Past JEE Question
Which of the following choices does not represent the same dimensions of the mentioned physical quanties?
a. Heat, potential energy and work
b. Electric flux, electric field, and electric dipole moment
c. Pressure, stress, and Young's modulus
d. Potential difference, voltage, and electromotive force of a galvanic cell.
(JEE 2005 screening)
Answer: b
Revision question 1
Find dimensions of a) work done by a force b) linear momentum, c)kinetic energy
Past JEE question
In the expression P = (α²/β)e-αZ/kθ, P is pressure, Z is distance, k is Boltzmann constant and θ is the temperature. The dimensional formula of β will be
a. M¹L³T-2
b. M¹L²T-2
c. M¹L¹T-1
d. M°L²T-1
Value of k = Boltzmann constant = 1.38*10-23 J/K
(JEE 2004 screening)
Saturday, May 16, 2009
IIT JEE Level Revision Questions - 3. Rest and Motion: Kinematics
Past JEE question
A small block slides without friction down an inclinded plane starting from the rest. If Sn is the distance travelled form time t = n to t = n+1, then Sn/Sn+1 is
a. (2n-1)/2n
b. (2n+1)/(2n-3)
c. (2n+1)/(2n+3)
d. 2n/(2n+1)
(JEE 2004, Screening)
answer: c
1982
1(ii)
A particle moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
a. zero
b. 1/SQRT(2) m/s² towards north-west
c. ½ m/s² toward north-west
d. ½ m/s² towards north
Answer (b)
The average acceleration must have an westward component and northward component to make the velocity towards east as zero and to make velocity towards north as 5 m/s
Hence it will act in the direction of northwest.
Westward component of average acceleration = (0-5)/10 = -½ m/s²
Northward component of average of acceleration = (5-0)/10 = ½ m/s²
Hence magnitude of average acceleration = Sqrt( (-1/2) ²+(1/2) ²) = 1/SQRT(2)
1983
1(i)
A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in the direction.
a. due north
b. 30° east of north
c 30° west of north
d. 60° east of north
Answer: ( c )
Since the man wants to reach the spot opposite to him, he has to swim to towards west of north to make use of water velocity in the direction west to east.
As he travels 10 metres/ per minute in that direction, he will have a moment of 5 metres/min in the west to east direction.
So angle is given by sin θ = 5/10 = ½
θ = 30°
4. State whether the statement is true or false
(i)
Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed. (neglect air resistance)
Answer: true
Both the balls thrown vertically with the same speed will reach equal height irrespective of their masses. These would also pass through the point of projection in their downward motion with the same speed when subject to acceleration due to gravity.
1984
State whether the statement is true or false
3(a)
A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.
Answer: true
At the top of the path, the vertical component of velocity is zero and the particle has only horizontal component. Hence the speed (the magnitude of the velocity) is minimum at the top of the path.
2004
A particle starts from rest. Its acceleration at the start is 10 m/s² and in 12 seconds the acceleration uniformly reduces to zero. The maximum speed of the particle will be
a. 20 m/s
b. 60 m/s
c. 550 m/s
d. 660 m/s
Revision question 1
A ball is dropped vertically from a height of d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting the subsequent motion and air resistance, (a) what will be the shape of velocity versus distance diagram? (b)What will be the shape of velocity versus time diagram.
Options for A
Linear or curvilinear
Options for B
Linear or curvilinear
(Based on Past JEE questions JEE 2000, screening)
Revision Problem 2
An aeroplane has to go from a point A to another point B, 500 km away 30° east of north. A wind is blowing due north at a speed of 20 m/s. the air-speed of the plane is 150 m/s.
(a) Find the direction in which the pilot should head the plane to reach the point B.
(b) Find the time taken by the plane to go from A to B.
