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AP EAMCET Engineering 17 Sep 2020 Shift 2 Paper
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© examsnet.com
Question : 13
Total: 160
A man, 2
m
tall, walks at the rate of 1
2
3
m
/
s
towards a street light which is 5
1
3
m
above the ground. The rate at which the length of his shadow is changing when he is 3
1
3
m
m away from the base of the light is _______
−
1
m
/
s
2
m
/
s
−
2
m
/
s
1
m
/
s
Validate
Solution:
A
B
=
street light,
C
=
Man.
From
∆
A
B
E
and
∆
D
C
E
,
A
B
C
D
=
A
E
C
E
=
A
C
+
C
E
C
E
⇒
A
B
C
D
=
A
C
C
E
+
1
⇒
A
C
C
E
=
A
B
C
D
−
1
⇒
A
C
C
E
=
5
1
3
2
−
1
=
5
3
⇒
C
E
=
3
5
A
C
Differentiating with respect to time we get,
d
d
t
C
E
=
3
5
(
d
d
t
A
C
)
....(i)
None
d
d
t
A
C
=
−
1
2
3
m
s
−
1
d
d
t
A
C
=
−
5
3
m
s
−
1
Negative sign as
A
C
decreases with time.
substituting in Eq. (i) we get.
Rate of change of shadow length
=
3
5
(
−
5
3
)
=
−
1
m
s
−
1
© examsnet.com
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