Subject

Physics

Class

JEE Class 12

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JEE Physics 2010 Exam Questions

Multiple Choice Questions

11.

In a series LCR circuit, R = 200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is

  • 305 W

  • 210 W

  • zero

  • 242 W


D.

242 W

The given circuit is under resonance as XL = XC. Hence, power dissipated in the circuit is 

P = V2/R = 242 W

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12.

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for the charge to reduce to one–fourth its initial value. Then the ratio t1/t2 will be

  • 1

  • 1/2

  • 1/4

  • 2


C.

1/4

straight U space equals space fraction numerator straight q squared over denominator 2 straight c end fraction
straight U space equals space straight U subscript max over 2
rightwards double arrow space straight q space equals space straight Q subscript straight o over straight v subscript 2
straight q space equals space straight Q subscript 0 straight e to the power of negative straight t divided by RC end exponent
In space straight q over straight Q subscript 0 space equals space minus straight t over RC semicolon
straight t space equals space RC space In straight Q subscript 0 over straight q
at space straight t subscript 1 space rightwards double arrow space straight q space equals space fraction numerator straight Q subscript 0 over denominator square root of 2 end fraction space straight t subscript 1 space equals space RC over 2 In space 2
at space straight t subscript 2 space rightwards double arrow space straight q space equals space straight Q subscript 0 over 4 space straight t subscript 2 space equals space 2 RC space In space 2
straight t subscript 1 over straight t subscript 2 space equals space 1 fourth
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13.

Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX' is given by


B.

The magnetic field in between because of each of the conductors will be in naturally opposite directions.
∴ Net magnetic field

straight B subscript in space between space space end subscript space equals space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space straight x end fraction space bold j with bold hat on top space plus space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space left parenthesis 2 straight d minus straight x right parenthesis end fraction left parenthesis negative straight j right parenthesis
space equals space fraction numerator straight mu subscript straight o straight i over denominator 2 space straight pi space straight x end fraction space open square brackets 1 over straight x minus fraction numerator 1 over denominator 2 straight d minus straight x end fraction close square brackets space left parenthesis space bold j with bold hat on top space right parenthesis
At x = d, B in between = 0
For x < d, B in between =bold j with bold hat on top
For x> d, B in between = left parenthesis bold minus bold j with bold hat on top right parenthesis
Towards x, net magnetic field will add up and direction will be  left parenthesis bold minus bold j with bold hat on top right parenthesis. Towards' x' b net magnetic field will add up and direction will be (bold j with bold hat on top)

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14.

Let there be a spherically symmetric charge distribution with charge density varying as  space straight rho left parenthesis straight r right parenthesis space equals space space straight rho subscript 0 space open parentheses 5 over 4 minus straight r over straight R close parentheses upto r = R, and ρ(r) = 0 for r > R, where r is the distance from the origin. The electric field at a distance r ( r < R) from the origin is given by

  • fraction numerator 4 space straight pi space straight rho subscript 0 straight r over denominator 3 straight epsilon subscript 0 end fraction space open parentheses 5 over 3 minus straight r over straight R close parentheses
  • fraction numerator begin display style space straight rho subscript 0 straight r end style over denominator begin display style 4 straight epsilon subscript 0 end style end fraction space open parentheses fraction numerator begin display style 5 end style over denominator begin display style 3 end style end fraction minus fraction numerator begin display style straight r end style over denominator begin display style straight R end style end fraction close parentheses
  • fraction numerator 4 space straight rho subscript 0 straight r over denominator 3 straight epsilon subscript 0 end fraction space open parentheses 5 over 3 minus straight r over straight R close parentheses
  • fraction numerator space straight rho subscript 0 straight r over denominator 3 straight epsilon subscript 0 end fraction space open parentheses 5 over 4 minus straight r over straight R close parentheses

B.

fraction numerator begin display style space straight rho subscript 0 straight r end style over denominator begin display style 4 straight epsilon subscript 0 end style end fraction space open parentheses fraction numerator begin display style 5 end style over denominator begin display style 3 end style end fraction minus fraction numerator begin display style straight r end style over denominator begin display style straight R end style end fraction close parentheses

