1. In a Young’s double slit experiment, the separation between the two slits is d and the wave length of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).
(A) If d = λ, the screen will contain only one maximum.
(B) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen.
(C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase.
(D) If the intensity of light falling on slit 2 is reduced so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase
2. Suppose the wave length λ and the double slit separation d in a Young’s double slit experiment are such that the 6th dark fringe is obtained at point P shown in the figure. The path difference (S2P – S1P) will be
(a) 5 λ
(b) 5 λ/2
(c) 6 λ
(d) 3 λ
(e) 11 λ/2
Visit for the answers
http://physicsplus.blogspot.com/2008/09/iit-jee-2008-question-mcq-on-youngs.html
Monday, September 29, 2008
Thursday, August 28, 2008
Spring - MQ
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. The maximum extension in the spring is
a. 4Mg/k
b. 2Mg/k
c. Mg/k
d. Mg/2k
(JEE 2002)
Answer (b)
Solution
Change in gravitational energy = Energy stored in the spring
Mgx = ½ kx²
=> x = 2Mg/k
a. 4Mg/k
b. 2Mg/k
c. Mg/k
d. Mg/2k
(JEE 2002)
Answer (b)
Solution
Change in gravitational energy = Energy stored in the spring
Mgx = ½ kx²
=> x = 2Mg/k
Kepler’s Laws - MQ
According to Kepler’s second law, the radius vector to a plant form the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of -----------------.
(JEE 1985)
Answer: Angular momentum
(JEE 1985)
Answer: Angular momentum
Tuesday, August 26, 2008
Collisions - Model questions - JEE
Collisions – Psat JEE Problems
1. A ball hits the floor and rebounds after an inelastic collision. In this case
a. the momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.
(More than one alternative may be correct)
JEE 1986
Answer ( c ) and (d)
Note: total kinetic energy is not conserved but total energy is conserved.
2. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .
(JEE 1987)
answer (3/2)mv²
Solution;
As the two fragments went in mutual perpendicular directions with velocity v the resultant momentum
2mv cos 45° = 2mv(1/√2) = √2 mv
if the velocity of fragment with 2m mass is V
then 2 mV =/√2mv
V = v/√2
Hence energy released = ½ mv² + ½ mv² + ½ *2 m (v/√2) ²
= (3/2)mv²
1. A ball hits the floor and rebounds after an inelastic collision. In this case
a. the momentum of the ball just after the collision is the same as that just before the collision.
b. the mechanical energy of the ball remains the same in the collision.
c. the total momentum of the ball and the earth is conserved.
d. the total energy of the ball and the earth is conserved.
(More than one alternative may be correct)
JEE 1986
Answer ( c ) and (d)
Note: total kinetic energy is not conserved but total energy is conserved.
2. A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m, are found to move with a speed v, each in mutually perpendicular directions. The total energy released in the process of explosion is _________ .
(JEE 1987)
answer (3/2)mv²
Solution;
As the two fragments went in mutual perpendicular directions with velocity v the resultant momentum
2mv cos 45° = 2mv(1/√2) = √2 mv
if the velocity of fragment with 2m mass is V
then 2 mV =/√2mv
V = v/√2
Hence energy released = ½ mv² + ½ mv² + ½ *2 m (v/√2) ²
= (3/2)mv²
Moment of inertia - Model Problems
1. A circular disc of radius r/3 is removed from the outer-edge of a bigger circular disc of mass 9M and radius r. the moment of inertia of the remaining portion of the disc about the centre O of the disc perpendicular to the plane of the disc is
a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²
JEE 2005
Answer (a)
Solution
Mass of the disc removed
Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²
Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M
Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²
Parallel axis theorem
Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes.
