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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2013 Exam Questions

Multiple Choice Questions

1.

A hoop of radius r and mass m rotating with an angular velocity ω0
is placed on a rough horizontal surface.The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?

  • 0/4

  • 0/3

  • 0/2

  • 0


C.

0/2

From conservation of angular momentum about any fixed point on the surfacemr squared straight omega subscript 0 space equals space 2 mr squared straight omega
therefore space straight omega space equals space straight omega subscript 0 over 2
therefore space straight V subscript CM space equals fraction numerator straight omega subscript 0 straight r over denominator 2 end fraction space

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2.

A projectile is given an initial velocity ofopen parentheses straight i with hat on top space plus 2 straight j with hat on top close parentheses straight m divided by straight s  where is along the ground and bold j with bold hat on top is along the vertical. If g = 10 m/s2, the equation of its trajectory is: 

  • y = x-5x2

  • y = 2x-5x2

  • 4y = 2x- 5x2

  • 4y = 2x-25x2


B.

y = 2x-5x2

Initial velocity, space straight V space equals space left parenthesis straight i with hat on top plus 2 straight j with hat on top right parenthesis space straight m divided by straight s
Magnitude of velocity,
straight v equals square root of left parenthesis 1 right parenthesis squared space plus left parenthesis 2 right parenthesis squared end root space equals space square root of 5 space straight m divided by straight s end root
the equation of trajectory of the projectile
straight y space equals space straight x space tan space straight theta space minus space fraction numerator gx squared over denominator 2 straight u squared end fraction left parenthesis 1 plus space tan squared straight theta right parenthesis
left square bracket because space tan space straight theta space equals space straight y over straight x space equals 2 over 1 space equals 2 right square bracket
straight y space equals space straight x.2 space minus space fraction numerator 10 left parenthesis straight x right parenthesis squared over denominator 2 left parenthesis square root of 5 right parenthesis squared end fraction left square bracket 1 plus left parenthesis 2 right parenthesis squared right square bracket
equals space 2 straight x minus fraction numerator 10 left parenthesis straight x squared right parenthesis over denominator 2 space straight x 5 end fraction left parenthesis 1 plus 4 right parenthesis space equals space 2 straight x minus 5 straight x squared

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3.

Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then

  • 0] = [M-1L-3T2A]

  • 0] = [M-1L-3T4A2]

  • 0] =[M-2L2T-1A-2]

  • 0] = [M-1L2T-1A2]


B.

0] = [M-1L-3T4A2]

From Coulomb's Law, F

straight F space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight R squared end fraction
straight epsilon subscript straight o space equals space fraction numerator straight q subscript 1 straight q subscript 2 over denominator 4 πFR squared end fraction

On Substituting the units, we get


straight epsilon subscript straight o space equals space fraction numerator straight C squared over denominator straight N minus straight m end fraction space equals space fraction numerator left square bracket AT right square bracket squared over denominator left square bracket MLT to the power of negative 2 end exponent right square bracket left square bracket straight L squared right square bracket end fraction space left parenthesis 4 straight pi space is space dimensionless right parenthesis
space equals space left square bracket straight M to the power of negative 1 end exponent straight L to the power of negative 3 end exponent straight T to the power of 4 straight A squared right square bracket

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4.

A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at the equilibrium position. The extension x0 of the spring when it is in equilibrium is

  • Mg over straight k
  • Mg over straight k open parentheses 1 minus LAσ over straight M close parentheses
  • Mg over straight k open parentheses 1 minus fraction numerator LAσ over denominator 2 straight M end fraction close parentheses
  • Mg over straight k open parentheses 1 plus LAσ over straight M close parentheses

C.

Mg over straight k open parentheses 1 minus fraction numerator LAσ over denominator 2 straight M end fraction close parentheses

At equilibrium ΣF = 0


kx subscript straight o space plus space open parentheses AL over 2 σg close parentheses minus Mg space equals 0
straight x subscript straight o space equals space Mg open square brackets 1 minus fraction numerator LAσ over denominator 2 straight M end fraction close square brackets


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5.

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and 2.2 × 1011N/m2 respectively?

  • 188.5 Hz

  • 178.2 Hz

  • 200.5 Hz

  • 770 Hz


B.

178.2 Hz

Fundamental space frequency space straight f space equals space fraction numerator 1 over denominator 2 l end fraction square root of straight T over straight mu end root
equals space fraction numerator 1 over denominator 2 l end fraction square root of straight T over Aρ end root
space equals equals space fraction numerator 1 over denominator 2 l end fraction square root of Stress over straight rho end root space equals fraction numerator 1 over denominator 2 space straight x 1.5 end fraction square root of fraction numerator 2.2 space straight x space 10 to the power of 11 space straight x space 10 to the power of negative 2 end exponent over denominator 7.7 space straight x space 10 cubed end fraction end root
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6.

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5s. In another 10s it will decrease to α times its original magnitude, where α equals

  • 0.7

  • 0.81

  • 0.729

  • 0.6


C.

