(1)

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Let $r$ be radius of circle.

Here, tension provides the required centripetal force,

           $\sum F_c=m{a}_c$

So, from initial condition,

            $T=mr{\omega }^2$                                      ...(i)

and, from final condition,

           $4T=mr\omega {\text{'}}^2$                                     ...(ii)

From eqn. (i) and (ii), we get

$\frac{T}{4T}=\frac{mr{\omega }^2}{mr\omega {\text{'}}^2}\Rightarrow \frac{1}{4}=\frac{{\omega }^2}{\omega {\text{'}}^2}\Rightarrow \omega {\text{'}}^2=4{\omega }^2$

$\omega \text{'}=2\omega =2\times 10=20\text{ rpm}$

(2)

When the bob is moving with speed $\text{'}\omega \text{'}$, then FBD is,

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Let $\text{'}r\text{'}$ be the radius of the circle.

Applying Newton’s second law on mass of the bob $\text{'}m\text{'}$ along centripetal direction, we have

$T=mr{\omega }^2$                                              ...(i)

When speed becomes $2\omega$, the FBD is

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$T\text{'}=mr{\left(2\omega \right)}^2=mr\left(4{\omega }^2\right)$

or $mr=\frac{T\text{'}}{4{\omega }^2}$                                    ...(ii)

Using value of equation (ii) in equation (i),

$T=\frac{T\text{'}{\omega }^2}{4{\omega }^2}=\frac{T\text{'}}{4}$

Hence, $T\text{'}=4T$

(2)

Direction of velocity is changing so it is not a constant and centripetal acceleration changes continuously as $\overrightarrow{v}$ is not constant and therefore $\overrightarrow{a}$ is also not constant.

(4)

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To keep the block stationary,

Frictional force $\ge$ Weight

 $\mu N\ge Mg$

Here, $N=M{\omega }^{2}r$

$r=1\text{m}, \mu =0.1$

For minimum $\omega ,$ $\mu M{\omega }^{2}r=Mg$

$\omega =\sqrt{\frac{g}{\mu r}}=\sqrt{\frac{10}{0.1\times 1}}=10$ $\text{rad}$ $s^{-1}$

(4)

$\text{ Centripetal force (}\frac{m{v}^2}{l}\text{) is provided by tension, so net force on the particle will be equal to tension }T.$

(4)

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For vertical equilibrium on the road,

        $N\cos \theta =mg+f\sin \theta \Rightarrow mg=N\cos \theta -f\sin \theta$            ...(i)

Centripetal force for safe turning,  $N\sin \theta +f\cos \theta =\frac{m{v}^2}{R}$        ...(ii)

From eqns. (i) and (ii), we get

$\frac{v^2}{Rg}=\frac{N\sin \theta +f\cos \theta }{N\cos \theta -f\sin \theta }\Rightarrow \frac{v_{\text{max}}^2}{Rg}=\frac{N\sin \theta +{\mu }_{s}N\cos \theta }{N\cos \theta -{\mu }_{s}N\sin \theta }$

$v_{\text{max}}=\sqrt{Rg\left(\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }\right)}$

(3)

Let $v$ be tangential speed of heavier stone. Then, centripetal force experienced by lighter stone is

          ${\left(F_c\right)}_{\text{lighter}}=\frac{m{\left(nv\right)}^2}{r}$

and that of heavier stone is ${\left(F_c\right)}_{\text{heavier}}=\frac{2m{v}^2}{(r/2)}$

But ${\left(F_c\right)}_{\text{lighter}}={\left(F_c\right)}_{\text{heavier}}$                              (given)

$\therefore$  $\frac{m{\left(nv\right)}^2}{r}=\frac{2m{v}^2}{(r/2)} \text{or,} n^2\left(\frac{m{v}^2}{r}\right)=4\left(\frac{m{v}^2}{r}\right)$

        $n^2=4 \text{or} n=2$