According to the parallel axis theorem
Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²
Moment of inertia of bigger disc before cutting the piece
= ½ (9M)(r²)
Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²
a. 4Mr²
b. (37/9)Mr²
c. (40/9)Mr²
d. 9Mr²
JEE 2005
Answer (a)
Solution
Mass of the disc removed
Total mass – M
Total area = π r²
Area of piece removed π (r/3) ²
Hence mass of the disc removed = M[π (r/3) ²]/( π r²) = M
Moment of inertia of removed about its own centre is = ½ (M)*(r/3) ²
Parallel axis theorem
Moment of inertia of a body about an axis parallel to its centre of mass is equal to
Moment of the body about the axis passing through its centre of mass plus Mr² (where M is the mass of body and r is the perpendicular distance between two axes.
According to the parallel axis theorem
Moment of inertia of the removed piece about an axis passing through the centre of the big disc = ½ (M)*(r/3) ² + M(2r/3) ²
=1/2(M) (r²/9) + M(4r²/9) = (½)(M) (r²/9 + 8r²/9) = (½)Mr²
Moment of inertia of bigger disc before cutting the piece
= ½ (9M)(r²)
Hence moment of inertia after removal of the piece
=( ½) (9M)(r²) - (½)Mr² = 4Mr²
Friday, August 15, 2008
Work and Energy Past JEE Questions
1. A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to
a. t1/2
b. t3/4
c. t3/2
d. t2
(1984)
2. A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is:
a. MgL
b. MgL/3
c. MgL/9
d. MgL/18
(1985)
3. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:
a. its velocity is constant.
b. its acceleration is constant
c. its kinetic energy is constant
d. it moves in a circular path
(1987)
4. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with t as ac = k²rt² where k is a constant. The power delivered to the particle by the force acting on it is
a. 2 πmk²r²t
b. mk²r²t
c. (mk4r²t5)/3
d. zero
(1994)
5. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of
a. 0.42 m from mass of 0.3 kg
b. 0.7 m form mass of 0.7 kg
c. 0.98 m from mass of 0.3 kg
d. 0.98 m from mass of 0.7 kg
(1995)
6. A force F = -K(yi + xj) (where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x axis to the point (a,0) and then parallel to the y axis to the point (a,a). The total work done by the force F on the particle is
a. -2Ka²
b. 2Ka²
c. -Ka²
d. Ka²
(1998)
solutions some time later
a. t1/2
b. t3/4
c. t3/2
d. t2
(1984)
2. A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is:
a. MgL
b. MgL/3
c. MgL/9
d. MgL/18
(1985)
3. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:
a. its velocity is constant.
b. its acceleration is constant
c. its kinetic energy is constant
d. it moves in a circular path
(1987)
4. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with t as ac = k²rt² where k is a constant. The power delivered to the particle by the force acting on it is
a. 2 πmk²r²t
b. mk²r²t
c. (mk4r²t5)/3
d. zero
(1994)
5. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of
a. 0.42 m from mass of 0.3 kg
b. 0.7 m form mass of 0.7 kg
c. 0.98 m from mass of 0.3 kg
d. 0.98 m from mass of 0.7 kg
(1995)
6. A force F = -K(yi + xj) (where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x axis to the point (a,0) and then parallel to the y axis to the point (a,a). The total work done by the force F on the particle is
a. -2Ka²
b. 2Ka²
c. -Ka²
d. Ka²
(1998)
solutions some time later
Thursday, August 14, 2008
JEE past questions - Centre of Mass
1. Two particles initially at rest move toward eahc other under a mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2 V, then the speed at the centre of mass of the system is
a. zero
b. 1.5 V
c. V
d. 3 V
(1982)
2. State whether the following statement is true or false.
Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2m/s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s, the velocity of the centre of mass is 0.75 m/s.
(1989)
a. zero
b. 1.5 V
c. V
d. 3 V
(1982)
2. State whether the following statement is true or false.
Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is 2m/s, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 m/s, the velocity of the centre of mass is 0.75 m/s.
