Questions on Electrostatics

The following question which appeared in IIT 1997 Entrance test paper checks whether you have a thorough understanding of the basic relation between the electric field and potential:
A non-conducting ring of radius 0.5m carries a total charge of 1.11×10^–10 C distributed non uniformly on its circumference producing an electric field E everywhere in space. The value of the line integral ∫ –E.dl between the limits l = ∞ to l = 0 ( l = 0 being the centre of the ring) in volts is
(a) +2 (b) –1 (c) –8 (d) zero
The line integral ∫ –E.dl between the limits l = ∞ and l = 0 gives the work done in bringing unit positive charge from infinity to the centre of the ring and therefore is equal to the electric potential at the centre of the ring. If the total charge on the ring is Q coulomb, the potential at the centre is V = (1/4πε₀)×Q/r where ‘r’ is the radius of the ring. Therefore, V = 9×10⁹ ×1.11×10^–10/0.5 = 2, very nearly. So, the correct option is (a).
Now, consider the following MCQ:
A dielectric slab (dielectric constant = K) of thickness ‘t’ is placed between the plates of a parallel plate air capacitor. If the capacitance of the capacitor is to be restored to the original value, the separation between the plates is to be increased by
(a) kt (b) k/t (c) t/k (d) t –(k/t) (e) t –(t/k)
When a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space are respectively q/ε₀A and q/Kε₀A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε₀A)(d–t)+(q/Kε₀A)t = (q/ε₀A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε₀A/[d–t+(t/K)] = ε₀A/[d–(t – t/K)].
Since the capacitance of the capacitor with air filling the entire space between the plates is ε₀A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).
The following MCQ appeared in Kerala Engineering Entrance 2003 question paper:
A parallel plate capacitor has a capacitance of 100 pF when the plates of the capacitor are separated by a distance of ‘t’. Then a metallic foil of thickness t/3 is introduced between the plates. The capacitance will then become
(a) 100 pF (b) (3/2)100 pF (c) (2/3)100 pF (d) (1/3)100 pF (e) (1/2)100 pF
As shown in the previous discussion, the capacitance of a parallel plate capacitor with a dielectric slab of thickness ‘t’ between the plates separated by a distance ‘d’ is given by,C = ε₀A/[d–t+(t/K)]. In the present problem, ‘d’ is to be replaced by ‘t’ and ‘t’ is to be replaced by t/3. Further, the dielectric constant K is to be replaced by ∞ since the dielectric constant of a conductor is infinite. The capacitance therefore becomes ε₀A/(t–t/3) = (3/2) ε₀A/t = (3/2)100 pF (since the original capacitance with air alone as the dielectric is ε₀A/t = 100 pF).

You will find more multiple choice questions with solution at physicsplus: Questions on Electrostatics

