NEET physics

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Question
CBSEENPH11020642

A ball is thrown vertically downwards from a height of 20m with an initial velocity vo, It collides with the ground, loses 50% of its energy in collision and rebounds to the same height. The initial velocity vo is,

(Take g = 10 ms-2)

  • 14 ms-1

  • 20 ms-1

  • 28 ms-1

  • 10 ms-1

Solution

B.

20 ms-1

Suppose a ball rebounds with speed v,

straight v space equals space square root of 2 gh space end root space equals space square root of 2 space straight x space 10 space straight x space 20 space end root space equals space 20 space straight m divided by straight s

Energy space of space straight a space ball space just space after space rebound comma

straight E space equals space 1 half space mv squared space equals space 200 space straight m

As comma space 50 space percent sign space of space energy space loses space in space collision space means space just space before space collision space energy space 400 space straight m.
According space to space law space of space conservation space of space energy comma space we space have
1 half space mv subscript 0 superscript 2 space space plus space mgh space equals space 400 space straight m

rightwards double arrow space 1 half mv subscript 0 superscript 2 space plus space straight m space straight x space 10 space straight x space 20 space equals space 400 space straight m space
rightwards double arrow space straight v subscript straight o space equals space 20 space straight m divided by straight s

Sponsor Area

Question
CBSEENPH11020645

A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth. Then,

  • the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant

  • the total mechanical energy of S varies periodically with time

  • the linear momentum of S remains constant in magnitude

  • the acceleration of S is always directed towards the centre of the earth

Solution

D.

the acceleration of S is always directed towards the centre of the earth

As we know that, the force on satellite is an only gravitational force which will always be towards the centre of the earth. Thus, the acceleration is S is always directed towards the centre of the earth

Question
CBSEENPH11020649

The heart of a man pumps 5 L of blood through  the arteries per minute at a pressure of 150 mm of mercury . If the density of mercury be 13.6 x 103 kg/ m3 and g = 10 m/s2, then the power of heart in watt is  

  • 1.70

  • 2.35

  • 3.0

  • 1.50

Solution

A.

1.70

 Given, pressure = 150 mm of Hg
Pumping rate of heart of a man
 = fraction numerator d straight V over denominator d straight t end fraction space equals space fraction numerator 5 space straight x space 10 to the power of negative 3 end exponent over denominator 60 end fraction space straight m cubed divided by straight s

Power space of space heart space equals space straight P space. dV over dt space equals space ρgh space. space dV over dt space left square bracket space straight p equals space ρgh right square bracket

rightwards double arrow space fraction numerator left parenthesis 13.6 space straight x space 10 cubed space kg divided by straight m cubed right parenthesis left parenthesis 10 right parenthesis space straight x 0.15 space straight x space 5 space straight x space 10 to the power of negative 3 end exponent space over denominator 60 end fraction space equals space 1.70 space straight W

Question
CBSEENPH11020651

An automobile moves on a road with a speed of 54 km h-1. The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3 kg m2 . If the vehicle is brought to rest in 15 s, the magnitude of average torque transmitted by its breaks to the wheel is

  • 6.66 kg m2s-2

  • 8.58 kg m2s-2

  • 10.86 kg m2s-2

  • 2.86 kg m2s-2

Solution

A.

6.66 kg m2s-2

As velocity of an automobile vehicle
straight v space equals space 54 space km divided by straight h space equals space 54 space straight x space 5 over 18 space equals space 15 space straight m divided by straight s
Angular space velocity space of space straight a space vehicle comma space straight v space equals space straight omega subscript straight o straight r

rightwards double arrow space straight omega subscript straight o space equals space straight v over straight R space equals space fraction numerator 15 over denominator 0.45 end fraction space equals space 100 over 3 space rad divided by straight s
So comma space angular space acceleration space of space an space antomobile
straight alpha space equals space fraction numerator increment straight omega over denominator straight t end fraction space equals space fraction numerator straight omega subscript straight t minus straight omega subscript straight o over denominator straight t end fraction space equals space fraction numerator 0 minus begin display style 100 over 3 end style over denominator 15 end fraction space equals space fraction numerator negative 100 over denominator 45 end fraction space rad divided by straight s squared
Thus comma space average space torque space transmitted space by space its space brakes space to space wheel
torque space equals space Iα
rightwards double arrow space 3 space straight x space 100 over 45 space equals space 6.66 space kg space straight m squared straight s to the power of negative 2 end exponent

Question
CBSEENPH11020654

straight A space force space straight F space equals space straight alpha stack space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 straight i with hat on top space minus space 6 space straight j with hat on top space minus 12 space straight k with hat on top. The value of α for which angular momentum about origin is conserved is
  • -1

  • 2

  • zero

  • 1

Solution

A.

-1

When the resultant external torque acting on a system is zero, the total angular momentum of a system is zero, the total angular momentum of a system remains constant.This is the principle of the conservation of angular momentum.
Given,
straight F space equals space straight alpha space straight i with hat on top space plus 3 space straight j with hat on top space plus space 6 space straight k with hat on top space is space acting space at space straight a space point space straight r equals space 2 space straight i with hat on top space minus 6 space straight j with hat on top space minus 12 space straight k with hat on top
As comma space angular space momentum space about space origin space is space consverved
straight i. straight e. comma
Torque space equals space constant
rightwards double arrow space Torque space equals space 0 space rightwards double arrow space straight r space xF space equals space 0

open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 cell negative 6 end cell cell negative 12 end cell row straight alpha 3 6 end table close vertical bar space equals space 0

rightwards double arrow space left parenthesis negative 36 space plus 36 right parenthesis straight i with hat on top space minus space left parenthesis 12 space plus space 12 straight alpha right parenthesis space straight j with hat on top space plus left parenthesis 6 space plus 6 straight alpha right parenthesis space straight k with hat on top space equals 0

rightwards double arrow space 0 space straight i with hat on top space minus space 12 space straight i with hat on top space left parenthesis space 1 plus straight alpha right parenthesis straight i with hat on top space plus space 6 space left parenthesis 1 space plus space straight alpha right parenthesis straight k with hat on top space equals 0

6 space left parenthesis 1 space plus straight alpha right parenthesis space equals 0 space rightwards double arrow space straight alpha space equals negative 1