Rotational Equilibrium

Assertion : The condition of equilibrium for a rigid body is Translational equilibrium Sigma F=0 and Rotational equilibrium Sigma tau=0 Reason : A rigid body musy be in equilibrium under the action of two equal and opposite forces.

Fixed Axis OF Rotation||Equilibrium Basic Problem(Rod Problem)||Pulley Problem

The optical rotation of the alpha -form of a pyranose is +150.7^(@) , that of the beta -form is +52.8^(@) . In solution an equilibrium mixture of these anomers has an optical rotation of +80.2^(@) . The precentage of the alpha -form in equilibrium mixture is :

Different Concepts on Calculation OF Equilibrium constant||Heterogeneous Equilibrium

What is work done on dipole to rotate it form stable equilibrium position (theta_(1) = 0^(@)) to unstable equilibrium (theta_(2) = 180^(@)) ?

The specific rotation of alpha- glucose is +112^(@) and beta- glucose is +19^(@) and the specific rotation of the constant equilibrium mixture is +52.7^(@) . Calculate the percentage composition of anomers (alpha and beta) in the equilibrium mixture.

The specific rotation of two glucose anomers are alpha = + 110^(@) and beta = 19^@ and for the constant equilibrium mixtures is +52.7^@ . Calculate the percentage compositions of the anomers in the equilibrium mixture.

How much of the alpha -anomer and beta -anomers are present in an equilibrium mixture with a specific rotation of +52.6^(@) ?

A short electric dipole of dipole moment P is placed in stable equilibrium in uniform electric field E as shown in the figure. The work done by the external agent to rotate the dipole slowly up to its unstable equilibrium will be