The following multiple choice questions involving friction were included in the AIPMT 2010 question paper:
(1) A block of mass m is in contact with the cart C as shown in the figure. The coefficient of static friction between the block and the cart is μ. The acceleration α of the cart that will prevent the block from falling statisfies
(1) α< g/μ
(2) α> mg/μ
(3) α> g/μm
(4) α≥ g/μ
When the cart moves forwards with acceleration α, the block of mass m presses against the vertical surface of the cart with a horizontal force of magnitude mα. Therefore, the normal force acting on the block is mαand the maximum frictional force trying to prevent the block from fallingis μmα. This must be greater than or equal to the weight mg of the block.
Therefore we have
μmα ≥ mg from which α≥ g/μ
(2) A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if
(1) r≥μg/ω2
(2)r=μgω2
(3) r< ω2/μg
(4) r≤μg/ω2
Let the mass of the coin be m. The coin will revolve with the record if the frictional force which supplies the centripetal force (required for the circular motion of the coin) is sufficient. Since the maximum value of the frictional force is μmg, we have
μmg ≥ mrω2
Thereforer≤μg/ω2.
You will find more questions (with solution) on friction on this site. One of the posts on friction ishere
Similar useful posts in this section can be seenhere
Today we will discuss three Multiple Correct Choice Type questions which appeared in the IIT-JEE 2009 question paper. Even though these questions are named ‘multiple correct choice’ type the number of correct choices can be one or more.These questions are simple, in line with the current trend, as you can see:
(1) Two metallic rings A and B, identical in shape and size but having different resistivities ρA and ρB, are kept on top of two identical solenoids as shown in the figure. When current I is switched on in both the solenoids in identical manner, the rings A and B jump to heights hA and hB respectively, withhA > hB. The possible relation(s) between their resistivities and their masses mA and mB is(are)
(A) ρA > ρB and mA = mB
(B) ρA < ρB and mA = mB
(C) ρA > ρB and mA > mB
(D) ρA < ρB and mA < mB
The voltages induced in the rings are equal. Since ring A jumps to greater height compared to ring B, the current in ring A must be greater. This means that the resistivity of ring A is less than that of ring B. SoρA < ρB.
Ring A can jump to a greater height if the mass of ring A is equal to or less than that of B. So the correct options are (B) and (D).
(2) A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then
(A) the intensity of the sound heard at the first resonance was more than that at the second resonance
(B)the prongs of the tuning fork were kept in a horizontal plane above the resonance tube
(C)the amplitude of vibration of the ends of the prongs is typically around 1 cm
(D)the length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength of the sound in air.
Option (A) is correct. (The second resonance will give sound of smaller intensity since the air column is longer and hence there is greater chance for the attenuation of the sound before and after the reflection at the water surface).
Option (B) is incorrect since the prongs are to be kept in a vertical plane to transfer energy efficiently into the resonance column.
Option (C) also is incorrect since the amplitude of vibration of the ends of the prongs is typically of the order of a millimetre only.
Option (D) is correct since the anti-node is obtained slightly above the end of the resonance tube and λ/4 = l + e where λ is the wave length of sound, l is the resonating length of the tube and e is the end correction.
(3) Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s)
(A) the angular momentum of the charge –q is constant
(B) the linear momentum of the charge –q is constant
(C) the angular velocity of the charge –q is constant
(D) the linear speed of the charge –q is constant.
The situation here is similar to that of a hydrogen atom (or, the motion of a planet around the sun) which is central field motion under an inverse square law force. The only correct option is (A) which high lights the constancy of angular momentum (Kepler’s law).
[Note that the torque is zero and hence the angular momentum is constant].
The following questions involving a spring-mass system are meant for checking whether you have a thorough understanding of the simple harmonic oscillations in which the restoring force is supplied by a spring:
(1) The frequency of vertical oscillations of a mass suspended at the end of a light spring is n. If the system is taken to a location where the acceleration due to gravity is reduced by 0.1%, the frequency of oscillation will be
(a) 1.01 n
(b) 0.99 n
(c) 1.001 n
(d) 0.999 n
(e) n
A spring-mass system (unlike the simple pendulum) does not require a gravitational force for oscillations since the restoring force required for oscillations is supplied by the elastic forces in the spring.
[Note that the period (T)of oscillations is given by T = 2π√(m/k) where m is the mass attached to the spring and k is the spring constant].
