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Standard XII
Physics
NCERT
Question
The potential energy for a force field
→
F
is given by U(x, y) = sin(x + y). Magnitude of the force acting on the particle of mass m at
(
0
,
π
4
)
is
1
√
2
1
√
2
0
A
1
√
2
B
0
C
1
D
√
2
Open in App
Solution
Verified by Toppr
→
F
=
(
x
,
y
)
=
−
∇
U
(
x
,
y
)
∇
U
(
x
,
y
)
=
∂
U
∂
x
ˆ
x
+
∂
U
∂
y
ˆ
y
=
cos
(
x
+
y
)
(
ˆ
x
+
ˆ
y
)
F
(
x
,
y
)
=
|
∇
U
(
x
,
y
)
|
F
(
0
,
π
4
)
=
|
∇
U
(
x
,
y
)
|
=
∣
∣
cos
(
π
4
)
∣
∣
|
ˆ
x
+
ˆ
y
|
F
(
0
,
π
4
)
=
1
√
2
√
1
2
+
1
1
=
1
u
n
i
t
Hence,
option
(
A
)
is correct answer.
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4
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→
F
is given by U(x, y) = sin(x + y). Magnitude of the force acting on the particle of mass m at
(
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