NEET physics

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Question
CBSEENPH11020740

If the lattice parameter for a crystalline structure is 3.6 A, then atomic radius in fcc crystal is

  • 1.18 A

  • 2.10 A

  • 2.92 A

  • 1.27 A

Solution

D.

1.27 A

Atomic radius is the distance between two nearest neighbours. For Fcc crystal 

Atomic space radius space equals space fraction numerator 1 over denominator 2 square root of 2 end fraction space straight x space lattice space parameter
or space space straight r space equals space fraction numerator straight a over denominator 2 square root of 2 end fraction
straight r equals space fraction numerator 3.6 over denominator 2 square root of 2 end fraction space straight A to the power of straight o space equals space 1.27 space straight A to the power of straight o

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Question
CBSEENPH11020741

If the error in the measurement of radius of sphere is 2%, then the error in the determination of volume of the sphere will be

  • 4%

  • 6%

  • 8%

  • 2%

Solution

B.

6%

Volume space of space straight a space sphere space equals space 4 over 3 straight pi space left parenthesis radius right parenthesis cubed
or space straight V space equals space 4 over 3 πR cubed
Taking space logarithium space on space both space sides comma space we space have space
log space straight V space equals space log space 4 over 3 space πR cubed
Taking space logarithium space on space both space sides comma space we space have
log space straight V space equals space log space 4 over 3 straight pi space plus 3 space log space straight R
Differentiating comma space we space get comma
fraction numerator increment straight V over denominator straight V end fraction space equals space 0 space plus fraction numerator 3 increment straight R over denominator straight R end fraction
Accordingly comma space fraction numerator increment straight R over denominator straight R end fraction space equals 2 percent sign
Thus comma space fraction numerator increment straight V over denominator straight V end fraction space equals space 3 space straight x space 2 percent sign space equals space 6 percent sign 

Question
CBSEENPH11020742

A roller coaster is designed such that riders experience " weightlessness" as they go round the top of a hill whose radius of curvature is 20 m.  The speed of the car at the top of the hill is between

  • 14 m/s and 15 m/s

  • 15m/s and 16 m/s

  • 16 m/s and 17 m/s

  • 13m/s and 17 m/s

Solution

A.

14 m/s and 15 m/s


Balancing the forces, we get
Mg-N = Mv2/R
For weightlessness, N = 0 
therefore, MV2/R = Mg
where R is the radius of curvature and v is the speeds of the car.
Therefore,
straight v space equals square root of Rg
putting space the space values comma space straight R space equals space 20 straight m comma space straight g space equals space 10.0 space straight m divided by straight s squared
So comma space straight v equals space square root of 20 space straight x space 10.0 end root space equals 14.14 space straight m divided by straight s squared
Thus, the speed of the car at the top of the hill is between 14 m/s and 15 m/s

Question
CBSEENPH11020743

The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is 

  • square root of 3 colon square root of 2
  • 1 colon square root of 2
  • square root of 2 colon 1
  • square root of 2 colon space square root of 3

Solution

B.

1 colon square root of 2

The square root of the ratio of the moment of inertia of rigid body nad its mass is called radius of gyration.
As in key idea, radius of gyration is given by
straight K space equals space square root of straight I over straight M end root
For space given space problem
straight K subscript disc over straight K subscript ring space equals space square root of straight I subscript disc over straight I subscript ring end root
But space straight I subscript disc space left parenthesis about space its space axis right parenthesis space equals space 1 half MR squared
and space straight I subscript ring space left parenthesis about space its space axis right parenthesis space equals MR squared
Where space straight R space is space the space the space radius space of space both space bodies.
Therefore comma space eq left parenthesis straight i right parenthesis space becomes
straight K straight K subscript disc over straight K subscript ring space equals space square root of fraction numerator begin display style 1 half end style MR squared over denominator MR squared end fraction end root space equals space 1 colon square root of 2

Question
CBSEENPH11020744

Water falls from a height of 60 m at the rate of 15 kg /s to operate a turbine. The losses due to frictional 

  • 8.1 kW

  • 10.2 kW

  • 12.3 kW

  • 7.0 kW

Solution

A.

8.1 kW

Power generated by the turbine is,
Pgenerated = Pinput  x 90/100

equals Mgh over straight t space straight x 90 over 100
Putting space the space given space values
straight M over straight t space straight x space 15 space kg divided by straight s comma space straight g space equals space 10 space straight m divided by straight s squared space comma space straight h space equals 60 space straight m
straight P subscript generated space end subscript space space equals space left parenthesis 15 space straight x space 10 space straight x 60 right parenthesis space straight x 90 over 100
space equals space 8.1 space kW