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A bullet of mass 20 g has an initial speed of 1 ms^{-1} , just before its starts penetrating a mud wall of thickness 20 cm . If the wall offers a mean resistance of 2.5 \times 10 ^{-2} N , the speed of the bullet after emerging from the other side of the wall is close to : 

 

  • Option 1)

    0.1 ms^{-1}

  • Option 2)

    0.7 ms^{-1}

  • Option 3)

    0.3 ms^{-1}

  • Option 4)

    0.4 ms^{-1}

Given ,   Option 1)Option 2)Option 3)Option 4)

The trajectory of a projectile near the surface of the earth is given as y=2x-9x^{2}. If it were launched at an angle \theta_{0} with speed v_{0} then \left ( g=10ms^{-2} \right ):

  • Option 1)

    \theta_{0}=\sin^{-1}\left ( \frac{2}{\sqrt{5}} \right ) and v_{0}=\frac{3}{5}ms^{-1}

  • Option 2)

    \theta_{0}=\cos^{-1}\left ( \frac{2}{\sqrt{5}} \right ) and v_{0}=\frac{3}{5}ms^{-1}

  • Option 3)

    \theta_{0}=\cos^{-1}\left ( \frac{1}{\sqrt{5}} \right ) and v_{0}=\frac{5}{3}ms^{-1}

  • Option 4)

    \theta_{0}=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right ) and v_{0}=\frac{5}{3}ms^{-1}

 
  Equation of path of a projectile - it is equation of parabola     Acceleratio due to gravity     initial velocity Angle of projection   - wherein Path followed by a projectile is parabolic is nature.     Given,       General Eqn.   Option 1)  and  Option 2)  and  Option 3)  and  Option 4)  and 

A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If  t_{1} and t_{2} are the values of the time taken by it to hit the target in two possible ways, the product t_{1}t_{2} is:

  • Option 1)

    R/2g

  • Option 2)

    R/4g

  • Option 3)

    R/g

  • Option 4)

    2R/g

 
  Horizontal Range - Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.       - wherein Special case of horizontal range For max horizontal range.     If the range is equal for both the target then the angle of projection may be different.   ;     ranges are  equal. Option 1) Option 2) Option 3) Option 4)

A particle starts from origin O from rest and moves with a uniform acceleration along the positive x-axis. Identify all figure that corectly represent the motion qualitatively. (a=acceleration, v=velocity,x=displacement,t=time)

  • Option 1)

    (B),(C)

  • Option 2)

    (A)

  • Option 3)

    (A),(B),(C)

  • Option 4)

    (A),(B),(D)

 
  1st equation or velocity time equation -   V = Final velocity u = Initial velocity A = acceleration T = time -     2nd equation or Position- time equation - Displacement Initial velocity acceleration time   -     a=Constant hence(A) v=at Hence(B)   Hence(D) Option 1) (B),(C) Option 2) (A) Option 3) (A),(B),(C) Option 4) (A),(B),(D)

Let \left | \overrightarrow{A_{1}} \right |=3,\left | \overrightarrow{A_{2}} \right |=5 and \left | \overrightarrow{A_{1}} +\overrightarrow{A_{2}} \right |=5 the value of \left | 2\overrightarrow{A_{1}} +3\overrightarrow{A_{2}} \right |\cdot \left | 3\overrightarrow{A_{1}} -2\overrightarrow{A_{2}} \right | is

  • Option 1)

    -106.5

  • Option 2)

    -99.5

  • Option 3)

    -112.5

  • Option 4)

    -118.5

 
  Scalar , Dot or Inner Product - Scalar product of two vector & written as is a scalar quantity given by the product of magnitude of & and the cosine of smaller angle between them. - wherein showing representation of scalar products of vectors.     ,  =     Option 1) -106.5 Option 2) -99.5 Option 3) -112.5 Option 4) -118.5

A particle is moving with speed \upsilon =b\sqrt{x} along the positive x-axis. Calculate the speed of the particle at timet=\tau(assume that the particle is at the origin at t=0).

 

  • Option 1)

    \frac{b^{2}\tau}{4}

     

     

     

  • Option 2)

    \frac{b^{2}\tau}{2}

  • Option 3)

    b^{2}\tau

  • Option 4)

    \frac{b^{2}\tau}{\sqrt{2}}

Option 1)      Option 2)Option 3)Option 4)

Two particles are projected from the same point with the same speed u such that they have the same rang R, but different maximum heights h_{1} and h_{2}. Which of the following is correct?

