Subject

Physics

Class

JEE Class 12

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JEE Physics 2011 Exam Questions

Multiple Choice Questions

11.

A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is

  • straight pi over 4 square root of LC
  • 2 straight pi space square root of LC
  • square root of LC
  • straight pi square root of LC

A.

straight pi over 4 square root of LC

As initially, charge is maximum
∴ q = qocos ωt
Current, 

683 Views

12.

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heat γ.It is moving with speed v and it suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by 

  • fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 γR end fraction Mv squared straight K
  • fraction numerator γMv squared over denominator 2 straight R end fraction minus straight K
  • fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 γR end fraction Mv squared straight K
  • fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 left parenthesis straight gamma plus 1 right parenthesis straight R end fraction Mv squared straight K

C.

fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 γR end fraction Mv squared straight K

As no heat is lost, therefore,
Loss of kinetic energy = Gain of internal energy of gas
i.e., 
1 half mv squared space equals space nC subscript straight v increment straight T
rightwards double arrow space 1 half mv squared space equals space straight m over straight M. fraction numerator straight R over denominator straight gamma minus 1 end fraction increment straight T
rightwards double arrow increment straight T space equals space fraction numerator Mv squared space left parenthesis straight gamma minus 1 right parenthesis over denominator 2 straight R end fraction straight K

776 Views

13.

A boat is moving due east in a region where the earth's magnetic field is 5.0 × 10-5 NA-1m-1 due north and horizontal. The boat carries a vertical aerial 2m long. If the speed of the boat is 1.50 ms-1, the magnitude of the induced emf in the wire of aerial is

  • 0.75 mV

  • 0.50 mV

  • 0.15 mV

  • 1 mV


C.

0.15 mV

Eind = B × v × l
= 5.0 × 10-5 × 1.50 × 2
= 10.0 × 10-5 × 1.5
= 15 × 10-5 vot.
= 0.15 mv

176 Views

14.

Water is flowing continuously from a tap having an internal diameter 8 × 10-3 m. The water velocity as it leaves the tap is 0.4 ms-1. The diameter of the water stream at a distance 2 × 10-1 m below the tap is close
to:

  • 7.5 x 10-3 m

  • 9.6x 10-3 m

  • 3.6x 10-3 m

  • 5.0x 10-3 m


C.

3.6x 10-3 m

From Bernoulli's theorem
ρgh space equals space 1 half straight rho left parenthesis straight v subscript 2 superscript 2 minus straight v subscript 1 superscript 2 right parenthesis
gh space equals space 1 half straight v subscript 1 superscript 2 space open parentheses open parentheses straight v subscript 2 over straight v subscript 1 close parentheses minus 1 close parentheses space left parenthesis therefore straight A subscript 1 straight v subscript 1 space equals straight A subscript 2 straight v subscript 2 right parenthesis
rightwards double arrow space space open parentheses straight A subscript 1 over straight A subscript 2 close parentheses squared space equals space 1 space plus space fraction numerator 2 hg over denominator straight v subscript 1 superscript 2 end fraction
rightwards double arrow open parentheses straight D subscript 1 over straight D subscript 2 close parentheses to the power of 4 space equals space 1 space plus space fraction numerator 2 gh over denominator straight v subscript 1 superscript 2 end fraction
rightwards double arrow space straight D subscript 2 space equals space straight D subscript 1 over open parentheses 1 plus begin display style fraction numerator 2 gh over denominator straight v subscript 1 superscript 2 end fraction end style close parentheses to the power of 1 divided by 4 end exponent space equals space fraction numerator 8 space straight x space 10 to the power of negative 3 end exponent over denominator open parentheses begin display style fraction numerator 1 plus space 2 space straight x space 10 space straight x 0.2 over denominator left parenthesis 0.4 right parenthesis squared end fraction end style close parentheses to the power of 1 divided by 4 end exponent end fraction
space equals space 3.6 space straight x space 10 to the power of negative 3 end exponent space straight m

494 Views

15.

