Fluid Mechanics – Multiple Choice Practice Questions for AP Physics C and IIT-JEE

Questions served for AP Physics C exam. and IIT-JEE are generally not as simple as the questions served for other degree entrance exams. You require more practice with tougher questions to obtain higher scores in these exams. Here are a few multiple choice practice questions on fluid mechanics:

[There will be 5 options for AP Physics C exam., but 4 options only for IIT-JEE]

(1) A bowl has a small hole at the centre of its bottom. Water poured into the bowl drains through the hole and the height of water column at any instant t is H. The radius of the free surface of water is R (fig.) at the instant t. If the time rate of decrease of the height of water column is constant, how is H related to R?

(a) H α R^1/2

(b) H α R

(c) H α R²

(d) H α R³

(e) H α R⁴

The time rate of decrease of the height of water column is dH/dt and we have

dH/dt = constant, as given in the question.

Since the velocity of the water flowing out (velocity of efflux) through the hole is √(2gH), the rate of flow is av = a√(2gH) where a is the area of the hole.

Therefore, we have

a√(2gH) = πR²(dH/dt)

Since a, g, π and dH/dt are costants, H α R⁴ [Option (e)].

(2) A jar of uniform area of cross section A has a hole of area a at its bottom. If there is water column of height H₁ in the jar initially, what time is required for the height to become H₂?

(a) (A/a) √(4/g) (√H₁ – √H₂)

(b) (A/a) √(2/g) (√H₁ – √H₂)

(c) (A/a) (4/g) (H₁ –H₂)

(d) (A/a) √(1/g) (√H₁ – √H₂)

(e) (a/A) √(4/g) (√H₁ – √H₂)

Rate of flow of water through the hole is av where v is the efflux velocity given by

v =√(2gx) where x is the height of water column.

If the height of water column changes by dx (fig.) in a time dt, we have

– Adx (= avdt) = a√(2gx) dt

[The negative sign shows that x decreases].

Therefore, dt = – (A/a)[1/√(2g)] (dx/x^1/2)

Integrating between the limits x = H₁ and x = H₂ we obtain the required time t.

Thus t = (A/a)[1/√(2g)] _H1∫^H2 (–x^–1/2)dx

Or, t = (A/a)[1/√(2g)] _H2∫^H1 (x^–1/2)dx

This gives t = (A/a) √(2/g) (√H₁ – √H₂)

(3) A sphere of wax (density 900 kgm^–3) has a volume of 20 cm³. Iron nails are pierced into it so that it just gets submerged in water. If the volume of the iron nails is negligible compared to the volume of the sphere of wax, what is the mass of the iron nails in the sphere?

(a) 0.001 kg

(b) 0.002 kg

(c) 0.01 kg

(d) 0.02 kg

(e) 0.09 kg

Since the sphere of wax containing the iron nails gets just submerged in water, the mean density of the sphere must be equal to the density of water (1000 kgm^–3).

Mass of wax in the sphere is 20×10^–6×900 kg.

If m represents the mass of the nails, we have

(20×10^–6×900 + m) /(20×10^–6) = 1000

This gives m = 2×10^–3 kg = 0.002 kg.

You will find a few more multiple choice questions (with solution) in this section here.

AP Physics & Degree Entrance Kinematics- Questions on One Dimensional Motion

The following questions numbered 1, 2 and 3 are based on the velocity-time graph of a particle in one dimensional motion shown in the adjoining figure.

(1) The displacement of the particle during the first second of its motion is nearly

(a) 0.25 m

(b) 0.5 m

(c) 1 m

(d) 1.5 m

(e) 2 m

The area under the velocity-time curve gives the displacement. The portion of the graph for the interval from zero to 1 second is straight and the area under the curve is triangular and is equal to (½)×1×4 = 2 m.

(2) The average accelerations during the 3^rd second and 7^th second respectively are nearly

(a) 0.5 ms^–2 and 2 ms^–2

(b) 1 ms^–2 and –2 ms^–2

(c) 1 ms^–2 and 2 ms^–2

(d) 1 ms^–2 and –2 ms^–2

(e) zero and –2 ms^–2

The 3^rd second is the interval from 2 seconds to 3 seconds and during this time the velocity increases from 7 ms^–1 to 8 ms^–1. The acceleration is therefore (8 – 7)/1 = 1 ms^–2.

The 7^th second is the interval from 6 seconds to 7 seconds and during this time the velocity decreases from 8 ms^–1 to 6 ms^–1.

The acceleration is therefore (6 – 8)/1 = – 2 ms^–2. [Option (b)]

(3) Which one among the following acceleration–time graphs most closely represents the motion of the particle?

The acceleration is positive and uniform initially. Afterwards the acceleration becomes zero (since the velocity remains constant) for some time and then becomes negative (since the velocity goes on decreasing) and uniform. The curve shown in (b) therefore represents the motion of the particle.

