(2)

Turning means motion on a curved path, which requires centripetal force. Bending of cyclist with respect to vertical direction provides the necessary centripetal force.

(4)

(1)

The net force on the particle directed towards the centre is $T$.

(2)

Centripetal force $=\frac{m{v}^2}{r}=\frac{100\times 10\times 10}{100}=100\text{\hspace{0.05em}N}$

Required frictional force to cross the turn, $=\mu mg=0.6\times 100\times 10=600\text{\hspace{0.05em}N}$

As the frictional force is greater than the centripetal force, so the rider will be able to cross the turn.

(1)

The object will slip if centripetal force $\ge$ force of friction.

       $mr{\omega }^2\ge \mu mg;$ $r{\omega }^2\ge \mu g$

$r{\omega }^2\ge \text{constant},$  $\text{or}$$\left(\frac{r_1}{r_2}\right)={\left(\frac{{\omega }_2}{{\omega }_1}\right)}^2$

$\frac{4\text{\hspace{0.05em}cm}}{r_2}={\left(\frac{{\displaystyle 2\omega }}{{\displaystyle \omega }}\right)}^2$

$\therefore$$r_2=1\text{\hspace{0.05em}cm}$

(3)

Here, $r=10\text{\hspace{0.05em}m},v=5{\text{\hspace{0.05em}m s}}^{-1},a_t=2{\text{\hspace{0.05em}m s}}^{-2}$

$a_r=\frac{v^2}{r}=\frac{5\times 5}{10}=2.5{\text{\hspace{0.05em}m s}}^{-2}$

The net acceleration is

$a=\sqrt{a_r^2+a_t^2}=\sqrt{{(2.5)}^2+2^2}=\sqrt{10.25}=3.2{\text{\hspace{0.05em}m s}}^{-2}$

(2)

Given:  $\overrightarrow{r}=A\cos \omega t\hat{i}+B\sin \omega t\hat{j}$

Velocity,  $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d}{dt}(A\cos \omega t\hat{i}+B\sin \omega t\hat{j})$

                  $=-A\omega \sin \omega t\hat{i}+B\omega \cos \omega t\hat{j}$

Acceleration, $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=-A{\omega }^2\cos \omega t\hat{i}-B{\omega }^2\sin \omega t\hat{j}$

$\overrightarrow{a}=-{\omega }^2[A\cos \omega t\hat{i}+B\sin \omega t\hat{j}]=-{\omega }^2\overrightarrow{r}$

The force acting on the particle is

$\overrightarrow{F}=m\overrightarrow{a}=m(-{\omega }^2\overrightarrow{r})=-m{\omega }^2\overrightarrow{r}$

(2)

Here, $r=3\text{\hspace{0.05em}m},\mu =0.1,g=10{\text{\hspace{0.05em}m s}}^{-2}$

The maximum speed with which a cyclist can take a turn without skidding is

$=\sqrt{\mu rg}=\sqrt{0.1\times \left(3\text{\hspace{0.05em}m}\right)\left(10{\text{\hspace{0.05em}m s}}^{-2}\right)}=\sqrt{3}{\text{\hspace{0.05em}m s}}^{-1}$

(3)

Here, $m=5\text{\hspace{0.05em}kg},r=10\text{\hspace{0.05em}m},T_{\max }=200\text{\hspace{0.05em}N}$

As  $T_{\max }=\frac{m{v}_{\max }^2}{r}$

$\therefore$$v_{\max }^2=\frac{T_{\max }\times r}{m}$

$\therefore$$v_{\max }^2=\frac{200\times 10}{5}=400$

$\Rightarrow$$v_{\max }=20{\text{\hspace{0.05em}m s}}^{-1}$

(1)

Here,  $v=720{\text{\hspace{0.05em}km h}}^{-1}=720\times \frac{5}{18}{\text{\hspace{0.05em}m s}}^{-1}=200{\text{\hspace{0.05em}m s}}^{-1}$