(Source: HC Verma, Prob. 49)
A small block slides without friction down an inclinded plane starting from the rest. If Sn is the distance travelled form time t = n to t = n+1, then Sn/Sn+1 is
a. (2n-1)/2n
b. (2n+1)/(2n-3)
c. (2n+1)/(2n+3)
d. 2n/(2n+1)
(JEE 2004, Screening)
answer: c
1982
1(ii)
A particle moving eastwards with a velocity 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is
a. zero
b. 1/SQRT(2) m/s² towards north-west
c. ½ m/s² toward north-west
d. ½ m/s² towards north
Answer (b)
The average acceleration must have an westward component and northward component to make the velocity towards east as zero and to make velocity towards north as 5 m/s
Hence it will act in the direction of northwest.
Westward component of average acceleration = (0-5)/10 = -½ m/s²
Northward component of average of acceleration = (5-0)/10 = ½ m/s²
Hence magnitude of average acceleration = Sqrt( (-1/2) ²+(1/2) ²) = 1/SQRT(2)
1983
1(i)
A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in the direction.
a. due north
b. 30° east of north
c 30° west of north
d. 60° east of north
Answer: ( c )
Since the man wants to reach the spot opposite to him, he has to swim to towards west of north to make use of water velocity in the direction west to east.
As he travels 10 metres/ per minute in that direction, he will have a moment of 5 metres/min in the west to east direction.
So angle is given by sin θ = 5/10 = ½
θ = 30°
4. State whether the statement is true or false
(i)
Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed. (neglect air resistance)
Answer: true
Both the balls thrown vertically with the same speed will reach equal height irrespective of their masses. These would also pass through the point of projection in their downward motion with the same speed when subject to acceleration due to gravity.
1984
State whether the statement is true or false
3(a)
A projectile fired form the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path.
Answer: true
At the top of the path, the vertical component of velocity is zero and the particle has only horizontal component. Hence the speed (the magnitude of the velocity) is minimum at the top of the path.
2004
A particle starts from rest. Its acceleration at the start is 10 m/s² and in 12 seconds the acceleration uniformly reduces to zero. The maximum speed of the particle will be
a. 20 m/s
b. 60 m/s
c. 550 m/s
d. 660 m/s
Revision question 1
A ball is dropped vertically from a height of d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting the subsequent motion and air resistance, (a) what will be the shape of velocity versus distance diagram? (b)What will be the shape of velocity versus time diagram.
Options for A
Linear or curvilinear
Options for B
Linear or curvilinear
(Based on Past JEE questions JEE 2000, screening)
Revision Problem 2
An aeroplane has to go from a point A to another point B, 500 km away 30° east of north. A wind is blowing due north at a speed of 20 m/s. the air-speed of the plane is 150 m/s.
(a) Find the direction in which the pilot should head the plane to reach the point B.
(b) Find the time taken by the plane to go from A to B.
(Source: HC Verma, Prob. 49)
IIT JEE Level Revision Questions - 6. Friction
Friction
1. fill blanks
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 of 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)
2. A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 m/s², the frictional force acting on the block is _________ .
(JEE 1984)
Answer: 5 N
Reason: The horizontal force acting on the cube due to the acceleration of the truck is
F = ma = 1 kg * (5 m/s²) = 5 N
As the block is not moving, a frictional force of 5 N is acting on it.
We can verify that the static friction can go up to a maximum value of 1 kg *(10 m/s²)*0.6 = 6 N if we take g as approximately equal to 10 m/s².
Revision Question 1.
A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, 9b) the lower block. Take g = 10 m/s².
1. fill blanks
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 of 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)
2. A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 m/s², the frictional force acting on the block is _________ .
(JEE 1984)
Answer: 5 N
Reason: The horizontal force acting on the cube due to the acceleration of the truck is
F = ma = 1 kg * (5 m/s²) = 5 N
As the block is not moving, a frictional force of 5 N is acting on it.
We can verify that the static friction can go up to a maximum value of 1 kg *(10 m/s²)*0.6 = 6 N if we take g as approximately equal to 10 m/s².
