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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2016 Exam Questions

Multiple Choice Questions

1.

A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

  • 0.75 mm

  • 0.80 mm

  • 0.70 mm

  • 0.50 mm


A.

0.75 mm

Given that the screw gauge has zero error.
So, least count of screw gauge = 0.5/50 mm

The thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line. we have

 =0 .50 mm + (25) x 0.5/50 mm
= 0.50 mm + 0.25 mm = 0.75 mm

971 Views

2.

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

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

straight v equals space square root of fraction numerator GM over denominator straight R plus straight h end fraction end root space equals space square root of GM over straight R end root space left parenthesis as space straight h less than less than straight R right parenthesis
Velocity space required space to space escape
1 half space mv squared space equals space fraction numerator GMn over denominator straight R plus straight h end fraction

straight v apostrophe space equals space square root of fraction numerator 2 GM over denominator straight R plus straight h end fraction end root space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space left parenthesis straight h less than less than straight R right parenthesis

therefore, the minimum increase in its orbital velocity required to escape from the Earth's Gravitational Field.

straight v apostrophe minus straight v space equals space square root of fraction numerator 2 GM over denominator straight R end fraction end root space minus square root of GM over straight R end root
space equals space square root of 2 gR end root minus square root of gR space equals space square root of gR space left parenthesis square root of 2 minus 1 right parenthesis

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3.

A pendulum clock loses 12 s a day if the temperature is 408C and gains 4 s a day if the temperature is 208C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively:

  • 25 C; α=1.85×10−5/ °C

  • 60 °C; α=1.85×10−4/ °C

  • 30°C; α=1.85×10−3/°C

  • 55°C; α=1.85×10−2/°8C


A.

25 C; α=1.85×10−5/ °C

Time period of pendulum,
straight T space equals space 2 straight pi square root of straight l over straight g end root
where comma space straight l space is space length space of space pendulum space and space straight g space is space acceleration space due space to space gravity
such space as space change space in space space time space period space of space straight a space pendulum comma
fraction numerator increment straight T over denominator straight T end fraction space equals space 1 half fraction numerator increment straight l over denominator straight l end fraction
When space clock space gains space 12 space straight s comma we space get
12 over straight T space equals space 1 half straight alpha space left parenthesis 40 minus straight theta right parenthesis
when space clock space loses space 4 straight s comma space we space get
4 over straight T space equals space 1 half straight alpha space left parenthesis straight theta minus 20 right parenthesis
Comparing space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
3 space equals space fraction numerator 40 minus straight theta over denominator straight theta minus 20 end fraction
rightwards double arrow space 30 minus 60 space equals space 40 minus 0
rightwards double arrow space 40 space equals space 100 space rightwards double arrow straight theta space equals space 25 to the power of straight o space straight C
substituting space the space value space of space straight theta space in space eq space left parenthesis straight i right parenthesis space we space have
12 over straight T space equals space 1 half straight alpha space left parenthesis 40 minus 25 right parenthesis
rightwards double arrow space substituting space the space value space of space straight theta space in space eq space left parenthesis straight i right parenthesis space we space have
12 over straight T space equals space 1 half straight alpha space left parenthesis 40 minus 25 right parenthesis
rightwards double arrow space fraction numerator 12 over denominator 24 space straight x space 3600 space end fraction space equals space 1 half straight alpha space left parenthesis 15 right parenthesis
straight alpha space equals space fraction numerator 24 over denominator 24 space straight x space 3600 space straight x space 15 end fraction
straight alpha space equals space 1.85 space straight x space 10 to the power of negative 5 end exponent divided by to the power of straight o straight C


Thus, the coefficient of linear expansion in pendulum clock = 1.85 x 10-5/C

838 Views

4.

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 :


  • 0.2 and 6.5 m

  • 0.2 and 3.5 m

  • 0.29 and 3.5 m

  • 0.29 and 6.5 m


C.

0.29 and 3.5 m

Energy lost over path PQ = μ mg cos θ x 4



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

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5.

