These are the practice questions with answers. Also see other numericals from books.

Fluid dynamics

1. Water flows through a horizontal pipe having a tapering bore. The velocity of water is 2

m/sec at the broader end and the pressure is 1 kilo-newton/m2 less at the narrow end. What is

the velocity of water at the latter end ?

Solution: Given: 𝑣1 = 2 𝑚/ sec , 𝑝1 = 𝑝 , 𝑣2 =? 𝑝2=(𝑝 − 103)𝑁/𝑚2 , Take density of water

ρ= 1000 𝑘𝑔/𝑚3.

From Bernoulli’s equation: 𝑝1

𝜌+

𝑣12

2=

𝑝2

𝜌+

𝑣22

2⇨

𝑝

𝜌+

22

2=

𝑝−103

𝜌+

1

2𝑣2

Solving, we get 𝑣2 = 6𝑚

𝑠.

1. A horizontal pipe of a non-uniform bore has water flowing through it such that the velocity

of flow is 40 cm/sec at a point where the pressure is 2 cm of mercury column. What is the pressure at a point where the velocity of flow is 60 cm/ sec? (Take g= 9.8 m/s2 and density

of water = 1 gm/c.c., density of mercury 13.6 gm/cm3 )

Given: 𝑣1 = 40𝑐𝑚

𝑠𝑒𝑐 , 𝑝1 = 2

𝑐𝑚

𝑠𝑒𝑐= 2 × 13.6 × 980. 𝑝2 =? , 𝑣2 = 60 𝑐𝑚 /𝑠𝑒𝑐. ρ = 1 gm /sec. g=

980 cm/sec2 . From Bernoulli’s theorem, 𝑝1

𝜌+

𝑣12

2=

𝑝2

𝜌+

𝑣22

2= 2 × 13.6 × 980 +

1

2402 = 𝑝2 +

1

2602 ⇨ 𝑝2 = 25650

𝑑𝑦𝑛𝑒𝑠

𝑐𝑚2 = 2565𝑁

𝑚2 (∵ 1 dyne/cm2= 10−1 N/m2 )

2. A venturimeter is connected to a horizontal main of radius 20 cm. If the radius of the throat

of the venturimeter be 15 cm and the difference of water level in the piezometer tubes be

10 cm, calculate the rate of flow of water per hour through the main.

Given : 𝑟1 = 20 𝑐𝑚 , 𝑟2 = 15 𝑐𝑚, ℎ = 10 𝑐𝑚 , 𝑔 = 980 𝑐𝑚

𝑠2 . Rate of water flowing is Q=

𝑎1𝑣1 = 𝑎1𝑎2√2𝑔ℎ

𝑎12−𝑎2

2 = 𝜋𝑟12𝜋𝑟2

2√2𝑔ℎ

𝑎12−𝑎2

2 = 3.14 × 3.14 × 400 × 225√2×980×10

3.14×400−3.14×225=

5.29 × 106 𝑐𝑚3

𝑠= 5.29 × 103 𝑙𝑖𝑡

𝑠𝑒𝑐= 19.04 × 106 𝑙𝑖𝑡

ℎ𝑜𝑢𝑟 . (Check it)

3. A flat plate of area 20 sq cm is placed on a horizontal surface coated with a layer of glycerine

1 mm thick. What force must be applied to the plate to keep it moving with a speed of 1

cm/sec over the horizontal surface ? (Coefficient of viscosity of glycerine=20 gm cm-1 sec-1).

Solution: A=20 cm2 , dx=1 mm=0.1 cm, dv= 1cm/sec, η=20 gm cm-1 sec-1.

According to Newton’s law of viscosity, viscous force F= 𝜂𝐴𝑑𝑣

𝑑𝑥= 20 × 20 ×

1

0.1=

4000 𝑑𝑦𝑛𝑒𝑠 = 0.04 𝑁.