Floating Bodies - Multiple Choice Questions

The weight of a floating body is equal to the weight of the displaced liquid. Questions based on this law of floatation can often be found in Medical and Engineering entrance test papers. Consider the following M.C.Q.:
A toy boat containing a piece of ice is floating in kerosene contained in a beaker. If the piece of ice is gently transferred to kerosene in the beaker, the level of kerosene in the beaker is
(a) lowered (b) raised (c) unchanged (d) first lowered and then raised (e) first raised and then lowered.
You should note that ice is denser than kerosene. So, on transferring the piece of ice in the toy boat to kerosene, it sinks, displacing a volume of kerosene equal to its own volume. When the piece of ice was in the toy boat, it could displace a greater volume of kerosene since the displaced kerosene should have the weight of the piece of ice. The level of kerosene is therefore lowered on transferring the ice to kerosene. The correct option is (a). If the ice melts without change of temperature, the correct option will still be (a), since the water so formed is certainly denser than kerosene and will sink, displacing a smaller volume of kerosene..
If you had a piece of iron instead of the piece of ice in the boat, the correct option would still be (a). But if you had a piece of cork instead, the level of kerosene will remain unchanged on transferring it to kerosene, since the density of cork is less than that of kerosene and it will float when overboard. (Inside the boat also it is a floating body and it will displace the same quantity of kerosene).
Now, consider the following question:
A toy boat containing an ice cube is floating in water contained in a beaker. The level of water in the beaker is noted. When the ice cube in the boat is gently transferred to water in the beaker, it melts without change of temperature. Then the level of water in the beaker is
(a) lowered (b) raised (c) unchanged (d) first raised and then lowered (e) first lowered and then raised.
The correct option here is (c) because the ice cube, while floating along with the boat, can displace a volume of water that has its own weight. On melting, the water produced will have the same volume and hence there is no change in the water level.
The boat in this problem is just a minor distraction. The answer is unchanged even if there is no boat and the ice cube is floating in water.
Now, suppose there is an iron nail on an ice cube floating in water contained in a beaker. If the ice melts without change of temperature, the level of water in the beaker will be lowered since the iron nail while floating along with ice can displace a greater volume of water.
A wooden log (density 700 kgm^-3) of mass 2100 kg floats in water. How much weight should be placed on it to make it just sink?
(a) 4800 kg (b) 2100 kg (c) 900 kg (d) 700 kg (e) 600 kg
The volume of the wooden log = 2100/700 =3 m³. While floating, the wooden log displaces water having weight 2100kg. Since the density of water is 1000 kgm^-3, the volume of the displaced water is 2100/1000 = 2.1 m³. The additional volume of water to be displaced by the wooden log on making it just sink is 3 – 2.1 = 0.9 m³. Weight of this additional volume of water = 0.9×1000 = 900 kg. This is the weight to be placed on the wooden log to make it just sink.

Multiple Choice Questions on Simple harmonic Motion

You will definitely find questions on simple harmonic motion in Medical and Engineering Entrance test papers as well as in GRE (Physics) question papers. You can find Multiple Choice Questions on Simple harmonic Motion with solution at physicsplus: Multiple Choice Questions on Simple Harmonic Motion . Additional questions with solution can be found here.

Multiple Choice Questions on One Dimensional Motion

Here is a question in kinematics which is a popular one and therefore requiring your attention:
The two ends of a train running with a constant acceleration passes a certain point with velocities v₁ and v₂. The velocity with which the middle point of the train passes the point is
(a) (v₁+v₂)/2 (b) √(v₁ ² +v₂²) (c) (v₁²+v₂²)/2 (d) (v₁+v₂)/√2

(e) √[(v₁²+v₂²)/2]
If ‘s’ is the length of the train, the velocity of the train changes from v₁ to v₂ when it moves through the distance ‘s’. Therefore we have,
v₂²- v₁² = 2as from which a = (v₂² - v₁² )/2s
If’ ’v’ is the velocity with which the mid point of the train passes the reference point, we have v² = v₁² +2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v = √[(v₁² +v₂ ² )/2].
The following question appeared in the Kerala Engineering Entrance Test paper of 2002:
A body dropped from a height ‘h’ with an initial velocity zero reaches the ground with a velocity 3km/hour. Another body of the same mass is dropped from the same height ‘h’ with an initial velocity 4km/hour. It will reach the ground with a velocity
(a) 3km/hour (b) 4km/hour (c) 5km/hour (d) 12km/hour (e) 8km/hour
We have, v² = u² + 2as with usual notations.
Therefore, 3² = 0 + 2gh for the first case and
v² = 4² + 2gh for the second case. These two equations yield the value v= 5km/hour. Note that we did not convert the velocities into m/s since the answer is required in km/hour. [Note that the answer is independent of the masses of the bodies].
Now consider the following simple question:
A particle starting from rest travels with uniform acceleration for t₁ seconds and then travels (continuously) with uniform retardation and comes to rest in another t₂ seconds. If the total distance traveled is ‘s’, the maximum velocity attained during the motion is
(a) s/(t₁ – t₂) (b) s/(t₁ + t₂) (c) 2s/(t₁ – t₂) (d) 2s/(t₁ + t₂) (e) s/t₁ + s/t₂
Since the acceleration and retardation are uniform, this can be easily solved using the concept of average velocity. If ‘v’ is the maximum velocity attained, the average velocity during the acceleration part as well as the deceleration part is v/2.
Therefore, s = (v/2)t₁+ (v/2)t₂ = (v/2)(t₁+t₂) from which v = 2s/(t₁+t₂).