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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2014 Exam Questions

Multiple Choice Questions

1.

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is (For steel Young’s modulus is 2×1011 N m–2 and coefficient of thermal expansion is 1.1×10–5 K–1 )

  • 2.2 x 108 Pa

  • 2.2 x 109 Pa

  • 2.2 x 107 Pa

  • 2.2 x 106 Pa


A.

2.2 x 108 Pa

According to Hooke's law, i.e,

Young apostrophe straight s space modulus space left parenthesis straight Y right parenthesis space equals space fraction numerator Tensile space stress over denominator Tensile space strain end fraction
So comma space straight Y space equals fraction numerator straight F divided by straight A over denominator increment straight L divided by straight L end fraction space equals space fraction numerator FL over denominator straight A increment straight L end fraction
If the rod is compressed, then compressive stress and strain appear. Their ratio Y is same as that for the tensile case. 
Give, length of a steel wire (L) = 10 cm,
Temperature (θ) = 100oC
As length is constant,
∴ Strain space equals space fraction numerator increment straight L over denominator straight L end fraction space equals space straight alpha increment straight theta
Now, pressure = stress =y x strain
[Given, Y = 2 x 1011N/M2 and α = 1.1 x 10-5 K-1]
= 2 x 1011 x1.1 x 10-5 x 100
=2.2 x 108

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2.

On the heating water, bubbles being formed at the bottom of the vessel detach and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r < < R and the surface tension of water is T, value of r just before bubbles detach is (density of water is ρw)

  • straight R squared square root of fraction numerator 2 straight rho subscript straight w straight g over denominator 3 straight T end fraction end root
  • straight R squared square root of fraction numerator straight rho subscript straight w straight g over denominator 6 straight T end fraction end root
  • straight R squared square root of fraction numerator straight rho subscript straight w straight g over denominator straight T end fraction end root
  • straight R squared space square root of fraction numerator 3 straight rho subscript straight w straight g over denominator straight T end fraction end root

A.

straight R squared square root of fraction numerator 2 straight rho subscript straight w straight g over denominator 3 straight T end fraction end root

The bubble will detach if, 
Buoyant force ≥ Surface tension force

4 over 3 πR cubed straight rho subscript straight w straight g space greater or equal than integral straight T space straight x space dl space sin space straight theta space


left parenthesis straight rho subscript straight w right parenthesis open parentheses 4 over 3 πR cubed close parentheses straight g greater or equal than left parenthesis straight T right parenthesis left parenthesis 2 πpr right parenthesis space sin space straight theta
sin space straight theta space equals space straight r over straight R
solving space straight r space equals space square root of fraction numerator 2 straight p subscript straight w straight R to the power of 4 straight g over denominator 3 straight T end fraction end root space equals space straight R squared square root of fraction numerator 2 straight rho subscript straight w straight g over denominator 3 straight T end fraction end root

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3.

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

  • square root of GM over straight R end root
  • square root of 2 square root of 2 GM over straight R end root
  • square root of GM over straight R left parenthesis 1 plus 2 square root of 2 right parenthesis end root
  • 1 half square root of GM over straight R left parenthesis 1 plus 2 square root of 2 right parenthesis end root

D.

1 half square root of GM over straight R left parenthesis 1 plus 2 square root of 2 right parenthesis end root

Net force acting on any one particle M,
 =fraction numerator GM squared over denominator left parenthesis 2 straight R right parenthesis squared end fraction space plus space fraction numerator GM squared over denominator left parenthesis straight R square root of 2 right parenthesis squared end fraction cos space 45 to the power of straight o space plus space fraction numerator GM squared over denominator left parenthesis straight R square root of 2 right parenthesis squared end fraction Cos space 45 to the power of straight o
space equals space GM squared over straight R squared open parentheses 1 fourth plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses


This force will equal to centripetal force
So, Mv squared over straight R space equals space GM squared over straight R squared open parentheses 1 fourth plus fraction numerator 1 over denominator square root of 2 end fraction close parentheses
straight v space equals space square root of fraction numerator GM over denominator 4 straight R end fraction left parenthesis 1 plus 2 square root of 2 right parenthesis end root space equals space 1 half square root of GM over straight R left parenthesis 2 square root of 2 plus 1 right parenthesis end root

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4.

When a rubber band is strecthed by a distance x, it exerts a restoring force of magnitude F = ax +bx2, where a and b are constants. The work done in stretching are unstretched rubber-band by L is

  • aL2 +bL2

  • 1 half left parenthesis aL squared plus bL cubed right parenthesis
  • aL squared over 2 space plus bL cubed over 3
  • 1 half space open parentheses aL squared over 2 plus bL cubed over 3 close parentheses

C.

aL squared over 2 space plus bL cubed over 3
straight U subscript straight f minus straight U subscript straight i space equals space minus straight W space equals space minus space integral subscript straight i superscript straight f straight F. dr
Given, F = ax +bx2
We know that work done in stretching the rubber band by L is 
|dW|= |Fdx|
vertical line straight W vertical line space equals space integral subscript 0 superscript straight L left parenthesis ax space plus bx squared right parenthesis dx
space equals space open square brackets ax squared over 2 close square brackets subscript straight O superscript straight L space plus space open square brackets bx cubed over 3 close square brackets subscript straight O superscript straight L
equals open square brackets aL squared over 2 minus fraction numerator ax space left parenthesis 0 right parenthesis squared over denominator 2 end fraction close square brackets space plus space open square brackets fraction numerator straight b space straight x space straight L cubed over denominator 3 end fraction minus fraction numerator straight b space straight x space left parenthesis 0 right parenthesis cubed over denominator 3 end fraction close square brackets
vertical line straight W vertical line space equals space aL squared over 2 plus bL cubed over 3
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5.