Three perfect gases at absolute temperature T1, T2 and T3 are mixed. The masses of molecules are m1,m2 and m3 and the number of molecules is n1,n2 and n3 respectively.Assuming no loss of energy, the final temperature of the mixture is

  • fraction numerator straight n subscript 1 straight T subscript 1 plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 over denominator straight n subscript 1 plus space straight n subscript 2 plus space straight n subscript 3 end fraction
  • fraction numerator straight n subscript 1 straight T subscript 1 superscript 2 space plus straight n subscript 2 straight T subscript 2 superscript 2 space plus straight n subscript 3 straight T subscript 3 superscript 2 over denominator straight n subscript 1 straight T subscript 1 space plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 end fraction
  • fraction numerator straight n squared subscript 1 straight T subscript 1 superscript 2 space plus straight n squared subscript 2 straight T subscript 2 superscript 2 space plus straight n squared subscript 3 straight T subscript 3 superscript 2 over denominator straight n subscript 1 straight T subscript 1 space plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 end fraction
  • fraction numerator open parentheses straight T subscript 1 plus straight T subscript 2 plus straight T subscript 3 close parentheses over denominator 3 end fraction

A.

fraction numerator straight n subscript 1 straight T subscript 1 plus straight n subscript 2 straight T subscript 2 space plus straight n subscript 3 straight T subscript 3 over denominator straight n subscript 1 plus space straight n subscript 2 plus space straight n subscript 3 end fraction

For adiabatic process i.e., no heat change

straight F over 2 straight n subscript 1 straight k subscript 1 straight T subscript 1 space plus fraction numerator begin display style straight F end style over denominator begin display style 2 end style end fraction straight n subscript 2 kT subscript 2 space plus fraction numerator begin display style straight F end style over denominator begin display style 2 end style end fraction straight n subscript 3 kT subscript 3 space
equals fraction numerator begin display style straight F end style over denominator begin display style 2 end style end fraction left parenthesis straight n subscript 1 plus straight n subscript 2 plus straight n subscript 3 right parenthesis kT
rightwards double arrow space straight T space equals space fraction numerator straight n subscript 1 straight T subscript 1 space plus straight n subscript 2 straight T subscript 2 plus straight n subscript 3 straight T subscript 3 over denominator space straight n subscript 1 plus straight n subscript 2 plus straight n subscript 3 end fraction

911 Views

16.

A current I flows in an infinitely long wire with cross-section in the form of a semicircular ring of radius R. The magnitude of the magnetic induction along its axis is

  • fraction numerator straight mu subscript 0 straight I over denominator 2 πR end fraction

  • fraction numerator straight mu subscript 0 straight I over denominator 2 πR end fraction
  • fraction numerator straight mu subscript 0 straight I over denominator 4 πR end fraction
  • fraction numerator straight mu subscript 0 straight I over denominator straight pi squared straight R end fraction

D.

fraction numerator straight mu subscript 0 straight I over denominator straight pi squared straight R end fraction
straight v equals space straight I over πR
dB space equals space open parentheses fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction close parentheses fraction numerator 2 straight I over denominator straight R end fraction
straight I space equals space λRdθ
therefore space straight B space equals space integral subscript negative straight pi divided by 2 end subscript superscript straight pi divided by 2 end superscript space dB space cos space straight theta
space equals space fraction numerator straight mu subscript 0 straight lambda over denominator 2 straight pi end fraction integral subscript negative straight pi divided by 2 end subscript superscript straight pi divided by 2 end superscript space cos space θdθ
space equals space fraction numerator straight mu subscript 0 straight lambda over denominator straight pi end fraction space equals space fraction numerator straight mu subscript 0 straight I over denominator straight pi squared straight R end fraction
1162 Views

17.