4. In an experiment with Poiseuille’s apparatus the following figures were obtained:

Volume of water issuing per min=7.08 c.c., Head of water =34.1 cm. Length of tube =56.45 cm.

Raius of the tube=0.0514 cm. Find the coefficient of viscosity.

Solution: Given Q=7.08 c.c per minute=7.08/60 c.c. per sec, Pressure difference P=34.1

cm=34.1×980×1 dynes/ cm2 (Using p=ρgh, where ρ be density of water), r=0.0514 cm, 𝑙 =

56.45 𝑐𝑚.

Then Poiseuille’s formula : 𝑄 =𝜋𝑃𝑟4

8𝜂𝑙 ⇨𝜂 =

𝜋𝑃𝑟4

8𝑄𝑙=

3.14×34.1×980×1×0.05144

8×7.08

60×56.45

=

0.014 𝐶. 𝐺. 𝑆. 𝑢𝑛𝑖𝑡(𝑃𝑜𝑖𝑠𝑒) = 0.0014 𝑆. 𝐼. 𝑢𝑛𝑖𝑡. (𝑘𝑔 𝑚−1𝑠−1).

5. Calculate the mass of water flowing in 10 minutes through a tube 0.1 cm in diameter, 40 cm

long. If there is a constant pressure head of 20 cm of water. The coefficient of viscosity of

water is 0.0089 C.G.S. units [Take density of water 1000 kg/𝑚3]

Solution: Given: radius of tube= 0.1/2 = 0.05 cm, length 𝑙 = 40 𝑐𝑚, Pressure head=20 cm

of water=20 × 1 × 981 dynes/ cm2, η=0.0089 C.G.S. units. Then Poiseuille’s formula ,

volume rate of flow of water given by

Q=𝜋𝑃𝑟4

8𝜂𝑙=

3.14×20×1×981×(0.05)4

8×0.0089×40= 0.135

𝑐.𝑐.

𝑠𝑒𝑐.

∴ volume of water flowing out in 10 minutes =0.135 ×10 ×60

Mass of water flowing out in 10 minutes= density × volume=1×0.135 ×10 ×60 =81

c.c.=81 ×10−6𝑚3.

6. A water drop of radius 0.01 cm is falling through air. Find its terminal velocity. Neglect air

density. (η for air is 1.8 × 10−4𝐶. 𝐺. 𝑆. units.

Solution: radius of water drop, r=0.001 cm, density of water= 1 gm/c.c., air density σ=0,

η=1.8 × 10−4 . Then from formula 𝑣 =2

9

(𝜌−𝜎)𝑟2𝑔

𝜂=

2

9

1×0.012×980

1.8 × 10−4 =1.2 cm /sec.

7. Two equal drops of water are falling through air with a steady terminal velocity of 5 cm/sec.

If the drops coalesce, what will be the new terminal velocity?

Solution: terminal velocity of small drops v=5 cm/sec. If they coalesce, let R be the radius

of larger drop, with r be the radius of small drop. Then

Volume of larger drop = 2 volume of small drop.

Or, 4

3𝜋𝑅3 = 2 ×

4

3 𝜋 𝑟3

Or, 𝑅 = 21/3𝑟.

Let terminal velocity for large drop be V , then

𝑣 =2

9

(𝜌−𝜎)𝑟2𝑔

𝜂

𝑎𝑛𝑑 𝑉 =2

9

(𝜌−𝜎)𝑅2𝑔

𝜂

Dividing both equations 𝑣

𝑉=

𝑟2

𝑅2

Or, 5

𝑉=

𝑟

22/3𝑟 ⇨V=22/3 × 5 = 7.93

𝑐𝑚

𝑠𝑒𝑐= 7.93 × 10−2 𝑚/𝑠 .

8. Determine the radius of the drop of water through air, if the terminal velocity of the drop is

1.2 cm/sec. Assume the coefficient of viscosity for air = 1.8 ×10−4 poise, and density of

air=1.21 x 10−3 gm/c.c.

