COMEDK Exam 2015 Solved Paper

Let AB be the ladder of length 5m.
We are given, ‌

dx

dt

=2m∕ secIn △ABC,
‌AB2=AC2+BC2
⇒(5)2=x2+h2 . . . (i)
Differentiating Eq. (i) w.r.t. t, we get
0‌=2x‌

dx

dt

+2h‌

dh

dt

⇒‌

dh

dt

‌=‌

−x

h

‌

dx

dt

. . . (ii)
Now, from Eq. (i), when x=4
‌h2=25−16
‌⇒h2=9⇒h=3m
‌‌ From Eq. (ii), we get ‌[‌

dh

dt

]=‌

−4

3

×2=‌

−8

3

‌∵ The negative sign shows the height decreases and decreasing rate is ‌

8

3

m∕ sec.