Apply shell theorem, the total charge upto distance r can be calculated as follows
dq space equals space 4 πr squared dr. straight rho
space equals space 4 πr squared. dr. straight rho subscript 0 space open square brackets 5 over 4 minus straight r over straight R close square brackets
left parenthesis because space dq space equals space ρdv right parenthesis
space equals space 4 πρ subscript 0 space open square brackets 5 over 4 straight r squared dr space minus straight r cubed over straight R dr close square brackets
space equals space integral space space dq space equals space straight q space equals space 4 πρ subscript 0 integral subscript 0 superscript straight r open parentheses 5 over 4 straight r squared dr minus straight r cubed over straight R dr close parentheses
space equals space 4 πρ subscript 0 space open square brackets 5 over 4 straight r cubed over 3 minus 1 over straight R straight r to the power of 4 over 4 close square brackets
As space the space electric space field comma space straight E space space equals kq over straight r squared
equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction 1 over straight r squared.4 πρ subscript 0 space open square brackets 5 over 4 open parentheses straight r cubed over 3 close parentheses minus fraction numerator straight r to the power of 4 over denominator 4 straight R end fraction close square brackets
straight E space equals space fraction numerator straight rho subscript 0 straight r over denominator 4 straight epsilon subscript 0 end fraction open square brackets 5 over 3 minus straight r over straight R close square brackets

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15.

In the circuit shown below, the key K is closed at t = 0. The current through the battery is

  • fraction numerator VR subscript 1 straight R subscript 2 over denominator square root of straight R subscript 1 superscript 2 plus straight R subscript 2 superscript 2 end root end fraction space at space straight t space equals space 0 space and space straight V over straight R subscript 2 space at space straight t space equals space infinity
  • straight V over straight R subscript 2 space at space straight t space equals space 0 space and space fraction numerator straight V left parenthesis straight R subscript 1 plus straight R subscript 2 right parenthesis over denominator straight R subscript 1 straight R subscript 2 end fraction space at space equals space infinity
  • straight V over straight R subscript 2 space at space straight t space equals space 0 space and space fraction numerator VR subscript 1 straight R subscript 2 over denominator square root of straight R subscript 1 superscript 2 space plus space straight R subscript 2 superscript 2 end root end fraction space at space straight t space equals space infinity
  • fraction numerator straight V space left parenthesis straight R subscript 1 space plus space straight R subscript 2 right parenthesis over denominator straight R subscript 1 straight R subscript 2 end fraction space at space straight t space equals space 0 space and space straight V over straight R subscript 2 space at space straight t space equals space infinity

B.

straight V over straight R subscript 2 space at space straight t space equals space 0 space and space fraction numerator straight V left parenthesis straight R subscript 1 plus straight R subscript 2 right parenthesis over denominator straight R subscript 1 straight R subscript 2 end fraction space at space equals space infinity

At t = 0, inductor behaves like an infinite resistance so , at t = 0,
i = V/R2
and at t = ∞, inductor behaves like a conducting wire i.e., resistance less wire
then, 
straight i space equals space straight V over straight R subscript eq space equals space fraction numerator straight V space left parenthesis straight R subscript 1 space plus space straight R subscript 2 right parenthesis over denominator straight R subscript 1 straight R subscript 2 end fraction

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16.

Two conductors have the same resistance a 0°C but their temperature coefficients o resistance are α1 and α2. The respective temperature coefficients of their series parallel combinations are nearly

  • fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space fraction numerator straight alpha subscript 1 space plus space straight alpha subscript 2 over denominator 2 end fraction
  • fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space straight alpha subscript 1 space plus straight alpha subscript 2
  • space straight alpha subscript 1 space plus straight alpha subscript 2 comma space fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction
  • straight alpha subscript 1 space plus straight alpha subscript 2 space comma space fraction numerator straight alpha subscript 1 straight alpha subscript 2 over denominator straight alpha subscript 1 space plus straight alpha subscript 2 end fraction

A.

fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space fraction numerator straight alpha subscript 1 space plus space straight alpha subscript 2 over denominator 2 end fraction

Re = Ro +Ro
2R(1 + αsΔT)
= R( 1 +α1ΔT) + R(1+α2ΔT)
= αs = α12 / 2
Rp = R x R / R+R
straight R over 2 left parenthesis space 1 space plus space straight alpha subscript straight p increment straight T right parenthesis space equals space fraction numerator straight R space left parenthesis 1 space plus straight alpha subscript 1 increment straight T right parenthesis space straight x space straight R left parenthesis space 1 space plus space straight alpha subscript 2 increment straight T right parenthesis over denominator straight R space left parenthesis 1 space plus straight alpha subscript 1 increment straight T right parenthesis space plus space straight R left parenthesis space 1 space plus space straight alpha subscript 2 increment straight T right parenthesis end fraction
fraction numerator begin display style 1 space plus space straight alpha subscript straight p increment straight T end style over denominator 2 end fraction space equals space open parentheses 1 space plus space straight alpha subscript 1 increment straight T close parentheses left parenthesis 1 space plus space straight alpha subscript 2 increment straight T right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket space 1 space plus space left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis increment straight T right square bracket to the power of negative 1 end exponent
space equals space left square bracket 1 space plus space left parenthesis straight alpha subscript 1 space plus straight alpha subscript 2 right parenthesis increment straight T right square bracket space left square bracket 2 space plus space open parentheses straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis increment straight T close parentheses to the power of negative 1 end exponent
space equals space 1 half space left square bracket space 1 space plus space left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis space increment straight T right square bracket space open square brackets 1 minus fraction numerator left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis over denominator 2 end fraction increment straight T close square brackets
fraction numerator 1 space plus space straight alpha subscript straight p increment straight T over denominator 2 end fraction space equals space open square brackets 1 plus fraction numerator begin display style left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis end style over denominator begin display style 2 end style end fraction increment straight T close square brackets
straight alpha subscript straight p space equals space fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction

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17.