(1989)
Saturday, August 9, 2008
Past JEE - MCQ-with Multiple Answers - Circular Motion
Circular Motion and Related Topics
1. A mass M is moving with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin:
a. is zero
b. remains constant
c. goes on increasing
d. goes on decreasing
(1985)
2. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:
a. its velocity is constant.
b. its acceleration is constant
c. its kinetic energy is constant
d. it moves in a circular path
(1987)
3. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R-5/2, then
a. T² is proportional to R
b. T² is proportional to R7/2
c. T² is proportional to R3/2
d. T² is proportional to R3.75
(1989)
1. A mass M is moving with a constant velocity parallel to the X-axis. Its angular momentum with respect to the origin:
a. is zero
b. remains constant
c. goes on increasing
d. goes on decreasing
(1985)
2. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:
a. its velocity is constant.
b. its acceleration is constant
c. its kinetic energy is constant
d. it moves in a circular path
(1987)
3. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to R-5/2, then
a. T² is proportional to R
b. T² is proportional to R7/2
c. T² is proportional to R3/2
d. T² is proportional to R3.75
(1989)
Past JEE MCQ - Circular Motion and Related Topics
Give the correct answer/choice
1. A thin circular ring of mass ‘M’ and radius ‘r’ is rotating about its axis with a constant angular velocity ω. Two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity
a. ω M/(M+m)
b. ω(M-2m)/(M+2m)
c. ω M/(M+2m)
d. ω (M+2m)/M
(1983)
2. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01R. The period of the second satellite is larger than the first one by approximately.
a. 0.7%
b. 1%
c. 1.5%
d. 3.0%
(1995)
3. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of
a. 0.42 m from mass of 0.3 kg
b. 0.70 m from mass of 0.7 kg
c. 0.98 m from mass of 0.3 kg
d. 0.98 m from mass of 0.7 kg
(1995)
4. If the distance between the earth and the sun were half its present value, the number of days in a year would have been
a. 64.5
b. 129
c. 182.5
d. 730
(1996)
5. A mass ‘m’ is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
a. is zero
b. remains constant
c. goes on increasing
d. goes on decreasing
(1997)
6. An artificial satellite moving in a circular orbit around the Earth has a total (kinetic + potential) energy E0. Its potential energy is
a. -E0
b. 1.5E0
c. 2E0
d. E0
(1997c)
7. A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane (x-y plane) along x-axis. The magnitude of angular momentum of disc about the origin O is
a. (1/2)MR² ω
b. MR² ω
c. (3/2) MR² ω
d. 2MR² ω
(1999)
8. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ, and gravity is neglected, then the time after which the bed starts slipping is
a. √(µ/ α)
b. µ/√( α)
c. 1/√(µ α)
d. infinitesimal
(2000)
9. A particle of charge q and mass m moves in a circular orbit of radius r with an angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on
a. ω and q
b. ω, q, and m.
c. q and m
d. ω and m.
(2000)
10. An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is 1/3. If the line joining the center of the hemispherical surface to the insect makes an angle α with the vertical, the maximum possible value of α is given by
a. cot α = 3
b. tan α = 3
c. se v = 3
d. cosec α = 3
(2001)
11.a geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. then, the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface ( Radius of earth = 6400 km) will approximately be
a. (1/2) hour
b. 1 hour
c. 2 hour
d. 4 hour
(2002)
12. Two identical balls, each of mass m, are connected by a rigid rod of length L. an impulse of ‘mv’ perpendicular to the rod is applied at one end of this system rest. The angular velocity produced in the system is
a. v/L
b. v/2L
c. v/3L
d. v/4L
(2003)
13. A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved?
a. inside the circle
b. outside the circle
c. center of the circle
d. on the circumference of the circle
(2003)
14. A steel ball is confined to rotate in a circular path with its speed decreasing due to friction. Which of the following statements is correct for the ball?
a. Its acceleration is towards the centre.
b. The ball moves towards the centre of circle in a spiral path.
c. Its angular momentum about the centre is conserved.
d. Only the direction of angular momentum is conserved.
