(n-1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector a with respect to the centre of the polygon. Find the position vector of centre of mass.

Case – I : When disk slides down

$t_1=\sqrt[]{\frac{2L}{gsin\theta }}..........\left(i\right)$              

Case – II : When disk rolls down

$t_2=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\beta }^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\left(\frac{k}{R}\right)}^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\left(\frac{R/\sqrt[]{2}}{R}\right)}^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+\frac{1}{2}}}}=\sqrt[]{\frac{4L}{3gsin\theta }}......\left(ii\right)$             

$\Rightarrow \frac{t_1}{t_2}=\sqrt[]{\frac{2L}{gsin\theta }}\times \sqrt[]{\frac{3gsin\theta }{4L}}=\sqrt[]{\frac{3}{2}}$              

This is a multiple choice type question as classified in NCERT Exemplar

a, b, d

a) according to the perpendicular axes theorem statement 1 is wrong

b) As z’|z so distance between them = a $\frac{\sqrt[]{2}}{2}=\frac{a}{\sqrt[]{2}}$

So according to parallel axes theorem I_z’=I_z+m (a/ $\sqrt[]{2}$ )²= I_z+ma²/2

Hence b is true

c) z’ is not parallel to z hence Parallel axes does not applied so statement is false

d) as x and y axes are symmetrical . hence I_x=I_y so d is true

d) as x and y axes are symmetrical . hence I_x=I_y so d is true

This is a multiple choice type question as classified in NCERT Exemplar

b, c

a) When r>r’

Torque about z-axis t=r $\times$ F

b) t’=r’ $\times F$ which is along negative z axis

c) t_z=Fr = magnitude of torque about z axis where r is perpendicular between F and z axis so torque along positive z axis is greater than negative z axis.

d) We are always calculating resultant torque about common axis. Hence total torque not equal to combination of torque along both axis of z, because they are not on common axis.

This is a multiple choice type question as classified in NCERT Exemplar

a, b, c, d

As we know torque = r $\times$ F = rFsin $\theta$

a) when forces act radially angle =0 hence torque =0

b) when forces are acting on the axis of rotation r=0 torque=0

c) when forces acting parallel to the axis of rotation angle =0 so torque =0

d) when torque by forces are equal and opposite torque net = t₁-t₂=0