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JEE Main 10 Apr 2019 Paper 1 Solved Paper
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© examsnet.com
Question : 1
Total: 90
PHYSICS
A particle of mass m is moving along a trajectory given by
x
=
x
0
+
a
cos
ω
1
t
y
=
y
0
+
b
sin
ω
2
t
The torque,acting on the particle about the origin, at t=0 is :
+
m
y
0
a
ω
1
2
^
k
m
(
−
x
0
b
+
y
0
a
)
ω
1
2
^
k
−
m
(
x
0
b
w
2
2
−
y
0
a
w
1
2
)
^
k
Zero
Validate
Solution:
→
r
=
(
x
o
+
a
cos
ω
1
t
)
i
+
(
y
0
+
b
sin
ω
2
t
)
j
;
at t=0
→
r
=
(
x
0
+
a
)
i
+
y
0
j
→
a
=
d
2
x
d
t
2
^
i
+
d
2
y
d
t
2
^
j
→
r
=
(
−
a
ω
1
2
cos
ω
1
t
)
^
i
+
(
−
b
ω
2
2
sin
ω
2
t
)
j
;
at t=0
→
a
=
−
a
ω
1
2
^
i
T
=
→
r
×
→
F
=
m
(
→
r
×
→
a
)
=
m
(
y
o
(
−
a
ω
1
2
)
−
k
)
=
m
y
0
a
ω
^
i
^
k
© examsnet.com
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