“Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages”
– Thomas A. Edison
Today I will give you a free response practice question on electromagnetic induction. You may try to answer this question within 15 minutes. Here is the question:
Two infinitely long straight parallel wires W1 and W2, separated by a distance ‘a’ in free space, carry equal currents I flowing in opposite directions as shown in the adjoining figure. A square loop PQRS of side ‘a’,made of nichrome wire of resistance ρ Ω per metre is arranged with its plane lying in the plane of the wires W1 and W2 so that the sides PQ and RS of the loop are parallel to the wiresW1 and W2. The side PQ of the loop is at a distance ‘a’ from the wire W2. Now, answer the following questions in terms of the given quantities and fundamental constants:
(a) Determine the magnetic flux density at a point midway between the wires W1 and W2.
(b)Determine the magnetic flux density at the mid point of the square loop PQRS.
(c) Calculate the magnetic flux through the loop PQRS.
(d) What is the average emf induced in the loop when the current through the wires is switched off in a time of 50 ms?
(e) When the current through the wires is switched off, it is found that at a certain instant t, the current decays at the rate of 40 As–1. Calculate the current induced in the loop PQRS at the instant t.
Indicate the direction of the current in the loop and justify your answer.
This question carries 15 points. Try to answer it. I’ll be back soon with a model answer for your benefit.
Essential points to be remembered in kinetic theory of gases were discussed in the post dated 13th March 2008. Questions on kinetic theory of gases were discussed subsequently. You can access all posts related to kinetic theory of gases by clicking on the label, ‘kinetic theory’ below this post. To access older posts you need to click on the ‘older posts’ button.
Today we will discuss a few more typical multiple choice questions on kinetic theory of gases:
(1) The root mean square (R.M.S.) speed v of the molecules of an ideal gas is given by the expressions,
v =√(3RT/M) and
v =√(3kT/m) where R is universal gas constant, T is the absolute (Kelvin) temperature, M is the molar mass, k is Boltzman’s constant and m is the molecular mass. The R.M.S. speed of oxygen molecules (O2) at temperature T1 is v1. When the temperature is doubled, if the oxygen molecules are dissociated into atomic oxygen, what will be R.M.S. speed of oxygen atoms? (Treat the gas as ideal).
(a) v1/2
(b) v1
(c) √2 v1
(d) 2v1
(e) 4v1
We have v1 =√(3RT1/M) or
v1 =√(3kT1/m)
On dissociation the molar mass as well as the molecular mass gets halved. Using the second equation, the R.M.S. speed v after dissociation is given by
v = √[3k×2T1/ (m/2)] = 2√(3kT1/m) = 2v1
(2) Four moles of an ideal diatomic gas is heated at constant volume from 20º C to 30º C. The molar specific heat of the gas at constant pressure (Cp) is 30.3 Jmol–1K–1 and the universal gas constant (R) is 8.3 Jmol–1K–1. The increase in internal energy of the gas is
(a) 80.3 J
(b) 303 J
(c) 332 J
(d) 880 J
(e) 1212 J
The increase in internal energy is MCv∆T where M is the mass of the sample of the gas, Cv is the specific heat at constant volume and ∆T is the rise in temperature of the gas. If we use the molar specific heat of the gas at constant volume for Cv, the number of moles in the sample of the gas is to be used in the place of M.
Now, Cv = Cp – R = 30.3 – 8.3 = 22 Jmol–1K–1.
Therefore, the increase in internal energy of the gas is 4×22×10 = 880 J.
(3) In the case of real gases, the equation of state, PV = RT (where P, V and T are respectively the pressure, volume and absolute temperature), is strictly satisfied only if corrections are applied to the measured pressure P and the measured volume V. The corrections for P and V arise respectively due to
(a) intermolecular attraction and the size of molecules
(b) size of molecules and expansion of the container
(c) expansion of the container and intermolecular attraction
(d) kinetic energy of molecules and collision of molecules
(e) intermolecular attraction and collision of molecules
In kinetic theory of gases it is assumed that there is no force between molecules But there is actually intermolecular attraction which reduces the pressure. So the correction for P arises due to intermolecular attraction.
The entire volume V of the container is not available for the molecules since the molecules have a finite size. The assumption (in kinetic theory) that the molecules are point masses without appreciable volume is incorrect. So the correction for V arises due to the size of molecules.
The correct option is (a).
(4) Gases exert pressure on the walls of the container because the gas molecules
(a) collide one another
(b) exert intermolecular attraction
(c) possess momentum
(d) expand on absorbing heat
(e) exert repulsive force
Because of the momentum of the gas molecules, they collide with the walls of the containing vessel and momentum transfer takes place, resulting in a force on the walls. Pressure is force per unit area. The basic reason for the pressure is the momentum of the gas molecules [Option (c)].
Questions involving friction were discussed on this site earlier. You can access them by clicking on the label ‘friction’ below this post or by trying a search for ‘friction’, using the search box provided on this page.
