MHCET 2011

Given :         Radius of the circular path         $$r = 20$$  $$m$$
                     Mass of the car        $$M = 1500kg$$
                     Speed        $$ v=12.5 $$  $$m/s$$
Let the coefficient of friction be  $$\mu$$.
The centrifugal force will balance the friction force so that the car does not slip.
$$\therefore$$     $$\dfrac{mv^2}{r}  =  \mu  mg$$                         $$\implies$$   $$\mu  = \dfrac{v^2}{rg}$$
$$\mu =  \dfrac{(12.5)^2}{20 \times 9.8}  =  0.8$$
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A car of mass $$1500$$ kg is moving with a speed of $$12.5$$ m/s on a circular path of radius $$20$$ m on a level road. What should be the coefficient of friction between the car and the road, so that the car does not slip?
A
$$0.2$$
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B
$$0.4$$
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C
$$0.6$$
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D
$$0.8$$
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Using            $$\vec{v}  =\vec{w}\times \vec{r} $$    $$\implies |\vec{v}| =  |w||r |sin\theta$$     where  $$\theta$$ is the angle between  $$\vec{w}$$  and  $$\vec{r}$$
$$\therefore$$   $$\theta = 90^0$$         $$\implies  sin\theta=  1$$
Thus     $$v = wr \times 1$$
$$\implies$$    $$w = \dfrac{v}{r}$$
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If the body is moving in a circle of radius $$r$$ with a constant speed $$v$$. Its angular velocity is:
A
$$\dfrac{v^2}{r}$$
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B
$$vr$$
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C
$$\dfrac{v}{r}$$
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D
$$\dfrac{r}{v}$$
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Given :    Mass of the ball      $$m = 0.25  kg$$
                Length of the string (radius of the circular motion)     $$l = 1.96$$  m  
                Maximum tension in the string        $$T = 25$$ N
Let the maximum speed with which the ball can be moved be  $$v$$.
The tension provides the necessary centripetal force required for the ball to move in circular path.
Using            $$T =\dfrac{mv^2}{l}$$
 $$\therefore$$  $$25  = \dfrac{0.25 \times v^2}{1.96}$$                  $$\implies  v^2 = 196$$
Thus        $$v = 14$$   $$m/s$$
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A ball of mass $$0.25$$ kg attached to the end of a string of length $$1.96$$ m is moving in a horizontal circle. The string will break if the tension is more than $$25$$ N. What is the maximum speed with which ball can be moved?
A
$$14$$ m/s
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B
$$3$$ m/s
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C
$$3.92$$ m/s
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D
$$5$$ m/s
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Let the initial velocity given to the sphere be  $$u$$. At the height of suspension  i.e  point B, its velocity becomes zero   i.e   $$v = 0$$
Work-energy theorem :             $$W_{all  forces} =  \Delta K.E$$
$$\therefore$$    $$ - mgl  = \dfrac{1}{2}mv^2  - \dfrac{1}{2}mu^2$$

$$- mgl  = 0  - \dfrac{1}{2}mu^2$$                         $$\implies   mgl  = \dfrac{1}{2}mu^2$$
$$\therefore$$    $$u  = \sqrt{2gl}$$
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A sphere is suspended by a thread of length $$l$$ . What minimum horizontal velocity has to be imparted the ball for it to reach the height of the suspension?
A
$$gl$$
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B
$$2\,gl$$
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C
$$\sqrt{gl}$$
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D
$$\sqrt{2\,gl}$$
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As the earth revolves around the sun, it experience the centrifugal force in a direction away from the centre of sun which tends to push it away from sun.
But the earth moves in a constant orbit, thus there must be a force in a direction towards the centre of sun which balances the centrifugal force.
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Which of the following is the evidence to show that there must be a force acting on earth and directed towards the sun?
A
Deviation of the falling bodies towards east
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B
Revolution of the earth around the sun
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C
Phenomenon of day and night
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D
Apparent motion of sun round the earth
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Acceleration due to gravity               $$g = \dfrac{GM}{R^2}$$
Mass of the planet,       $$M = Vd   = d \times \dfrac{4}{3}\pi R^3$$          where  $$d$$ is the density of the planet
$$\implies$$      $$g  =  \dfrac{4}{3}\pi dGR$$              (for earth)