Revision Question 1.
A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, 9b) the lower block. Take g = 10 m/s².
IIT JEE Level Revision Questions - 7. Circular motion
Past JEE Questions
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
Revision Questions
1. A circular road of radius r is banked for a speed of v = 40 km/hr. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.
a. The car cannot make a turn without skidding.
b. If the car turns at a speed less than 40 km/hr, it will slip down.
c. If the car turns at the correct speed of 40 km/hr, the force by the road on the car is equal to mv²/r.
d. If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than mv²/r.
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
Revision Questions
1. A circular road of radius r is banked for a speed of v = 40 km/hr. A car of mass m attempts to go on the circular road. The friction coefficient between the tyre and the road is negligible.
a. The car cannot make a turn without skidding.
b. If the car turns at a speed less than 40 km/hr, it will slip down.
c. If the car turns at the correct speed of 40 km/hr, the force by the road on the car is equal to mv²/r.
d. If the car turns at the correct speed of 40 km/hr, the force by the road on the car is greater than mg as well as greater than mv²/r.
IIT JEE Level Revision Questions - 8. 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)
7. 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
Revision Questions
1. A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring. (Source HC Verma, prob. 44)
2. The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity of √(10gl) where l is the length of the pendulum. Find the tension in the string when
(a) the string is horizontal
(b) the bob is at its highest point and
(c) the string makes an angle of 60° with the upward vertical
(Source HC Verma Prob 53)
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)
7. 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
Revision Questions
1. A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved. Find the spring constant of the spring. (Source HC Verma, prob. 44)
2. The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity of √(10gl) where l is the length of the pendulum. Find the tension in the string when
(a) the string is horizontal
(b) the bob is at its highest point and
(c) the string makes an angle of 60° with the upward vertical
(Source HC Verma Prob 53)
IIT JEE Level Revision Questions - 9. Centre of mass, linear momentum, collision
Past JEE Questions
Centre of mass. Linear momentum collisions fill blanks
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 _________ .
1982
1(iv)
Two particles initially at rest move toward each other under a mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, then the speed at the centre of the mass of the system is
a. zero
b. 1.5 V
c. V
d. 3 V
Answer (a)
Velocity of centre of mass due to motion of particles due to internal forces is zero.
1986
1.(ii)
Multiple alternatives are correct. Find them
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.
Answer; © and (d)
In an inelastic collision, momentum of the system is conserved. Also the total energy of the system is conserved. But total kinetic energy is not conserved.
1(iii)
A shell is fired from a cannon with a velocity V (m/s) at an angle θ with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon and the speed (in m/s) of the other piece immediately after the explosion is:
Answer: (a)
At the highest point, the shell has only horizontal component of velocity. When the particle breaks into pieces, the momentum of the system is conserved. As one of the components retraces back its horizontal component at the time of explosion is V cos θ.
The systems momentum at the time of explosion is mV cos θ. The momentum of the piece that retraced the path is (m/2)V cos θ.
Therefore mV cos θ = -(m/2)V cos θ + (m/2)V’
(m/2)V’ = 3(m/2) V cos θ
V’ = 3 V cos θ.
1987
2(iii)
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 ---------------.
Answer: (3/2)mv²
The momentum of the fragment with mass 2m has to be equal to the sum of momentums of the other two fragments.
As the magnitude of momentum of other two vectors is mv and the angle between them is 90°, the resultant vector will have the magnitude 2mv cos 45° .
2mv cos 45° = [2/SQRT(2)] mv
2mv’ = [2/SQRT(2)] mv
v’ = v [1/SQRT(2)]
So energy released in the explosion = sum kinetic energies of the three fragments
(½)mv² + (1/2)mv² +(1/2) 2m [(v(1/SQRT(2)] ²
mv²+mv²(1/2) = (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
Centre of mass. Linear momentum collisions fill blanks
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 _________ .