(2005)
1. A thin circular ring of mass ‘M’ and radius ‘r’ is rotating about its axis with a constant angular velocity ω. Two objects, each of mass m, are attached gently to the opposite ends of a diameter of the ring. The wheel now rotates with an angular velocity
a. ω M/(M+m)
b. ω(M-2m)/(M+2m)
c. ω M/(M+2m)
d. ω (M+2m)/M
(1983)
2. A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01R. The period of the second satellite is larger than the first one by approximately.
a. 0.7%
b. 1%
c. 1.5%
d. 3.0%
(1995)
3. Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of
a. 0.42 m from mass of 0.3 kg
b. 0.70 m from mass of 0.7 kg
c. 0.98 m from mass of 0.3 kg
d. 0.98 m from mass of 0.7 kg
(1995)
4. If the distance between the earth and the sun were half its present value, the number of days in a year would have been
a. 64.5
b. 129
c. 182.5
d. 730
(1996)
5. A mass ‘m’ is moving with a constant velocity along a line parallel to the x-axis, away from the origin. Its angular momentum with respect to the origin
a. is zero
b. remains constant
c. goes on increasing
d. goes on decreasing
(1997)
6. An artificial satellite moving in a circular orbit around the Earth has a total (kinetic + potential) energy E0. Its potential energy is
a. -E0
b. 1.5E0
c. 2E0
d. E0
(1997c)
7. A disc of mass M and radius R is rolling with angular speed ω on a horizontal plane (x-y plane) along x-axis. The magnitude of angular momentum of disc about the origin O is
a. (1/2)MR² ω
b. MR² ω
c. (3/2) MR² ω
d. 2MR² ω
(1999)
8. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration α. If the coefficient of friction between the rod and the bead is µ, and gravity is neglected, then the time after which the bed starts slipping is
a. √(µ/ α)
b. µ/√( α)
c. 1/√(µ α)
d. infinitesimal
(2000)
9. A particle of charge q and mass m moves in a circular orbit of radius r with an angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on
a. ω and q
b. ω, q, and m.
c. q and m
d. ω and m.
(2000)
10. An insect crawls up a hemispherical surface very slowly. The coefficient of friction between the insect and the surface is 1/3. If the line joining the center of the hemispherical surface to the insect makes an angle α with the vertical, the maximum possible value of α is given by
a. cot α = 3
b. tan α = 3
c. se v = 3
d. cosec α = 3
(2001)
11.a geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. then, the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface ( Radius of earth = 6400 km) will approximately be
a. (1/2) hour
b. 1 hour
c. 2 hour
d. 4 hour
(2002)
12. Two identical balls, each of mass m, are connected by a rigid rod of length L. an impulse of ‘mv’ perpendicular to the rod is applied at one end of this system rest. The angular velocity produced in the system is
a. v/L
b. v/2L
c. v/3L
d. v/4L
(2003)
13. A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved?
a. inside the circle
b. outside the circle
c. center of the circle
d. on the circumference of the circle
(2003)
14. A steel ball is confined to rotate in a circular path with its speed decreasing due to friction. Which of the following statements is correct for the ball?
a. Its acceleration is towards the centre.
b. The ball moves towards the centre of circle in a spiral path.
c. Its angular momentum about the centre is conserved.
d. Only the direction of angular momentum is conserved.