Let us leave the velocity time graph here and consider a couple of different questions:

(4) Two boys running at uniform speeds v₁ and v₂ respectively along a straight line path in opposite directions get 9 m closer each second. While running along the same direction with their speeds reduced by 50%, they get 0.5m closer each second. The speeds v₁ and v₂ are respectively

(a) 6 ms^–1 and 3 ms^–1

(b) 5 ms^–1 and 4 ms^–1

(c) 5 ms^–1 and 4.5 ms^–1

(d) 5 ms^–1 and 3.5 ms^–1

(e) 5.5 ms^–1 and 3.5 ms^–1

This is a simple question involving relative velocity. Wile running along opposite directions, we have

v₁+ v₂ = 9

While running along the same direction, we have

v₁/2 – v₂/2 = 0.5, from which v₁– v₂ = 1

Solving the above equations, we obtain v₁ = 5 ms^–1 and v₂ = 4 ms^–1

(5) The rear end of a train running on a straight track with uniform acceleration has velocities 6 ms^–1 and 10 ms^–1 respectively when passing points A and B in its path. The velocity of the rear end midway between these points is approximately

(a) 7 ms^–1

(b) 7.5 ms^–1

(c) 8ms^–1

(d) 8.2 ms^–1

(e) 8.4 ms^–1

We have v² = u² + 2as where u and v are the initial ans final velocities respectively, a is the acceleration and s is the displacement.

If the distance between A and B is s, we have

10² = 6² + 2as from which 2as = 64

If v₁ is the velocity midway between A and B we have

v₁² = 6² + 2a(s/2) = 6² + 32 = 68

Therefore, v₁ = 8.2 ms^–1 nearly.

You will find more questions (with solution) on one dimensional motion and other sections at AP Physics Resources

AP Physics Exam Resources- Two Questions (MCQ) on Moment of Inertia

Multiple choice questions discussed on this site will be useful for entrance examinations for admission to various degree courses including professional courses. They will be suitable for those preparing for AP Physics Examination, as can be judged by working out the following two questions:

(1) Three circular discs of radii R, R and 2R are cut from a metallic sheet of uniform thickness and the smaller discs are placed symmetrically on the larger disc as shown in the figure. If the mass of a smaller disc is M, the moment of inertia of the system about an axis at right angles to the plane of the discs and passing through the centre of the larger disc is

(a) 5MR² (b) 7MR² (c) 9MR²

(d) 11MR² (e) 12MR²

The mass of the larger disc is 4M (since its radius is twice that of the smaller disc) and its moment of inertia is (4M)×(2R)²/2 = 8MR².

The moment of inertia of each smaller disc about the axis passing through the centre of the larger disc (as given by the parallel axis theorem) is MR²/2 + MR² = 3MR²/2.

Note that the moment of inertia is a scalar quantity. Therefore, the total moment of inertia of the system of three discs is 8MR² + 2×3MR²/2 = 11MR².

(2) A small body of regular shape made of iron rolls up with an initial velocity ‘v’ along an inclined plane. It reaches a maximum height of 7v²/10g where ‘g’ is the acceleration due to gravity. The body is a

(a) ring (b) disc (c) solid sphere

(d) hollow sphere (e) cylindrical rod

The initial kinetic energy of the body is ½ Mv² + ½ I ω² where M is its mass and I is its moment of inertia about its axis (of rolling). The first term is its translational kinetic energy and the second term is its rotational kinetic energy.

Since the entire kinetic energy is used in gaining gravitational potential energy, we have

½ Mv² + ½ I ω² = Mgh where ‘h’ is the maximum height reached.

Therefore, ½ Mv² + ½ I v²/R² = Mg×7v²/10g, from which

I = (2/5)MR².

The body is therefore a solid sphere.

Multiple Choice Questions on Newton’s Laws of Motion

You will have to apply Newton’s Laws of Motion in different branches of Physics, but you will find questions specifically meant for checking your understanding of these laws in AP Physics Examination, Graduate Record Examination (GRE) and Medical and Engineering Entrance Examinations.

Consider the following MCQ based on impulse:
The force ‘F’ acting on a particle of mass ‘m’ is indicated by a force-time graph (Fig.). The momentum received by the particle during the time from zero to 8 s is

(a) 24 Ns (b) 20Ns (c) 12Ns (d) 6Ns (e) zero

The area under the force-time graph gives the impulse imparted to the particle. Impulse is a vector quantity and so you must consider its sign while adding the areas. The impulse received from zero to 2 seconds is positive and is equal to the area of the triangle, which is 6 Ns. The impulse received during the time from 2 s to 4 s is the area of the rectangle, which is – 6 Ns. The impulse received during the time from 4 s to 8 s is the area of the larger rectangle, which is 12 Ns.

Hense the net impulse received during the time from zero to 8 seconds is 6 –6 +12 = 12 Ns.

Since the impulse is equal to the change of momentum, the correct option is (c).

Now consider the following question:

A boy caught a ball of mass 200g moving with a speed of 30 ms^–1. If the catching process be completed in 0.1 s, the force of impact exerted by the ball on the hands of the boy is

(a) 60 N (b) 40 N (c) 30 N (d) 20 N (e) 10 N

We have force F = dp/dt where ‘dp’ is the change in linear momentum during the time dt.

