Physics Musing Problem Set 28 8
JEE Workouts 12
Core Concept 18
Physics Musing Solution Set 27 21
Exam Prep 2016 23
JEE Accelerated Learning Series 31
Brain Map 46
JEE Advanced Practice Paper 2016 58
Ace Your Way CBSE XI 64
Thought Provoking Problems 71
Ace Your Way CBSE XII 74
You Ask We Answer 82
Live Physics 83
Crossword 85
CONTENTS
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single oPtion correct tyPe
1. A metal ring of initial radius r and cross-sectional area A is fitted onto a wooden disc of radius R > r. If Young’s modulus of the metal is Y, then the tension in the ring is
(a) AYR
r (b) YrAR
(c) AY R r(r − ) (d) Y R r(Ar− )
2. A piece of pure gold (r = 19.3 g cm–3) is suspected
to be hollow from inside. It weighs 38.250 g in air and 33.865 g in water. The volume of the hollow portion in gold is
(a) 1.982 cm3 (b) 2.403 cm3
(c) 3.825 cm3 (d) 4.385 cm3
3. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats g. It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by (a) 2((gg+−21))RMv2 (b) (g2g−1R)Mv2
(c) gMv2R2 (d) (g −12R )Mv2
4. In the figure shown, there are 10 cells each of emf e and internal resistance r. The current through resistance
R is
(a) zero (b) e/r
(c) 3e/r (d) 4e/r
5. A rectangular loop with a
sliding connector of length 1 m is situated in a uniform magnetic field of 2 T perpendicular to the plane of loop. Resistance of connector is 2 W.
Two resistances of 6 W and 3 W are connected as shown in the figure. The external force required to keep the connector moving with a constant velocity 2 m s–1 is
(a) 2 N (b) 1 N
(c) 4 N (d) 6 N
6. A thin lens of refractive index 1.5 and focal length in air 20 cm is placed inside a large container containing two immiscible liquids as shown in figure.
If an object is placed at an infinite distance close to principal axis, the distance between two images will be
(a) 25 cm (b) 40 cm
(c) 65 cm (d) 85 cm
P
hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material.In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You.
The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue.
We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.
PHYSICS
PHYSICS
MUSING
MUSING
Solution Senders of Physics Musing set-27
1. Manmohan Krishna (Bihar) 2. Anubhav Jana (WB)
3. Shiekh Md. Shakeel Hassan (Assam) 4. Swati Shah (Rajasthan)
set-26 1. Md. Samim Jahin (Assam) 2. Deep Anand Basumatary (Assam) 3. Harsimran Singh (Punjab) 4. Sayantan Bhanja (WB)
7. The figure shows several
equipotential lines.
Comparing between
points A and B, choose the best possible statement.
(a) The electric field has a greater magnitude at point A and is directed to left.
(b) The electric field has a greater magnitude at point A and is directed to right.
(c) The electric field has a greater magnitude at point B and is directed to left.
(d) The electric field has a greater magnitude at point B and is directed to right.
8. A light wire AB of length 10 cm can slide on a vertical frame as shown in figure. There is a film of soap solution trapped between the frame and the wire.
Find the mass of the load W that should be suspended from the wire to keep it in equilibrium. Neglect friction. Surface tension of soap solution is 25 dyne cm–1. (Take g = 10 m s–2)
(a) 0.25 g (b) 0.50 g
(c) 2.50 g (d) 5.00 g
ParagraPh tyPe
Read the given paragraph and answer question number 9 and 10.
Consider the situation shown in figure in which a block A of mass 2 kg is placed over a block B of mass 4 kg.
The combination of the blocks
are placed on an inclined plane of inclination 37° with horizontal. The system is released from rest. (Take g = 10 m s–2 and sin 37° = 0.6)
9. The coefficient of friction between block B and inclined plane is 0.4 and in between the two blocks is 0.5. Then
(a) Both blocks will move but block A will slide over the blocks B.
(b) Both blocks will move together. (c) None of them will move. (d) Only block A will move.
10. The frictional force acting between the blocks will be (a) 8 N (b) 6.4 N (c) 4 N (d) zero nn
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� CLASSROOM STUDY MATERIAL Your logo here1. Assume ABCDEF to be a regular hexagon. Choose the correct statements.
(a) ED DB BE + + = 0 (b) FE BC = E D C F A B (c) AD = 2 FE (d) DC = −AF
2. A spring mass system is hanging from the ceiling of an elevator in equilibrium as shown in figure. The elevator suddenly starts accelerating upwards with acceleration a, consider all the statements in the reference frame of elevator and choose the correct one(s).
(a) The frequency of oscillation is 1 2π
k m.
(b) The amplitude of the resulting SHM is ma
k .
(c) The amplitude of resulting SHM is m g a
k
( + ).
(d) Maximum speed of block during oscillation is
m k a .
3. An ideal gas has molar heat capacity at constant pressure CP = 5R
2 .. The gas is kept in a cylindrical vessel fitted with a piston which is free to move.
Mass of the frictionless piston is 9 kg. Initial volume of the gas is
0.0027 m3 and cross-section area of the piston is 0.09 m2. The initial temperature of the gas is 300 K.
m k
Atmospheric pressure P0 = 1.05 × 105 N m–2. An amount of 2.5 × 104 J of heat energy is supplied to the gas, then
(a) Initial pressure of the gas is 1.06 × 105 N m–2. (b) Final temperature of the gas is 1000 K. (c) Final pressure of the gas is 1.06 × 105 N m–2. (d) Work done by gas is 9.94 × 103 J.
4. A man has fallen into a ditch of width d and two of his friends are slowly
pulling him out using a light rope and two fixed pulleys as shown in figure. Assume both the friends apply forces of equal magnitude. Choose the correct statements.
(a) The force exerted by both the friends decreases as the man moves up.
(b) The force applied by each friend is
mg
h d h
4 2+4 2when the man is at depth h.
(c) The force exerted by both the friends increases as the man moves up.
(d) The force applied by each friend is mgh d2+h2
when the man is at depth h.
