JEE Main 17 March 2021 Shift 1 Solved Paper

Given, F=4

∧

i

−3

∧

j

Resolving the components of position vector in horizontal and vertical direction.
∴ Horizontal component, rx=r‌cos‌θ
=10‌cos‌60∘ [∵cos‌60∘=‌

1

2

]
=10×‌

1

2

=5 units
Vertical component, ry=r‌sin‌θ
=10‌sin‌60∘  [∵sin‌60∘=‌

√3

2

]
=10×‌

√3

2

=5√3 units
∴ Position vector w.r.t. O,rO=5

∧

i

+5√3

∧

j

and position vector w.r.t. Q,
rQ=−5

∧

i

+5√3j
Torque at point P w.r.t. O,
τO=rO×F=(5

∧

i

+5√3

∧

j

)×(4

∧

i

−3‌

∧

j

)
=(−15−20√3)

∧

k

=(15+20√3)(−

∧

k

)
Torque at point P w.r.t. Q,
τQ=rQ×F=(4

∧

i

−3

∧

j

)×(−5

∧

i

+5√3

∧

j

)
=(−15+20√3)

∧

k

=(15−20√3)(−

∧

k

)