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UPSC NDA Math Model Paper 1
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© examsnet.com
Question : 3
Total: 120
Forces
3
∧
i
+
2
∧
j
+
5
∧
k
and
2
∧
i
+
∧
j
−
3
∧
k
are acting on a particle and displace it from the point
2
∧
i
−
∧
j
−
3
∧
k
to the point
4
∧
i
−
3
∧
+
7
k
The work done by the force is:
18 units
30 units
24 units
36 units
Validate
Solution:
Concept:
If two points
A
and
B
have position vectors
→
A
and
→
B
respectively, then the vector
→
A
B
=
→
B
−
→
A
For two vectors
→
A
and
→
B
at an angle θ to each other:
Dot Product is defined as:
→
A
.
→
B
=
|
→
A
|
|
→
B
|
c
o
s
θ
Resultant Vector is equal
→
A
+
→
B
.
Work: The work (W) done by a force
(
→
F
)
in moving (displacing) an object along a vector
→
D
is given by:
W
=
→
F
.
→
D
=
|
→
F
|
|
→
D
|
c
o
s
θ
Calculation:
Let's say that the forces acting on the particle are
→
F
1
=
3
∧
i
+
2
∧
j
+
5
∧
k
and
→
F
2
=
2
∧
i
+
∧
j
−
3
∧
k
∴
The resulting force acting on the particle will be
→
F
=
→
F
1
+
→
F
2
⇒
→
F
=
(
3
∧
i
+
2
∧
j
+
5
∧
k
)
+
(
2
∧
i
+
∧
j
−
3
∧
k
)
⇒
→
F
=
5
∧
i
+
3
∧
j
+
2
∧
k
.
Since the particle is moved from
2
∧
i
+
2
∧
j
−
3
∧
k
to the point
4
∧
i
−
3
∧
j
+
7
∧
k
, the displacement vector
→
D
will be:
→
D
=
(
4
∧
i
−
3
∧
j
+
7
∧
k
)
−
(
2
∧
i
+
∧
j
−
3
∧
k
)
⇒
→
D
=
2
∧
i
−
2
∧
j
+
10
∧
k
And finally, the work done W will be:
W
=
→
F
.
→
D
=
(
5
∧
i
+
3
∧
j
+
2
∧
k
)
.
(
2
∧
i
−
2
∧
j
+
10
∧
k
)
⇒
W
=
(
5
)
(
2
)
+
(
3
)
(
−
2
)
+
(
2
)
(
10
)
⇒
W
=
10
−
6
+
20
=
24
units
© examsnet.com
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