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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2004 Exam Questions

Multiple Choice Questions

1. 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


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
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2.

Which of the following statements is false for a particle moving in a circle with a constant angular speed?

  • The velocity vector is tangent to the circle.

  • The acceleration vector is tangent to the circle.

  • The acceleration vector points to the centre of the circle.

  • The velocity and acceleration vectors are perpendicular to each other.


B.

The acceleration vector is tangent to the circle.

For a particle moving in a circle with constant angular speed, the velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of the circle or is always along the radius of the circle. Since, the tangential vector is perpendicular to radial vector, therefore, velocity vector will be perpendicular to the acceleration vector. But in no case acceleration vector is tangent to the circle.
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3.

Two masses m1 = 5 kg and m2 = 4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift free to move?
(g = 9.8 m/s2 )

  • 0.2 m/s2

  • 9.8 m/s2

  • 5 m/s2

  • 4.8 m/s2


A.

0.2 m/s2


On release, the motion of the system will be according to the figure.
m1g - T = m1a ...... (i)  and
T - m2g = m2a ...... (ii)

On solving;

straight a space equals space open parentheses fraction numerator straight m subscript 1 minus straight m subscript 2 over denominator straight m subscript 1 plus straight m subscript 2 end fraction close parentheses space straight g space........ space left parenthesis iii right parenthesis

Here, m1 = 5 kg, m2 = 4.8 kg, g = 9.8 m/s2

straight a space equals space open parentheses fraction numerator 5 minus 4.8 over denominator 5 plus 4.8 end fraction close parentheses space straight x space 9.8
space equals space fraction numerator 0.2 over denominator 9.8 end fraction space straight x space 9.8
space equals space 0.2 space straight m divided by straight s squared

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4.

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


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

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5.

A machine gun fires a bullet of mass 40 g with a velocity 1200 ms−1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?

  • one 

  • Four

  • Two

  • Three


D.

Three

The force exerted by machine gun on man's hand in firing a bullet = change in momentum per second on a bullet or rate of change of momentum

space equals open parentheses 40 over 1000 close parentheses space straight x space 1200 space equals space 48 space straight N

The force exerted by man on machine gun = 144 N Hence, number of bullets fired =144/48 = 3

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6.

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


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

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7.

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.


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

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8.

A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

  • 7.2 J

  • 3.6 J

  • 120 J

  • 1200 J 


B.

3.6 J

Mass per length
= M/L
= 4/2 = 2 kg/m
The mass of 0.6 m of chain = 0.6 x 2 = 1.2 kg
The centre of mass of hanging part = 0.6 +0 /2 = 0.3 m
Hence, work done in pulling the chain on the table
W =mgh
= 1.2 x 10 x 0.3
= 3.6 J

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9.

An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

  • 20 m

  • 40 m

  • 60 m

  • 80 m


D.

80 m

Third equation of motion gives
v2 = u2 + 2as ⇒ 2
s ∝ u (since v = 0)
where a = retardation of body in both the cases

therefore, straight s subscript 1 over straight s subscript 2 space equals space fraction numerator straight u subscript 1 superscript 2 over denominator straight u subscript 2 superscript 2 end fraction .... (i)

Here, s1 = 20 m, u1 = 60 km/h, u2 = 120 km/h. Putting the given values in eq. (i), we get

20 over straight s subscript 2 space equals space open parentheses 60 over 120 close parentheses squared
space straight s subscript 2 space equals space 20 space straight x space open parentheses 120 over 60 close parentheses squared
space equals space 20 space straight x space 4 space
space equals space 80 space straight m

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10.

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


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
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