A thin semi–circular ring of radius r has a positive charge q distributed uniformly over it. The net field straight E with rightwards arrow on top at the center O is

  • fraction numerator straight q over denominator 4 straight pi squared straight epsilon subscript 0 straight r squared end fraction space bold j with hat on top
  • negative fraction numerator straight q over denominator 4 straight pi squared straight epsilon subscript 0 straight r squared end fraction
  • negative fraction numerator straight q over denominator 2 straight pi squared straight epsilon subscript 0 straight r squared end fraction straight j with hat on top
  • fraction numerator straight q over denominator 2 straight pi squared straight epsilon subscript 0 straight r squared end fraction bold space bold j with bold hat on top

C.

negative fraction numerator straight q over denominator 2 straight pi squared straight epsilon subscript 0 straight r squared end fraction straight j with hat on top
straight E space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q over straight r squared fraction numerator sin space straight pi divided by 2 over denominator straight pi divided by 2 end fraction

space equals space fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight q over straight r squared 2 over straight pi
straight E with rightwards arrow on top space equals space minus space fraction numerator straight q over denominator 2 straight pi squared straight epsilon subscript 0 straight r squared end fraction straight j with hat on top
797 Views

18.

A rectangular loop has a sliding connector PQ of length

  • straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 6 straight R end fraction comma space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals space minus straight I subscript 2 space equals space fraction numerator straight B space calligraphic l straight v over denominator straight R end fraction comma space straight I space equals fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction comma space straight I space equals space fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction
  • straight I subscript 1 space equals straight I subscript 2 space equals space straight I space equals space fraction numerator straight B space calligraphic l straight v over denominator straight R end fraction

C.

straight I subscript 1 space equals space straight I subscript 2 space space equals space fraction numerator straight B space calligraphic l straight v over denominator 3 straight R end fraction comma space straight I space equals space fraction numerator 2 straight B space calligraphic l straight v over denominator 3 straight R end fraction

A moving conductor is equivalent to a battery of emf
= vBI
Equivalent circuit I = I1 + I2


Applying Kirchhoff's law
I1R + IR -vBl = 0 .... (i)
I2R + IR -vBl = 0 .... (ii)
Adding Eqs. (i) and (ii), we get
2IR + IR = 2vBI
I = 2VBI/3R
I1 = I2 = VBI/3R

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19.

The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by y= 0.02(m) sin open square brackets 2 straight pi open parentheses fraction numerator straight t over denominator 0.04 space left parenthesis straight s right parenthesis end fraction minus fraction numerator straight x over denominator 0.50 space left parenthesis straight m right parenthesis end fraction close parentheses close square brackets The tension in the string is 

  • 4.0 N

  • 12.5

  • 0.5 N

  • 6.25 N


D.

6.25 N

The tension in the string
on comparing with the standard equation,
y = a sin (ω t = kx)

T = μv2 = μω2/k2 = 0.04

 = 0.04 (2π/0.004)2/(2π/0.50)2
 = 6.25

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20.

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30º with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. If density of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is

  • 4

  • 3

  • 2

  • 1


C.

2

From FBD of sphere, using Lami theorem


F/Mg = Tan θ .... (i)
when suspended in liquid as θ remains unchanged,

therefore space fraction numerator straight F apostrophe over denominator mg space open parentheses 1 minus begin display style straight rho over straight d end style close parentheses end fraction space equals space tan space straight theta space... left parenthesis ii right parenthesis
Using space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
straight F over mg space equals space fraction numerator straight F apostrophe over denominator mg space open parentheses 1 minus begin display style straight rho over straight d end style close parentheses end fraction comma space where space straight F apostrophe space equals space straight F over straight K
therefore space space straight F over mg space equals space fraction numerator straight F over denominator mgK space open parentheses 1 minus begin display style straight rho over straight d end style close parentheses end fraction
space or
straight K space equals space fraction numerator 1 over denominator 1 minus begin display style straight rho over straight d end style end fraction space equals space fraction numerator 1 over denominator open parentheses 1 minus begin display style fraction numerator 0.8 over denominator 1.6 end fraction end style close parentheses end fraction space equals space 2

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