\}\] \[\Rightarrow \]               \[\frac{dl}{dt}=\frac{2}{3}\times 30\text{m}/\min =20\text{m}/\min \]