5. Two balls are thrown from an inclined plane at angle of projection a with the plane, one up the incline and other down the incline as shown in figure. (Here, T stands for total time of flight).
d
Which of the following are correct? (a) h h v g 1 2 0 2 2 2 = = sin cos α q (b) T T1= 2 =2vg0cossinqα (c) R2 – R1 = g(sinq) T12 (d) vt1 =vt2
6. Two identical buggies move one after other due to inertia (without friction) with the same velocity v0.
A man of mass m rides the rear buggy. At a certain moment, the man jumps into the front buggy with a velocity u relative to his buggy. If mass of each buggy is equal to M and velocity of buggies after jumping of man are vrear and vfront. Then
(a) v v m m Mu rear = 0+ + (b) v v m m Mu rear = 0− + (c) v v mM m M u front = 0+( + )2 (d) vfront v m MmM u ( ) = − + 0 2
7. A spherical body of radius R rolls on a horizontal surface with linear velocity v. Let L1 and L2 be the
magnitudes of angular momenta of the body about centre of mass and point of
contact P respectively. Then (here K is the radius of gyration about its geometrical axis)
(a) L2 = 2L1 if radius of gyration K = R
(b) L2 = 2L1 for all cases
(c) L2 > 2L1 if radius of gyration K < R
(d) L2 > 2L1 if radius of gyration K > R
8. Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a
string. They are fully immersed in a fluid of density dF. They get arranged
into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if (a) dA < dF (b) dB > dF
(c) dA > dF (d) dA + dB = 2dF
9. A body of mass m is attached to a spring of spring constant k which hangs from the ceiling of an elevator at rest in equilibrium. Now the elevator starts accelerating upwards with its acceleration varying with time as a = pt + q, where p and q are
A B
positive constants. In the frame of elevator,
(a) The block will perform SHM for all value of p and q.
(b) The block will not perform SHM in general for all value of p and q except p = 0.
(c) The block will perform SHM provided for all value of p and q except p = 0.
(d) The velocity of the block will vary simple harmonically for all value of p and q.
10. A string of mass m is fixed at both its ends. The fundamental mode of string is excited and it has an angular frequency w and the maximum displacement amplitude A. Then
(a) The maximum kinetic energy of the string is
EK = 1mA
4 2 2w .
(b) The maximum kinetic energy of the string is
EK = 1mA
2 2 2w .
(c) The mean kinetic energy of the string averaged over one periodic time is <EK > = 1mA
4 2 2w .
(d) The mean kinetic energy of the string averaged over one periodic time is <EK > =1mA
8 2 2w .
11. A bottle is kept on the ground as shown in the figure. The bottle can be modelled as having two cylindrical zones. The lower zone of the bottle has a cross-sectional radius of
R 2 and is filled with
honey of density 2r. The upper zone of the bottle is filled with the water of density r and has a cross-sectional radius R. The height of the lower zone is
H while that of the upper
zone is 2H. If now the honey and the water parts
are mixed together to form a homogeneous solution, then
(Assume that total volume does not change)
(a) The pressure inside the bottle at the base will remain unaltered.
(b) The normal reaction on the bottle from the ground will remain unaltered.
(c) The pressure inside the bottle at the base will increase by an amount 12rgH .
(d) The pressure inside the bottle at the base will
decrease by an amount 1
4
12. A particle moving with kinetic energy 3 J makes an elastic collision (head-on) with a stationary particle which has twice its mass. During impact
(a) The minimum kinetic energy of system is 1 J (b) The maximum elastic potential energy of the
system is 2 J.
(c) Momentum and total energy are conserved at every instant.
(d) The ratio of kinetic energy to potential energy of the system first decreases and then increases.
13. Two blocks A and B each of mass m are connected
by a massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block C, moving on the floor with a speed v along the line joining A and B, collides with A. Then (a) The maximum compression of the spring is
v m k/ .
(b) The maximum compression of the spring is
v m k/2 .
(c) The kinetic energy of A-B system at maximum compression of the spring is zero.
(d) The kinetic energy of A-B system at maximum compression of the spring is mv2/4.
14. Three planets of same density and with radii R1, R2 and R3 such that R1 = 2R2 = 3R3, have gravitational fields on the surfaces E1, E2, E3 and escape velocities
v1, v2, v3 respectively. Then (a) EE1 2 1 2 = (b) EE1 3 =3 (c) vv1 2 = 2 (d) v v13 1 3 =
15. Water is flowing smoothly through a closed pipe system. At one point A, the speed of the water is 3.0 m s–1 while at another point B, 1.0 m higher, the speed is 4.0 m s–1. The pressure at A is 20 kPa when the water is flowing and 18 kPa when the water flow stops. Then
(a) the pressure at B when water is flowing is 6.7 kPa.
(b) the pressure at B when water is flowing is 8.2 kPa.
(c) the pressure at B when water stops flowing is 10.2 kPa.
(d) the pressure at B when water stops flowing is 8.2 kPa.
solutions
1. (a, b, c, d) : As ABCDEF is a regular hexagon, \ FE BC AD = , =2FE DC , = − AF
Also, from triangle law of vector addition,
ED DB BE
+ + = 0.
2. (a, b, d) : As it is a isochronous system
\ υ= π 1 2 k m
From the reference frame of elevator, A ma= k
v A k m ma k m k a max =w = = 3. (a, c) : CP =5R CV = R 2 3 2 , ∆ ∆ W Q n C C nC P V p = ( − ) = − =1 3 5 2 5 \ ∆W =2∆Q = × × = 5 2 2 55 104 104 . J
Pressure is constant and equal to
P P mg A = 0+ =1 05 10× 5+ ×9 10 0 09 . . = 1.06 × 105 N m–2 4. (b, c) : From figure, sinq = + h h2 d2 4 As man moves slowly 2T sin q = mg
T = mg
2sinq
As man moves upward, q becomes small \ sin q decreases ⇒ T increases T mg h h d mg d h h = × + = + 2 4 4 4 2 2 2 2
5. (a, b, c) : Maximum height of projectile on an
inclined plane, h v g h 1 0 2 2 2
max =( sin )cosaq = max
⇒ (a) is correct Time of flight
T v
g T
1= 2 0cossinqa = 2 ⇒ (b) is correct T T mg d/2 h
where a = angle of projection from inclined plane q = angle of inclination of surface.
R1 v0 T1 1g T12
2
=( cos )a − sinq
(Range upward the inclined plane)
R2 v0 T2 1g T22
2
=( cos )a + sinq
(Range downward the inclined plane) ⇒ (R2 – R1) = g sin q T12 ⇒ (c) is correct
vt1 and vt2 are the velocities of the particles at their maximum height. Let the particles reach their maximum heights at time t1 and t2 respectively.
