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JEE Main 16 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 5
Total: 90
A mosquito is moving with a velocity
v
=
(
0.5
t
2
∧
i
+
3
t
∧
j
+
9
∧
k
)
m
∕
s
and accelerating in uniform conditions. What will be the direction of mosquito after 2s?
[16 Mar 2021 Shift 2]
tan
−
1
(
2
3
)
from
X
-axis
tan
−
1
(
2
3
)
from
γ
-axis
tan
−
1
(
5
2
)
from
Y
-axis
tan
−
1
(
5
2
)
from
X
-axis
None of the above
Validate
Solution:
Given, velocity of mosquito,
v
=
(
0.5
t
2
∧
i
+
3
t
∧
j
+
9
∧
k
)
m
∕
s
At
t
=
2
s
,
v
=
0.5
(
2
)
2
∧
i
+
3
(
2
)
∧
j
+
9
∧
k
=
2
∧
i
+
6
∧
j
+
9
∧
k
∴
Angle along
X
-axis,
cos
α
=
2
√
2
2
+
6
2
+
9
2
=
2
√
121
=
2
11
or
α
=
cos
−
1
(
2
11
)
Similarly, angle along
Y
-axis,
cos
β
=
6
11
⇒
β
=
cos
−
1
(
6
11
)
Angle along
Z
-axis,
cos
γ
=
9
11
⇒
γ
=
cos
−
1
(
9
11
)
∴
Angle made by the direction of motion of mosquito will be
cos
−
1
(
2
11
)
from
X
-axis
=
tan
−
1
√
117
2
cos
−
1
(
6
11
)
from
Y
-axis
=
tan
−
1
√
85
2
cos
−
1
(
9
11
)
from
Z
-axis
=
tan
−
1
(
√
40
9
)
© examsnet.com
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