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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2011 Exam Questions

Multiple Choice Questions

1.

A pulley of radius 2m is rotated about its axis by a force F = (20t - 5t2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, the number of rotations made by the pulley before its direction of motion if reversed is

  • more than 3 but less than 6

  • more than 6 but less than 9

  • more than 9

  • less than


A.

more than 3 but less than 6

To reverse the direction
integral τdθ space equals 0
straight tau space equals space left parenthesis 20 space straight t minus 5 straight t squared right parenthesis 2 space equals space 40 straight t minus 10 straight t squared
straight alpha space equals space straight tau over straight I space equals space fraction numerator 40 straight t minus 10 straight t squared over denominator 10 end fraction space equals space 4 straight t minus straight t squared
straight omega space equals space integral subscript 0 superscript straight t space αdt space equals space 2 straight t squared minus straight t cubed over 3
straight omega space is space zero space at
2 straight t squared minus straight t cubed over 3 space equals space 0
straight t cubed space equals space 6 straight t squared
straight t space equals space 6 space sec
straight theta space equals space integral ωdt
space equals space integral subscript 0 superscript 6 left parenthesis 2 straight t squared minus straight t cubed over 3 right parenthesis dt
open square brackets fraction numerator 2 straight t cubed over denominator 3 end fraction minus straight t to the power of 4 over 12 close square brackets subscript 0 superscript 6 space equals space 216 space open square brackets 2 over 3 minus 1 half close square brackets space equals space 36 space rad.
No space of space revolution space fraction numerator 36 over denominator 2 straight pi end fraction space Less space than space 6

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2.

A thin horizontal circular disc is rotating about a vertical axis passing through its centre. An insect is at rest at a point near the rim of the disc. The insect now moves along a diameter of the disc to reach its other end. During the journey of the insect, the angular speed of the disc 

  • remains unchanged

  • continuously decreases

  • continuously increases 

  • first increases and then decreases


D.

first increases and then decreases


From angular momentum conservation about the vertical axis passing through centre. When the insect is coming from circumference to center. Moment of inertia first decreases then increase. So angular velocity increase than decrease. 
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3.

Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

  • -4Gm/r

  • -6Gm/r

  • -9Gm/r

  • zero


C.

-9Gm/r

GM over straight x squared space equals space fraction numerator straight G left parenthesis 4 straight m right parenthesis over denominator left parenthesis straight r minus straight x squared right parenthesis end fraction
1 over straight x space equals space fraction numerator 2 over denominator straight r minus straight x end fraction
straight r minus straight x space equals space 2 straight x
3 straight x space equals space straight r over 3


= -3Gm/r - 6Gm/r = -9Gm/r


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4.

Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d < < l) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate.As a result the charges approach each other with a velocity v. Then as a function of distance x between them

  • v ∝ x-1

  • v ∝ x1/2

  • v ∝ x

  • v ∝ x-1/2


D.

v ∝ x-1/2



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5.

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

  • g

  • 2/3g

  • g/3

  • 3/2g


B.

2/3g


mg - T = ma
TR space equals space fraction numerator mR squared straight alpha over denominator 2 end fraction
straight T space equals space mRα over 2 space equals space ma over 2
mg space minus ma over 2 space equals space ma
fraction numerator 3 space ma over denominator 2 end fraction space equals space mg
straight a space equals space fraction numerator 2 straight g over denominator 3 end fraction
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6.

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is

  • straight pi straight v to the power of 4 over straight g squared
  • straight pi over 2 straight v to the power of 4 over straight g squared
  • straight pi straight v squared over straight g squared
  • straight pi straight v squared over straight g

C.

straight pi straight v squared over straight g squared

Maximum range of water coming out of the fountain

straight R subscript straight m space equals space straight v squared over straight g
therefore comma space Total space area space around space fountain
space straight A space equals space πR subscript straight m superscript 2 space equals straight pi straight v to the power of 4 over straight g squared

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7.

The transverse displacement y(x,t) of a wave on a string is given by
straight y left parenthesis straight x comma straight t right parenthesis space equals space straight e to the power of negative left parenthesis ax squared plus bt squared space plus space 2 square root of ab space xt right parenthesis end exponent
This represents a

  • wave moving in +x direction with speed square root of straight a over straight b end root

  • wave moving in -x direction with speed square root of b over a end root

  • standing wave frequency square root of straight a

  • Standing wave of frequency square root of 1 over straight b end root


B.

wave moving in -x direction with speed square root of b over a end root

Given wave is,
straight y space left parenthesis straight x comma straight t right parenthesis space equals space straight e to the power of negative left parenthesis ax squared space plus bt squared space plus 2 square root of ab space xt right parenthesis end exponent
space equals space straight e to the power of negative left parenthesis square root of ax space plus square root of bt right parenthesis squared end exponent
It space is space straight a space function space of space type space straight y space equals space straight f left parenthesis ωt space plus kx right parenthesis
space equals space straight f open parentheses straight t minus straight x over straight v close parentheses
Where space straight v space is space the space wave space velocity.
therefore space straight y space left parenthesis straight x comma straight t right parenthesis space represents space wave space travelling space along space minus straight x space direction
Speed space of space wave space space equals space straight omega over straight k space equals space fraction numerator square root of straight b over denominator square root of straight a end fraction space equals space square root of straight b over straight a end root

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8.

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly. (Surface tension of soap solution = 0.03 Nm-1)

  • 4π mJ 

  • 0.2π mJ

  • 2π mJ

  • 0.4π mJ


D.

0.4π mJ

Work done = Change in surface energy
⇒ W = 2T x 4π (R22-R12)
 = 2 x 0.03 x 4π [ (5)2-(3)2] x 10-4
 = 0.4 π mJ

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9.

An object moving with a speed of 6.25m/s, is decelerated at a rate given by dv/dt =- 2.5√v, where v is the instantaneous speed. The time taken by the object, to come to rest, would be

  • 2 s

  • 4 s

  • 8 s

  • 1 s


A.

2 s

dv over dt space equals space minus space 2.5 square root of straight v
fraction numerator dv over denominator square root of straight v end fraction space equals space minus 2.5 dt
integral subscript 6.25 end subscript superscript 0 straight v to the power of negative 1 divided by 2 end exponent space dv space equals space minus 2.5 integral subscript 0 superscript straight t dt
minus 2.5 left square bracket straight t right square bracket subscript 0 superscript straight t space equals space left square bracket 2 space straight v to the power of 1 divided by 2 end exponent right square bracket subscript 6.25 end subscript superscript 0
straight t space equals 2 straight s
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10.

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm
Circular scale reading: 52 division
Given that 1 mm on the main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is

  • 0.052 cm

  • 0.026 cm

  • 0.005

  • 0.52


A.

0.052 cm

Least count of screw gauge =1/100 mm = 0.01 mn
Diameter - Divisions on cirular scale × least count + main scale reading = 52 × 1/100 + 0
100 = 0.52 mm
diameter = 0.052 cm

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