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be:

  • left parenthesis 92 space plus-or-minus 2 right parenthesis straight s
  • left parenthesis 92 space plus-or-minus 5 right parenthesis straight s
  • left parenthesis 92 space plus-or-minus space 1.8 right parenthesis straight s
  • left parenthesis 92 space plus-or-minus 3 right parenthesis straight s

A.

left parenthesis 92 space plus-or-minus 2 right parenthesis straight s

Arithmetic mean time of an oscillating simple pendulum

fraction numerator straight capital sigma space straight x subscript straight i over denominator straight N end fraction space equals space fraction numerator 90 space plus 91 space plus 92 space plus 95 over denominator 4 end fraction space equals space 92 space straight s
Mean deviation of a simple pendulum

equals fraction numerator straight capital sigma space vertical line space begin display style straight x with minus on top end style space minus space straight x subscript straight i vertical line over denominator straight N end fraction space equals space fraction numerator 2 space plus space 1 space plus 3 space plus space 3 space plus space 0 over denominator 4 end fraction space equals space 1.5
Given, minimum division in the measuring clock, i.e. simple pendulum = 1s. Thus, the reported mean time of an oscillating simple pendulum = left parenthesis space 92 space plus-or-minus space 2 right parenthesis straight s

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6.

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:

  • turn left

  • turn right

  • go straight

  • turn left and right alternately


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.

763 Views

7.

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:

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

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

equals space straight L space equals space straight r space straight x space straight p space equals space straight r subscript straight p space sin space straight theta space straight n with hat on top space or space straight L space equals space straight r space left parenthesis straight p right parenthesis straight n with hat on top
When space straight a space particle space is space moving space from space straight D space to space straight A
straight L space equals space fraction numerator straight R over denominator square root of 2 end fraction space mv space left parenthesis negative straight k with hat on top right parenthesis
straight A space particle space is space moving space from space straight A space to space straight B
straight L space equals space fraction numerator straight R over denominator square root of 2 end fraction space mv space left parenthesis negative straight k with hat on top right parenthesis
and space it space moves space from space straight C space to space straight D
space straight L space equals space open parentheses fraction numerator straight R over denominator square root of 2 end fraction plus straight a close parentheses mv space left parenthesis straight k with hat on top right parenthesis
For space straight B space to space straight C comma space we space have space
straight L space space equals space open parentheses fraction numerator straight R over denominator square root of 2 end fraction plus straight a close parentheses mv space left parenthesis straight k with hat on top right parenthesis


1517 Views

8.

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn=constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively):

  • straight n space equals straight C subscript straight p over straight C subscript straight v
  • straight n equals space fraction numerator straight C minus space straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction
  • straight n equals fraction numerator straight C subscript straight p minus straight C over denominator straight C minus straight C subscript straight v end fraction
  • straight n space equals space fraction numerator straight C minus straight C subscript straight v over denominator straight C minus straight C subscript straight p end fraction

B.

straight n equals space fraction numerator straight C minus space straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction

For the polytropic process, specific heat for an ideal gas.
straight C space equals space fraction numerator straight R over denominator 1 minus straight n end fraction space plus space straight C subscript straight v
therefore comma space fraction numerator straight R over denominator 1 minus straight n end fraction space plus straight C subscript straight v space equals space straight C
rightwards double arrow space fraction numerator straight R over denominator 1 minus straight n end fraction space equals space straight C minus straight C subscript straight v
rightwards double arrow fraction numerator straight R over denominator straight C minus straight C subscript straight v end fraction equals space 1 minus straight n space left parenthesis where comma space straight R thin space equals space straight C subscript straight p minus straight C subscript straight v right parenthesis
rightwards double arrow space fraction numerator straight C subscript straight p minus straight C subscript straight v over denominator straight C minus straight C subscript straight v end fraction space equals space 1 minus straight n
straight n space equals space 1 minus space fraction numerator straight C subscript straight p minus straight C subscript straight v over denominator straight C minus straight C subscript straight v end fraction space rightwards double arrow space straight n space equals space fraction numerator straight C minus straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction
Thus comma space number space of space moles space straight n space is space given space by space
straight n space equals space fraction numerator straight C minus straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction

700 Views

9.

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


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 

628 Views

10.

A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms−2 )

  • 2 straight pi square root of 2 straight s
  • 2s

  • 2 square root of 2 straight s
  • square root of 2 straight s

C.

2 square root of 2 straight s

A uniform string of length 20 m is suspended from a rigid support. Such that the time taken to reach the support.

straight T equals mgx over straight l
So comma space velocity space at space point space straight P space equals space square root of fraction numerator begin display style mgx over straight l end style over denominator straight m divided by straight l end fraction end root
straight v equals space square root of gx
dx over dt space equals space square root of gx
integral subscript 0 superscript 20 fraction numerator dxt over denominator square root of straight x end fraction space equals space integral subscript 0 superscript straight t square root of straight g space dt
left square bracket 2 square root of straight x right square bracket subscript 0 superscript 20 space equals space square root of 10 straight t
square root of 20 space equals space square root of 10 straight t
straight t space equals space 2 square root of 2 space straight s 

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