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Manipal Entrance Test 2013 Physics Paper
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© examsnet.com
Question : 32
Total: 60
A man crosses a 320 m wide river perpendicular to the current in 4 min. If in still water he can swim with a speed
5
3
times that of the current, then the speed of the current, in mm
−
1
in is
30
40
50
60
Validate
Solution:
v
r
2
=
v
m
2
−
v
2
v
=
320
4
m/min = 80 m/min
v
m
=
5
3
v
r
v
r
2
=
t
5
3
(
v
r
)
2
−
(
80
)
2
16
9
v
r
2
=
(
80
)
2
v
r
=
60
m/min
u
2
,
u
0
form canours linear
© examsnet.com
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