Solution: We have formula for terminal velocity: 𝑣 =2

9

(𝜌−𝜎)𝑟2𝑔

𝜂 , where r=radius =?, density

of air σ= 1.21 x 10−3 gm/c.c. , 𝑣 = 1.2cm

sec, η = 1.8 × 10−4 poise, g=980 cm/s2 , density of

water ρ=1 gm/c.c.

Then, 1.2 =2

9

(1−1.21 x 10−3)𝑟2×980

1.8 ×10−4 ⇨𝑟 = 9.965 × 10−4cm.

Rotational motion

1. Two spheres are identical in mass and volume but one is hollow and the other is solid. How will

you identify them experimentally? Solution: To identify we let to slide both in an inclined plane. The acceleration down the plane

will be then 𝑎 = (𝑅2

𝐾2+𝑅2) 𝑔 sin 𝛳…. (i)

For solid cylinder 𝐾2 =2

5𝑅2 , while that of hollow sphere 𝐾2 =

2

5(

𝑅5−𝑟5

𝑅3−𝑟3) =2

5 𝑅5

𝑅3 (1−

𝑟5

𝑅5

1−𝑟3

𝑅3

)

Since 𝑟5

𝑅5 <𝑟3

𝑅3 , 1 −𝑟5

𝑅5 is greater than 1 −𝑟3

𝑅3 . Hence K for hollow sphere is greater . Hence the

acceleration from equation (i) of solid sphere is greater that means solid comes quicker on the

ground. In this way we identify whether the spheres are hollow or solid.

2. A wheel of radius 6 cm is mounted so as to rotate about a horizontal axis through its center. A

string of negligible mass, wrapped round its circumference, carries a mass of 200 gm attached to

its free end. When let fall, the mass descends through 100 cm in the first 5 seconds. Calculate

the angular acceleration of the wheel and its moment of inertia. Take g=9.8 m/s2 Solution: Given radius R=6 cm= 0.06 m, mass m= 200 gm= 0.2 kg, descended distance h=100

cm= 1 m, time t= 5 sec. Angular acceleration α=? moment of inertia 𝐼 =?

From 𝑠 = 𝑢𝑡 +1

2𝑎 𝑡2= 0 +

1

2𝑎 𝑡2 ⇨ 𝑎 =

1

52 = 0.04m

𝑠2 . Also v = at = 0.04 × 5 =

0.2m

s. Now angular velocity ω =

v

R=

0.2

0.06= 3.33

rad

sec.

Also a = αR which gives α =𝑎

𝑅=

0.04

0.06=

2

3rad/sec2. Now, loss in P. E = gain in K. E.

So, mgh =1

2𝐼𝜔2 +

1

2𝑚𝑣2 ⇨ 𝑚𝑔ℎ −

1

2𝑚𝑣2 =

1

2𝐼𝜔2

Hence 𝐼 =2𝑚𝑔ℎ−𝑚𝑣2

𝜔2 =2×0.2×9.8×1−0.2×0.22

3.332 = 0.35 𝑘𝑔𝑚2.

3. A solid sphere of mass 100 gm and radius 2.5 cm rolls without sliding with a uniform velocity of

10 cm/sec along a straight line on a smooth horizontal table. calculate its total energy. Solution: mass m=100 gm =0.1 kg, radius R=2.5 cm= 0.025 m, velocity v= 10 cm/sec= 0.1 m/sec.

Total energy of rolling 𝐸 =1

2 𝑚𝑣2 (1 +

𝐾2

𝑅2).

For solid sphere moment of inertia I = m𝐾2 =2

5𝑚𝑅2 ⇨ 𝐾2 =

2

5𝑅2.

Hence 𝐸 =1

2 0.1 × 0.12 × (1 +

2

5) = 7 × 10−4joule.

4. A symmetrical body is rotating about its axis of symmetry. Its moment of inertia about the axis

of rotation being 1 kg/m2 and its rate of rotation 2 rev/sec. (a)What is its angular momentum?

(b) What additional work will have to be done to double its rate of rotation?

Solution: Moment of inertia, 𝐼 = 1 𝑘𝑔/𝑚2, frequency 𝑓 = 2rev

sec, angular momentum j =?

𝑗 = 𝐼𝜔 = 𝐼2𝜋𝑓 = 1 × 2𝜋 × 2 = 4π kg m2/sec

The work done W= ½ 𝐼𝜔2 =1

2× 1 × (2𝜋 × 2)2= 8𝜋2joule.

If frequency is doubled , 𝑊′ =1

2𝐼(2𝜋 × 2 × 2)2 = 32 𝜋2joule. Hence additional work=

32𝜋2 − 8𝜋2 = 24 𝜋2joule.

5. A solid cylinder (a) rolls (b) slides from rest down an inclined plane. Neglect friction and compare

the velocities in both cases when the cylinder reaches the bottom of the inclined plane. Solution: when a cylinder rolls down the inclined plane , its acceleration will be

𝑎 = (𝑅2

𝐾2+𝑅2) 𝑔 sin 𝛳 , where 𝐾2 =𝑅2

2 [about its own axis]. Hence 𝑎 =

2

3 𝑔 sin 𝛳 . Now

velocity 𝑣12 = 0 + 2𝑎𝑠 =

4

3 𝑔𝑠 sin 𝛳 , where initial velocity u is zero and 𝑠 be the length of the

inclined plane. For the cylinder sliding, we have acceleration 𝑎 = 𝑔 sin 𝛳 and velocity 𝑣22 = 0 +

2𝑎𝑠 = 2 𝑔𝑠 sin 𝛳 . Hence 𝑣2

𝑣1= √

6

4= 1.225. ⇨𝑣2 = 1.225 × 𝑣1

6. A uniform sphere of mass 2 kg and radius 10 cm is released from rest on an inclined plane which

makes 300 angle with the horizontal. Deduce (i) its angular acceleration (ii) linear acceleration

along the plane and (iii) kinetic energy as it travels 2 meters along the plane. Take g=9.8 m/s2.

Solution: Mass of the sphere m= 2 kg, radius R=10 cm= 0.1 m, angle of inclination ϴ=300. Length

of travel along inclined plane 𝑠 = 2 m.

Linear acceleration 𝑎 = (𝑅2

𝐾2+𝑅2) 𝑔 sin 𝛳 where 𝐾2 =2𝑅2

5 [about its diameter].

Hence 𝑎 =5

7 𝑔 sin 𝛳 =

5

7× 9.8 × sin 300 = 3.5 𝑚/𝑠2. Since 𝑎 = 𝛼𝑅 ⇨ 𝛼 =

𝑎

𝑅=

3.5

0.1=

35 𝑟𝑎𝑑/𝑠𝑒𝑐2.

Gain in kinetic energy= loss in potential energy

Or, 𝐾. 𝐸 = 𝑚𝑔𝑠 sin 𝛳 = 2 × 9.8 × 2 × sin 300 = 19.6 joule.

7. A flywheel of mass 10kg and radius 20cm is mounted on an axle of mass 8 kg and radius 5 cm. A

rope is wound round the axle and carries a weight of 10kg. The flywheel and the axle are set into

rotation by releasing the weight. Calculate (i) the angular velocity and the kinetic energy of the

wheel and the axle and (ii) the velocity and kinetic energy of weight when the weight has

descended 20 cm from its original position. (g=9.8 m/s2)

Solution: mass of the weight W=10 kg, mass of the flywheel M=10 kg, mass of the axle m=8 kg,

height through which weight is descended h=20 cm=0.2 m, let I be the moment of inertia of

flywheel and axle with radius of the wheel R=20 cm=0.2 m, radius of the axle r= 5 cm=0.05 m.

Here I = M (𝑅2+𝑟2

2) +

1

2𝑚𝑟2 [∵ flywheel is taken similar to annular disc and axle is taken as

solid cylinder and moment of inertia of flywheel is about the axis passing through its centre and

perpendicular to its plane, moment of inertia of axle is about the axis of cylindrical symmetry]

Or, I = 10 (0.22+0.052

2) +

1

2× 8 × 0.052 = 0.212 + 0.01 = 0.222 kg 𝑚2.

Linear velocity of the weight (W), 𝑣 = 𝑟𝜔 = 0.05𝜔 , where ω be the angular velocity of

wheel and axle. According to conservation of energy

Loss of potential energy by weight= gain in kinetic energy of wheel and axle.

Or, 𝑊gℎ =1

2𝐼𝜔2 +

1

2 𝑊𝑣2

Or, 10 × 9.8 × 0.2 =1

2× 0.222 × 𝜔2 +

1

2× 10 × (0.05𝜔)2

Or, 0.1235𝜔2 = 19.6 ⇨ 𝜔 = 12.59 rad/sec .

Kinetic energy of weight= ½ W𝑣2 =1

2 × 10 × (0.05 × 12.59)2 = 1.40 joule.

Velocity 𝑣 = 𝑟𝜔 = 0.05 × 12.59 = 0.63m

s.

8. A flywheel in the form of a solid circular disc of mass 500 kg and radius 1 meter is rotating ,

making 120 rev/min. Compute the kinetic energy and the angular acceleration if the wheel

is brought to rest in 2 seconds, friction is neglected.

Solution: mass m=500 kg , radius R=1m, frequency f = 120rev

min ⇨ angular speed 𝜔0 =

2π f = 2 × 3.14 ×120

60= 4π rad/sec.

Final angular speed =0, time to brought rest (t)= 2 sec.

Hence 0 = 𝜔0 + 𝛼𝑡

Or, 𝛼 = −4𝜋

2= −2

rad

𝑠𝑒𝑐2 , negative sign indicates the retardation.

Moment of inertia I =m𝑅2

2=

1

2× 500 × 12 = 250

kg

𝑚2

Kinetic energy E =1

2𝑚 𝜔0

2 =1

2 × 500 × (4𝜋)2 = 3.94 × 104 joule.

9. Calculate the Coriolis force on a mass of 50 gm placed at a distance of 10 cm from the axis of

a rotating system if the angular speed of the frame is 10 rad/sec.

Solution: Mass m= 50 gm= 50 ×10−3𝑘𝑔, distance r= 10 cm = 10 ×10−2𝑚, angular speed

ω= 10 rad/sec. The Coriolis force is given by F= 2mωv = 2mωrω = 2mr𝜔2 = 2 × 50 ×

10−3 × 10 × 10−2 × 102 = 10−3N.

10. An annular disc of mass 0.2 kg and radii 0.2 m and 0.25 m rolls such that the centre has

velocity of 0.5 m/s. Calculate its kinetic energy.

Hint: Total kinetic energy 𝐸 = kinetic energy due to rotation(𝐸1) + kinetic energy due to linear motion(𝐸2).

=1

2 𝑀𝑣2 +

1

2𝐼𝜔2, where 𝐼 =

1

2 𝑀(𝑅2

2 + 𝑅12) =

1

2 × 0.2 × (0.22 + 0.252) = 1.025 ×

10−2 kg m2and 𝜔 =𝑣

𝑅2=

0.5

0.25= 2

rad

s

⇨E =1

2 × 0.2 × 0.52 +

1

2× 1.025 × 10−2 × 22 =0.0455 J.

Gravity

1. Show that the intensity and potential at any point on the surface of the earth are g

and g R respectively, assuming the earth to be a uniform sphere.

Solution: Let the earth be a uniform sphere of radius R and mass M. Then

gravitational intensity at any point on its surface , E= −𝑀

𝑅2 𝐺

But gravitational force on mass m is given by F=−𝐺𝑀𝑚

𝑅2

Now gravitational intensity is force per unit mass i.e. E=𝐹

𝑚= −𝐺

𝑀

𝑅2 = 𝑔

Also potential at a point on the surface is given by V= − (𝑀

𝑅𝐺) = − (

𝑀

𝑅2 𝐺) 𝑅 = 𝑔𝑅

2. A smooth straight tunnel is bored through the earth and a small particle is allowed to

move in it from a positon of rest.

Find the periodic time of one

vibration. Given that G== 6.67 ×

10−11 𝑁𝑚2𝑘𝑔−2 , and mean

density of earth =5.6 gm/c.c.

Solution :

Let a tunnel AB be bored through

the earth, fig 2.7and let a particle

of mass m be placed in it a P such

that its distance from the center O

of the earth is r and that from the

midpoint C of the of the tunnel and

let the angle OPC be ɵ.

Gravitational force on the particle

of it is given by

F=

−𝑚(𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝ℎ𝑒𝑟𝑒 𝑜𝑓 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟)𝐺

𝑟2

Figure 2.7 : calculation of time period of the particle inside the tunnel .

Where negative sign indicates the force is directed towards O, the center of the

sphere.

Or, F= −𝑚(

4

3𝜋𝑟3𝜌)

𝑟2 𝐺 = −4

3𝜋𝑟𝜌𝑚𝐺.

Component of force acting along PC

= 𝐹 cos ɵ = −4

3π𝑟ρ𝑚𝐺 cos ɵ = −

4

3π𝑟ρ𝑚𝐺

𝑥

𝑟= −

4

3π𝑟ρ𝑚𝐺𝑥.

Now acceleration of the particle towards C , (a)=𝑓𝑜𝑟𝑐𝑒

𝑚𝑎𝑠𝑠= −

4

3𝜋𝜌𝑚𝐺𝑥

𝑚= −

4

3𝜋𝜌𝐺𝑥

Hence a= −𝜇𝑥 ,where 4

3𝜋𝜌𝐺 = 𝜇 𝑏𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

Hence

a∝ −𝑥

Hence, acceleration is directly proportional to its displacement from C and is directed

towards it. It therefore, executes a simple harmonic motion about C. Its time period is

given by

T= 2𝜋√𝑙

𝜇= 2𝜋√

3

4𝜋𝜌𝐺= √

3𝜋

𝜌𝐺= √

3𝜋

5.6×103 ×6.67×10−11 = 5023.1 𝑠𝑒c.

3. One uniform shell of mass m2 lies inside and concentric with a larger uniform shell of

mass m1. Obtain the gravitational

field due to the system (i) at a point

out side the two shells (ii) at a point

in the space between two shell (iii) at

a point inside the smaller shell.

Solution : Fig 2.8 shows the two shells

with their common center O ,and let

points P1, 𝑃2, 𝑎𝑛𝑑 𝑃3 lie outside the

two shells, in between the two shells

and inside the inner shell at distances

r1, r2 , and r3 respectively.

(i)Gravitational intensity at a point P

at distance r1=Sum of intensities due

to two shell

Or, E=−𝑚1𝐺

𝑟12 + (−

𝑚2

𝑟12 ) 𝐺 =

− (𝑚1+𝑚2

𝑟12 ) 𝐺

(ii) For point P2 lies between the two shells: since P2 lies inside the outer shell , so

intensity due to it is zero, but there is contribution from the inner shell i.e.

E= 0 + (−𝑚2

𝑟22 ) 𝐺 = (−

𝑚2

𝑟22 )

(iii) Since P3 lies inside both the shell, total gravitational intensity at this point is equal

to zero.

(4). Obtain value of escape velocity for an atmospheric particle above the surface of

Figure 2.8: calculation of gravitational field at different points of the shell.

earth.[Given mass of earth Me= 5.98 × 1024 𝑘𝑔, Radius R=6.37 × 106𝑚 , 𝐺 = 6.67 ×

10−11 𝑁𝑚2/𝑘𝑔2]

Solution : 𝑣𝑒 = √2𝑀𝐺

𝑅′ , 𝑅′ = 𝑅 + 1000 𝐾𝑚 = 7.37 × 106𝑚 Hence, 𝑣𝑒 = √2𝑀𝐺

𝑅′ =

1.04 × 104𝑚/𝑠𝑒𝑐.

(5). Estimate the mass of sun , assuming orbit of the earth round the sun to be circle. The

distance between sun and earth is 1.49 ×1013 cm and G= 6.68 × 10-12 SI unit.

Solution:

Since gravitational force between sun and earth is equal to the centripetal force

𝐺𝑀𝑚

𝑟2 =𝑚𝑣2

𝑟 ,

𝑂𝑟, M=𝑣2𝑟

𝐺 , where v=

2𝜋𝑟

𝑇.

𝑂𝑟, M=4𝜋2𝑟3

𝑇2𝐺 =1.972 ×1022 kg.

(6). Calculate the gravitational self energy of (i) the sun (ii) the earth-sun system, given that

the mass of the sun = 2 x 1030 kg and its radius = 7×108 meters, mass of the earth = 6 ×1024

kg and mean earth- sun distance = 1.5 ×108 km. Take G= 7×10-11 N-𝑚2/𝑘𝑔2.

Solution:

Self energy of the sun, 𝐸𝑠 = − 35

𝑀𝑠2

𝐺𝑅𝑠

,where 𝑀𝑠 is mass of the sun= 2 x 1030 kg, 𝑅𝑠 is the

radius of the sun. Self energy of the earth-sun system , 𝐸𝑒𝑠 = −𝑀𝑠𝑀𝑒

𝑟𝑒𝑠𝐺, where 𝑀𝑒 be the

mass of earth=6 ×1024 kg, 𝑟𝑒𝑠 be the mean distance between sun and earth=1.5 ×108 km=1.5

×1011 m. Putting all these values , we get 𝐸𝑠 = −2.4 × 10−41joules and Ees = −5.6 ×

1033joules.

(7). Calculate the period of revolution of Neptune round the sun given that the diameter of

the orbit is 30 times the diameter of the earth orbit round the sun, both orbits being assumed

to be circular.

Solution: Let 𝑟1 and 𝑟2 be the radii of the orbit of earth and Neptune respectively. Then

𝜏22

𝜏12 = (

𝑟2

𝑟1)

3 , where

𝑟2

𝑟1= 30, 𝜏1 = 1 year.

⇨𝜏2

1= 303/2

∴ 𝜏2 = 164.3 years.

Elasticity

1. Find the amount of work done in twisting a steel wire of radius 1.00 mm and length 25 cm through

an angle of 450. Given η for steel =8×1010𝑁/𝑚2

Solution : radius r= 1 mm =1×10−3𝑚 , length L= 25 cm = 25× 10−2𝑚, η =8×1010𝑁/𝑚2,

Angle, ɵ=450 =𝜋𝑐

180× 45 = 0.785 𝑟𝑎𝑑𝑖𝑎𝑛.

Now, work done in twisting the wire through angle ɵ , W= ½ Cɵ2 ,

where C is twisting couple per unit twist= 𝜋𝑟4𝜂

2𝐿=

3.14×(1×10−3)4×8×1010

2×25× 10−2 =0.50 N-m .

Hence W= ½ ×0.5 ×0.7852 = 0.154 𝐽𝑜𝑢𝑙𝑒.

2. A wire 300 cm long and 0.625 square cm in cross-section is found to stretch 0.3 cm under a tension

of 1200 kg. What is the Young’s modulus for the material of the wire.