Today we will discuss some more multiple choice practice questions on friction:
(1) A uniform chain is placed on a rough horizontal table so that one end of the chain hangs down over the edge of the table. When 20% of the length of the chain hangs over the edge, it starts sliding. What is the coefficient of static friction between the chain and the table?
(a) 0.1
(b) 0.15
(c) 0.25
(d) 0.35
(e) 0.5
When the chain just begins to slide, the frictional force between the chain and the table is maximum and is called limiting frictional force which is equal to μsN where μs is the coefficient of static friction between the chain and the table and N is the normal force.
Here N = 0.8mg and μsN = 0.2mg where m is the mass of the chain.
[Note that the weight of 20% of the length of the chain (= 0.2mg) balances the frictional forceμsN].
Therefore, μs×0.8 mg = 0.2 mg from which μs = 0.25
(2) A wooden block is placed on a horizontal surface and a horizontal force equal to the limiting frictional force is applied on it. If the coefficient of static friction and the coefficient of kinetic friction are respectively 0.5 and 0.4 and the acceleration due to gravity is 10 ms–2, the acceleration of the wooden block is
(a) 1 ms–2
(b) 1.1 ms–2
(c) 1.4 ms–2
(d) 1.5 ms–2
(e)zero
The limiting (or, maximum) frictional force is μs mg where μs is the coefficient of static friction, and mg is the weight of the wooden block. When the block moves along the horizontal surface, the frictional force acting is reduced to kinetic frictional force and is equal to μk mg where μk is the coefficient of kinetic friction.
[If the frictional force were not reduced, the block would have moved with uniform velocity since there would be no net force].
The net force acting on the block is μs mg – μk mg and the acceleration a of the block is given by,
a = (μs mg – μk mg)/m = (μs – μk) g = (0.5– 0.4) × 10 = 1 ms–2
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(3) A slab of mass 50 kg is placed on frictionless horizontal floor and a block of mass 10 kg is placed on the slab as shown. The coefficient of static friction and the coefficient of kinetic friction between the slab and the block are respectively 0.4 and 0.2 and the acceleration due to gravity is 10 ms–2. A horizontal force equal to 30 N acts on the block. The acceleration of the slab will be
(a) 0.1 ms–2
(b) 0.2 ms–2
(c) 0.3 ms–2
(d) 0.5 ms–2
(e) 1.2 ms–2
The limiting frictional force (μsN) between the block and the slab is given by
μsN = μsMg = 0.4×10×10 = 40 N
The block will not slide along the slab since the applied force (= 30 N) is less than the limiting frictional force. So the slab and the block will move together, with acceleration a given by
a = 30 N/(50+10)kg = 0.5 ms–2.
[Since there is no relative motion between the slab and the block, the coefficient of kinetic friction does not play any role in the above problem. It just serves the purpose of a distraction].
The above questions may be ‘enjoyed’ by AP Physics B & Caspirants.The following questions are specifically for AP Physics Caspirants:
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(4) A slab A of mass M = 6 kg is placed on rough horizontal floor. The coefficient of kinetic friction between the floor and the slab is 0.1. A block B of mass m = 4 kg is placed on the slab (Fig.). The coefficient of static friction between the slab and the block is 0.2. When a horizontal force of magnitude F is applied on the slab, the block B just begins to slide along the slab A. What is the value of F? (Take g = 10 ms–2).
(a) 40 N
(b) 30 N
(c) 15 N
(d) 10 N
(e) 8 N
The limiting frictional force between the slab and the block is μsmg where μs is the coefficient of static friction between the slab and the block.
When the force F is applied on the slab, the entire system containing the slab and the block must move with an acceleration a so that the inertial force (= ma) on the block just balances the frictional force.
[The inertial force must be infinitesimally greater than the frictional force for the slipping to occur].
Therefore, we have μsmg = ma so that a = μsg = 0.2×10 = 2 ms–2
If the floor were smooth, the force to be applied on the slab to attain the slipping condition would have been a(M+m).
But since the floor is rough, the applied force has to overcome the force of kinetic friction [μk(M+m)g] between the floor and the slab.
Therefore, F = μk(M+m)g + a(M+m)
Thus F = (M+m)( μkg + a) = 10(1+2) = 30 N.
[The above question could be made a little more difficult if the coefficient of static friction (say, 0.25) between the floor and the slab also is given (to distract you). You should remember that once the slab slides along the floor, the friction called into play is kinetic friction and hence your answer will be unchanged].
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(5) A man of mass M = 80 kg tries to pull down a tree using a rope. He exerts the pulling force at an angle of 30º with the horizontal (Fig.). If the coefficient of static friction between the man’s feet and the ground is 0.6 and the gravitational acceleration is 10 ms–2, the maximum pulling force he can exert (before he slips) is approximately
(a) 50 N
(b) 100 N
(c) 200 N
(d) 400N
(e) 600 N
If the man exerts too much pulling force his feet will slip and hence the maximum pulling force is determined by the coefficient of friction and the angle of pull. If T is the maximum tension produced in the rope, the maximum pulling force, F = T.
The vertical component of T which acts upwards reduces the normal force N and we have
N = Mg – T sin 30º = 80×10 – T/2
The maximum frictional force = μsN = 0.6×(800 – T/2).