$\theta =15°,g=10{\text{\hspace{0.05em}m s}}^{-2};\text{ As }\tan \theta =\frac{v^2}{rg}$

$\therefore$$r=\frac{v^2}{\tan \theta g}=\frac{{\left(200{\text{\hspace{0.05em}m s}}^{-1}\right)}^2}{\tan 15°\times 10{\text{\hspace{0.05em}m s}}^{-2}}=14815\text{\hspace{0.05em}m}=14.8\text{\hspace{0.05em}km}$

(1)

At the lowest point, $mg$ acts downwards and $T_1$ upwards so that net force $=mg-T_1$.

At the highest point, both $mg$ and $T_2$ act downwards so that net force $=mg+T_2$.

Hence, option (1) is correct.

(1)

Here, $\mu =0.15$,

$\nu =33\frac{1}{3}\text{\hspace{0.05em}rpm}=\frac{100}{3}\text{\hspace{0.05em}rpm}=\frac{100}{3\times 60}\text{\hspace{0.05em}rps}=\frac{5}{9}\text{\hspace{0.05em}rps}$

$\therefore$$\omega =2\pi \nu =2\times \frac{22}{7}\times \frac{5}{9}=\frac{220}{63}{\text{\hspace{0.05em}rad s}}^{-1}$

The coin will revolve with the disc, if the force of friction is enough to provide the necessary centripetal force.

i.e.,  $\frac{m{v}^2}{r}\le \mu mg$

As $v=r\omega$,

$\therefore$$m{\omega }^{2}r\le \mu mg$

For the given $\mu$ and $\omega$, the condition is $r\le \frac{\mu g}{{\omega }^2}$    $\cdots \left(i\right)$

As  $\frac{\mu g}{{\omega }^2}=\frac{0.15\times 10}{{{\displaystyle (}{\displaystyle \frac{{\displaystyle 220}}{{\displaystyle 63}}}{\displaystyle )}}^2}\approx 12\text{\hspace{0.05em}cm}$

For coin A, $r=4\text{\hspace{0.05em}cm}$.

The above condition is satisfied, therefore coin A will revolve with the disc.

For coin B, $r=14\text{\hspace{0.05em}cm}$.

The above condition is not satisfied, therefore coin B will not revolve with the disc.

Note: We have nothing to do with the radius of the disc.

(2)

Here, $R=300\text{\hspace{0.05em}m},\theta =15°,g=10{\text{\hspace{0.05em}m s}}^{-2},\mu =0.2$

The optimum speed of the car to avoid wear and tear is given by

$v=\sqrt{Rg\tan \theta }$$=\sqrt{300\times 10\times \tan 15°}=\sqrt{810}=9\sqrt{10}{\text{\hspace{0.05em}m s}}^{-1}$

(3)

The maximum permissible speed is given by

$v_{\max }=\sqrt{\frac{Rg(\mu +\tan \theta )}{1-\mu \tan \theta }}$$=\sqrt{\frac{300\times 10\times (0.2+0.27)}{1-0.2\times 0.27}}=38.6{\text{\hspace{0.05em}m s}}^{-1}$

(4)

Force will be same for all the surfaces.

(3)

Here, mass of the man, M = 55 kg

As the man is standing stationary w.r.t. the belt, acceleration of man = acceleration of belt.

$\therefore$    Acceleration of man, $a=1$ $\mathrm{m}{\mathrm{s}}^{-2}$

$\therefore$    Net force on the man, $F=Ma$$=\left(55\text{\hspace{0.05em}kg}\right)\left(1{\text{\hspace{0.05em}m s}}^{-2}\right)=55\text{\hspace{0.05em}N}$

(4)

Here, Mass of helicopter, $M=2000\text{\hspace{0.05em}kg}$

Mass of the crew and passengers, $m=500\text{\hspace{0.05em}kg}$

Vertically upwards acceleration, $a=15{\text{\hspace{0.05em}m s}}^{-2}$, $g=10{\text{\hspace{0.05em}m s}}^{-2}$

(i) Force on the floor by the crew and passengers

    $=m(g+a)$$=500\text{\hspace{0.05em}kg} (10+15){\text{\hspace{0.05em}m s}}^{-2}=12500\text{\hspace{0.05em}N}=1.25\times {10}^4\text{\hspace{0.05em}N}$

     It acts vertically downwards.