Therefore the change in ‘g’ does not affect the frequency and the correct option is (e).
(2) One end of a light spring is fixed to the ceiling and a mass M is suspended at the other end. When an additional mass m is attached to the mass M, the additional extension in the spring is e. The period of vertical oscillation of the spring-mass system now is
(a) 2π√[(M+m)e/mg]
(b) 2π√(me/mg)
(c) 2π√[(M+m) /mge]
(d) 2π√[me/(M+m)g]
(e)2π√(M/mge)
The period (T)of oscillations is given by T = 2π√[(M+m)/k] where k is the spring constant.
Since an additional weight mg attached to the spring produces an additional extension e, the spring constant k = mg/e.
Therefore, period of oscillations T = 2π√[(M+m)/(mg/e)] = 2π√[(M+m)e/mg], as given in option (a).
(3) The period of vertical oscillations of a mass M suspended using a light spring of spring constant k is T. The same spring is cut into three equal parts and they are used in parallel to suspend the mass M as shown in the adjoining figure. What is the new period of oscillations?
(a) T
(b) 3T
(c) 9T
(d) T/9
(e)T/3
The original period of oscillation T is given by
T = 2π√(M/k)
When the spring is cut into three equal parts, each piece has spring constant 3k.
[Since the length of each piece is reduced by a factor three, the extension for a given applied force will be reduced by a factor three so that the spring constant (which is the ratio of force to extension) will become three times].
Since the three pieces are connected in parallel the effective spring constant of the combination is 3k+3k+3k = 9k. The new period of oscillations T1is given by
T1= 2π√(M/9k) = T/3 since T = 2π√(M/k)
You will find some multiple choice questions with solution in this sectionhere.
Today we will discuss two multiple choice questions from modern physics. These questions are from the section ‘electrons and photons’ and they were included in the EAMCET (Medical) 2009 question paper:
(1) In Millikan’s oil drop experiment, a charged oil drop of mass 3.2×10–14 kg is held stationary between two parallel plates 6 mm apart, by applying a potential difference of1200 V between them. How many electrons does the oil drop carry?
(1) 7
(2) 8
(3) 9
(4) 10
The electric field ‘E’ between the plates is V/d where V is the potential difference and d is the separation between the plates.
ThereforeE = 1200/(6×10–3) = 2×105 volt per metre.
The electric force on the drop is Eq where ‘q’ is the charge on the drop. Since the drop is stationary, the electric force balances the weight mg of the drop.
Since the electronic charge is 1.6×10–19 coulomb, the number of excess electrons carried by the drop is (1.6×10–18) /(1.6×10–19) = 10.
(2) Electrons accelerated by a potential of ‘V’ volts strike a target material to produce ‘continuous X-rays’. Ratio between the de Broglie wave length of the electrons striking the target and the shortest wave length of the ‘continuous X-rays’ emitted is
(1) h/√(2Vem)
(2) (1/c) √(2m/Ve)
(3) (1/c) √(Ve/2m)
(4) he/√(Ve/2m)
The de Broglie wave length λ of the electron is given by
λ = h/p where h is Planck’s constant and p is the momentum of the electron.
The shortest wave length λmin of the ‘continuous X-rays’ produced by the striking electrons is given by
hc/λmin = Ve where c is the speed of light in free space and e is the electronic charge.
[Note that we have equated the entire energy of the electron to the energy of the X-ray photon for obtaining minimum wave length (or, the maximum frequency νmax):
eV = hνmax = hc/λmin].
Therefore, λmin = hc/Ve.
The ratio of wave lengths required in the problem is
λ /λmin = (h/p)/(hc/Ve ) = Ve/pc………..(i)
The kinetic energy of the electron is eV and the momentum p of the electron is related to its kinetic energy as
The following question appeared in Kerala Engineering Entrance 2009 Examination question paper:
The truth table for the following logic circuit is
The top AND gate gets inputs (NOT A) and B. The bottom AND gate gets inputs A and (NOT B). The outputs of the two AND gates are respectively [(NOT A) AND B] and [A AND (NOT B)]. These are the inputs for the final OR gate. The output of the final OR gate is [(NOT A) AND B] OR [A AND (NOT B)]. This is the Boolean expression for the XOR gate (exclusive OR gate, also termed EXOR gate). The truth table for the XOR gate is given in option (A).
Even if you don’t know the XOR gate and its truth table, you can substitute values 0 and 1 for the inputs A and B and find the output in each trial. You can easily arrive at the answer given in option (A).
