Class 11 NEET Physics Work, Energy and Power Questions

Given that
Mass of rod = 500 g; Length of rod = 200 cm
Rod will be in equilibrium, when net torque about point O will be zero.
Torque at point O due to 2 kg mass

→

τ

=

→

r

×

→

F

=rF‌sin‌θ(

^

n

)
τ1=20×20×10−2×sin‌90°(

^

k

) =4Nm(

^

k

)
Torque due to mass of rod :
τ2=5×60×10−2×sin‌90°(−

^

k

) =3Nm(−

^

k

)
Torque due to mass m
τ3=mg×120×10−2×sin‌90°(−

^

k

) =12mNm(−

^

k

)
Net torque about point O will be zero
So

→

τ

1+

→

τ

2+

→

τ

3=0
⇒4−3−12m=0
⇒12m=1
m=

1

12

kg