Answer:

A particle has a displacement of 40M along OX in the first 5s and 30m perpendicular to OX in

the next 5s. The magnitude of average velocity of the particle during the TIME internal of 10s is

a) 6 ms–1 b) 7 ms–1 c) 8 ms–1 d) 5 ms–1 [ A uniform metre scale has two weights of 10 gf and 8 gf suspended at the 10 cm and 80 cm marks respectively. If the metre scale itself weights 50 gf, find where MUST the weight be, so that the metre scale stays balanced?.

At the balancing condition, Anticlock wise moment must be equal to clock wise moment.

Clockwise moment, about the balancing point is

= moment by 80 gf

=80gf×(80−X)cm

=6400gfcm−80Xgfcm.....(1)

Anticlock wise moment about the balancing point is

= moment by 10 gf + moment by 50 gf

=10gf×(X−10)+50gf×(X−50)

=10Xgfc,−100gfcm+50Xgfcm−2500gfcm.....(2)

At the balancing condition, Clock wise at equilibrium = ANTI clock wise moments.

6400−80X=100X−100+50X−2500

6400+2600=140X⇒140X=8000

⇒X=1408000=64.28

Scale is balanced at 64.28 cm from the beginning.