Problem Based Learning in Basic Physics - VII
A. K .Mody
C-14 Beverly Hills, Lam Road, Devlali, Nashik 422401, Maharashtra
H. C. Pradhan
HBCSE, TIFR, V. N. Purav Marg,
Mankhurd
Mumbai – 400 088
[In this article- seventh in the series of articles we
present problems for a problem based learning course from the area of material
properties. We present the learning objectives in this area of basic physics
and what each problem tries to achieve with its solution. ]
In this article, seventh in the series of Problem Based Learning in Basic Physics, we present problems on material properties. Methodology and philosophy of selecting these problems are already discussed. (Pradhan 2009, Mody 2011)
To review methodology in brief, we
note here that this PBL (Problem Based Learning) starts after students have
been introduced to formal structure of Physics. Ideally
students would attempt only main problem. If they find it difficult, then
depending upon their area of difficulty, right auxiliary problem have to be
introduced by teacher who is expected to be a constructivist facilitator.
Teacher may choose as per his/her requirement or may construct questions on the
spot to guide student to right idea and method.
Material Properties: Elasticity of solids and fluids
as well as mechanics of fluids
Learning
objectives:
1. To
understand the role that property of elasticity of solid, liquid and gases plays
within material according to Hooke’s law along with laws of physics and how it
works in known situations.
2. To
understand behavior of fluids according to Archimedes principle, equation of
continuity and Bernoulli’s principle.
Problems:
1. A sphere of mass 1 kg is suspended at
the end of 2 m long steel wire whose other end is fixed to the ceiling. The
wire has a cross-sectional area of 1 mm2. The sphere is raised by
some height and then dropped to give sudden jerk to the wire. Find the maximum
height to which the ball can be raised so that wire doesn’t break. [Given: Ysteel
= 2×1011 N/m2 and breaking stress for steel =
1.1×109 N/m2]
Tasks
involved in this problem are:
a.
To relate spring
constant with elastic modulus.
b.
To apply conservation
of energy to the wire to find its extension assuming the model of elastic
spring.
c.
To calculate stress
based on extension of the wire.
d.
To estimate maximum
height that will not allow stress to exceed is breaking limit.
Fig.
P(2)
2.
A
rod of length 1.05 m having negligible mass is supported at its ends by two
wires, A of steel and B of aluminium of equal lengths as shown in figure. The
cross-section areas of wires A and B are 1.0 mm2 and 2.0 mm2
respectively. At what point along the rod should a mass m be suspended
in order to produce (a) equal stress and (b) equal strains in bith the wires.[Given:
Ysteel = 20×1010 N/m2 Yaluminium
= 7×1010 N/m2]
[NCERT
XI]
Tasks
involved in this problem are:
a. To
apply condition of equilibrium under given situation.
b. To
calculate position where mass to be hanged to meet the required condition.
3.
A
wire of density 9 gm/cm3 is stretched between two clamps 100 cm
apart, while subjected to an extension of 0.05 cm. What is the lowest frequency
of transverse vibrations in the wire, assuming Young’s modulus of the material
to be 9×1011 dyne/cm2. [JEE 1975]
Tasks involved
in this problem are:
a. To calculate tension due to stretching of
wire.
b.
To calculate fundamental frequency of vibration based on this tension.
Fig.
P(4)
4.
A
rail track made of steel having length 10 m is clamped on a railway line at its
two ends. On a summer day due to rise in temperature by 20oC, it is
deformed as shown in the figure. Find x, (displacement of the centre) if
αsteel = 1.2×10 – 5 /oC and Ysteel
= 20×1010 N/m2]
[NCERT EP XI]
Tasks
involved in this problem are:
a. To
calculate increase in length in accordance with property of thermal expansion.
b. To
find geometric deformation.
Fluids:
5. A
piece of brass (alloy of copper and zinc) weighs 12.9 g in air. When completely
immersed in water it weighs 11.3 g. What is the mass of copper contained in the
alloy? Specific gravity of copper and zinc, are 8.9 and 7.1 respectively. [JEE
1966]
[Note:
This is the problem similar to what Archimedes must have faced when he
was assigned the task to estimate gold in the crown.]
Tasks
involved in this problem are:
a. To
frame the equations for total mass and loss of weight.
b. To
realize that it is volume of the body that decides the weight of the liquid
displaced and hence loss of weight.
c. Solve
the two equations for volume of one of
the constituent and hence estimate the masses.
[Here auxiliary
problems are given to assist in realizing formation of simple equation for loss
of mass and how it relates to volume of the object.]