(2005)
Friday, August 8, 2008
Past JEE Fill in the Blanks – Circular Motion and Related Chapters
1. A particle moves in a circle of radius R. In half the period of revolution, its displacement is ___________ and the distance covered is _______________. (1983)
2. The numerical value of the angular velocity of rotation of the earth should be __________ rad/s in order to make the effective acceleration due to gravity equal to zero at the equator. (1984)
3. A uniform cube of side ‘a’ and mass ‘m’ rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimum value of F for which the cube begins to tip about an edge is _________. (Assume that the cube does not slide). (1984)
4. According to Kepler’s second law, the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of ____________. (1985)
5. A spotlight S rotates in horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light P moves along the wall at distance of 3 m. The velocity of spot P when θ = 45° is ___________ m/s. (1987)
6. A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5R from the surface of the earth is ___________ hours. (1987)
7. A cylinder of mass ‘M’ and radius ‘R’ is resting on a horizontal platform (which is parallel to the x-y plane) with its axis fixed along the y-axis and free to rotate about its axis. The platform is given a motion in the x-direction given as x = A cos ωt. There is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder during its motion is ___________________. (1988)
8. A stone of mass ‘m’ tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Arn where A is a constant, r is the instantaneous radius of the circle, and n = _____________. (1993)
9. The ratio of Earth’s orbital angular momentum (about the Sun) to its mass is 4.4*1015 m²/s. The area enclosed by the Earth’s orbit is approximately ___________ m². (1997 cancelled)
2. The numerical value of the angular velocity of rotation of the earth should be __________ rad/s in order to make the effective acceleration due to gravity equal to zero at the equator. (1984)
3. A uniform cube of side ‘a’ and mass ‘m’ rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point that is directly above the centre of the face, at a height 3a/4 above the base. The minimum value of F for which the cube begins to tip about an edge is _________. (Assume that the cube does not slide). (1984)
4. According to Kepler’s second law, the radius vector to a planet from the sun sweeps out equal areas in equal intervals of time. This law is a consequence of the conservation of ____________. (1985)
5. A spotlight S rotates in horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light P moves along the wall at distance of 3 m. The velocity of spot P when θ = 45° is ___________ m/s. (1987)
6. A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5R from the surface of the earth is ___________ hours. (1987)
7. A cylinder of mass ‘M’ and radius ‘R’ is resting on a horizontal platform (which is parallel to the x-y plane) with its axis fixed along the y-axis and free to rotate about its axis. The platform is given a motion in the x-direction given as x = A cos ωt. There is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder during its motion is ___________________. (1988)
8. A stone of mass ‘m’ tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Arn where A is a constant, r is the instantaneous radius of the circle, and n = _____________. (1993)
9. The ratio of Earth’s orbital angular momentum (about the Sun) to its mass is 4.4*1015 m²/s. The area enclosed by the Earth’s orbit is approximately ___________ m². (1997 cancelled)
Circular Motion and Related Chapters - Past JEE Questions - TRUE or FALSE
State whether the following are True or False
1. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°. When its angular displacement is 20°, the tension in the string is greater than mg cos 20°. (1984)
2. It is possible ot put an artificial statellite into an orbit in such a way that it will always remain directly over New Delhi. (1984)
3. Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails. (1985)
4. A triangular plate (ABC) of uniform thickness an density is made to rotate about an axis perpendicular to the plane of the paper and (a) passing through A and (b) passing through B, by the application of the same force, F at D(mid point of A). The angular acceleration in both the cases will be the same. (1985).
5. A thin uniform disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω. Another disc of the same dimension but of mass M/4 is placed gently on the first disc coaxially. The angular velocity of the system now is 2ω/√(5). (1986)
1. A simple pendulum with a bob of mass m swings with an angular amplitude of 40°. When its angular displacement is 20°, the tension in the string is greater than mg cos 20°. (1984)
2. It is possible ot put an artificial statellite into an orbit in such a way that it will always remain directly over New Delhi. (1984)
3. Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails. (1985)
4. A triangular plate (ABC) of uniform thickness an density is made to rotate about an axis perpendicular to the plane of the paper and (a) passing through A and (b) passing through B, by the application of the same force, F at D(mid point of A). The angular acceleration in both the cases will be the same. (1985).