Given :          $$d' = d$$   and    $$R' = 0.2 R$$
$$\therefore$$        $$g'  =\dfrac{4}{3}\pi d'GR'     = \dfrac{4}{3}\pi d G (0.2 R)$$
$$\implies$$    $$g' = 0.2 g$$
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If the density of a small planet is the same as that of earth, while the radius of the planet is $$0.2$$ times that the earth, the gravitational acceleration on the surface of that planet is:
A
$$0.2$$ g
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B
$$0.4$$ g
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C
$$2$$ g
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D
$$4$$ g
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Given  :             $$\dfrac{M_E}{M_M}  = 81$$             and                $$\dfrac{R_E}{R_M}  = 3.5$$
Escape velocity             $$v_e  = \sqrt{\dfrac{2GM}{R}}$$
$$\implies$$    $$\dfrac{v_{e_E}}{v_{e_{M}}}  =  \sqrt{\dfrac{M_E}{M_M} \times  \dfrac{R_M}{R_E}}$$

  $$\dfrac{v_{e_E}}{v_{e_{M}}}  =  \sqrt{81 \times \dfrac{1}{3.5}}   = 4.81$$
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The mass of earth is $$81$$ times that of the moon and the radius of the earth is $$3.5$$ times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will be:
A
$$0.2$$
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B
$$2.57$$
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C
$$4.81$$
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D
$$0.39$$
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Initial speed         $$\nu_i  =1200$$  $$r.p.m     =  \dfrac{1200}{60}$$  $$r.p.s   =    20$$  $$r.p.s$$
$$\therefore$$      $$w_i = 2\pi  \nu_i  =  2\pi \times  20 = 40 \pi   $$    $$rad/s$$
Final angular velocity    $$w_f= 0$$ 
Using         $$w_f^2  - w_i^2   =  2\alpha S$$
$$\therefore$$    $$0 - (40\pi)^2  = 2 (-4) S$$        $$\implies$$   $$S = 200  \pi^2$$  $$rad$$
Number of revolution            $$N = \dfrac{200  \pi^2}{2\pi}  = 100 \pi  = 314$$
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A wheel has a speed of $$1200$$ revolution per minute and is made to slow down at a rate of $$4\;rad/s^2$$. The number of revolution it makes before coming to rest is
A
$$143$$
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B
$$272$$
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C
$$314$$
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D
$$722$$
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A disc is composed of rings and moment of inertia of disk can be found by using integral of moment of inertia of concentric elemental rings. 
Moment of inertia of continuous body is found by using $$I=\int R^2 dm$$. 
For increasing the moment of inertia of a non-uniform disc, it is hence desired that the mass density is more in the exterior parts of the disc. 
Thus, B is the correct option. 
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A circular disc is to be made by using iron and aluminium, so that it acquires maximum momentum of inertia about its geometrical axis. It is possible with
A
Iron and aluminium layers in alternate order
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B
Aluminium at interior and iron surrounding it
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C
Iron at interior and aluminium surrounding it
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D
Either (a) or (c)
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Work done              $$W = \dfrac{1}{2}kx^2$$
where  $$k$$ and $$x$$ are the spring constant and extension of the spring respectively.
$$\implies  W  \propto    k$$                 (for same $$x$$ )
$$\therefore$$   $$k_A>k_B$$   $$\implies   W_A >W_B$$
Thus work required to stretch the spring  $$A$$ is more.
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Two springs have spring constant $$k_A$$ and $$k_B$$ and $$k_A > k_B$$. The work required to stretch them by same extension will be:
A
more in spring A
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B
more in spring B
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C
equal in both
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D
nothing can be said
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