The initial momentum of the ball is 0.2×30 = 6 kgms^–1 and the final momentum is zero so that dp = 6.

Therefore, F = 6/0.1 = 60 N.

The following MCQ which appeared in Karnataka CET 2002 question paper is meant for checking your grasp of the law of conservation of linear momentum:

A projectile is moving at 20 ms^–1at its highest point, where it breaks into two equal parts due to an internal explosion. One part moves vertically up at 30 ms^–1 with respect to the ground. Then the other part will move at

(a) 30 ms^–1 (b) 50 ms^–1 (c) 10√3 ms^–1 (d) 20 ms^–1

If the mass of the projectile is ‘m’, the mass of each fragment after the explosion is m/2. The momentum of the part which moved upwards is (m/2)×30 =15m. This is shown as vector OA in the figure.

The momentum of the projectile at the highest point of its path is 20m and its direction is horizontal. This is shown as vector OC.

The momentum of the other part (after the explosion) is (m/2)×v where ‘v’ is its velocity. This is shown as vector OB in the figure.

Since the momentum is conserved, the total final momentum, which is the vector sum of the momentum vectors OA and OB must be equal to the initial momentum, represented by the vector OC.

Evidently, OB = √(OA² + OC²).

Or, mv/2 = √[(15m)² + (20m)²] = 25m.

This gives v = 50 ms^–1.

Magnetic Force on Moving Charge- Kerala Medical Entrance 2007 Question

The following MCQ appeared in Kerala Medical Entrance 2007 question paper:

A proton with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the proton is (Take mass of proton to be 1.6×10^–27 kg)

(a) 3×10^–12 N (b) 8×10^–10 N (c) 8×10^–12 N (d) 2×10^–10 N (e) 3×10^–10 N

The velocity ‘v’ of the proton is to be found first using the expression for kinetic energy E (in joule). Note that E = 2 MeV = 2×10⁶×1.6×10^–19 joule.

We have E = ½ mv² from which v = √(2E/m) = [2×2×10⁶×1.6×10^–19/(1.6×10^–27)]^1/2 = 2×10⁷ ms^–1.

Substituting this value of ‘v’ in the expression for magnetic force (F = qvB), we obtain

F = 1.6×10^–19×2×10⁷×2.5 = 8×10^–12 N.

Note: In the above question we did not take the relativistic increase of mass of the proton into consideration. Since the energy is 2 MeV only and the proton is is fairly heavy, the relativistic increase of mass will be about 0.2% only and you will obtain the velocity of the proton of 2 MeV energy as 1.995×10⁷ ms^–1. So, the magnetic force will be slightly reduced.

You should be aware of the relativistic increase in mass when you deal with questions like the above one, especially if you are preparing for GRE Physics Exam or AP Physics Exam in which you can expect questions of the type given below:

An electron with energy of 2 MeV enters a uniform magnetic field of 2.5 T normally. The magnetic force on the electron is nearly (Take the rest mass of electron to be 9.1×10^–31 kg)

(a) 1.17×10^–10 N (b) 8×10^–10 N (c) 3.35×10^–10 N (d) 8×10^–12 N (e) 3.14×10^–14 N

If you calculate the velocity of the electron without considering the relativistic increase in mass, as we did in the previous question, you will get (nearly) v = 8.39×10⁸ ms^–1 and the magnetic force F = 3.35×10^–10 N, nearly. But, the velocity is more than the velocity of light in free space and therefore is absurd. So, (c) is not the correct option.

If m₀ is the rest mass of the electron and ‘m’ is its mass while moving with kinetic energy E (= 2 MeV), we have

(m – m₀)c² = E, from which m = m₀ + E/c². The energy E is to be substituted in joule in this equation

Therefore, m = 9.1×10^–31 + (2×10⁶ ×1.6×10^–19)/(3 ×10⁸)² = 9.1×10^–31 +3.55×10^–30 = 4.46×10^–30 kg. Note that the mass of the electron has become nearly five times its rest mass.

But m = m₀/√(1 – v²/c²) so that v = c√[1 – (m₀/m)²] = 3×10⁸×√[1 – (9.1×10^–31 /4.46×10^–30)²] = 2.93×10⁸ ms^–1

The magnetic force on the electron is qvB = 1.6×10^–19×2.93×10⁸×2.5 = 1.17×10^–10 N.

AP (Advanced Placement) Physics Examinations

Many among the visitors of this site might have already noted that the posts here are useful for preparing for the AP (Advanced Placement) Physics Exams. Even though you find multiple choice questions here, you will definitely find the posts useful for answering the free-response section of the exams since the questions are answered with all necessary theoretical details. In many cases the discussions touch even minute details. This is done for helping students who are just above average. The needs of those who appear for the AP Physics Exams will be considered while discussing questions here. You may make use of the facility for comments for communications in this regard.