JEE physics

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Question
CBSEENPH11020250

A particle of mass m is moving along the side of a square of side ‘a’, with a uniform speed v in the x-y plane as shown in the figure:
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Which of the following statements is false for the angular momentum → L about the origin?

  • bold L space equals space minus fraction numerator mv over denominator square root of 2 end fraction space straight R space straight k with hat on top comma space when space the space paticle space is space moving space from space straight A space to space straight B
  • straight L space equals space mv space open parentheses fraction numerator straight R over denominator square root of 2 end fraction space plus straight a close parentheses space bold k with bold hat on top comma space when space the space particle space is space moving space from space straight B space to space straight C
  • straight L space equals space mv space open parentheses fraction numerator straight R over denominator square root of 2 end fraction space minus straight a close parentheses space straight k with hat on top comma space when space the space particle space is space moving space from space straight B space to space straight C
  • bold L space equals space minus fraction numerator mv over denominator square root of 2 end fraction space straight R space straight k with hat on top comma space when space the space paticle space is space moving space from space straight D space to space straight A

Solution

B.

straight L space equals space mv space open parentheses fraction numerator straight R over denominator square root of 2 end fraction space plus straight a close parentheses space bold k with bold hat on top comma space when space the space particle space is space moving space from space straight B space to space straight C

For a particle of mass, m is moving along the side of a square a. Such that
Angular momentum L about the origin
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Sponsor Area

Question
CBSEENPH11020251

A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals µ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction µ and the distance x(=QR), are, respectively close to :


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  • 0.2 and 6.5 m

  • 0.2 and 3.5 m

  • 0.29 and 3.5 m

  • 0.29 and 6.5 m

Solution

C.

0.29 and 3.5 m

Energy lost over path PQ = μ mg cos θ x 4
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Energy lost over path QR = μ mgx
i.e μ mg cos 30°  x 4 = μ mgx  (∴ θ = 30°)
straight x space equals space 2 square root of 3 space equals space 3.45 space straight m
From Q to R energy loss is half of the total energy loss.
i.e μ mgx = mgh/2
μ = 0.29
The values of the coefficient of friction μ and the distance x (=OR) are 0.29 and 3.5 m

Question
CBSEENPH11020252

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms−2:

  • 2.45 ×10−3 kg

  • 6.45 x×10−3 kg

  • 9.89 ×10−3 kg

  • 12.89 ×10−3 kg

Solution

D.

12.89 ×10−3 kg

Given potential energy burnt by lifting weight
= mgh = 10 x 9.8 x 1 x 1000 = 9.8 x 104 J
If mass lost by a person be m, then energy dissipated
 = m x 2 x 38 x 107 J /10
⇒ m = 5 x 10-3 x 9.8 / 3.8
= 12.89 x 10-3 kg 

Question
CBSEENPH11020253

A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

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  • turn left

  • turn right

  • go straight

  • turn left and right alternately

Solution

A.

turn left

As, the wheel rolls forward the radius of the wheel, decreases along AB hence for the same number of rotations it moves less distance along AB, hence it turns left.

Question
CBSEENPH11020254

A satellite is revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R ; h<<R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)

  • square root of 2 gR end root
  • square root of gR
  • square root of gR divided by 2 end root
  • square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

Solution

D.

square root of gR space left parenthesis square root of 2 space minus 1 right parenthesis

Given, a satellite is revolving in a circular orbit at a height h from the Earth's surface having radius of earth R, i.e h <<R
Orbit velocity of a satellite
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therefore, the minimum increase in its orbital velocity required to escape from the Earth's Gravitational Field.
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