The following question appeared in Kerala Medical Entrance 2009 Examination question paper:
If the two inputs of a NAND gate are shorted, the gate is equivalent to
(a) XOR
(b) OR
(c) NOR
(d) NOT
(e) AND
If both inputs of a two input NAND gate are shorted, both inputs get the same signal.Because of the NOT operation at the output side of the NAND, the output is the complement of the input. The circuit therefore behaves as a NOT gate [Option (d)].
I give below 4 questions which appeared in Karnataka Common Entrance Test (CET) question papers of the years 2009, 2008, 2007 and 2006 respectively:
(1) In the following combination of logic gates, the outputs A, B and C are respectively
(1) 0, 1, 0
(2) 1, 1, 0
(3) 1, 0, 1
(4) 0, 1, 1
In circuit (A), the output of the input NAND gate is 0 since both inputs are 1. After inversion it becomes 1 and the final OR gate produces output 1.
In circuit (B), the output NAND gate gets 1 and 0 as its inputs and hence the final output is 1.
In the circuit (C), the NOR gate at the input side supplies 0 as one input to the AND gate at the output side. Hence the final output is 0.
Therefore, the correct option is (2).
(2) To get an output Y = 1 from the circuit shown, the inputs A, B and C must be respectively
(1) 0, 1, 0
(2) 1, 0, 0
(3) 1, 0, 1
(4) 1, 1, 0
In all cases the OR gate will provide a high (level 1) input to the AND gate at the output side. The AND gate should have both inputs at level 1 in order to have its output at 1. This is possible only if input C is 1. So the correct option is (3).
(3) The truth table given below is for
(A and B are the inputs, yis the output)
(1) NAND
(2) XOR
(3) AND
(4) NOR
The output y is the complement of AND operation and hence the answer is NAND [Option (1)].
(4) Identify the logic operation performrd by the cicuit given below
(1) NOT
(2) AND
(3) OR
(4) NAND
If you know De Morgan’s theorem, you will immediately obtain the answer as AND. The NOR gate at the output side gets inputs which are complement of A and complement of B. According to De Morgan’s theorem,(NOT A OR NOT B) is the same as [NOT (A AND B)]. Since the output gate is not an OR gate, but a NOR gate, the output is (A AND B).
Therefore, given circuit performs AND operation [Option (2)].
[If you do not know DeMorgan’s theorem, you may try substituting values 0 and 1 for A and B to obtain the answer].
Here are two questions from wave optics. The first question is meant for testing your understanding of coherent sources and interference while the second one is meant for testing your knowledge of the interference pattern produced by Young’s double slit.
(1) Two coherent sources emitting light of wave length λ and amplitude A produce an interference pattern on a screen. The maximum intensity of the interference pattern is I. If the light sources are not coherent but the wave length and amplitude are λ and A respectively, the maximum intensity on the screen will be
(a) I/4
(b) I/2
(c) I
(d) √I
(e) 2√I
At the interference maximum the amplitudes of the light waves get added. Therefore, the resultant amplitude at the interference maximum is A + A = 2A.
The intensity is directly proportional to the square of the amplitude so that we have
Iα 4A2
Or, I = k×4A2……………(i)
where k is the constant of proportionality
Since the amplitude of each light wave is A, the intensity (I0) of light produced by each source is given by
I0= k×A2
This gives I0 = I/4 from (i).
Each source will produce uniform intensity I0 everywhere on the screen and the resultant intensity everywhere will be 2I0 (since there are two sources) and we have
2I0 = 2×I/4 = I/2
The correct option is (b).
(2) The angular width of the interference fringes obtained in a double slit experiment using light of wave length λ is found to be θ. If the entire experimental arrangement is immersed in water having refractive index 4/3, the angular fringe width will be
(a) θ/4
(b) θ/3
(c) θ
(d) 4θ/3
(e) 3θ/4
The angular fringe width is λ/d where d is the separation between the slits. Therefore we have
θ =λ/d.
When the arrangement is submerged in water of refractive index n, the wave length of light is reduced to λ' given by
λ'= λ/n = 3λ/4 since n = 4/3
The angular fringe width now becomes θ'=λ'/d = 3λ/4d = 3θ/4.
(1) A block of mass m is in contact with the cart C as shown in the figure. The coefficient of static friction between the block and the cart is μ. The acceleration α of the cart that will prevent the block from falling statisfies
