JEE Main 2019MathematicsApplication of DerivativesEasyMCQ

JEE Main 2019 — Application of Derivatives Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25cm/sec , then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is:

Choose an option

Show full solutionCorrect option: D
Correct answer
D253

Step-by-step explanation


dydt=-25 cm/sec , dxdt=?
Now, x2+y2=22=4
differentiating w.r.t. t both side
2xdxdt+2ydydt=0 x2+y2=4 when y=1 ⇒x2=3⇒x=3
⇒dxdt=-yxdydt
⇒dxdt 3,1=-13×-25=253 cm/sec

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Application of Derivatives chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, covering the Application of Derivatives chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.