(ii) Action of the rotor of the helicopter on the surrounding air

     $=(M+m)(g+a)$$=(2000+500)\text{\hspace{0.05em}kg} (10+15){\text{\hspace{0.05em}m s}}^{-2}=62500\text{\hspace{0.05em}N}=6.25\times {10}^4\text{\hspace{0.05em}N}$

      It acts vertically downwards.

(iii) Force on the helicopter due to the surrounding air is equal and opposite to the action of the rotor of the helicopter on the surrounding air.

      $\therefore$  Force on the helicopter due to the surrounding air $=6.25\times {10}^4$ N

      It acts vertically upwards.

(3)

Here, $v=20{\text{\hspace{0.05em}m s}}^{-1},a=2{\text{\hspace{0.05em}m s}}^{-2},g=10{\text{\hspace{0.05em}m s}}^{-2}$

The coin will fall back into the person's hand after $t$ s.

$\therefore$$t=\frac{2v}{a+g}=\frac{2\times 20{\text{\hspace{0.05em}m s}}^{-1}}{(2+10){\text{\hspace{0.05em}m s}}^{-2}}=\frac{40}{12}\text{\hspace{0.05em}s}=\frac{10}{3}\text{\hspace{0.05em}s}$

(3)

The reading on the scale is a measure of the force on the floor by the person. By Newton’s third law this is equal and opposite to the normal force N on the person by the floor.

$\therefore$   When the lift is ascending upwards with an acceleration of 9 $\mathrm{m}{\mathrm{s}}^{-2}$, then

N − 50 × 10 = 50 × 9  or  N = 50 × 10 + 50 × 9 = 50(10 + 9) = 950 N

$\therefore$   The reading of weighing machine is 95 kg.

(2)

As the system is in equilibrium,

$\therefore$    $T=W$                 $\cdots \left(i\right)$

and       $T=mg\sin 30°$  $\cdots \left(ii\right)$

From (i) and (ii), we get

$W=mg\sin 30°=\left(100\text{\hspace{0.05em}N}\right)\times \frac{1}{2}=50\text{\hspace{0.05em}N}$

(4)

Here, $m_1=10\text{\hspace{0.05em}kg},m_2=20\text{\hspace{0.05em}kg},F=600\text{\hspace{0.05em}N}$

Let $T$ be tension of the string and $a$ be common acceleration of the system.

$\therefore$$a=\frac{F}{m_1+m_2}=\frac{600\text{\hspace{0.05em}N}}{10\text{\hspace{0.05em}kg}+20\text{\hspace{0.05em}kg}}=\frac{600}{30}{\text{\hspace{0.05em}m s}}^{-2}=20{\text{\hspace{0.05em}m s}}^{-2}$

When a force $F$ is applied on 10 kg block, then the tension in the string is

$T=m_{2}a=\left(20\text{\hspace{0.05em}kg}\right)\left(20{\text{\hspace{0.05em}m s}}^{-2}\right)=400\text{\hspace{0.05em}N}$

(2)

When a force $F$ is applied on 20 kg block, then the tension in string is

$T=m_{1}a=\left(10\text{\hspace{0.05em}kg}\right)\left(20{\text{\hspace{0.05em}m s}}^{-2}\right)=200\text{\hspace{0.05em}N}$

(1)

The free body diagram of 3 kg block is as shown in the figure (a).

The equation of motion of 3 kg block is

$T_2-3g=3a$

$T_2=3(a+g)=3(2+10)=36\text{\hspace{0.05em}N}$          $\cdots \left(i\right)$

The free body diagram of 5 kg is as shown in the figure (b).

