A meterstick is supported at each side by a spring scale. A heavy mass is then hung on the meterstick so that the spring scale on the left hand side reads four times the value of the spring scale on the right hand side. If the mass of the meterstick is negligible compared to the hanging mass, how far from the right hand side is the large mass hanging. (A)25 cm (B) 50 cm (C)67 cm (D)75 cm (E) 80 cm in

80 cm

Explanation:

Suppose the large mass is at point A which is at length 'x' meter  from right, Since it is on meter stick so the length on left will be (1-x) meter .

Since the system is in equilibrium position

clock wise torque around point A= anti clock wise torque around point A

force on left side × distance from point A on left side = force on right side × distance from point A on right side

Since left side is 4 times more reading - it means four times force of tension is acting there as compared to right side !

Let force be F

then,

4F × (1-x)=F × x

==> 4(1-x)=x

==> 4-4x=x

==> 4 =5x

==> x= 4/5=0.8 meter = 0.8 × 100 cm = 80 cm