TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

 A boy standing on a moving truck throws a projectile such that he is able to catch it back after the truck has moved 100 m. if the truck is moving horizontally along a straight line with a constant speed 30 m/s, at what speed (relative to the truck) must the projectile is thrown.(Assume, g=10 m/s²)

Answer:

Explanation:

Given, velocity of truck, v=30 m/s

 time taken by truck to move a distance of 100 m

 $t= \frac{100}{30}\Rightarrow$  $t= \frac{10}{3} s$

 If $v_{1}$  be the velocity of projectile in upward

direction then time taken by the projectile to reach  at maximum height is $t_{1}$ , then

    $0=v_{1}-gt$,

 $v_{1}=gt_{1}$

But, $t_{1}=\frac{t}{2}=\frac{\frac{10}{3}}{2}=\frac{5}{3}s$

 $\therefore$   $v_{1}=g\times \frac{5}{3}=\frac{10 \times 5}{3}=\frac{50}{3} m/s$

An iron rod of length 1.5 m lying on a horizontal table is pulled up from one end along a vertical line so as to move it with a constant velocity 3 m/s. while the other end of the rod slides along the floor. After how much time the speed of the end sliding  on the floor equals to the speed of the end being pulled up

Answer:

Explanation:

 According to question,

 Given,

AB is an iron rod whose length is given by

   l=1.5 m

 velocity of end A,

   v=3 m/s

Let t be the time when both the ends have same speed. since end A has always a constant velocity, therefore the velocity of end B at point B' is also become 3 m/s.

 In the position of B'A' of the rod, i.e, after time t,

 $(l-x)^{2}+(3t)^{2}=l^{2}$

 $(1.5-x)^{2}+9t^{2}=(1.5)^{2}$

 $2.25+x^{2}-3x+9t^{2}=2.25$

 $x^{2}-3x+9t^{2}=0$   ............(i)

 As rod is moving with a constant  velocity then from equation of the motion,

 $x=ut+\frac{1}{2}at^{2}$

 x=3t     .............(ii)

 [ $\because$  velocity ,u=3 m/s and accleration , a=0]

 From Eqs.(i) and (ii)

$(3t)^{2}-3(3t)+9t^{2}=0$

 $9t^{2}-9t+9t^{2}=0$

 $18t^{2}-9t=0 \Rightarrow t= \frac{1}{2}$

 A particle covers a distance from A to B over a period of time; the distance versus time plot is shown below. Then which of the following is true for the motion of the particle?

Answer:

Explanation:

 The distance versus time plot for movement of a particle from A and B is shown  in the diagram

 From graph , it is slope of distance-time graph for position A to C and D to B is always positive, therefore both the average speed and instantaneous speed are not always zero, therefore, option (a) is not correct.

 Average velocity= Total displacement from A to B / Total time interval

 Since, total displacement depends on straight distance between initial and final points, hence, the displacement between A to B always non zero.  therefore the average speed is always non zero.

 Since, no distance is travelled between point C to D is zero. Hence, instantaneous speed at every instant between C to D is zero.

Hence, option (b)  is correct and option (c) and (d) are incorrect.

A current carrying conductor obeys Ohm's law  (V= RI). If the current passing through the conductor is  $l=(5\pm0.2)$A  and voltage   developed  is  $V=(60\pm6)$ V , then find the percentage  of error is resistance, R

Answer:

Explanation:

Given, current passing through  the conductor

  $I= (5\pm0.2)A, \triangle l=0.2 A$

 Voltage developed,

 $V=(60 \pm  6) V$

 $\Rightarrow$     $\triangle V=6V$

 By Ohm's law,

   V=RI

 $R=\frac{V}{I}$

 $\therefore$    $\frac{\triangle  R}{R}=\frac{\triangle V}{V}+\frac{\triangle I}{I}$

                             =$\frac{6}{60}+\frac{0.2}{5}$

 = 0.1+0.04

  $\frac{ \triangle R}{R}=0.14$

 $\therefore$     $\frac{ \triangle R }{R} \times 100=0.14 \times$ 100=14%

Identify the correct option

Answer:

Explanation:

 The properties of the four basic forces are given in the following table,

 The above table suggests that the range of weak nuclear force is smaller than the range of strong nuclear force, gravitation and electromagnetic force.

 Hence, option (c) is correct

• Physics - General questions