The equation of motion of 5 kg block is

$T_1-T_2-5g=5a$

$T_1=5(a+g)+T_2$

      $=5(2+10)+36=96\text{\hspace{0.05em}N}$         $\text{(Using (i))}$

(1)

(3)

If $a$ is downward acceleration of 4 kg block, the upward acceleration of 1 kg block must be $2a$.

Let $T$ be tension in each part of string. The equation of motion on 4 kg block is

$4g-2T=4a$                     $\cdots \left(i\right)$

The equation of motion on 1 kg block is

$T-1g=1\times 2a$                $\cdots \left(ii\right)$

or  $2T-2g=4a$              $\cdots \left(iii\right)$

Adding (i) and (iii), we get

$2g=8a$  or  $a=\frac{g}{4}$

$\therefore$   Acceleration of 1 kg block

$=2a=\frac{g}{2}\text{ upwards.}$

(2)

Let $a$ be the common acceleration of the system and $T$ be tension of the string.

The equations of motion of two blocks are

$T-8g=8a$            $\cdots \left(i\right)$

and  $12g-T=12a$    $\cdots \left(ii\right)$

Adding (i) and (ii), we get

$4g=20a$     or     $a=\frac{g}{5}$

(1)

Here, mass of monkey, $m=40\text{\hspace{0.05em}kg}$

Maximum tension the rope can stand, $T=500\text{\hspace{0.05em}N}$

Tension in the rope will be equal to apparent weight of the monkey $\left(R\right)$.

The rope will break when $R$ exceeds $T$.

(1)   When the monkey climbs up with an acceleration $a=5{\text{\hspace{0.05em}m s}}^{-2}$

        $R=m(g+a)=40(10+5)=600\text{\hspace{0.05em}N}$          $\therefore$$R>T$

        Hence, the rope will break.

(2)    When the monkey climbs down with an acceleration $a=5{\text{\hspace{0.05em}m s}}^{-2}$

         $R=m(g-a)=40(10-5)=200\text{\hspace{0.05em}N}$          $\therefore$$R<T$

         Hence, the rope will not break.

(3)    When the monkey climbs up with a uniform speed $v=5{\text{\hspace{0.05em}m s}}^{-1}$, its acceleration $a=0$.

         $\therefore$$R=mg=40\times 10=400\text{\hspace{0.05em}N}$             $\therefore$  $R<T$

         Hence, the rope will not break.

(4)     When the monkey falls down the rope freely under gravity,

          $a=g$      $\therefore$$R=m(g-a)=m(g-g)=0$

          Hence, the rope will not break.

(4)

(1)

Here, $m_1=40\text{\hspace{0.05em}kg}$

$m_2=30\text{\hspace{0.05em}kg},\theta =30°$

Let $T$ be the tension in the string and $a$ be the acceleration of the system.

Their equations of motion are

$m_{1}g\sin 30°-T=m_{1}a$            $\cdots \left(i\right)$

$T-m_{2}g\sin 30°=m_{2}a$           $\cdots \left(ii\right)$

Adding (i) and (ii), we get

$(m_1+m_2)a=(m_1-m_2)g\sin 30°$

Substituting the given values, we get

$(40+30)a=(40-30)\times 9.8\times \frac{1}{2}=49$

$\Rightarrow$$a=\frac{49}{70}=0.7{\text{\hspace{0.05em}m s}}^{-2}$

(1)

Case I:

Using Newton's second law of motion,

$T_1-mg=ma$

$T_1=m(g+a)$          $\cdots \left(i\right)$

Case II:

Using Newton's second law of motion,

$mg-T_2=ma$

$T_2=m(g-a)$          $\cdots \left(ii\right)$

From eqn. (i) and (ii),

$\frac{T_1}{T_2}=\frac{g+a}{g-a}=\frac{9.8+4.9}{9.8-4.9}=\frac{3}{1}$