TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

A solid spherical ball rolls on a horizontal surface at 10 m/s and continuous to roll up  on an inclined surface as shown in the figure.If  the mass of balls is 11 kg and frictional losses are negligible, the value of h where the ball stop and starts rolling down the inclination is (Assume , g =10 m/s² )

Answer:

Explanation:

Key idea  For rolling motion , the rotational linetic energy is given by expression k

$KE= \frac{1}{2} mv^{2}\left(1+\frac{k^{2}}{R^{2}}\right)$

 gIven, rolling velocity on horizontal surface. 

v_H=  10 m/s  , mass of ball , m=11kg and g=10 m/s²

Kinetic energy of rotation,

$KE= \frac{1}{2} \times 11\times (1.0)^{2}\left(1+\frac{\frac{2}{5}R^{2}}{R^{2}}\right)$

  ( $\because$     For solid spherical ball,    $k=\sqrt{\frac{2}{5}}R$ )

 $\Rightarrow$       $=11 \times 50 \times \frac {7}{5}$J

 By applying the law  pof energy comservation,

 $ KE_{i}+v_{i}= KE_{f}+v_{f}$

$\Rightarrow$      $11\times50\times\frac{7}{5}+0=0+mgh$

 $\Rightarrow$      $h= \frac{50}{10} \times \frac{7}{5}$=7 m

hence, the correct option is (c)

A mass of 2kg initially at a height of  1.2 m above an uncompressed spring with spring constant $2 \times 10^{4} $ N/m, is released from rest to fall on the spring. Taking the acceleration due to gravity  as 10 m/s² and neglecting the air resistance, the compression of the spring in mm is 

Answer:

Explanation:

 Key idea    when  a mass m is fall from height h on an uncompressed spring then the maximum compression is given by the relation,

 mgh=$\frac{1}{2}kx^{2}$

(principle of energy conservation)

Given, mass of body , m=2kg, height,h=1.2m , spring constant , k= 2 x 10⁴ N/m and acceleration due to gravity , g-10 m/s²

 Then,   $x=\sqrt{\frac{2mgh}{k}}\Rightarrow x=\sqrt{\frac{2\times2\times1.2\times10}{2\times 10^{4}}}$

 $\Rightarrow$       =$4.89 \times 10^{-2} m$   $\approx$ 50mm

So, the correct option is  (c)

A particle of mass m is moving along a circle of radius R such that its tangential acceleration  a_t varies with distance covered x as

a_t =ax² where $\alpha$   is a constant. The kinetic energy,K of the particle varies with the distance as K= $\beta x^{c}$, where $\beta$  and c are constants. The values of $\beta$ and c are

Answer:

Explanation:

 Given, mass of a particel =m

 tangential acceleration of particel ,  $a_{t}=ax^{2} $ and kinetic energy , K=$\beta x^{c}$

Here, as we know that torque,  $\tau=F.R=I\alpha$

$\Rightarrow$     $F.R=I\frac{a_{1}}{R}$              $\left(\because \alpha=\frac{a}{R}\right)$

$\Rightarrow$      $F.R= mR^{2}\frac{aX^{2}}{R}$       ($\because   I= MR^{2}$)

 $\Rightarrow$    $F=max^{2}$  ............(i)

 Hence, the work done is displacement  of dx,

$W=\int f.dx=\int max^{2} dx$

 W=$\frac{1}{3}max^{3}$  ......(iii)

 from work -energy theorm,

  $W= \triangle  KE$

 here, W= work done and $\triangle KE$ = change in kinetic energy,

 $\frac{max^{3}}{3}=\beta x^{2} $     $ (\because K=\beta x^{2})$

 Now, from the above equation , we get

 $\beta =\frac{ma}{3}$ and c=3

 Hence , the correct option is (a)

 A ball is projected vertically up from ground. Boy A standing at the window of first floor of a nearby building observes that the time interval  between the ball crossing him while going up and the ball crossing him while going down is 2s . Another boy B  standing  on the second  floor notices that time interval between the ball passing him twice, during up motion  and down  motion is 1s . Calculate the difference between  the vertical positions  of boy B and boy A (Assume , acceleration  due to gravity ,g=10 m/s²)

Answer:

Explanation:

 The figure given below , shows the vertical motion of ball,

Here, ball crosses the 1st floor in 2s, so it goes up in 1s and goes down in next one second.

So,   $v-v_{A}=-gt$

 $\Rightarrow$    0-$v_{A}=-10(1) \Rightarrow   v_{A}=10m/s$

$\because$ Boy, B observed the ball crossing him while up and down motion in 1s, So , time taken by the ball to reached the floor B from A is 0.5s

So,    s=$v_{A}t- \frac{1}{2} gt^{2}$

=$10 \times 0.5 -\frac{1}{2}\times 10 \times (0.5)^{2}$

 =5-1.25=3.75 m

 Hence, the correct option is (b)

Consider a car initially at rest, starts to move along a straight road first with acceleration 5 m/s² , then with uniform velocity and finally, decelerating at 5 m/s², before coming to a stop. Total time taken from start to end is t=25 s. If the average  velocity during that time is 72 km/hr, the car moved with uniform velocity for a time of 

Answer:

Explanation:

 Given , acceleration of car, a= 5 m/s²,

 deceleration of   car a=5 m/s² , total time taken from start of end is, t=25s and average velocity of car

   $v_{avg}$=72 km/hr=20 m/s   ( 1$\frac{km}{hr}=\frac{5}{18}m/s$)

Since, $v_{avg}$= total displacement/total time taken

 2t, total time taken by the car during acceleration  and deceleration,

 $v_{avg}=20=\frac{d_{1}+d_{(25-2t)}+d_{t}}{25}$

     $=\frac{2d_{1}+d_{(25-2t)}}{25}$

 Since, $d_{t}=0+\frac{1}{2}{at^{2}}=\frac{1}{2}at^{2}=\frac{5}{2}t^{2} $   and  

    $d_{(25-2t)}= v_{uni.}(25-2t)$

where, $v_{uni.} $=5t

now,   $v_{avg}=20=\frac{2\left(\frac{5}{2}t^{2}\right)+5t(25-2t)}{25}$

$\therefore$       $20 \times 25 =5t^{2}+5t(25-2t)$

$\Rightarrow$  500=$5t^{2}+125t-10t^{2}$

 $\Rightarrow$  $t^{2}-25t+100=0$

 so, it gives t=20 and 5s

 Hence , the time of uniform motion,

 $t_{20}=25-2t=25-2 \times 20=-15  s$

                                   ($\because$ Not possible )

 or $t_{5}=25-10=15s$

So, the correct option is (a)

• Physics - General questions