TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

How much pressure (in atm) is needed to compress a sample for water by 0.4 % ?  (Assume , Bulk modulus of water = $2.0 \times 10^{9}$ Pa)

Answer:

Explanation:

Given, the percentage  change in volume =0.4%

 The change in volume of water is given by 

$\triangle V= -\frac{p.V}{B}$

 where , p= pressure applied on the sample of the matter.

 Given,

 $\frac{ \triangle  V}{V}$= -0.4%= $\frac{0.4}{100}$

 Pressure  required , p=   $B\left(-\frac{\triangle V}{V}\right)$

  $=\frac{0.4}{100} \times 2\times 10^{9} P_a= \frac{0.4}{100}\times\frac{2\times 10^{9}}{101325}atm=79 atm\approx 80 atm$

 Hence, 80 atm pressure is needed to compress a sample of water by 0.4%

Consider a spherical planet which  is rotating about its axis such that the speed of a point on its equator is v and the effective acceleration due to gravity on the equator is $\frac{1}{3}$ of its value at the poles. What is the escape velocity for a particular at the pole of this planet

Answer:

Explanation:

We know that the escape velocity of a particle from the surface of a planet is given by

$v_{e}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2gR^{2}}{R}}$            [ $\because GM= gR^{2}$]

     $=\sqrt{2gR}$

 where, g= acceleration due to gravity, and R= radius of the planet

 Given, the velocity at equator , $v_{E}= V$

 and  acceleration due to gravity at equator

 $g_{E}= \frac{1}{3}g_{p}$

$\therefore$     $g_{p}= 3 g_{E}$.........(i)

 The escape velocity of a particle at equator,

$V_{E}=\sqrt{2g_{E}R}$  ..........(ii)

 and at poles,

$V_{P}=\sqrt{2g_{P}R}$

 $=\sqrt{2\times 3g_{E}R}=\sqrt{3}\sqrt{2g_{E}R}$             [From Eq.(i)]

 $=\sqrt{3}V_{E}$                    [From Eq.(ii)]

 = $\sqrt {3}$ v                    [ $\because v_{E}=V$]

 Hence, $\sqrt{3}$ v  is the escape velocity for a particle  at the pole of this planet.

A vertical spring mass system has the same line period as simple pendulum undergoing small oscillations .Now , both of them are put in an elevator going downwards with an  acceleration 5 m/s² . The ratio of time period of the spring mass system to the time period of the pendulum is (Assume, acceleration due to gravity , g= 10 m/s²)

Answer:

Explanation:

 We know that time period of a spring mass system

  $T_{1}=2\pi\sqrt{\frac{m}{k}}$

 where, m= mass of body

 and k= force constant of the spring

 Time period of simple pendulum

 $T_{2}=2\pi\sqrt{\frac{l}{g}}$

 According to the  question, initially time period of a spring mass system. $T_{1}$= time period of simple pendulum , $T_{2}$

 $2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{l}{g}}$

 $\sqrt{\frac{m}{k}}=\sqrt{\frac{l}{10}}$    .........(i)   [ $\because   g=10 m/s^{2}$]

 when both of them are put in an elevator going downwards with an acceleration 5 m/s² , then no effect occurs on the time period of spring mass system.

 i.e,  $T_{1}'=T_{2}=2\pi\sqrt{\frac{m}{k}}$

 But time period of simple pendulum changes in elevator and is given by

 $T_{2}'=2\pi\sqrt{\frac{l}{g_{eff}}}$ , where $g_{eff}$= g-a = effective

 acceleration of the pendulum when elevator is accelerating  downwards with a m/s²

$T_{2}'=2\pi\sqrt{\frac{l}{g-5}}=2\pi \sqrt{\frac{l}{5}}$

$\therefore$     $\frac{T_{1}'}{T_{2}'}=\frac{2\pi\sqrt{\frac{m}{k}}}{2\pi\sqrt{\frac{l}{5}}}=\frac{\sqrt{l/10}}{\sqrt{l/5}}$                             [From Eq. (i)]

$\therefore$      $\frac{T_{1}'}{T_{2}'}=\frac{1}{\sqrt{2}}$

Hence, the ratio of time period of the spring mass system to the time period of the pendulum is   $\frac{1}{\sqrt{2}}$

A  ball of mass 2kg is thrown from a tall building with velocity, v=$ (20 m/s)\hat{i}+(24 m/s)\hat{j}$ at time t=0 s. Change in the potential energy of the ball

after, t=8 s is ( The ball is assumed to be in air during its motion between 0s and 8 s , $\hat{i}$  is along the horizontal and $\hat{j}$ is along the vertical direction. (Take $g=10 m/s^{2}$)

Answer:

Explanation:

 Given , velocity of the ball , v=$ (20 \hat{i}+24\hat{j})$ m/s and mass of the ball, m=2kg

 From the equation of motion , when a ball is thrown, 

  v=u-gt   .........(i)

 Substituting the given values in Eq.(i) , we get

 v= $ (20 \hat{i} +24 \hat{j})-(10 \hat{j})8$

  = $20 \hat{i}+24 \hat{j}-80 \hat {j}= 20 \hat{i}-56 \hat{j}$ .....(ii)

 From the law of conservation of energy , change in potential energy of the ball= change in kinetic energy of the ball

 $\Rightarrow$    $\triangle PE$=   $\frac{1}{2}mu^{2}-\frac{1}{2}mv^{2}$

= $\frac{1}{2}m(u^{2}-v^{2})$

 $=\frac{2}{2}(u.u-v.v)$

  =$[(20 \hat{i}+24 \hat{j}).(20 \hat{i}+24 \hat{j})- [(20 \hat{i}-56 \hat{j}).(20 \hat{i}-56 \hat{j})]$

 = $ (20)^{2}+(24)^{2}-(20)^{2}-(56)^{2}$

 = $(24)^{2}-(56)^{2}$= 576-3136

  =-2560=-2.56 kJ

Ball-I  is dropped from the top of a building from rest . At the same moment, ball -2 is thrown upward towards ball -1 with a speed 14 m/s from a point 21 m below the top of building . How far  will the ball -1 have dropped when it passes ball-2, (Assume , acceleration due to gravity  , g=10 $m/s^{2}$)

Answer:

Explanation:

Suppose that both the balls meet at a distance h from ball -1 after time t from the start as shown in the figure,

 For downward motion of ball-1, from second equation  of the motion

  $h= ut+ \frac{1}{2} gt^{2}$ ......(i)

 h= $0 \times t + \frac {1}{2} \times 10 t^{2}$

                                                 [ $\because u=0$]

 h= $5 t^{2}$  ..........(ii)

 For upward motion of ball -2 , from second equation of the motion.

 $h= vt- \frac{1}{2}g t^{2}$ ........(iii)

 Given, speed of ball -2 , v=14 m/s , acceleration due to gravity ,$ g=10 m/s^{2}$

 Substituting these values in Eq.(iii) , we ger

 $21-h=14 t -\frac{1}{2} \times 10 t^{2}$

$\Rightarrow$    $21-h=14t-5t^{2}$  ............(iv)

 Adding Eq.(ii) and (iv) , we get

 21=14 t

 $\Rightarrow$     $ t= \frac{3}{2}s =1.5 s$

 $\therefore$   Eq.(ii)  , we get

  h= $ 5 \times  (1.5)^{2}$= 11.25 m

   = $\frac{45}{4}$m

 Therefore , the ball-1 will have dropped  $\frac{45}{4}$ m  when it passes ball-2

• Physics - General questions