NEET physics

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Question
CBSEENPH11020547

The dimensions of left parenthesis straight mu subscript straight o straight epsilon subscript straight o right parenthesis to the power of negative 1 divided by 2 end exponent are

  • [L1/2 T-1/2]

  • [L-1T]

  • [LT-1]

  • [L1/2 T1/2]

Solution

C.

[LT-1]

left parenthesis straight mu subscript straight o straight epsilon subscript straight o right parenthesis to the power of negative 1 divided by 2 end exponent space is space velocity space of space light
Hence space straight c space equals space fraction numerator 1 over denominator square root of straight mu subscript straight o straight epsilon subscript straight o end root end fraction space equals space left square bracket LT to the power of negative 1 end exponent right square bracket

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Question
CBSEENPH11020548

A stone is dropped from a height h. It hits the ground with a certain momentum p. If the same stone is dropped from a height 100% more than the previous height, the momentum when it hits the ground will change by

  • 68%

  • 41%

  • 200%

  • 100%

Solution

B.

41%

velocity space straight v space equals space square root of 2 gh end root space space space space space space space.... left parenthesis straight i right parenthesis
and space momentum space straight p space equals space mv space space space space... space left parenthesis ii right parenthesis
From space left parenthesis Eqs. right parenthesis space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space have
space straight p proportional to square root of straight h
Here comma
straight p subscript 2 over straight p subscript 1 space equals space square root of straight h subscript 2 over straight h subscript 1 end root
therefore space fraction numerator straight p subscript 2 over denominator straight p subscript 1 space end fraction space equals square root of fraction numerator 2 straight h over denominator straight h end fraction space end root space equals square root of 2

straight p subscript 2 space equals 1.414 straight p subscript 1
percent sign change space equals space fraction numerator straight p subscript 2 minus straight p subscript 1 over denominator straight p subscript 1 end fraction space straight x space 100 space equals space 41 percent sign

Question
CBSEENPH11020549

A car of mass m starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude Po. The instantaneous velocity of this car is proportional to

  • t2Po

  • t1/2

  • t-1/2

  • bevelled fraction numerator straight t over denominator square root of straight m end fraction

Solution

B.

t1/2

Power Po = Fv
equals open parentheses straight m dv over dt close parentheses straight v
equals space mv dv over dt
Integrating space both space sides
integral straight P subscript straight o dt space equals integral mv space dv
straight P subscript straight o straight t space equals space mv squared over 2
straight v squared space equals space fraction numerator 2 straight P subscript straight o straight t over denominator straight m end fraction
straight v proportional to space straight t to the power of 1 divided by 2 end exponent

Question
CBSEENPH11020550

A car of mass m is moving on a level circular track of radius R. If straight mu subscript straight s represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by

  • square root of straight mu subscript straight s mRg end root
  • square root of Rg divided by straight mu subscript straight s end root
  • square root of mRg divided by straight mu subscript straight s end root
  • square root of straight mu subscript straight s Rg end root

Solution

D.

square root of straight mu subscript straight s Rg end root

In this condition, centripetal force is equal to static frictional force between road and tyres, 
so
straight mu subscript straight s space mg space equals mv squared over straight R
straight v subscript max space equals space square root of straight mu subscript straight s Rg end root

Question
CBSEENPH11020551

A circular platform is mounted on the frictionless vertical axle. Its radius R =2 m and its moment of inertia about the axle is 200 kg m2. It is initially at rest. A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms-1 relative to the ground. Time taken by the man to complete one revolution is

  • π sec

  • 3π/2 sec

  • 2π sec

  • π/2 sec

Solution

A.

π sec

For conservation of angular momentum
Iω space equals space mvr
200 space straight x space straight omega space equals space 50 space straight x space 2 space straight x space 1
space straight omega space equals space 1 half space rad divided by straight s
straight v equals space rω space equals space 1 space straight m divided by straight s
straight T equals space fraction numerator 2 πr over denominator 1 minus left parenthesis negative 1 right parenthesis end fraction space equals space fraction numerator 2 πr over denominator 2 end fraction space equals space πr space sec