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
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  • 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 WiredFaculty
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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
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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
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On Substituting the units, we get

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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
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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.
∴ WiredFaculty