0.729

Amplitude of damped oscillator
straight A space equals space straight A subscript 0 straight e to the power of negative fraction numerator bt over denominator 2 straight m end fraction end exponent
After space 5 straight s comma space space space 0.9 space straight A subscript straight o space equals space straight A subscript straight o straight e to the power of negative fraction numerator straight b left parenthesis 5 right parenthesis over denominator 2 straight m end fraction end exponent space space.. space left parenthesis straight i right parenthesis space
After space 10 space more space second comma
straight A space equals space straight A subscript straight o straight e to the power of negative straight b fraction numerator 15 over denominator 2 straight m end fraction end exponent
straight A space equals space straight A subscript straight o open parentheses straight e to the power of negative straight b fraction numerator 15 over denominator 2 straight m end fraction end exponent close parentheses cubed space space... space left parenthesis ii right parenthesis
From space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma
straight A space equals space 0.729 space straight A subscript straight o space equals space αA subscript 0
Hence comma space straight alpha space equals space 0.729

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7.

What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

  • 5GmM/6R

  • 2GmM/3R

  • GmM/2R

  • GmM/3R


A.

5GmM/6R

From conservation of energy,
Total energy at the planet = Total energy at the altitude
negative GMm over straight R space plus space left parenthesis KE right parenthesis subscript surface space equals space minus fraction numerator GMm over denominator 3 straight R end fraction space plus space 1 half mv subscript straight A superscript 2 space... space left parenthesis straight i right parenthesis
In its orbit the necessary centripetal force provided by gravitational force.
∴ therefore space fraction numerator mv subscript straight A superscript 2 over denominator left parenthesis straight R space plus 2 straight R right parenthesis end fraction space equals space fraction numerator GMm over denominator left parenthesis straight R space plus 2 straight R right parenthesis squared end fraction
straight v subscript straight A superscript 2 space equals space fraction numerator GM over denominator 3 straight R end fraction space... space left parenthesis ii right parenthesis
From space eq space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get
left parenthesis KE right parenthesis subscript surface space equals space 5 over 6 GMm over straight R

514 Views

8.

A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is

  • fraction numerator 2 Bωl cubed over denominator 2 end fraction
  • fraction numerator 3 Bωl cubed over denominator 2 end fraction
  • fraction numerator 4 Bωl squared over denominator 2 end fraction
  • fraction numerator 5 Bωl squared over denominator 2 end fraction

D.

fraction numerator 5 Bωl squared over denominator 2 end fraction


de space equals space straight B space left parenthesis ωx right parenthesis. dx
straight e space equals space Bω integral subscript 2 straight L end subscript superscript 3 straight L end superscript xdx
space equals fraction numerator 5 BωL squared over denominator 2 end fraction
1106 Views

9.

This question has Statement I and Statement II. Of the four choices given after the Statements, choose the
one that best describes the two Statements.
Statement – I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as f open parentheses 1 half mv squared close parentheses comma space then space straight f space equals space open parentheses fraction numerator straight m over denominator straight M plus straight m end fraction close parentheses
Statement – II : Maximum energy loss occurs when the particles get stuck together as a result of the collision.

  • Statement – I is true, Statement – II is true, Statement – II is a correct explanation of Statement – I.

  • Statement – I is true, Statement – II is true, Statement – II is not a correct explanation of Statement – I.

  • Statement – I is true, Statement – II is false.

  • Statement – I is false, Statement – II is true 


D.

Statement – I is false, Statement – II is true 

Before collision, the mass is m and after collision, the mass is m+M
therefore, Maximum energy loss
fraction numerator straight p squared over denominator 2 straight m end fraction minus fraction numerator straight p squared over denominator 2 left parenthesis straight m plus straight M right parenthesis end fraction
space equals space fraction numerator straight p squared over denominator 2 straight m end fraction open square brackets fraction numerator begin display style straight m end style over denominator straight m plus straight M end fraction close square brackets space space space space
space space space open square brackets because KE space equals space fraction numerator straight p squared over denominator 2 straight m end fraction close square brackets
equals space 1 half mv squared open curly brackets fraction numerator straight m over denominator straight m plus straight M end fraction close curly brackets
open square brackets straight f space equals space fraction numerator straight m over denominator straight m plus straight M end fraction close square brackets

398 Views

10.

The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is

  • povo

  • open parentheses 13 over 2 close parentheses straight p subscript 0 straight v subscript 0
  • open parentheses 11 over 2 close parentheses straight p subscript straight o straight v subscript 0
  • 4povo


B.

open parentheses 13 over 2 close parentheses straight p subscript 0 straight v subscript 0

Heat is extracted from the source in path DA and AB is

space straight Q space equals space 3 over 2 straight R open parentheses fraction numerator straight P subscript 0 straight V subscript 0 over denominator straight R end fraction close parentheses space plus 5 over 2 straight R open parentheses fraction numerator begin display style 2 straight P subscript 0 straight V subscript 0 end style over denominator straight R end fraction close parentheses space equals space 13 over 2 straight P subscript 0 straight V subscript 0

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