A body mass 1 kg is thrown upwards with velocity 20 ms-1. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction?  (g = 10 ms-2)

  • 20 J

  • 30 J

  • 40 J

  • 40 J


A.

20 J

The energy lost due to air friction is equal to the difference of initial kinetic energy and final potential energy.

Initially, body possess only kinetic energy and after attaining a height the kinetic energy is zero.

Therefore, loss of energy = KE - PE

=

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The dependence of acceleration due to gravity g on the distance r from the centre of the earth assumed to be a sphere of radius R of uniform density is as shown in figure below


The correct figure is

  • (4) 

  • (1)

  • (2)

  • (3)


A.

(4) 



So, the correct figure will be

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The figure shows elliptical orbit of a planet m about the sun S. The shaded are SCD is twice the shaded area SAB. If t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then,

  • t1 > t2

  • t1 =4 t2

  • t1 = 2t2

  • t1 = 2t2


C.

t1 = 2t2

Apply Kepler's law of area fo planetary motion.

The line joining the sun to the planet sweeps out equal areas time interval ie, areal velocity is constant.


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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

  • 16 m/s and 17 m/s


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,

Thus, the speed of the car at the top of the hill is between 14 m/s and 15 m/s
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Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at a temperature  to C, the power received by a unit surface (normal to the incident rays) at a distance R from the centre of the sun is :

  • fraction numerator 4 πr squared σt to the power of 4 over denominator straight R squared end fraction
  • fraction numerator straight r squared straight sigma left parenthesis straight t plus 273 right parenthesis to the power of 4 over denominator 4 πR squared end fraction
  • fraction numerator 16 space straight pi squared straight r squared σt to the power of 4 over denominator straight R squared end fraction
  • fraction numerator 16 space straight pi squared straight r squared σt to the power of 4 over denominator straight R squared end fraction


D.

fraction numerator 16 space straight pi squared straight r squared σt to the power of 4 over denominator straight R squared end fraction

From Stefan's law, the rate at which energy is radiated by sun at its surface si 

P = σ x 4 πr2T4

[Sun is a perfectly black body as it emits radiations of all wavelengths and so for it e =1]
The intensity of this power at earth's surface (under the assumption R >>ro) is

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