Hence, 0 = (v0 sin a) – (g cos q) t1
⇒ t =v g 1 0cossinqa Similarly, sin cos . t v g 2= 0 aq Hence, t2 = t1
Hence, vt1 = v0 cos a – (g sin q) t1
vt2 = v0 cos a + (g sin q) t2
⇒ vt1 ≠vt2
6. (b, c) : As no external force is applied to the system
v0 v0
v v0
u v+
Conserving linear momentum of man and rear buggy, (M + m)v0 = Mv + m(v + u) ⇒ = − + v v mu M m 0 = vrear
Conserving linear momentum of man and front buggy, m(u + v) + Mv0 = (M + m)v′ m u v mu M m Mv M m v + − + 0 + 0 =( + ) ′ (M m v) Mmu ( ) M m M m v + + + = + ′ 0 v v Mmu M m ′ = + + 0 ( )2 = vfront 7. (a, d)
8. (a, b, d) :Let V be the volume of each sphere and T is the tension in the string.
For the string to be taut,
dFVg > dAVg or dF > dA (a) is correct and dBVg > dFVg or dB > dF A B T d VgA d VF g d VB g T d VF g (b) is correct For an equilibrium dFVg + dFVg + T = T + dAVg + dBVg or dA + dB = 2dF (d) is correct
9. (c, d) : In the frame of elevator
mg ma kx m d x dt + − = 22 ⇒ = − − + d x dt k m x m g ak 2 2 ( ) or d x ( ) dt k m x m g pt qk 2 2 = − − + +
There is a term involving t on R.H.S., this does not represent S.H.M. unless p = 0
Differentiating with respect to time
d x dt k m dx dt mp k 3 3 = − − or d vdt mk v mpk 2 2 = − −
Thus the velocity of the block will vary simple harmonically.
10. (a, d) : Let the displacement of the string be given by
y x t A x
L t
( , )= sinπ cos(w +d)
where d is a phase factor. So the transverse velocity is given by v x t y y A x L t ( , )= ∂ sin sin( ) ∂ = −w + π w d
The maximum kinetic energy is equal to the string’s total energy of oscillation. Note that all points of the string achieve their maximum kinetic energy at the same instant of time, where y = 0 for all x. Since
dm = mdx where m =m
L is the mass per unit
length of the uniform string. The maximum kinetic energy,
E y t dm K = ∂∂
∫
maximumof 1 2 2 = ∂ ∂ ∫
maximumof 12 2 0 m y y dx LThe maximum value of ∂∂yy
2 , occurs when sin2 (wt + d) = 1 Hence E A x L dx K L =m w
∫
π 2 2 20sin2The integral sin2
0
πx
L dx
L
∫
over the half-cycle hasthe average value of L2
Hence, EK =1A L= mA 2 2 1 4 2 2w m 2 2w \ (a) is correct
The mean kinetic energy of the string averaged over one periodic time is obtained by integrating the time dependent factor sin2 (wt + d) over one period, 0 to T.
Now since sin (2 )
0 2
wt d dt T
T
+ =
∫
The mean kinetic energy of the string averaged over one periodic time is
<E > = E
∫
+ = T t dt E K K K T sin (2 ) 0 2 w d = = 1 2 1 4 1 8 2 2 2 2 mA w mA w \ (d) is correct11. (b, c) : Initial pressure at the bottom = rg × 2H + 2r × g × H = 4rgH
Final density of the homogeneous mixture
= × × + × × × + × = r r r A H A H A H A H 2 2 2 2 2 3 2 Final pressure at the bottom = 3
2 3
9 2 r× ×g H= rgH
12. (a, b, c, d) : In a head on elastic collision between two particles, the kinetic energy becomes minimum and potential energy becomes maximum at the instant when they move with a common velocity. The momentum and energy are conserved at every instant.
Let m and u be the mass and initial velocity of the first particle, 2m be the mass of second particle and
v be the common velocity.
Then, 1
2mu = J ; mu2 3 =(m+2m v) orv u= 3 Minimum kinetic energy of system
=1 =
2 3 3 1
2
( )m u J
Maximum potential energy of system = 2 J
13. (b, d) : After collision of C with A, let velocity acquired by A and B be v′ and spring gets compressed by length x. Using law of conservation of linear momentum, we have
mv = mv′ + mv′
or v′ = v/2
Using law of conservation of mechanical energy, we have 1 2 1 2 1 2 1 2 2 2 2 2 mv = mv′ + mv′ + kx or mv2 m v 2 m v 2 kx2 2 2 = + + or mv2 kx2 2 = or x v= 2mk 1 2/ \ (b) is correct
At maximum compression of the spring, the kinetic energy of A-B system will be
=1 ′ + ′ = = = 2 1 2 2 4 2 2 2 2 2 mv mv mv m v mv \ (d) is correct 14. (b, c) : E=GMR2 = G R R 4 3 3 2 π r or E ∝ R E E R R R R 1 2 1 2 2 2 2 2
= = = \ (a) is not correct. E E R R R R 1 3 1 3 3 3 3 3 = = = \ (b) is correct. Escape velocity, v= 2GMR = 2RG4 R 3π r3 = 8 3π rR G or v ∝ R2 v v R R 1 2 1 2 2 = = \ (c) is correct. v v R R 1 3 1 3 3 = = \ (d) is not correct.
15. (a, d) : Let P1, h1, and v1 and P2, h2 and v2 represent
the pressures, heights and velocities of flow at the two points respectively. According to the Bernoulli’s theorem P1 gh1 1 v12 P2 gh2 v22 2 1 2 +r + r = +r + r ...(i) Putting v1 = 3.0 m s–1, v2 = 4.0 m s–1, (h2 – h1) = 1 m, P1 = 20 kPa we get, P2 20 103 9 8 1 103 3 2 9 16 10 = + × − + − × − . ( ) [ ] = 20 – 9.8 – 3.5 = 6.7 kPa
Also when the flow stops, v1 = v2 = 0 and then from (i),
P2 = 18 – 9.8 = 8.2 kPa
Pressure inside a liquid
We choose atmospheric pressure = P0
At a depth h below the free surface, pressure = P.
h P0
P
To find P, we choose a liquid column of height h and cross-sectional area A.
Since the liquid column is unaccelerated,
P0A + mg = PA
⇒ P P mg= +
A
0 =P0+r( )AhA g
⇒ P = P0 + rgh
The additional pressure with respect to atmospheric pressure is known as Gauge pressure.
Hence we conclude that pressure changes by an amount rgh on moving through a distance h vertically.
Note that this result has been derived from equilibrium of the liquid column. Hence if the container or liquid was vertically accelerated, it would not be applicable. In such cases if the container is vertically accelerated, say upward with a, then
(Fnet)upward direction = ma
⇒ (PA) – (P0A + mg) = ma
⇒ P P m= + +
A g a
0 ( )
⇒ P = P0 + r(g + a)h
\ We replace g with g eff where,
geff = + −g ( )a
\ P = P0 + rgeff h
So, it is interesting to see that we can also have a situation that all points inside a liquid irrespective of
their location, will have same pressure as atmospheric if
geff = 0, as in case of free fall.
Measurement of atmospheric pressure (P0)
We take a tub filled partially with mercury and a tube completely filled with mercury. We seal the mouth of the tube and invert it upside down with the mouth inside the mercury in the tub.
The liquid in the tube drops down a little, creating almost vacuum in the upper closed end of the tube as shown.
At equilibrium,
PA + rgh = PB = PC = P0
⇒ rgh = P0 [ PA = 0]
Hence, measuring the length of the liquid column in the tube, P0 can easily be calculated.
Taking P0 = 1 atm ≈ 105 N m–2
For Hg, r = 13.6 g cm–3, h comes out to be almost
760 mm or 76 cm.
Hence you would often see that pressure is given in terms of length and not N m–2 or pascal. If instead
of Hg, some other liquid is used, to find the height of liquid risen we can easily use,
r1h1 = r2h2
Supposedly, we keep this set-up in an upward accelerating frame, then how will h change?
Clearly, in such case also, atmospheric pressure does not change, we need to change g with geff.
\ P1 = rgh = rgeff h′ = r(g + a)h′ \ ′ = + h g g a h
\ h′ < h ⇒ height of liquid column decreases. Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
Pressure difference in a horizontally accelerated container
In such case, clearly, the pressure at same horizontal level would not be same. To know the exact relation we consider a thin horizontal liquid column of length l as drawn. Hence from free body diagram,
(P2A – P1A) = ma
⇒ (P2 – P1)A = (rAl)a
⇒ P2 = P1 + ral
\ DP = ral along horizontal, as DP = rgh along vertical.
Vertically pressure increases in the direction of gravity, horizontally acceleration increases opposite to the direction of acceleration.
Let us calculate the inclination of free surface with horizontal now.
We can find pressure at B from A as well as C. PB = PA + rgh = PC + raL ⇒ tanq h= = L a g
This could also have been found out using the fact that liquids cannot tolerate tangential force on its surface. Hence the free surface should be perpendicular to geff
as below.
geff = + −g ( )a
[We revert the acceleration of container and add it vectorially to g ]
Let us apply this to a more complicated situation. Assume a container falling down on a smooth inclined plane.
We have to find a.
Clearly, a = gsinq downwards Hence, geff = + −g ( ) a g gcos gsin a g= sin parallel to incline gcos = geff
Since, geff is perpendicular to inclined plane, hence free
surface of liquid is parallel to inclined plane. Hence a = 0°
U-tubes
Let us consider a U-tube filled with two immiscible liquids on two limbs. Note that,
PC = PF but
PA ≠ PD
PB ≠ PE,
even though (A, D) and (B, E) pair of points are at same level. Since if we move up from C to B and F to E, the change in height is same but we have different density of liquids hence r1h ≠ r2h.
If difference in height of free surface of liquids is to be calculated in terms of H, PC = PF ⇒ P0 + r1gh1 = P0 + r2 gH ⇒ h1= 2H 1 r r \ = − Dh 1 2 H 1 r r Archimedes’ principle
If an object is submerged inside a liquid, partially or completely, it experiences an upward force by the liquid due to pressure difference along the vertical column of the liquid, which is equal to the weight of liquid displaced by the object. To prove this, let us imagine an object of cross-sectional area A and height H partially submerged till height h as shown here.
H h l s FBD of object mg P A0 ( +P0 lgh A)
\ By definition, force of upthrust (also known as buoyant force) is
Fup = (P0 + rl gh)A – P0A
= rl (hA)g = rlVsub g
Note : This result is applicable only if the container is
vertically unaccelerated, else we need to replace g with
geff in the result.
Now, let us see what would the condition of floatation for the object be.
For equilibrium, Fup = mg ⇒ rl(Ah)g = rs(AH)g ⇒ h = ≤ ≤ H sl h H r r 1 [ for floatation]
\ If rs ≤ rl, the object floats, else it sinks. Hence it
does not matter how heavy an object is for floating, what matters is how dense the object is!
Note here, that the fraction of submerged portion, h
H is
independent of g, hence even in accelerated containers, this same fraction will be submerged.
Now, let us seen an application. Suppose a helium filled balloon
He air
is floating in air (their densities given as rHe and rair) with a
string tied to a box as shown here.
Now, if the box is accelerated towards right with acceleration a, we have to find the direction and angle with the vertical in which the string gets deflected at equilibrium.
One would be tempted to say that the string will deflect towards left due to the pseudo force.
But there is a basic point, one is missing here. As we saw force of upthrust being generated due to pressure difference along vertical column of liquid, similar to it a side thrust can also be generated if pressure difference is created along horizontal column and on similar approach it can be proved.
Fside thrust = rl Vsub a
Hence, since, rair > rHe,
so side thrust will be greater than the pseudo force, so the balloon deflects towards right.
HeVg T HeVa airVa airVg T (air–He)Va (air–He)Vg \ = − − = tan ( ) ( ) q r r rairair rHeHe Va Vg a g
Force exerted by a liquid on a vertical surface
Let us assume, that the wall of a dam of width w stops a liquid of height h from flowing. It obviously is experiencing force due to the liquid’s pressure.
To find this, let us consider a horizontal strip of height
dy at a depth y below the free surface.
The force experienced by this surface due to pressure of liquid is
dF = (rgy)(dyw)
(P0 is not considered since
it is due to atmosphere) ⇒ F= ∫ = ∫dF hw gydyr 0 = rgh w 2 2
\ Net horizontal force on vertical surface = rg h hw2( )
= pressure at centroid of submerged portion × area of submerged portion Force exerted by liquid on a horizontal surface
Let us consider a horizontal face of area A at a depth h below the free surface.
We have to find the force exerted by the liquid only. Net vertical force = PA
= (rgh)A = r(Ah)g
= weight of liquid column above its surface
Solution Set-27
1. (d) :Here, d = 0.5 mm = 0.5 × 10–3 m
D = 0.5 m
l = 500 nm = 500 × 10–9 m
The distance of third maxima from the second minima on the other side is
= 9
2b (where is the fringe width)b = 9 2 lDd = × × × × × − − 9 500 10 0 5 2 0 5 10 9 3 . . = 2.25 × 10–3 m = 2.25 mm
2. (c) : During motion of the particle, total mechanical energy remains constant.
At the surface of earth, total mechanical energy is Ei = −GmMR + 12mv02 = −GM + R2 mR mv0 2 1 2 = −gmR+ mv g GM= R 1 2 02 2
Total mechanical energy at height h = R is
E GmM
R mv
gmR mv
f = − 2 +12 2= − 2 +12 2
According to law of conservation of mechanical energy, Ei = Ef ∴ −gmR+1mv = −gmR+ mv 2 2 1 2 0 2 2 or −2gR v+ 02= −gR v ∴+ 2 v= v02−gR 3. (a) : If we consider the cylindrical surface to be a
ring of radius R, there will be an induced emf due to changing field. E dl d dt A dBdt ⋅ = − = −
∫
f ⇒ E R = −A dB= − ⇒ = − dt R dBdt E R dB dt (2 ) 2 2 π π∴ Force on the electron
F Ee eR dB dt = − = 2 ⇒ Acceleration =12eRm dB dt
As the field is increasing being directed inside the paper, hence there will be anticlockwise induced current (in order to oppose the cause) in the ring
(assumed). Hence there will be a force towards left on the electron.
4. (c) : Total time of flight is T = 4 s and if u is its initial speed and q is the angle of projection. Then
T u
g
=2 sinq=4
or usinq = 2g ...(i)
After 1 s velocity vector makes an angle of 45° with horizontal i.e.,
vx = vy
ucosq = usinq – gt
ucosq = usinq – g ( t =1s)
ucosq = 2g – g (Using (i))
or ucosq = g ...(ii)
Squaring and adding (i) and (ii), we get
u2sin2q + u2cos2q = (2g)2 + (g)2
u2 = 5g2 = 5(10)2 m2 s–2 ∴ u = 22.36 m s–1
Dividing (i) by (ii), we get,
u u g g sin cos q q= = 2 2 tanq = 2 or q = tan–1(2) 5. (d) :Here, T l g = 2π ...(i)
When lift is accelerated upwards with acceleration
a, let time period becomes T2. Then
T l
g a
2=2π + ...(ii)
Dividing Eq. (i) by Eq. (ii), we get
2 1 1 2 = + = + g a g a g /
Squaring both sides, we get 4 1= + a
g or a = 3g
6. (c) : Lengths of the two inclined planes are
l1 h l h
1 2 2
= =
sinq and sinq
Accelerations of the block down the two planes are
a1 = g sinq1 and a2 = g sinq2
As l1 1a t1 12and l2 a t2 22 2 1 2 = = ∴ l = = = × l a t a t t t a l a l g g 1 2 1 12 2 22 22 12 1 2 2 1 1 2 1 2 or sin sin sin sin q q q q ∴ t = t21 12 sin sin q q 7. (a) : Given : f = at2 + bt
ε=df= + = + dt d dt(at2 bt) 2at b Current flowing, I=| |Rε =2at bR+ Average emf = = + = +
∫
∫
∫
ε τ τ τ τ τ dt dt at b dt a b 0 0 0 2 ( )Total charge flowing,
q Idt at b R dt a b R =
∫
=∫
(2 + ) = + 0 2 0 τ τ τ τ8. (a) : Free body diagram for m For m,
mg – T = m 2a … (i)
N = ma … (ii)
Free body diagram for M For M 2T – N = Ma …(iii) On solving, a mg M m = + 2 5 ( ) ∴ Net acceleration of m, a a a a mg m M m= 4 2+ 2 = 5 =(52 5+ ) 9. (b) : tan sin cos a q q = + B A B ...(i)
where a is the angle made by the vector (A B + with ) A.
Similarly, tan sin
cos b q q = − B A B ...(ii)
where b is the angle made by the vector(A B − )with A.
Note that the angle between Aand ( )−B is (180° – q). Adding (i) and (ii), we get
tan tan sin
cos sin cos a b q q q q + = + + − B A B B A B = − + + + − AB B AB B A B A B
sin sin cos sin sin cos
( cos )( cos ) q q q q q q q q 2 2 = − 2 2 AB2 2 A B sin ( cos ) q q 10. (a) : tan sin cos a q q = + B A B tan sin cos . a q q = + 1 …(i) m mg 2a N T a tan sin cos a q q 2 2 2 = + a a a tan sin ( cos ) a q q 2 = 2+ …(ii) tan tan tan a a a = − 2 2 1 2 2 sin cos sin cos sin ( cos ) q q q q q q 1 2 2 1 2 2 2 + = ⋅ + − + On solving, cosq = − 1 2 ∴ q = 120° nn Winners (October 2015)
Basheer Mazahar (Varanasi) sidharth sankar sahu (Odisha) Ayushi Tripathi (Delhi)
solution senders (september 2015) Anoop Jain (Delhi)
sanchit Mehta (Haryana) Viraj Thapa (Assam)
solution of october 2015 crossword
1 S K E W R A Y K A O N I D Y N A M O A T D F R M S Z E N I T H U C L E A C A N L R A Y S Y C T O C T F L I P F L O P D R Y C E L L P F L A V O R R I P P L E I T S U N D R Y I C E F O R C E T E N S I M E T E R C LL O Y D M I R R O R E R OF S S I L F U E L S M I L K Y W A Y T O U R M A L I N E E R W E I G H T O E A N E C H O I C G A T E S I N A L M C H I N C L U D 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Mechanical ProPerties of fluids
1. A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 103 kg m–3. If outer diameter and the density of the bowl are 1 m and 2 × 104 kg m–3 respectively, then the inner diameter of the bowl will be
(a) 0.94 m (b) 0.96 m
(c) 0.98 m (d) 0.99 m
2. A body of density r is dropped from rest at a height
h into a lake of density s, where s > r. Neglecting
all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface.
(a) s rh− (b) hrs
(c) s rhr− (d) s rhs−
3. Water in a vessel of uniform cross-section escapes through a narrow tube at the base of the vessel. Which of the following graphs represents the variation of the height h of the liquid with time t? (a) h t (b) h t (c) h t (d) h t
4. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is
13.6 g cm–3 and the angle of contact of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is
(a) 1 : 0.15 (b) 1 : 3
(c) 1 : 6.5 (d) 1.5 : 1
5. A metallic sphere of mass M falls through glycerine with a terminal velocity v. If we drop a ball of mass 8M of same metal into a column of glycerine, the terminal velocity of the ball will be
(a) 2 v (b) 4 v
(c) 8 v (d) 16 v
6. A cylindrical drum, open at the top, contains 15 L of water. It drains out through a small opening at the bottom. 5 L of water comes out in time t1, the
next 5 L in further time t2 and the last 5 L in further time t3. Then
(a) t1 < t2 < t3 (b) t1 > t2 > t3 (c) t1 = t2 = t3 (d) t2 > t1 = t3
7. A sealed tank containing a liquid of density r moves with a horizontal acceleration a, as shown in figure. The difference in pressure between the points A and
B is h A B l a (a) h r g + l r a (b) h r g – l r a (c) h r g (d) l r a
8. The surface area of air bubble increases four times when it rises from bottom to top of a water tank where the temperature is uniform. If the
chapterwise McQs for practice
atmospheric pressure is 10 m of water, the depth of the water in the tank is
(a) 30 m (b) 40 m
(c) 70 m (d) 80 m
9. A block of wood floats in water with 45
th
of its volume submerged, but it just floats in a liquid. What is the density of liquid?
(a) 750 kg m–3 (b) 800 kg m–3
(c) 1000 kg m–3 (d) 1250 kg m–3
10. A spherical ball is dropped in a long column of viscous liquid. Which of the following graphs represent the variation of
(i) gravitational force with time (ii) viscous force with time
(iii) net force acting on the ball with time?
F P Q R t (a) Q, R, P (b) R, Q, P (c) P, Q, R (d) R, P, Q
11. Water is flowing through a horizontal pipe. If at one point pressure is 2 cm of Hg and velocity of flow of the liquid is 32 cm s–1 and at another point, velocity of flow is 40 cm s–1, the pressure at this point is
(a) 1.45 cm of Hg (b) 1.98 cm of Hg
(c) 1.67 cm of Hg (d) 1.34 cm of Hg
12. The rate of flow of glycerine of density 1.25 × 103 kg m–3 through the conical section of a pipe, if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is 10 N m–2, is
(a) 5.28 × 10–4 m3 s–1 (b) 6.28 × 10–4 m3 s–1 (c) 7.28 × 10–4 m3 s–1 (d) 8.28 × 10–4 m3 s–1 13. Water rises in a capillary tube to a height h. Choose
the false statement regarding a capillary rise from the following.
(a) On the surface of Jupiter, height will be less than h.
(b) In a lift, moving up with constant acceleration, height is less than h.
(c) On the surface of the moon, the height is more than h.
(d) In a lift moving down with constant acceleration, height is less than h.
14. A candle of diameter d is floating on a liquid in a cylindrical container of diameter D(D>>d) as shown in figure. If it is burning at the rate of 2 cm h–1, then the top of the candle will
L L d D (a) remain at the same height (b) fall at the rate of 1 cm h–1 (c) fall at the rate of 2 cm h–1 (d) go up at the rate of 1 cm h–1
15. A frame made of metallic wire enclosing a surface area A is covered with a soap film. If the area of the frame of metallic wire is reduced by 50%, the energy of the soap film will be changed by
(a) 100% (b) 75% (c) 50% (d) 25%
therMal ProPerties of Matter
16. The plots of intensity versus wavelength for three black bodies at temperatures T1, T2 and T3
respectively are as shown. Their temperatures are such that T1 T3 T2 I (a) T1 > T2 > T3 (b) T1 > T3 > T2 (c) T2 > T3 > T1 (d) T3 > T2 > T1
17. A solid copper sphere (density r and specific heat capacity c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required (in ms) for the temperature of the sphere to reach 100 K is
(a) 807 r crs (b) 807 r csr (c) 277 r crs (d) 277 r crs
18. A clock with an iron pendulum keeps correct time at 15°C. What will be the error in time per day, if the room temperature is 20°C?
(The coefficient of linear expansion of iron is 0.000012°C–1.)
(a) 2.6 s (b) 6.2 s
(c) 1.3 s (d) 3.1 s
19. A body cools from 60°C to 50°C in
10 min. If room temperature is 25°C, temperature of body at the end of next 10 min will be
(a) 38.5°C (b) 40°C
(c) 45°C (d) 42.8°C
20. Two rods of same length and material transfer a given amount of heat in 12 s, when they are joined end to end (i.e., in series). But when they are joined in parallel, they will transfer same heat under same conditions in
(a) 24 s (b) 3 s
(c) 48 s (d) 1.5 s
21. A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt should be
(a) 450 (b) 900
(c) 225 (d) 1800
22. A body in laboratory takes 4 min to cool from 61°C to 59°C. If the laboratory temperature is 30°C, then the time taken by it to cool from 51°C to 49°C is
(a) 4 min (b) 6 min
(c) 8 min (d) 5 min
23. The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm. The radiation intensity for the star is (Take s = 5.67 × 10–8 W m–2 K–4, Wien’s constant, b = 2898 mm K)
(a) 5.67 × 108 W m–2 (b) 5.67 × 1012 W m–2 (c) 5.67 × 107 W m–2 (d) 5.67 × 1014 W m–2
24. The reading of Centigrade thermometer coincides
with that of Fahrenheit thermometer in a liquid. The temperature of the liquid is
(a) –40°C (b) 0°C
(c) 100°C (d) 300°C
25. For a black body at temperature 727 °C, its radiating power is 60 W and temperature of surrounding
is 227°C. If the temperature of the black body is changed to 1227°C, then its radiating power will be
(a) 120 W (b) 240 W
(c) 304 W (d) 320 W
26. A metal plate 4 mm thick has a temperature difference of 32°C between its faces. It transmits 200 kcal h–1 through an area of 5 cm2. Thermal conductivity of the material is
(a) 58.33 W m–1 °C–1 (b) 33.58 W m–1 °C–1 (c) 5 × 10–4 W m–1 °C–1 (d) 5 × 10–2 W m–1 °C–1
27. A 2 kg copper block is heated to 500°C and then it is placed on a large block of ice at 0°C. If the specific heat capacity of copper is 400 J kg–1°C–1 and latent heat of fusion of water is 3.5 × 105 J kg–1, the amount of ice that can melt is
(a) (7/8) kg (b) (7/5) kg
(c) (8/7) kg (d) (5/7) kg
28. The maximum wavelength of radiation emitted at 2000 K is 4 mm. What will be the maximum wavelength emitted at 2400 K?
(a) 3.3 mm (b) 0.66 mm
(c) 1 m (d) 1 mm
29. The net rate at which heat is lost by a body due to radiation does not depend upon
(a) temperature of the body
(b) temperature of the surroundings (c) material of the body
(d) nature of its surface
30. We plot a graph, having temperature in °C on x-axis and in °F on y-axis. If the graph is straight line, then it
(a) passes through origin (b) intercepts the positive x-axis (c) intercepts the positive y-axis
(d) intercepts the negative axis of both x-and
y-axis
solutions
1. (c) : Let D1 be the inner diameter of the hemispherical bowl. As bowl is just floating so 4 3 1 2 1 2 10 4 3 1 2 2 2 10 3 3 3 1 3 4 p × × = p − × × . D ( )
or 1 2 10 2 10 1 3 4 13 . × × = − D ⇒ D1= −1 1 220. =1 3/ 18 820. 1 3/ On solving, we get D1 = 0.98 m
2. (c) : The speed of the body just before entering the liquid is u= 2 . The buoyant force Fgh B of
the lake, i.e., upward thrust of liquid on the body is greater than the weight of the body W, since s > r. If V is the volume of the body and a is the acceleration of the body inside the liquid, then
FB – W = ma or sVg – rVg = rVa or (s – r)g = ra or a=(s r−r )g
Using the relation, v2 = u2 + 2as, we have 0=( 2gh)2−2g(s r− )s
r or s= −s rhr
3. (a) : Let dV be the decrease in volume of water in vessel in time dt. Therefore rate of decrease of water in vessel = rate of water flowing out of narrow tube
So, −dV = − dt P P r l p h ( 1 2) 4 8 But, P1 – P2 = hrg \ −dVdt =p r(h g rhl) =(prhl Agr× ) (× ×h A) 4 4 8 8
where h × A = volume of water in vessel at a time t = V
\ dV= −pr8hgrlA4 ×V dt= −lV dt
or dVV = − ldt
where, pr8hgrlA4 = = constantl
Integrating it within the limits, as time changes from 0 to t, volume changes from V0 to V.
or ln VV t
0 = − l
or V = V0–lt
where, V0 = initial volume of water in vessel = Ah0
Therefore,
h = h0e–lt
Thus, the variation of h and t will be represented by exponential curve as given by (a).
4. (c) : As, h= 2 cosSr gr θ \ S=2 cos orhr grθ S∝coshrθ \ SSw hh Hg = × × 1 2 2 1 1 2 cos cosθθ rr = −( . )103 42 ×coscos1350° ×° 13 61. = 1 ×3 42.0 0 70713 6. . =6 51. 5. (b) : As, M=43p rr3 and 8M=43p rR3 , So, R3 = 8r3 ⇒ R = 2r Now v ∝ r2 so, v v R r r r 1= 2 2 2 4 = = or v1 = 4 v
6. (a) : If h is the initial height of liquid in drum above the small opening, then velocity of efflux,
v= 2 . As the water drains out, h decreases, gh
hence v decreases. This reduces the rate of drainage of water. Due to which, as the draining continues, a longer time is required to drain out the same volume of water. So, clearly, t1 < t2 < t3.
7. (a) : Since points A and C are in the same horizontal line but separated by distance l and liquid tank is moving horizontally with acceleration a, hence
PC – PA = l r a or PC = PA + l r a h A B l a C
Points B and C are vertically separated by h \ PB – PC = h r g or PB – (PA + l r a) = h r g or PB – PA = h r g + l r a 8. (c) : Surface area, A = 4pr2 or r= A 4 1 2 p / Volume V=4 r = A =kA 3 4 3 4 3 3 2 3 2 p p p / / where ( ) / constant 4 3 1 4 3 2 p p × = =k
Using Boyle’s law, we have P1V1 = P2V2 or P PV ( )/ / V h kA kA 2 1 1 2 13 2 23 2 10 = = + or ( ) ( ) / / P h A A h 2 1 2 3 2 3 2 10 10 1 4 = + = + =108+h As P2 = 10 m of water, so 10 10 8 80 10 = +h or = +h or h = 70 m 9. (b) : Density of water =
Density of solid ×total volume of solid immersed volume \ r1 r 4 r 5 5 4 = × V = V
And density of liquid, r2 =rV =r
V \ r1=5r2 4 or r2 4r1 kg m 3 5 4 5 1000 800 = = × = −
10. (c) : Gravitational force remains constant on the falling spherical ball. It is represented by straight line P. The viscous force (F = 6 phrv) increases as the velocity increases with time. Hence, it is represented by curve Q. Net force = gravitational force – viscous force. As viscous force increases, net force decreases and finally becomes zero. Then the body falls with a constant terminal velocity. It is thus represented by curve R.
11. (b) : As per Bernoulli’s theorem,
P1 1 v12 P2 v22 2 1 2 + r = + r or P P1 2 1 (v22 v12) 2 − = r − or P P1 2 1 [ 2 2] dyne 2 2 1 40 32 288 − = × − = cm− = × = 288 13 6 980. 0 02. cm of Hg \ P2 = P1 –0.02 = 2 – 0.02 = 1.98 cm of Hg
12. (b) : According to continuity equation,
v v A A 2 1 1 2 2 2 0 1 0 04 25 4 = = × × = p p ( . ) ( . ) ...(i)
According to Bernoulli’s equation for horizontal tube,
P1 1 v12 P2 v22 2 1 2 + r = + r i e v. ., 22−v12 =2(P P1− 2) r i e v. ., v ( ) ( . ) 22 12 2 103 3 2 2 1 25 10 16 10 − = × × = × − m s ...(ii)−
Substituting the value of v2 from equation (i) in (ii)
(6.25v1)2 – v12 = 16 × 10–3,
i.e., v1 = 0.02 m s–1
So rate of flow through the tube
= A1v1 (= A2v2)
= p × (0.1)2 × 0.02 = 6.28 × 10–4 m3 s–1 13. (d)
14. (b) : Volume of candle = area × length = p d × L
2 2
2
Weight of candle = weight of liquid displaced
Vrg = V′rg or pd2 L r pd2 L r 4 ×2 4 = × ′ ⇒ ′ = r r 1 2 ...(i)
Since, candle is burning at the rate of 2 cm h–1, then after an hour, candle length is 2L – 2
\ (2L – 2)r = (L – x)r′ \ ′ = − − r r L x L 2( 1) or 1 2 2= 1 − − L x L ( ) (using (i)) \ x = 1 cm
Hence, it also decreases 1 cm outside, so, it falls at the rate of 1 cm h–1.
15. (c) : Surface energy = surface tension × surface area or U = S × 2A
New surface energy, U1 S 2 A S A
2 = × = × % decrease in surface energy
=U U− × = − × = U SA SA SA 1 100 2 2 100 50 % % %
16. (b) : According to Wien’s law, lm ∝ 1 .T
From the figure, (lm)1 < (lm)3 < (lm)2, therefore
17. (b) : dT dt
A
mcJ T T
= s ( 4− 04)
Here, fall in temperature of body
dT = (200 – 100) = 100 K, temperature of surrounding T0 = 0 K, initial temperature of body T = 200 K
100 4 4 3 200 0 2 3 4 4 dt r r cJ = s p − p r ( ) \ dt r cJ= r × − =r c⋅ × − s r s 48 10 4 2 48 10 6s . 6 (As J = 4.2) = 7 80 r cr s ms 18. (a) : Here, DT = 20 – 15 = 5°C a = 0.000012°C–1 = 12 × 10–6 °C–1 Time lost per day = 1
2a( )DT ×86400s
= ×1 × − × ×
2 12 10 6 5 86400s
= 2.592 s ≈ 2.6 s
19. (d) : According to Newton’s law of cooling
T T t K T T T 2 1 2 1 0 2 − = + − \ − = + − = 60 50 10 60 50 2 25 30 K K or K = 1 30 For next 10 min
50 10 50 2 25 2 60 − = + − = = T K T K T T or 70T = 3000, T =300= ° 7 42 8. C 20. (b) : D D D D t Q x KA T = ( ) ( )
When two rods of same length are joined in parallel
A′ = 2A and Dx′ =Dx 2 \ Dt′ =Dt = × = 4 1 4 12s 3s
21. (d) : For a spherical black body of radius r at T K, Power radiated = energy radiated per second
P = 4pr2 (sT4) \ P = P r T r T 2 1 22 24 12 14 P2 2 4 450 1 2 2 4 = ×( ) = P2 = 4 × 450 = 1800 watt
22. (b) : According to Newton’s law of cooling
T T t K T T T 1 2 1 2 0 2 − = + − \ 61 59 4 61 59 2 30 30 − = + − = K K or 60 K = 1 ⇒ K = 1/60 Now 51 49− = 51 49+2 −30 20 = t K K =20 1×60 = 1 3 \ t = 3 × 2 = 6 min 23. (a) : Here, lm = 289.8 nm = 289.8 × 10–9 m s = 5.67 × 10–8 W m–2 K–4 b = 2898 mm K = 2898 × 10–6 m K If T is the temperature of star, then according to Wien’s law, lm T = b
T b m = = × × = − − l 2898 10 289 8 10 10 6 9 4 . From Stefan’s law, E = sT4
= 5.67 × 10–8 × (104)4 = 5.67 × 108 W m–2
24. (a) : Temperature on Celsius scale and Fahrenheit scale are related as
TC −0=TF− 100
32 180
If the temperature is T at which the readings of two scales coincide, then from
T T T 100 32 180 40 = − \ = − ° C 25. (d) : Radiating power, P∝(T4−T04)
Here, T is temperature of the black body and T0 is
temperature of surrounding. \ = − − P P T T T T 2 1 24 04 14 04 \ = − − P2 4 4 4 4 60 1500 500 1000 500 ( ) ( ) ( ) ( ) or P2 80 15 60 = × = 320 W 26. (a) : Here, Dx = 4 mm = 4 × 10–3 m, A = 5 cm2 = 5 × 10–4 m2, DT = 32°C Heat transmit per hour
D D Q t =200 kcal h−1 = × × × − 200 1000 4 2 60 60 . J s 1= 233.33 J s–1 As DDQt KA TDD x =
\ Thermal conductivity of material K Q t A T x = D D D D / ( / ) or K = × × × × − − 233 33 4 10 5 10 32 3 4 . = 58.33 W m–1°C–1
27. (c) : Let x kg of ice melts. Using law of calorimetry,
heat lost by copper = heat gained by ice \ 2 × 400 × (500 – 0) = x × 3.5 × 105
or x = × ×
× 2 400 500
3 5 10. 5 = 87kg
28. (a) : According to Wien’s displacement law lmaxT = constant l l max max 2 1 1 2 = T T l l m max max . 2 2 4 2000 2400 4 2024 3 3 = or = × = m 29. (c) 30. (c) : As 100C = −F18032 \ F =9C+ 5 32
Thus the graph between °C and °F is a straight line with positive intercept on y-axis as shown in the figure above.
T
here are plenty of technologies available in the market to treat water. But most
of them are capable of treating only tap water a
nd are either membrane-based or use chemicals
. In remote villages or areas hit by natural disasters, po
nds and wells still remain the primary sou
rce of water, but this is not fit for drinking.
Traditional water-trea
tment technologies are not designed to treat very dirty or waste
water. Membranes need to be changed and chemicals m
ust be disposed of in a safe manner. At the Flexible Ele
ctronics Lab at the IISc, we have been working on the id
ea of developing a more efficient water tre
atment technology since 2010; a small, portable, maybe even a ha
nd-held device that could convert very dirty w
ater into drinkable water. It was only a year later that we seriously sta
rted working on the project. We used electric fields to achieve the objective. When an electric field is established
in a container with water, the impurities and small pa
rticles get polarised and are then attracted to each oth
er. Smaller particles coalesce into larger granu
les while remaining suspe nded in water. These granules can the
n be filtered easily, using a normal tea filter or
a clean cloth. For our tests, we used water from
Mavalipura, near Bangalore, where the w
ater quality is really bad, and the results we achieved were
satisfactory. There was no new phys
ics to the experiment that we performed, as the problem
was essentially an engineering
issue. The key was to u
nderstand how electric fields permeate ion rich fluids and how
they interact with neutral impurities. This understa
nding helped us identify the field strengths neede
d to maximise this interaction and then use this conc
ept to improve the filtration process.
We have managed to develop a t
echnology that can convert very dirty water — from pon
ds, lakes, wells and even waste water — into drinkable quality. Unlike in the traditional filtration technologies, there is zero wastage
of water. In fact, even the waste w
ater can be reused. The entire instrument can be encased
in a one-litre bottle and needs very little po
wer to run, either a battery or hand-held dynamo will serve the
purpose. The instrument can filter one liter of w
ater in a maximum time of about 10 minu
tes, though there have been occasions where one litre of water
has been treated within three minutes.
So far, we have tested the newly
developed water purifying system in labo
ratories and now we are trying to take the technology to the field.
We estimate that a bottle filter could cost a
bout Rs 1,000, but the prices will come down substantially as we in
crease the capacity of the instrument. The new filter and
our ideas on how to introduce the technology to so
ciety was recently awarded the first place at a compe
tition organised by Google in Zurich, Switzerland. Courtesy : Indian
express
One d rop
at a time
:
A fast, cheap
and
effective
water p
urifier
Sanjiv Sambandan, Flexible electronics Lab, Dept. of Instrumentationand Applied Physics, IISC , bangalore
Finding a way to replace membrane, chemical-based
purifiers with electric fields to filter very dirty water
and make it fit for drinking