  • Option 1)

    R^{2}=4h_{1}h_{2}

     

  • Option 2)

    R^{2}=16h_{1}h_{2}

  • Option 3)

    R^{2}=2h_{1}h_{2}

  • Option 4)

    R^{2}=h_{1}h_{2}

 
R is the same for  and  angle of projection  ,     ,         Option 1)   Option 2) Option 3) Option 4)

The position vector of a particle changes with time according to the relation \vec{r}(t)=15t^{2}\:\hat{i}+(4-20t^{2})\:\hat{j}. What is the magnitude of the accelaration at t=1 ?

  • Option 1)

    40

  • Option 2)

    25

  • Option 3)

    100

  • Option 4)

    50

     Option 1)Option 2)Option 3)Option 4)

The position of a particles as a function of time t , is given by

x(t)=at+bt^{2}-ct^{3}

where a,b,and c are constants . When the particle attains zero acceleration , then its velocity will be :

  • Option 1)

    a+\frac{b^{2}}{4c}

  • Option 2)

    a+\frac{b^{2}}{3c}

  • Option 3)

    a+\frac{b^{2}}{c}

  • Option 4)

    a+\frac{b^{2}}{2c}

 
Given  Option 1) Option 2) Option 3) Option 4)

The stream of a river is flowing with a speed of 2\: km/h. A swimmer can swim at a speed of 4\: km/h . What should be the direction of the swimmer with respect to the flow of the river to cross the river straight ?

 

  • Option 1)

    90^{\circ}         

  • Option 2)

    150^{\circ}

  • Option 3)

    120^{\circ}

  • Option 4)

    60^{\circ}

 
         Now, the angle with the downstream is             Option 1)           Option 2) Option 3) Option 4)

Ship A is sailing towards north-east with velocity \vec{v}= 30 \widehat{i}+50\widehat{j} km/hr where \widehat{i} points east and \widehat{j}, north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in :

  • Option 1)

    2.6 hrs.

  • Option 2)

    2.2 hrs.

  • Option 3)

    4.2 hrs.

  • Option 4)

    3.2 hrs.

 Option 1)2.6 hrs.Option 2)2.2 hrs.Option 3)4.2 hrs.Option 4)3.2 hrs.
Screenshot_2019-03-02-15-10-57-209_com.android.chrome.png Two guns A and B can fire bullets at speeds 1 km/s and 2km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by two guns , on the ground is:
  450 Sin 2(45) nge = R — U •sin•2f-3 R —K [12 pnd A IIR A1 [14 2 1 16
Screenshot_2019-03-02-15-01-13-335_com.android.chrome.png A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v , the total area around the fountain that gets wet is
Horizontal Range - Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits. - wherein Special case of horizontal range For max horizontal range.    
15498700639711664720689.jpg The magnitude of displacement and average velocity of the top of the seconds hand of a clock of dial radius R 25 cm during t 10s.
@Aditi  For 60 sec total theta is 360 So for 10 sec theta is 60 now we find displacement which is equal to (1/4) m and avarage velocity=d/t=(1/40)m/s
IMG_20190120_100813.jpg A stone falling from the top of a vertical tower has descended x metres when another is let fall from a point? y metres below the top. if they fall from rest and reach the ground together, show that the height of the tower is
Let A and B two stone as shown in figure and B just fell when A has descended at a distance x from the top of tower   similarly      
A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after which the splash is heard is given by
@Suraj raj Time taken for thrown ball to reach the bottom  = Time taken by sound to reach the top  = h/v  Thus total time =

A particle just clears a wall of height b at a
distance a and strikes the ground at a distance c
from the point of projection. The angle of
projection is

  • Option 1)

    tan^-(\frac{bc}{a*(c-a)})

  • Option 2)

    tan^-(\frac{bc}{a })

  • Option 3)

    tan^-(\frac{b}{a*c})

  • Option 4)

    45^0

 

2

A body falling from rest under gravity passes a certain point P. It was at a distance of 400 m from P, 4s prior to passing through P. If g = 10m/s^2 , then the height above the point P from where the body began to fall is

  • Option 1)

    720

  • Option 2)

    900

  • Option 3)

    320

  • Option 4)

    680

 
Option 1) 720 Option 2) 900 Option 3) 320 Option 4) 680
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