A mass M, attached to a horizontal spring, executes SHM with an amplitude A1. When the mass M passes through its mean position than a smaller mass m is placed over it and both of them move together with amplitude A2. The ratio of (A1/A2) is

  • fraction numerator straight M space plus straight m over denominator straight M end fraction
  • open parentheses fraction numerator straight M over denominator straight M plus straight m end fraction close parentheses to the power of 1 divided by 2 end exponent
  • fraction numerator straight M over denominator straight M plus straight m end fraction

C.

At mean position, Fnet = 0
Therefore, By conservation of linear momentum,
Mv1 = (M+m)v2
1A1 = (M+m)ω2A2
⇒ rightwards double arrow space open parentheses fraction numerator straight M over denominator straight M plus straight m end fraction close parentheses space equals space fraction numerator straight omega subscript 2 straight A subscript 2 over denominator straight omega subscript 1 straight A subscript 1 end fraction
But space straight omega subscript 1 space equals space square root of straight k over straight M end root space and space straight omega subscript 2 space equals space square root of fraction numerator straight k over denominator straight M plus straight m end fraction end root
On space solving comma space we space get space straight A subscript 1 over straight A subscript 2 space equals space square root of fraction numerator straight m plus straight M over denominator straight M end fraction end root

1444 Views

18.

Two particles are executing simple harmonic motion of the same amplitude Aand frequency ω along the x - axis. Their mean position is separated by distance X0 (X0 >A). If the maximum separation between them is (X0 + A), the phase difference between their motion is

  • π/3

  • π/4

  • π/6

  • π/2


A.

π/3

x1 = A sin( ωt + Φ1)
x2 = A sin( ωt + Φ2)
straight x subscript 1 space minus straight x subscript 2 space equals space straight A space open square brackets 2 sin open square brackets ωt space plus space fraction numerator straight ϕ subscript 1 plus straight ϕ subscript 2 over denominator 2 end fraction close square brackets sin open square brackets fraction numerator straight ϕ subscript 1 minus straight ϕ subscript 2 over denominator 2 end fraction close square brackets close square brackets
straight A space equals space 2 straight A space sin space open parentheses fraction numerator straight ϕ subscript 1 minus straight ϕ subscript 2 over denominator 2 end fraction close parentheses
fraction numerator straight ϕ subscript 1 minus straight ϕ subscript 2 over denominator 2 end fraction space equals space straight pi over 6
straight ϕ subscript 1 space equals space straight pi over 3

1715 Views

19.

100g of water is heated from 30°C to 50°C ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/Kg/K)

  • 8.4 kJ

  • 84 kJ

  • 2.1 kJ

  • 4.2 kJ


A.

8.4 kJ

ΔQ = M,S,ΔT
= 100 × 10-3 × 4.184 × 20 = 8.4 × 103
ΔQ = 84 kJ, ΔW = 0
ΔQ = ΔV + ΔW
ΔV = 8.4 kJ

188 Views

20.

A Carnot engine operating between temperatures T1 and T2 has efficiency 1/6. When T2 is lowered by 62 K, its efficiency increases to 1/3. Then T1 and T2 are, respectively

  • 372 K and 330 K

  • 330 K and 268 K

  • 310 K and 248 K

  • 372 K and 310 K


D.

372 K and 310 K

The efficiency  is given by,

straight eta subscript 1 space equals space 1 space minus straight T subscript 2 over straight T subscript 1
rightwards double arrow space 1 over 6 space equals space 1 space minus space straight T subscript 2 over straight T subscript 1
rightwards double arrow space straight T subscript 2 over straight T subscript 1 space equals space 5 over 6
straight eta subscript 2 space equals space 1 minus fraction numerator straight T subscript 2 minus 62 over denominator straight T subscript 1 end fraction space space.... space left parenthesis straight i right parenthesis
1 third space equals space 1 minus fraction numerator straight T subscript 2 minus 62 over denominator straight T subscript 1 end fraction space... space left parenthesis ii right parenthesis
On space solving space Eqs space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
straight T subscript 1 space equals space 372 space straight K space and space straight T subscript 2 space equals space 310 space straight K

763 Views

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