JEE physics

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Question
CBSEENPH11020285

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

Solution

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

Sponsor Area

Question
CBSEENPH11020286

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 

Solution

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

Question
CBSEENPH11020287

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]

Solution

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

Question
CBSEENPH12039450

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

Solution

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

Question
CBSEENPH11020290

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

Solution

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