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If   gas molecules each of mass Kg collide over the area of   in unit second with  Speed , the pressure exerted will be?

  • Option 1)

    3Nm^{2}

  • Option 2)

    2 \ Nm^2

  • Option 3)

    4Nm^{2}

  • Option 4)

    6Nm^{2}

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

A ball is thrown upward with an initial velocity V_{o} from the surface of the earth.

The motion of the ball is affected by a drag force equal to m\gamma v^{2} ( where m is 

mass of the ball, v is its instataneous velocity and \gamma is a constant). Time taken

by the ball to rise to its zenith is:

  • Option 1)

    \frac{1}{\sqrt{\gamma g}}\tan^{-1}(\sqrt{\frac{\gamma }{g}}V_{o})

  • Option 2)

    \frac{1}{\sqrt{\gamma g}}\sin^{-1}(\sqrt{\frac{\gamma }{g}}V_{o})

  • Option 3)

    \frac{1}{\sqrt{\gamma g}}\ln(1+\sqrt{\frac{\gamma }{g}}V_{o})

  • Option 4)

    \frac{1}{\sqrt{2\gamma g}}\tan^{-1}(\sqrt{\frac{2\gamma }{g}}V_{o})

 
  Newton's 2nd Law -     Therefore   - wherein  in C.G.S & S.I Force can be defined as rate of change of momentum.     =>  =>  Option 1) Option 2) Option 3) Option 4)

Two blocksA and B of masses m_{A}=1Kgand  m_{B}=3Kg re kept on the table as shown in figure . The coefficient of friction between A and B is 0,2 .and between B and the surface of the table is also 0.2 The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is :\left [ Take\, \, g=10m/s^{2} \right ]

  • Option 1)

    8N

  • Option 2)

    16N

  • Option 3)

    40N

  • Option 4)

    12N

for 1 Kg block Now,  Option 1)Option 2)Option 3)Option 4)

 

Light is incident normally on a completely absorbing surface with an energy flux of  25 Wcm ^{-2} . If the surface has an area of 25cm ^{-2} ,The momentum transferred to the surface in 40 min time duration will be : 

 

  • Option 1)

    6.3 \times 10 ^ { -4} Ns

  • Option 2)

    1.4 \times 10 ^ { -6} Ns

  • Option 3)

    5.0 \times 10 ^ { -3} Ns

  • Option 4)

    3.5 \times 10 ^ { -6} Ns

Pressure = I/C  Force = Pressure * Area = I/C * A  momentum transferred =  Option 1)Option 2)Option 3)Option 4)

A man ( mass=50 kg ) and his son ( mass=20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70ms^{-1} with respect to the man. The speed of the man with respect to the surface is :

  • Option 1)

    0.47ms^{-1}

  • Option 2)

    0.14ms^{-1}

  • Option 3)

    0.28ms^{-1}

  • Option 4)

    0.20ms^{-1}

 

Option 4

 

A uniform rod of length l is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely ?

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

  Centripetal Force - F = Centripetal force  Angular velocity n = frequency - wherein Force acts on the body along the radius and towards centre.                                                                                                                         This equation follows graph(1)   Option 1) Option 2) Option 3) Option 4)

A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F=20N, making an angle of 30^{\circ} with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is \mu =0.2. The difference between the accelerations of the block, in case (B) and case (A) will be : (g=10ms^{-2})

 

  • Option 1)

    0.4 ms^{-2}

  • Option 2)

    3.2 ms^{-2}

  • Option 3)

    0.8 ms^{-2}

  • Option 4)

    ms^{-2}

 
  Newton's 2nd Law -     Therefore   - wherein  in C.G.S & S.I Force can be defined as rate of change of momentum.                                                                                                                                                                                                                                                                     Option...

