The following questions appeared in Karnataka CET 2010 question paper:
(1) A battery of e.m.f. E has internal resistance ‘r’. A variable resistance R is connected to the terminals of the battery. A current I is drawn from the battery. V is the terminal P.D. If R alone is graually reduced to zero, which of the following best describes I and V?
(a) I approaches zero, V approaches E
(b) I approaches E/r, V approaches zero
(c) I approaches E/r, V approaches E
(d) I approaches infinity, V approaches E
The current I is given by
I = E/(r+R)
When R is reuced to zero, I = E/r
Since the terminal P.D. is givenby V = ER/(r+R), we have V = 0 when R = 0.
The correct option is (b).
(2) Three voltmeters A, B and C having resistances R, 1.5 R and 3 R respectively are used in a circuit as shown. When a P.D. is applied between X and Y, the readings of the voltmeters are V1, V2 andV3 respectively. Then
(a) V1 = V2 = V3
(b) V1 <>2 = V3
(c) V1 > V2 > V3
(d) V1 > V2 = V3
The effective resistance of the voltmeters B and C in parallel is 1.5R× 3R/(1.5R+3R) = R.
Therefore, the same P.D. exists across A and the parallel combination of B and C and thevoltmeters show the same reading [Option (a)].
Today we will discuss two questions on surface tension which appeared in EAMCET 2010 question papers.
The following multiple choice question appeared in EAMCET 2010 (Engineering) question paper:
The excess pressure inside a spherical soap bubble of radius 1 cm is balance by a column of oil (Sp. gravity 0.8) 2 mm high. The surface tension of oil is
(1) 3.92 N/m
(2) 0.0392 N/m
(3) 0. 392 N/m
(4) 0.00392 N/m
The excess pressure ∆P insie a spherical bubble of radius r is given by
∆P = 4 T/r where T is the surface tension.
The pressure p exerted by a liquid column of height h is given by
p = hρg where ρ is the density of the liquid.
Therefore we have
4 T/r = hρg from which T= rhρg/4
Substituting for known values, T = (1×10–2×2×10–3×0.8×103×10)/4 = 0.004 N/m [Option(4)].
[If you substitute g = 9.8 ms–2 instead of 10 ms–2 (as we did above for convenience), you will get the answer as 0.00392 N/m].
Here is the question which appeared in EAMCET 2010 (Agriculture and Medicine) question paper:
A spherical liquid drop of diameter D breaks up to n identical spherical drops. If the surface tension of the drop is ‘σ’, the change in energy in this process is
(1) πσD2(n1/3 – 1)
(2) πσD2(n2/3 – 1)
(3) πσD2(n – 1)
(4) πσD2(n4/3 – 1)
The surface energy (E1) of a drop of radius R is given by
E1 =4πR2σ
Or E1 = πD2σ where D is the diameter of the drop.
When a drop of radius R breaks into n identical droplets, the radius r of each droplet is given by (on equating the volumes)
(4/3) πR3 = n×(4/3) πr3
Therefore, r = R/n1/3
The total surface energy (E2) of all the n droplets is given by
Any intelligent fool can make things bigger and more complex…
It takes a touch of genius and a lot of courage to move in the opposite direction.
– Albert Einstein
Here are two questions (MCQ) on friction, which may easily tempt you to answer incorrectly:
(1) Three identical wooden blocks A, B and C, each of mass 1 kg, are placed on a rough horizontal surface as shown. A horizontal force F acts on the block A and the system of masses moves with an acceleration of 2 ms–2.If the coefficient of friction between the blocks and the horizontal surface is 0.3 and the gravitational acceleration is 10 ms–2, the force exerted by block A on block B is
(a) 2 N
(b) 4 N
(c) 6 N
(d) 8N
(e) 10 N
The block A pushes the blocks B and C and produces an acceleration in them. But when the blocks A and B move, an opposing frictional force Ff is produced. The opposing frictional force is given by
Ff = μmg where μ is the coefficient of friction, m is the total mass of blocks A and B and g is the gravitational acceleration.
Therefore, Ff =0.3×2×10 = 6 N.
The resultant force acting on block B is FB– Ff where FB is the force exerted by block A on block B. Therefore, we have
FB– Ff = ma = 2×2 since the resultant force moves blocks A and B with an acceleration of 2 ms–2.
Therefore, FB– 6 = 4 so that FB = 10 N.
(2) A block of mass Mslides down with uniform velocity from the top A to the bottom B of a rough inclined plane of angle θ and height H. If the coefficient of kinetic friction between the block and the plane is μk,the work done against friction during the motion of the block is
(a) MgH sin θ
(b) μkMgH sin θ
(c) μkMgH/ sin θ
(d) μkMgH
(e)MgH
When the block reaches the bottom of the incline, the gravitational potential energy of the block decreases by MgH. If the plane were smooth, the kinetic energy of the block would have increased by an equal amount. Since the kinetic energy of the block remains unchanged, it follows that the loss in the gravitational potential energy is equal to thework done against friction [Option (e)].
