CBSE Questions for Class 11 Engineering Physics Motion In A Plane Quiz 9 - MCQExams.com

Centripetal acceleration and tangential acceleration is normal to each other.

Centripetal acceleration, $${{a}_{c}}=\dfrac{{{v}^{2}}}{r}=\dfrac{{{20}^{2}}}{50}=8\,m{{s}^{-2}}$$

Tangential acceleration, $${{a}_{t}}=0$$

Hence, magnitude of acceleration is $$8\,m{{s}^{-1}}$$ 

The acceleration will be directed towards the centre of the circular loop.

angular velocity = 2πf = 2 x 22/7 x 14/25 = 88/25 rad/s

Centripetal acceleration = rω² = 0.8 m x 88 x 88 / 225 = 27.53 m/s²

LetAngle of projection for A=30oAngle of projection for B=60oRequired condition:Distance traveled by A−Distance traveled by B=x[u3√]cos30o×t−ucos60o×t=x[u3√](3√2)×t−u(12)×t=xut[32−12]=xut=xt=xu

Moment of inertia of uniform rod of length(l) -

$$I=\dfrac { M{ l }^{ 2 } }{ 12 } $$

- wherein

About axis passing through its center & perpendicular to its length.

Law of conservation of angular moment -

$$\vec { r } =\dfrac { d\vec { L }  }{ dt } $$

- wherein

If net torque is zero

i.e $$\dfrac { d\vec { L }  }{ dt } =0,\quad $$

$$\overrightarrow { L } =$$ constant

angular momentum is conserved only when external torque is zero .

 Centre of mass from system from 0

$$=>\dfrac { 8m\times 0\quad +m(\dfrac { L }{ 3 } )-\quad 2m(\dfrac { L }{ 6 } ) }{ 8m+m+2m } =0$$

So,centre of mass is at zero.

From conservation of ang.momentum; $${ L }_{ i }={ L }_{ f }$$

$${ L }_{ i }=m(2v)\times (\dfrac { L }{ 3 } )+\quad 2mv\times (\dfrac { L }{ 6 } )=mvL$$       

$${ L }_{ f }=[(8m)\times \dfrac { { L }^{ 2 } }{ 12 } +\quad m{ (\dfrac { L }{ 3 } ) }^{ 2 }+2m{ (\dfrac { L }{ 6 } ) }^{ 2 }]\omega $$

$$=[\dfrac { 2 }{ 3 } m{ L }^{ 2 }+\dfrac { m{ L }^{ 2 } }{ 9 } +\dfrac { m{ L }^{ 2 } }{ 18 } ]\omega $$

$$=(\dfrac { 12+2+1 }{ 18 } )m{ L }^{ 2 }\omega =\frac { 5 }{ 6 } m{ L }^{ 2 }\omega $$

$$\dfrac { 5 }{ 6 } m{ L }^{ 2 }\omega =mvl,$$

$$\therefore \omega =\dfrac { 6v }{ 5L } $$

Correct answer: Option A

Hint: Differentiation of displacement gives velocity and differentiation of velocity gives acceleration with respect to time.

Explanation of correct option:

Given: Displacement - Time relation

$$x = {(t + 5)^{ - 1}}$$...................(i)

Step 1:Differentiating both side w.r.t. $$'t'$$ of equation (i)

We get,

$$\dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}{(t + 5)^{ - 1}}$$

$$\therefore\dfrac{dx}{dt} = ( - 1){(t + 5)^{ - 2}}(1 + 0)$$

$$\dfrac{dx}{dt}$$ is velocity.

$$ \Rightarrow v =  - {(t + 5)^{ - 2}}$$...................(ii)

Step 2:Differentiating both side w.r.t. $$'t'$$ of equation (ii)

We get,

$$\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}[ - {(t + 5)^{ - 2}}]$$

$$\therefore \dfrac{dv}{dt}$$ =  $$- ( - 2){(t +5)^{ - 3}}.(1)$$

$$\dfrac{dv}{dt}$$ is acceleration.

Formatting equation (iii) to out the value of equation (ii)

$$ \Rightarrow a = 2 \times \left\{ { - {{(t + 5)}^{ - 2}}} \right\}\left\{ { - {{(t + 5)}^{ - 2 \times \dfrac{1}{2}}}} \right\}$$

$$ \Rightarrow a = 2v \times {v^{\frac{1}{2}}}$$

$$\therefore a = 2{v^{\frac{3}{2}}}$$

As $$2$$ is constant so $$a$$ varies with $$v$$ such as

$$a \propto {v^{\frac{3}{2}}}$$ i.e. $$(acceleration) \propto (velocity)^{\frac{3}{2}}$$

Hence, Acceleration of a particle is proportional to  $$ (velocity)^{\frac{3}{2}}$$

When a move in circular motion, centripetal force will act on the body toward center.

When body moves with uniform angular velocity in circular path.

Then,

(i)                  Centripetal force, $${{F}_{c}}=m\omega {{l}^{2}}$$

(ii)                Speed of the particle, $$v=\omega l$$

(iii)               Because it is horizontal circle and there is no gravity. Therefore there is no force in vertical direction.

$${\textbf{Step1: Identify the values that are given.}}$$

               $${\text{Radius = R}}$$

               $${\text{Static friction = }}\mu $$

               $${\text{Now, the centripetal force is providing by frictional force between the road and tyre}}$$

$${\textbf{Step2: Find v. }}$$

               $$\text{The centripetal force is given by } \dfrac{{m{v^2}}}{R} $$

               $$\text{and the frictional force is  }  = \mu mg $$

               $$\text{For avoiding car to skid the cetripetal force must be equal to frictional force } $$

               $$\dfrac{{m{v^2}}}{R} = \mu mg$$

               $$v = \sqrt {\mu Rg} $$

               $${\text{Hence,this is the required solution}}$$

$${\textbf{Hence,}}$$ $${\textbf{the final answer is }}$$ $$\mathbf {v =} $$ $$\mathbf { { \sqrt {\mu Rg} .}}$$