$$v = \sqrt {gr} = \sqrt {10 \times 40} = 20\,m{s^{ - 1}}$$

According to given question, a uniform circular disc of radius $$50\,cm$$  at rest is free to turn about an axis having perpendicular to its plane and passes through its centre. This situation can be shown by the figure given below :

$$\eqalign{ & \therefore {\text{Angular}}\,{\text{acceleration,}}\,\alpha = 2\,rad\,{s^{ - 2}}\,\,\left( {{\text{given}}} \right) \cr & {\text{Angular}}\,{\text{speed,}}\,\omega = \alpha t = 4\,rad\,{s^{ - 1}} \cr & \because {\text{Centripetal}}\,{\text{acceleration,}}\,{a_c} = {\omega ^2}r = {\left( 4 \right)^2} \times 0.5 \cr & = 16 \times 0.5 \cr & {a_c} = 8\;m/{s^2} \cr} $$
$$\because $$ Linear acceleration at the end of $$2\,s,$$
$$\eqalign{ & {a_t} = \alpha r = 2 \times 0.5 \cr & \Rightarrow {a_t} = 1\;m/{s^2} \cr} $$
Therefore, the net acceleration at the end of $$2.0\,s$$  is given by
$$\eqalign{ & a = \sqrt {a_c^2 + a_t^2} \cr & a = \sqrt {{{\left( 8 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {65} \cr & \Rightarrow a \approx 8\;m/{s^2} \cr} $$

Net force on particle in uniform circular motion is centripetal force $$\left( {\frac{{m{v^2}}}{l}} \right)$$  which is provided by tension in string so the net force will be equal to tension i.e., $$T.$$

The angle of banking $$\tan \theta = \frac{{{v^2}}}{{rg}}$$
$${\text{Given,}}\,\theta = {45^ \circ }$$
Radius of banked curve road
$$\eqalign{ & r = 90\,m\,{\text{and}}\,g = 10\,m/{s^2} \cr & \Rightarrow \tan {45^ \circ } = \frac{{{v^2}}}{{90 \times 10}} \cr & v = \sqrt {90 \times 10 \times \tan {{45}^ \circ }} \cr & = \sqrt {90 \times 10 \times 1} \cr & = 30\,m/s \cr} $$

On a banked road, $$\frac{{V_{\max }^2}}{{Rg}} = \left( {\frac{{{\mu _s} + \tan \theta }}{{1 - {\mu _s}\tan \theta }}} \right)$$
Maximum safe velocity of a car on the banked road
$${V_{\max }} = \sqrt {Rg\left[ {\frac{{{\mu _s} + \tan \theta }}{{1 - {\mu _s}\tan \theta }}} \right]} $$

Torque $$\left( \tau \right)$$ acting on a body and angular acceleration $$\left( \alpha \right)$$ produced in it are related as $$\tau = I\alpha $$
Consider a hollow cylinder, around which a rope is wounded as shown in the figure.

Torque acting on the cylinder due to the force $$F$$ is $$\tau = Fr$$
Now, we have $$\tau = I\alpha $$
where, $$I$$ = moment of inertia of the cylinder about the axis through the centre $$ = m{r^2}$$
$$\alpha $$ = angular acceleration
$$\eqalign{ & \Rightarrow \alpha = \frac{\tau }{I} = \frac{{Fr}}{{m{r^2}}} = \frac{F}{{mr}} = \frac{{30}}{{3 \times 40 \times {{10}^{ - 2}}}} \cr & = \frac{{100}}{4} = 25\,rad/{s^2} \cr} $$

Position vector of the particle is given by $$r = \cos \omega t\hat x + \sin \omega t\hat y$$
where $$\omega $$ is a constant.
Velocity of the particle is
$$\eqalign{ & v = \frac{{dr}}{{dt}} = \frac{d}{{dt}}\left( {\cos \omega t\,\hat x + \sin \omega t\,\hat y} \right) \cr & = \left( { - \sin \omega t} \right)\omega \hat x + \left( {\cos \omega t} \right)\omega \hat y \cr & = - \omega \left( {\sin \omega t\,\hat x - \cos \omega t\hat y} \right) \cr} $$
Acceleration of the particles
$$\eqalign{ & a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left[ { - \omega \sin \omega t\,\hat x + \omega \cos \omega t\,\hat y} \right] \cr & = - {\omega ^2}\cos \omega t\,\hat x - {\omega ^2}\sin \omega t\,\hat y \cr & = - {\omega ^2}\left( {\cos \omega t\,\hat x + \sin \omega t\widehat y} \right) \cr & \Rightarrow a = - {\omega ^2}r = {\omega ^2}\left( { - r} \right) \cr} $$
Assuming the particle as $$P,$$ then its position vector is directed as shown in the diagram.

Therefore, acceleration is directed towards $$- r$$  that is towards $$"O"$$  (origin).
$$\eqalign{ & {\text{Now, we have}}\,v \cdot r = - \omega \left( {\sin \omega t\,\hat x - \cos \omega t\,\hat y} \right)\left( {\cos \omega t\,\hat x + \sin \omega t\,\hat y} \right) \cr & = - \omega \left[ {\sin \omega t \cdot \cos \omega t + 0 + 0 - \sin \omega t \cdot \cos \omega t} \right] \cr & = - \omega \left[ 0 \right] = 0\,\,\,\left[ {\because \hat x \bot \hat y} \right] \cr & v \bot r \cr} $$
Thus, velocity is perpendicular to $$r.$$

For upper half of inclined plane
$${v^2} = {u^2} + 2a\frac{S}{2} = 2\left( {g\sin \theta } \right)\frac{S}{2} = gS\sin \theta $$
For lower half of inclined plane
$$\eqalign{ & 0 = {u^2} + 2g\left( {\sin \theta - \mu \cos \theta } \right)\frac{S}{2} \cr & \Rightarrow - gS\sin \theta = gS\left( {\sin \theta - \mu \cos \theta } \right) \cr & \Rightarrow 2\sin \theta = \mu \cos \theta \cr & \Rightarrow \mu = \frac{{2\sin \theta }}{{\cos \theta }} = 2\tan \theta \cr} $$

$$\eqalign{ & N\cos \theta = mg\,\,{\text{and}}\,\,N\sin \theta = m{\omega ^2}r \cr & \therefore \tan \theta = \frac{{{\omega ^2}r}}{g}\,......\left( {\text{i}} \right) \cr & {\text{Given}}\,y = {x^2} \cr & \therefore \frac{{dy}}{{dx}} = 2x \cr & {\text{or}}\,\,\tan \theta = 2 \times 1 = 2\,......\left( {{\text{ii}}} \right) \cr} $$
From above equations, we get
$$\omega = \sqrt {2g} \,\,\left( {r = 1\;m} \right)$$

Given; speed $$= 10\,m/s;$$   radius $$r = 10\,m$$
Angle made by the wire with the vertical $$\tan \theta = \frac{{{v^2}}}{{rg}} = \frac{{{{10}^2}}}{{10 \times 10}} = 1 \Rightarrow \theta = {45^ \circ } = \frac{\pi }{4}$$