Today we will discuss two multiple choice questions on optics, which were included in the IIT-JEE 2010 question paper. The first question is single correct choice type mcq where as the second one is multiple correct choice type mcq.
(1) A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is
(A) virtual and at a distance of 16 cm from the mirror
(B) real and at distance of 16 cm from the mirror
(C) virtual and at a distance of 20 cm from the mirror
(D) real and at a distance of 20 cm from the mirror
The lens has focal length f = 15 cm and the object O is placed at distance 30 cm which is equal to 2f. Therefore a real image I1 will be formed at distance 2f (= 30 cm) on the other side of the lens, if the mirror is absent.
Since the image I1 is at 20 cm from the plane mirror, a real image I2 would be formed (by the reflection of the rays at the plane mirror) at distance 20 cm from the mirror. But the returning rays are further refracted by the lens and the final image is formed at I3. The image I2 serves as the object for the lens for the formation of the final image I3.
As is clear from the figure, the object distance u3 for the formation of the final image is 10 cm. Therefore we have
1/v3– 1/10 = 1/15
[We have used the law of distances 1/v– 1/u = 1/f, in accordance with the Cartesian sign convention]
Therefore, 1/v3 = 1/15 + 1/10
This gives v3 = 6 cm.
The image is real and its distance from the plane mirror is (6+10) cm = 16 cm [Option (b)].
(2) A ray OP of monochromatic light is incident on the face AB of prism ABCD near the vertex B at an incident angle of 60º (see figure). If the refractive index of the material of the prism is √3 , which of the following is (are) correct ?
(A) The ray gets totally internally reflected at face CD
(B) The ray comes out through face AD
(C) The angle between the incident ray and the emergent ray is 90º
(D) The angle between the incident ray and the emergent ray is 120º
When the angle of incidence (at P) is 60º, the angle of refraction r is 30º since the refractive index n of the material of the prism is given by
n = sin i/sin r
Or, √3 = sin 60º/sin r so that sin r = (sin 60º)/√3 = ½ from which r = 30º
From the quadrilateral PABQ (fig.) it follows that angle PQB = 45º. Therefore the angle of incidence of the ray at the point Q is 45º. But this is greater than the critical angle for the interface.
[If θc is the critical angle for the interface, we have 1/sin θc = n =√3so thatsin θc = 1/√3. Since sin45º = 1/√2 it follows that 45º > θc]
The ray incident at Q is therefore totally reflected and it is incident at an angle of 30º at the point R. The angle of refraction at R is 60º and hence the angle between the incident ray and the emergent ray is 90º.
Options A, B and C are therefore correct.
[If you know the action of the prism commonly used in the constant deviation spectrograph, you will be able to answer this question immediately].
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., but4 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) Hα R1/2
(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/x1/2)
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 getssubmerged 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.
Today we will discuss three multiple choice questions which appeared in IIT-JEE 2010 question paper:
(1)Consider a thin square sheet of side L and thickness t, made of a material of resistivity ρ. The resistance between two opposite faces, shown by the shaded areas in the figure is
(A) directly proportional to L
(B) directly proportional to t
(C) independent of L
(D) independent of t
We have R = ρL/A where A is the cross section area which is equal to L t.
Therefore, R = ρL/Lt = ρ/t, which is independent of L [Option (C)].
(2) Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with increase in temperature. If at room temperature, 100 W, 60 W and 40 W bulbs have filament resistances R100, R60 and R40 respectively, the relation between these resistances is :
(A) 1/R100 = 1/R40 + 1/R60
(B) R100 = R40 + R60
(C) R100 > R60 > R40
(D) 1/R100 > 1/R60 > 1/R40
The power P dissipated in a resistance R is given by
P = V2/R where V is the voltage applied across the resistance.
This shows that the resistance of the filament is smaller if the wattage of the incandescent lamp is greater. In other words, the reciprocal of the filament resistance is greater if the wattage of the incandescent lamp is greater. This is true at all temperatures including room temperature. Therefore we have
1/R100 > 1/R60 > 1/R40
(3)To verify Ohm's law, a student is provided with a test resistor RT, a high resistance R1, a small resistance R2, two identical galvanometers G1 and G2, and a variable voltage ource V. The correct circuit to carry out the experiment is
To verify Ohm’s law, the current through RT is to be measured by connecting an ammeter inseries with it and the voltage across RT is to be measured by connecting a voltmeter inparallel with it. One galvanometer is to be converted into an ammeter by connecting the small resistance R2 across it. The other galvanometer is to be converted into a voltmeter by connecting the high resistance R1 in series with it. So the correct circuit to carry out the experiment is the one shown in option (C).
You will find some multiple choice questions (with solution) on direct current circuits here.