1982
1(iv)
Two particles initially at rest move toward each other under a mutual force of attraction. At the instant when the speed of A is V and the speed of B is 2V, then the speed at the centre of the mass of the system is
a. zero
b. 1.5 V
c. V
d. 3 V
Answer (a)
Velocity of centre of mass due to motion of particles due to internal forces is zero.
1986
1.(ii)
Multiple alternatives are correct. Find them
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.
Answer; © and (d)
In an inelastic collision, momentum of the system is conserved. Also the total energy of the system is conserved. But total kinetic energy is not conserved.
1(iii)
A shell is fired from a cannon with a velocity V (m/s) at an angle θ with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon and the speed (in m/s) of the other piece immediately after the explosion is:
Answer: (a)
At the highest point, the shell has only horizontal component of velocity. When the particle breaks into pieces, the momentum of the system is conserved. As one of the components retraces back its horizontal component at the time of explosion is V cos θ.
The systems momentum at the time of explosion is mV cos θ. The momentum of the piece that retraced the path is (m/2)V cos θ.
Therefore mV cos θ = -(m/2)V cos θ + (m/2)V’
(m/2)V’ = 3(m/2) V cos θ
V’ = 3 V cos θ.
1987
2(iii)
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 ---------------.
Answer: (3/2)mv²
The momentum of the fragment with mass 2m has to be equal to the sum of momentums of the other two fragments.
As the magnitude of momentum of other two vectors is mv and the angle between them is 90°, the resultant vector will have the magnitude 2mv cos 45° .
2mv cos 45° = [2/SQRT(2)] mv
2mv’ = [2/SQRT(2)] mv
v’ = v [1/SQRT(2)]
So energy released in the explosion = sum kinetic energies of the three fragments
(½)mv² + (1/2)mv² +(1/2) 2m [(v(1/SQRT(2)] ²
mv²+mv²(1/2) = (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
IIT JEE Level Revision Questions - 10. Rotational mechanics
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
IIT JEE Level Revision Questions - 16. Sound waves
JEE Question
Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have the maximum amplitude of vibration is _____________ m. (JEE 1984 screening)
Answer 1/8
Revision Question 1
The equation of a travelling sound wave is y = 6.0 sin (600 t - 1.8 x) where y is measured in 10^-5 m. t in second and x in metre.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.
(b) Find the ratio of the velocity amplitude of the particles to the wave speed.
Source: HC Verma, Prob NO. 8)
Sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have the maximum amplitude of vibration is _____________ m. (JEE 1984 screening)
Answer 1/8
Revision Question 1
The equation of a travelling sound wave is y = 6.0 sin (600 t - 1.8 x) where y is measured in 10^-5 m. t in second and x in metre.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.
(b) Find the ratio of the velocity amplitude of the particles to the wave speed.
Source: HC Verma, Prob NO. 8)
IIT JEE Level Revision Questions - 44. X-Rays
Past JEE Question
To produce charateristic 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 __________.
(The binding energy of the innermost electron in tungsten is 40 Kev).(JEE 1983 Screening)
Answer: 40 kV, 40 kV
Revision question 1
Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic K or L X-ray.
The energy levels of the tungsten atom with an electron knocked out are as follows:
Cell---------------- Energy
containing---------- in
vacancy-------------keV
K -----------------69.5
L------------------11.3
M------------------ 2.3
(Reference: Problem No. 21 in HC Verma)
To produce charateristic 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 __________.
(The binding energy of the innermost electron in tungsten is 40 Kev).(JEE 1983 Screening)
Answer: 40 kV, 40 kV
Revision question 1
Find the maximum potential difference which may be applied across an X-ray tube with tungsten target without emitting any characteristic K or L X-ray.
The energy levels of the tungsten atom with an electron knocked out are as follows:
Cell---------------- Energy
containing---------- in
vacancy-------------keV
K -----------------69.5
L------------------11.3
M------------------ 2.3
(Reference: Problem No. 21 in HC Verma)
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