Auxilliary
Problems:
A.
What percentage of
volume of ice remains submerged while it is floating on the water surface? Ice
has specific gravity is 0.91.
B.
A piece of copper
having an internal cavity weighs 264 g in air and 221 g in water. Find the
volume of the cavity. Density of copper = 8.8 g cm – 3. [JEE
1963]
6. A
boat carrying number of large stones is floating in a water tank that is about
to overflow. If the stones are unloaded into the water tank, what will happen
to water level in water tank? Will water in the tank over flow? Give scientific
explanation based on Archimedes principle.
Tasks involved
in this problem are:
This
is not necessarily a mathematical problem. However one can write conditions to
arrive at the conclusion. One can also argue qualitatively.
7.
A cylindrical vessel of
height 500 mm has an orifice (small hole) at its bottom. The orifice is
initially closed and water is filled in it up to height H. Now the top
is completely sealed with a cap and the orifice at the bottom is opened. Some
water comes out from the orifice and the water level in the vessel becomes
steady with height of water column being 200 mm. Find the fall in the height
(in mm) of water level due to opening of the orifice. [Take atmospheric
pressure = 1.0×105 N/m2, density of water = 1000 kg/m3
and g = 10 m/s2. Neglect any effect of surface tension.] [JEE
2009]
Tasks
involved in this problem are:
a. To
realize and write equation for the condition that has to be satisfied once the
orifice is opened for liquid to stop flowing out.
b. To
understand the thermodynamic condition of air trapped above the water.
c. To
solve the two conditions to estimate the loss of height of water.
Fig.
P(8)
8. Water
in a cylinder is being rotated at constant angular speed ω about the
axis of the cylinder (see the figure). Ignore effect of surface tension and
find the equation of the surface of water. Can you identify the surface?
Tasks
involved in this problem are:
a.
To find the net force direction on a mass
element on the liquid surface so that surface assumes position perpendicular to
the force.
b.
Geometrically finding
the angle to arrive at equation of the surface.
Fig.
P(9)
9.
A non-viscous liquid of
constant density 1000 kg/m3 flows in a streamline motion along a
tube of variable cross-section. The tube is kept inclined in the vertical plane
as shown in the figure. The area of cross-section of the tube at two points P
and Q at heights of 2 metre and 5
metre are respectively 4×10 – 3m2 and 8×10 – 3m2.
The velocity of the liquid at point P is 1
m/s. Find the work done per unit volume by the pressure and the gravity forces
as the fluid flows from point P to Q. [JEE 1997]
Tasks
involved in this problem are:
a. Apply
equation of continuity to estimate speed of the fluid at the other end.
b. Calculate
pressure difference at the two ends
using Bernoulli’s principle.
c. To
estimate the work/energy accordingly.
Fig.
P(10)
10.
A uniform wire having
mass per unit length λ is placed over a liquid surface. The wire causes
the liquid to depress by y (y << a) as shown in the figure.
Find the surface tension of the liquid. Neglect end effect. [JEE 2004]
Tasks
involved in this problem are:
a. To
understand using free body diagram as how does the force of surface tension
support this object to float on the liquid surface.
b. To
understand how and why small angle approximation have to be applied to solve
the equations obtained in part ‘a’ to get the expression for surface
tension.
Fig.
P(11)
11.
A soap bubble having
surface tension T and radius R is formed on a ring of radius b (b << R).
Air is blown inside tube with velocity v as shown. The air molecules collide perpendicularly with the wall of
the bubble and stops. Calculate the radius at which the bubble separates from
the ring.
[JEE 2003]
Tasks
involved in this problem are:
a.
To apply Bernoulli’s
principle inside and outside the bubble.
b.
To use excess pressure
inside the bubble concept to understand why bubble would separate from the
ring. [Remember this is a popular toy based problem which children play with.]
12.
Eight droplets of
mercury, each of radius 1 mm, coalesce into a single drop. Find (i) the radius
of the single drop formed (assuming all the droplets and the drop to be
spherical in shape), (ii) the change in the surface energy of the mercury drop,
(iii) the change in the temperature of the mercury. [Surface tension of mercury
= 0.465 J/m2, Density of
mercury = 13.6×103 kg/m3, Specific heat of mercury = 140
J Kg - 1 K - 1] [Ans:
(i) 2 mm (ii) 2.337×10 – 5 J
(iii) 3.663×10 – 4 K]
Tasks
involved in this problem are:
a.
To understand that
volume /mass of the liquid drop/s remain same when they combine or breakup.
b.