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90o angle at centre. Radius joining their interface makes an angle α with vertical. ratio d1/d2 is

  • fraction numerator 1 plus sin space straight alpha over denominator 1 minus sin space straight alpha end fraction
  • fraction numerator 1 plus space cosα over denominator 1 minus cosα end fraction
  • fraction numerator 1 plus space tan space straight alpha over denominator 1 minus tan space straight alpha end fraction
  • fraction numerator 1 plus cos space straight alpha over denominator 1 minus cos space straight alpha end fraction

C.

fraction numerator 1 plus space tan space straight alpha over denominator 1 minus tan space straight alpha end fraction
straight P subscript straight A space equals space straight P subscript straight B
straight P subscript 0 space plus space straight d subscript 1 straight g end subscript straight R space left parenthesis Cos space straight alpha space minus sin space straight alpha right parenthesis space equals straight P subscript 0 space plus straight d subscript 2 straight R space left parenthesis Cos space straight alpha plus sin space straight alpha right parenthesis
rightwards double arrow space straight d subscript 1 over straight d subscript 2 space equals space fraction numerator cos space straight alpha space plus sin space straight alpha over denominator cos space straight alpha minus sin space straight alpha end fraction space equals space fraction numerator 1 plus space tan space straight alpha over denominator 1 minus space tanα end fraction
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6.

From the tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H,u and n is

  • 2gH = n2u2

  • gH = (n-2)2u2

  • 2gH = nu2(n-2)2u2

  • gH = (n-2)2u2


C.

2gH = nu2(n-2)2u2

Time is taken to reach the maximum height t1 = u/g

If t2 is the time taken to hit the ground,
i.e, negative straight H space equals space ut subscript 2 minus 1 half gt subscript 2 superscript 2
straight t subscript 2 space equals space nt subscript 1
minus straight H space equals space straight u nu over straight g minus 1 half straight g fraction numerator straight n squared straight u squared over denominator straight g squared end fraction
minus straight H space equals space nu squared over straight g minus 1 half fraction numerator straight n squared straight u squared over denominator straight g end fraction
2 gH space equals nu squared left parenthesis straight n minus 2 right parenthesis

740 Views

7.

A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?

  • A meter scale.

  • A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

  • A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.

  • A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.


B.

A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.

If student measures 3.50 cm, it means that there is an uncertainty of order 0.01 cm
For vernier scale with 1 MSD = 1mm
and 9 MSD = 10 VSD
LC of vernier scale with 1 MSD - 1 VSD
1 over 10 open parentheses 1 minus 9 over 10 close parentheses
space equals space 1 over 100 cm

450 Views

8.

A bob of mass m attached to an inextensible string of length l is suspended from a vertical support. The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension

  • angular momentum is conserved

  • angular momentum changes in magnitude but not in the direction

  • angular momentum changes in direction but not in magnitude

  • angular momentum changes both in direction and magnitude


C.

angular momentum changes in direction but not in magnitude

542 Views

9.

A block of mass m is placed on a surface with a vertical cross-section given by y = x3/6. If the coefficient of friction is 0.5, the maximum height above ground at which the block can be placed without slipping is 

  • 1 over 6 straight m
  • 2 over 3 straight m
  • 1 third straight m
  • 1 half straight m

A.

1 over 6 straight m

A block of mass m is placed on a surface with vertical cross section, then


tan space straight theta space equals space dy over dx space fraction numerator straight d open parentheses begin display style straight x cubed over 6 end style close parentheses over denominator dx end fraction space equals space straight x squared over 2
At limiting equilibrium, we get
μ = tan θ
0.5 = x2/2
⇒ x2 =1
⇒ x = ±1
Now, putting the value of x in y = x3/6, we get
When x =1

straight y space equals space fraction numerator left parenthesis 1 right parenthesis cubed over denominator 6 end fraction space equals 1 over 6
When x =-1

straight y space equals space fraction numerator left parenthesis negative 1 right parenthesis cubed over denominator 6 end fraction space equals fraction numerator negative 1 over denominator 6 end fraction
So, the maximum height above the ground at which the block can be placed without slipping is 1/6m.

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10.

A mass ‘m’ is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?

  • 2g/3

  • g/2

  • 5g/6

  • g


B.

g/2

For the mass m,
mg-T = ma


As we know, a = Rα    ... (i)
So, mg-T = mRα
Torque about centre of pully
T x R = mR2α ...... (ii)
From Eqs. (i) and (ii), we get 
a = g/2
Hence, the acceleration of the mass of a body fall is g/2.

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