JEE Physics

Question 1

Which one of the following represents the correct dimensions of the coefficient of viscosity?

  • ML−1 T−2

  • MLT−1

  • ML−1 T−1

  • ML−2 T−2

Solution

C.

ML−1 T−1

straight eta space equals space fraction numerator straight F over denominator straight A left parenthesis increment straight V subscript straight x divided by increment straight z right parenthesis end fraction
therefore space Dimensions space of space straight eta
space equals space fraction numerator Dimensions space of space force over denominator Dimensions space of space area space straight x space Dimensions space of space velocity space gradient end fraction
space equals space fraction numerator left square bracket MLT to the power of negative 2 end exponent right square bracket over denominator left square bracket straight L squared right square bracket left square bracket straight T to the power of negative 1 end exponent right square bracket end fraction space equals space left square bracket ML to the power of negative 1 end exponent straight T to the power of negative 1 end exponent right square bracket
Question 2

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

  • x2

  • ex

  • x

  • logex

Solution

A.

x2

In this problem acceleration (a) is given in terms of displacement (x)  to determine the velocity with respect to position or displacement we have to apply integration method.
From given information a =-kx, where a is acceleration, x is displacement and k is proportionality constant. 
vdx over dx space equals space minus space kx space open square brackets therefore space dv over dt space equals space fraction numerator begin display style dv end style over denominator dx end fraction open parentheses dx over dt close parentheses space equals space straight v dv over dx close square brackets
straight v space dv space equals space minus kx space dx
Let for any displacement from 0 to x , the velocity changes from vo to v
integral subscript straight v subscript straight o end subscript superscript straight v space vdv space equals space minus integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space fraction numerator straight v squared minus straight v subscript 0 superscript 2 over denominator 2 end fraction space equals space minus space kx squared over 2
rightwards double arrow space straight m space open parentheses fraction numerator straight v squared minus space straight v subscript 0 superscript 2 over denominator 2 end fraction close parentheses space equals space fraction numerator negative mkx squared over denominator 2 end fraction
increment straight K space proportional to space straight x squared

Question 3

A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?

  • h/9 metres from the ground

  • 7h/9 metres from the ground

  • 8h/9 metres from the ground

  • 17h/18 metres from the ground.

Solution

C.

8h/9 metres from the ground


second law of motion
straight s space equals ut space plus space 1 half space plus gT squared
or space straight h equals 0 plus 1 half gT squared space left parenthesis because space straight u equals 0 right parenthesis
therefore space straight T equals square root of open parentheses fraction numerator 2 straight h over denominator straight g end fraction close parentheses end root
At space straight t space equals space straight T over 3 straight s comma
straight s space equals space 0 plus space 1 half straight g space open parentheses straight T over 3 close parentheses squared
rightwards double arrow space straight s space equals space 1 half space straight g. straight T squared over 9
rightwards double arrow space straight s space equals space straight g over 18 space straight x space fraction numerator 2 straight h over denominator straight g end fraction space space space space open parentheses therefore space equals space square root of fraction numerator 2 straight h over denominator straight g end fraction end root close parentheses
therefore s = h/9 m
Hence, the position of ball from the ground= h- h/9 = 8h/9 m
Question 4

If space straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top space equals space straight B with rightwards arrow on top space straight x space straight A with rightwards arrow on top then the angle between A and B isπ
  • π

  • π/3

  • π/2

  • π/4

Solution

A.

π

open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses space equals space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis
rightwards double arrow space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses minus space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses plus space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis space equals space 0
open square brackets therefore space left parenthesis straight B with rightwards arrow on top space straight x space straight A with rightwards arrow on top right parenthesis space equals space minus space left parenthesis straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top right parenthesis close square brackets
2 space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses space equals 0
rightwards double arrow space space 2 AB space sin space straight theta space equals space 0
sin space straight theta space equals space 0 space space left square bracket space because space vertical line straight A with rightwards arrow on top vertical line space equals space straight A space not equal to 0 comma space vertical line straight B with rightwards arrow on top vertical line space equals space straight B not equal to space 0 right square bracket
straight theta space equals space 0 space or space straight pi space
Question 5

A projectile can have the same range R for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

  • 1/R2

  • 1/R

  • R

  • R2

Solution

C.

R

We know in advance that range of projectile is same for complementary angles i.e. for θ and (900 - θ )
straight T subscript 1 space equals space fraction numerator 2 straight u space sin space straight theta over denominator straight g end fraction

straight T subscript 2 space equals space fraction numerator 2 straight u space sin space left parenthesis 90 to the power of 0 minus straight theta right parenthesis over denominator straight g end fraction space equals space fraction numerator 2 space straight u space cos space straight theta over denominator straight g end fraction
and space straight R space equals space fraction numerator straight u squared space sin space 2 straight theta over denominator straight g end fraction
Therefore comma space straight T subscript 1 space straight T subscript 2 space equals space fraction numerator 2 space straight u space sin space straight theta over denominator straight g end fraction space straight x space fraction numerator 2 space straight u space cos space straight theta over denominator straight g end fraction
space equals space fraction numerator 2 straight u squared space left parenthesis 2 space sin space straight theta space cos space straight theta right parenthesis over denominator straight g squared end fraction
space equals space fraction numerator 2 straight u squared space left parenthesis sin space 2 straight theta right parenthesis over denominator straight g squared end fraction
space equals space 2 straight R divided by straight g
space equals space straight T subscript 1 straight T subscript 2 space proportional to space straight R