5. A thin uniform disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω. Another disc of the same dimension but of mass M/4 is placed gently on the first disc coaxially. The angular velocity of the system now is 2ω/√(5). (1986)
Sunday, June 22, 2008
X-Rays - 6
1. Which one of following statements is Wrong in the context of X-rays generated from a X- ray tube ?
(A) Wavelength of characteristic X-rays decreases when the atomic number of the target increases
(B) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target
(C) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube
(D) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube
(JEE 2008)
Answer (B)
Reason:m (λc) cuttoff wavelength is given as
λc = hc/K.E. = hc/eV
Hence λc does not depend on atomic number of target but depends on potential difference between cathode and anode (target)
(A) Wavelength of characteristic X-rays decreases when the atomic number of the target increases
(B) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target
(C) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube
(D) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube
(JEE 2008)
Answer (B)
Reason:m (λc) cuttoff wavelength is given as
λc = hc/K.E. = hc/eV
Hence λc does not depend on atomic number of target but depends on potential difference between cathode and anode (target)
X-Rays - 7
1. The shortest wavelength of x-rays emitted from an x-ray tube depends on
a. current in the tube
b. the voltage applied to the tube
c. the nature of gas in the tube
d. the atomic number of the target material
More than one alternative possible
(JEE 1982)
answer: (b)
The wave length of emitted X-rays depends on the voltage applied to the x-ray tube.
2. To produce characteristic X-rays using a tungsten target in an X-ray generator, the accelerating voltage should be greater than ---------------- volts and the energy of the characteristic radiation is ---------------- eV.
(The binding energy of the innermost electron in tungsten is 40 KeV)
(JEE 1983)
answer: 40 kilo volts, 40 KeV
Energy of the accelerating electron should be equal to the binding energy of the electron in tungsten
Hence E = eV = 40 keV
V = 40 kV
Energy of the characteristic radiation is equal to the characteristic kinetic energy of bombarding electrons.
3. A potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation
a. the intensity increases
b. the minimum wave length increases
c. the intensity remains unchanged
d. the minimum wave length decreases
Answer: (d)
(JEE 1988)
λmin = hc/eV
If V is increased, minimum wave length will decrease
a. current in the tube
b. the voltage applied to the tube
c. the nature of gas in the tube
d. the atomic number of the target material
More than one alternative possible
(JEE 1982)
answer: (b)
The wave length of emitted X-rays depends on the voltage applied to the x-ray tube.
2. To produce characteristic X-rays using a tungsten target in an X-ray generator, the accelerating voltage should be greater than ---------------- volts and the energy of the characteristic radiation is ---------------- eV.
(The binding energy of the innermost electron in tungsten is 40 KeV)
(JEE 1983)
answer: 40 kilo volts, 40 KeV
Energy of the accelerating electron should be equal to the binding energy of the electron in tungsten
Hence E = eV = 40 keV
V = 40 kV
Energy of the characteristic radiation is equal to the characteristic kinetic energy of bombarding electrons.
3. A potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation
a. the intensity increases
b. the minimum wave length increases
c. the intensity remains unchanged
d. the minimum wave length decreases
Answer: (d)
(JEE 1988)
λmin = hc/eV
If V is increased, minimum wave length will decrease
Nucleus - 6
1. The nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this figure and choose the correct choice(s) given below.
(A) Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy
(B) Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will release energy
(C) Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments
(D) Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments
(JEE 2008)
[Ans. B,D]
Reason: Energy is released when binding energy per nucleon increases in fusion of two nuclei of mass number from 51 to 100, final nuclei has mass number 102 to 200 where B.E./A is greater similarly in fission of 200 to 260 final mass no becomes 100 to 130.
(A) Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy
(B) Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will release energy
(C) Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments
(D) Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments
(JEE 2008)
[Ans. B,D]
Reason: Energy is released when binding energy per nucleon increases in fusion of two nuclei of mass number from 51 to 100, final nuclei has mass number 102 to 200 where B.E./A is greater similarly in fission of 200 to 260 final mass no becomes 100 to 130.
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