To understand need for
energy in breaking or release of energy when drops combine.
c.
To realize what happens
to excess (surface) energy in accordance with energy conservation principle.
Solutions:
1. Let
the sphere be raised to height h. When it reaches the bottom, the
gravitational P.E. at the bottom point gets converted in to elastic P.E. due to
extension of the wire.
i.e., mgh = ½ kx2 where x
is the extension of the wire. [Here
from Solution of P(2)
above.]
This strain causes
. To prevent breaking this has to be less than breaking
stress SB. Which gives the condition
.
Fig.
P(2)
2. Let
TA and TB represent tensions in
string A (steel) and B (aluminum) respectively, then
(a) For
equal stress :
and for torque around
point of suspension of weight to be in equilibrium, TA x = TB
(L – x)
Solving
the above two gives
(b) For
equal strain : ΔlA = ΔlB which gives
and for torque around
point of suspension of weight to be in equilibrium, TA x = TB
(L – x)-
Solving
the two equations gives
3. For
a wire extended under tension,
and frequency of
vibrating string
where m is mass
per unit length of the wire.
In terms of given
quantities, m = ρA which gives
Fig.
P(4)
4. Referring
to the figure given and applying Pythagoras,
(neglecting higher
order term of ΔL we get x2 = ½ LΔL,
Which gives
5. Let
1 denote Copper and 2 denote zinc.
The given information than gives m1
+m2 = ρ 1V1 + ρ2V2 =
12.9 gm
and
ρW (V1 + V2) = 1.6 gm
Substituting appropriate specific
gravities (note ρW = 1) , we get m1 = 7.61
gm
Auxilliary
Solutions:
A. For
the floating ice, let V be the volume of ice and V’ be the volume
of the submerged portion.
Weight of the ice =
weight of the water displaced
\
ρiceVg = ρwaterV’g
\
B.
Let V be the volume of the
metal piece and V0 be that of cavity.
Then we have ρ(V – V0)
= 264 and ρWV =
264 - 221 = 43
Solving which we get V0
= 13 cm3.
6. Volume of water displaced = (weight of the
stone)/(ρwaterg)
and
volume of stone = ρstoneVstoneg
this gives
and since ρwater
< ρstone we get Vstone < Vwater
The tank will not over flow.
Fig.
P(7)
7. Initially the pressure of air above the
liquid is PA : the atmospheric pressure
after the orifice is opened, the pressure of air above the liquid is P
= PA – hρg
and also PA (L
– H) A = P (L – h) A where A is the area of
cross-section of the cylinder and
L =500 mm and h = 200 mm. finding P and solving the equations
we get H = 206 mm so
that H - h = 6 mm.
Fig.
P(8)
8. Referring
to figure above,
for a small element of
mass m at (x, y)
Integration
yields,
, which is equation of parabola.
Fig.
P(9)
9. From
equation of continuity vpAp = vqAq that gives vq = ½ m/s.
Work done by gravity = ρgh
= 29.4×103 J/m3
From Bernoulli’s
equation we have Pp + ½ ρvp2 = Pq
+ ½ ρvq2 + ρgh
This gives work by
pressure force = Pp – Pq = ρgh - (½
ρvp2 - ½ ρvq2) = 29.03×103
J/m3
Fig.
P(10)
10. Referring
to the figure, when the needle is in equilibrium,
and 2Tsinθ =
λg
in small angle
approximation : tanθ ≈ sinθ
wihich gives
Fig.
P(11)
11. Referring
to figure : and using Bernoulli’s equation for outside the bubble and inside
the bubble, we get Pout +
½ ρvout2 = Pin + ½ ρvin2
where T is the
surface tension and vin = 0 yields
12. When
drops coalesce into one, the total volume remains constant : i.e.,
which gives R =
2r = 2 mm
Also loss of surface
energy results in rise in temperature of the drop.
i.e., ΔUsurface = msΔθ here
which gives Δθ
= 3.663×10 – 4 K
References:
- Mody A. K. & Pradhan H. C., ‘Problem Based Learning in Basic Physics – I, School Science 49 (3) Sept 2011
- Pradhan H.C. & Mody A. K., ‘Constructivism applied to physics teaching for capacity building of undergraduate students’, University News, 47 (21) 4-10, (2009)
- JEE - Joint Entrance Examination for admission to IIT
- Physics Textbook for class XI part-I, NCERT New Delhi (2006)
- NCERT EP XI : Physics Exemplar Problems class XI, NCERT New Delhi (2009)
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