The horizontal component of the maximum tension T = T cos 30º = T×(√3)/2
The frictional force should balance the horizontal component of tension so as to prevent the man from slipping.
Therefore we have
0.6×(800 – T/2) = T ×(√3)/2
Or, 960 – 0.6 T = 1.732 T so that T = 960/2.332 = 400 N, approximately
AP Physics C aspirants are expected to have a fairly good idea regarding vector methods in Physics. AP Physics B aspirants also should have some idea of vectors. Today we will discuss some multiple choice practice questions on vectors.
(1) The magnetic flux density B at a point P is 0.5 tesla. What is the maximum number of components into which the vector B can be resolved?
(a) 1
(b) 2
(c) 3
(d) 6
(e) infinite
The resultant of all possible components should make the vector B of the given direction and of magnitude 0.5 tesla. We can imagine an infinite number of components in various directions to combine and produce the field B and hence the correct option is (e).
(2) The force F acting on a particle has its line of action (direction) lying in the XY plane and is inclined at an angle θ with the x-axis. What is the z-component of the force F?
(a) zero
(b) F cos θ
(c) F sin θ
(d) F/cos θ
(e) F/sin θ
Since the direction of the vector F is lying in the XY plane it has no z-component. The correct option is (a).
(3) The resultant of two forces F1 and F2 of the same magnitudeF has magnitude F itself. The angle between F1 and F2 is
(a) 30º
(b) 45º
(c) 60º
(d) 90º
(e) 120º
The magnitude F of the resultant force is given by
F = √(F12 + F22 + 2 F1 F2 cosθ) where θ is the angle between the forces F1 and F2.
Since the magnitudes F1 and F2 of the two forces are the same and equal to F, we have
F = √(F2 + F2 + 2 F F cosθ)
Squaring, F2 = F2 + F2 + 2 F2cosθ
This gives cosθ = – ½ so that θ = 120º
[The adjoining figure will be useful to understand how the two forces produce the resultant satisfying the given conditions].
(4) The resultant of two forces is 40 N and the smaller force, which has magnitude 30 N, is normal to the resultant. The larger force is
(a) 40 N
(b) 45 N
(c) 50 N
(d) 60 N
(e) 80 N
The forces and their resultant are shown in the adjoining figure.
Since the resultant R of magnitude 40 N is perpendicular to the forceF1 of 30 N, we have (from the right angled triangle),
F2 = √(302 + 402) = 50 N.
(5) Rain drops are falling vertically with a speed of 10 ms–1. A boy holding an umbrella runs southward with a speed of 5 ms–1. What is the direction in which he should hold his umbrella so that he will not get drenched?
(a) At an angle of tan–1(1/2) with the vertical, towards north
(b) At an angle of tan–1(1/2) with the vertical, towards south
(c) At an angle of tan–1(2) with the vertical, towards south
(d) At an angle of 30º with the vertical, towards north
(e)At an angle of 30º with the vertical, towards south
In the figure the velocities of the boy and rain drops are represented respectively by the vectors OA and OB.
The relative velocity of the rain drops with respect to the boy = (velocity of rain drops) – (velocity of boy).
The quantity on the right hand side is the vector sum of the velocity of rain drops and the negative of the velocity (reversed velocity shown by vector OC) of the boy. This is given by the vector OD which makes an angle θ with the vertical and is given by
tan θ = 5/10 = ½.
Therefore, the boy should hold his umbrella at an angle of tan–1(1/2) with the vertical, towards south.
The following multiple choice questions are specifically meant for AP Physics C aspirants:
(6) Suppose that the ground coincides with the XZ plane of a right handed Cartesian coordinate system and i, j, k are unit vectors along x, y, z directions respectively. An inclined plane of inclination 60º is placed on the ground and a force F = – 20 j N is applied at a point on the inclined plane. What is the component of this force parallel to the inclined plane?
(a) 20/√3 N
(b) 10 N
(c) 10√3 N
(d) 5√3 N
(e) Zero
The force F of magnitude 20 N is directed vertically downwards (along the negative y-direction).
With reference to the figure, you can easily see that the component of F parallel to the inclined plane is 20 sin 60º = 10√3 N.
(7) The magnitude of the area of a parallelogram formed by the vectors A = i + 2 j + 2 k metre and B = 2 i– 2 j + k metre as adjacent sides is (i, j, k are unit vectors)
(a) √78 m2
(b) 9 m2
(c) √91 m2
(d) 11 m2
(e) 12 m2
The magnitude of the area of the parallelogram is equal to the magnitude of the vector product (cross product) of the vectors A and B.
[Remember that area is a vector].
Area = A ×B = (i + 2 j + 2 k) × (2 i– 2 j + k)
Or, Area = –2 k –j – 4 k + 2i + 4 j + 4 i
= 6 i + 3 j – 6 k
The magnitude of area vector = √(62 + 32 + 62) = √81 = 9 m2
You will find similar useful questions (with solution) on vectors here.
is the question:
