question_answer

A ladder 5 m in length is resting against vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of \[1.5\,m/\sec \]. The length of the highest point of the ladder when the foot of the ladder  \[4.0\,m\] away from the wall decreases at the rate of [Kurukshetra CEE 1996]

A)            2 m/sec

B)            3 m/sec

C)            2.5 m/sec

D)            1.5 m/sec

Correct Answer: A

Solution :

           According to fig. \[{{x}^{2}}+{{y}^{2}}=25\]                         .....(i)                    Differentiate (i) w.r.t. t, we get                    \[2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\]                                  ?..(ii)                    Here \[x=4\] and \[\frac{dx}{dt}=1.5\]                    From (i), \[{{4}^{2}}+{{y}^{2}}=25\Rightarrow y=3\]                    \[\therefore \] From (ii), \[2(4)(1.5)+2(3)\frac{dy}{dt}=0\]                    So, \[\frac{dy}{dt}=-2m/\sec \]                    Hence, length of the highest point decreases at the rate of 2m/sec.