Question

A long one meter wooden scale of uniform mass density, is allowed to rotate about an axis passing through its center normal to its length[50th cm], an object of weight X gram weight is hanged at 10th cm of the scale this weight is balanced by the known weight of 200 gram weight hanged at 90th cm of the same scale the weight X in gram weight is

Answer 1

Explanation:

Buoyancy force will act on the center of submerged part of plank ,

distance from O is d=

2cosθ

0.5

distance of center of mass from is a=1/2=0.5m.

let the cross section be A and density of water be ρ

weight of plank is w=1×A×0.5ρ×g,}

writing torque about O

we get Bdcosθ=W×0.5×cosθ

we get cos

2

θ=1/2

so

cos

2

θ

1

=2

Answer 2

Given:

A long one meter wooden scale of uniform mass density, is allowed to rotate about an axis passing through its center normal to its length[50th cm], an object of weight X gram weight is hanged at 10th cm of the scale this weight is balanced by the known weight of 200 gram weight hanged at 90th cm of the same scale.

To find:

The value of X gram?

Calculation:

In-order to attain rotational equilibrium the net torque experienced by the wooden scale should be zero with respect to an axis passing through the centre of the scale.

• Now , force of 200 gm is acting on 90 cm , i.e. it is (90 - 50) = 40 cm from centre.

• Similarly , force of X gram is acting on 10 cm, i.e. it is (50 - 10) = 40 cm from centre.

So, we can say :

$\therefore \sf{ \sum( \tau) = 0}$

$= > \sf{ (Xg \times 40) - (200g \times 40)= 0}$

$= > \sf{ Xg \times 40 = 200g \times 40}$

$= > \sf{ X = 200 \: grams}$

So, value of X is 200 grams.