A smooth wire of length 2\pi r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed \omega about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then value of \omega ^{2} is equal to:

 

  • Option 1)

    \frac{\sqrt{3}g}{2r}

  • Option 2)

    2g/(r\sqrt{3})

  • Option 3)

    (g\sqrt{3})/r

  • Option 4)

    2g/r

from geometry  equating (1) and (2) Option 1)Option 2)Option 3)Option 4)

A spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths l_{1} and l_{2} where, l_{1}=nl_{2} and n is an integer. The ratio k_{1}/k_{2} of the corresponding force constants, k_{1} and k_{2} will be:

 

  • Option 1)

    n

     

     

     

  • Option 2)

    \frac{1}{n^{2}}

  • Option 3)

    \frac{1}{n}

  • Option 4)

    n^{2}

{  }Option 1)n      Option 2)Option 3)Option 4)

A wedge of mass M=4m lies on a frictionless plane . A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge . The maximum height climbed by the particle on the wedge is given by :

  • Option 1)

    \frac{v^{2}}{g}

  • Option 2)

    \frac{2v^{2}}{7g}

  • Option 3)

    \frac{2v^{2}}{5g}

  • Option 4)

    \frac{v^{2}}{2g}

 
Applying  Conservation of linear momentum Applying conservation of mechanical energy Option 1) Option 2) Option 3) Option 4)

A ball is thrown vertically up (taken as + z-axis) from the ground . The correct momentum -height (p-h) diagram is :

  • Option 1)

        

  • Option 2)

      

  • Option 3)

      

  • Option 4)

 
As we know  According to above equation the graph of p v/s h  is given by Option 1)      Option 2)    Option 3)    Option 4)

If 10^{22} gas molecules each of mass 10^{-26} kg collide with a surface (perpendicular to it) elastically per second over an area 1\; m^{2} with a speed 10^{4}\; m/s, the pressure exerted by the gas molecules will be of the order of :

  • Option 1)

    2\; N/m^{2}

  • Option 2)

    10^{4}\; N/m^{2}

  • Option 3)

    10^{16}\; N/m^{2}

     

  • Option 4)

    10^{8}\; N/m^{2}

 
    For 1 particle Momentum  = change for n particle For unit second Option 1) Option 2) Option 3)   Option 4)

Meu = tank theta please explain this

 

Here we have the FBD of the block on the wedge. By the figure,
a book of weight 20N is pressed between two hands and each hand excerts a force of 40N if the book just starts to slides down.coefficient of frction
Weight of book = Frictional force between hands and book mg = μN 20 = μ(40 + 40) μ = 20 / 80 μ = 0.25 Coefficient of friction is 0.25
IMG_20190307_110207.jpg Arrangement of two block system is as shown in the figure . mass of block A is •m A 5K g and mass of B is 10Kg A constant force F = 100 N applied on upper block A . Friction between A and B is and b/w B and ground surface is smooth , then find work done by frictional force on block B in t = 2 sec if system starts from rest smooth
 For block A  N = mg     =   For block B 
IMG_20190307_105330.jpg A block of mass m = 4 Kg is given an initial velocity v = 20 m/s up to the plane as shown in the figure . find the work done by frictional force 0.5 and g = 10 m/s before the block stop for a while
   Now             
6,7 & 8

6) Three blocks A,B and C are suspended as shown in figure.  Mass of each block A and C is m, if system is in equilibrium and mass of B is M then

a)M=2m b) M<2m c) M>2m d) M=m

7) In the arrangement shown in figure pullys are massless and frictionless and threads are light and inextensible. Block of mass m1 will remain at rest if

a) 1/m1=1/m2 + 1/m3 b) 4/m1= 1/m2 + 1/m3 c) m1=m2+m3 d) 1/m3= 2/m2 + 3/m1

8) A uniform rope of length L and mass M is placed on a smooth fixed wedge as shown. Both end of rope are at same horizontal level. The rope is initially released from rest, then the magnitude of initial acceleration is

 

The answer to question  number 8 is explained below:

15513417365571608153671.jpg

one slug is equivalent to 14.6 Kg. A force of 10 Pound is applied on a body of 1Kg. The acceleration of the body is

Just convert Pound force into newton 1 pound  = 4.448 N Now apply F = ma

1550991201052683533745.jpg

The solution of the above question is mentioned below:   

Ans. pls

Free body diagram (FBD) for a car which is moving on a frictionless circular road has an angle \theta with the horizontal. Choose the correct FBD for car and specify reasons to choose it.

 Resolve force vectors into components that are parallel to and perpendicular to the acceleration of the object. In the case of a vehicle driving around a banked curve the acceleration is horizontal, towards the center so it's the normal force (and friction if any) that you would resolve into components.
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