Solution :

Given: Length of wire L = 300 cm = 300 ×10−2m. Area A= 0.625 ×10−4𝑚2. Elongation l= 0.3cm =

0.3 ×10-2 m.

Tension F= 1200 Kg weight= 1200 ×9.8 =11760 N.

Then Y= 𝐹𝐿

𝐴𝑙=

11760×300×10−2

0.625×10−4×0.3 ×10−2=1.88×1011 N/𝑚2

3. Find the ratio of the adiabatic and isothermal elasticities of a gas.

Solution: If a pressure dP is applied to a gas having original volume V then, let dV the decrease

in volume , then strain produced is −𝑑𝑉

𝑉 . Hence Bulk modulus

K=𝑠𝑡𝑟𝑒𝑠𝑠

𝑠𝑡𝑟𝑎𝑖𝑛= −

𝑑𝑃𝑑𝑉

𝑉

= −𝑉𝑑𝑃

𝑑𝑉,which is the elasticities of gas.

If the gas is compressed under isothermal conditions, i.e. if the temperature remains constant

then, PV=k(constant)

Or, P=𝑘

𝑉, … (1)

Differentiating both side with respect to V we get, dP

dV= −

𝑘

𝑉2

Or, VdP

dV= −𝑉

𝑘

𝑉2 = −𝑘

𝑉 …. (2)

Hence from (1) and (2)

−VdP

dV = P .Hence isothermal elasticities of gas is equal to the pressure P

Now to calculate the adiabatic elasticities , we have adiabatic equation P𝑉𝛾 = 𝑘.

Or, P=𝑘

𝑉𝛾 … . . (3). Differentiating both side with respect to V we get, 𝑑𝑃

𝑑𝑉= −𝛾𝑘𝑉−𝛾−1 = −𝛾𝑘𝑉−𝛾𝑉−1

Or, −𝑉𝑑𝑃

𝑑𝑉= 𝛾

𝑘

𝑉𝛾 = 𝛾𝑃 ,from equation (3).

Hence adiabatic elasticities is γP

Now the ratio of two elasticities becomes 𝛾𝑃

𝑃= 𝛾.

4. Find the greatest length of a steel wire that can hang vertically without breaking.( Breaking

stress for steel= 7.9 x 108 N/m2 , Density of steel =7.9 x 103kg/m3)

Solution: When steel wire is hanging, then its weight = mg= ρLAg , where ρ be the density , L is

length , A is cross-section . Then stress = Force/area = ρLAg/A=L ρ g.

According to question,

L ρ g=breaking stress = 7.9 x 108 N/m2

Or, L=7.9 x 108

7.9×103×10= 104𝑚

5. A wire 4 meters long and 0.3 mm in diameter is stretched by a force of 80 N . If the extension in

length amounts to 1.5 mm , calculate the energy stored in the wire. If we take another wire of

cross section 1 sq.mm and length 2 meters , then calculate the work done in stretching a wire

through 0.1 mm if young's modulus for the material of the wire is 2 x 1011 N/m2..

Solution : Force = 80 N and extension (l)= 0.1mm= 0.1 ×10−3𝑚 𝑚. Then energy stored = ½ F x l =

½ ×80×1.5 𝑥 10−3= 0.06 Joule. For second case , cross-sectional area A= 1 mm2=10-6 m2. Length

L= 2m, extension l= 0.1 mm =

0.1 ×10-3 m. Young modulus Y= 𝐹/𝐴

𝑙/𝐿= 2 × 1011.

Or, F= 𝑌𝐴𝑙

𝐿 . Now work done W=

1

2𝐹 × 𝑙 =

1

2

𝑌𝐴𝑙

𝐿× 𝑙 =

1

2

2 ×1011×10−6

2(0.1 × 10−3)2 = 5 ×

10−4Joule.

6. A metal wire of length 3 meters and diameter 1mm is stretched by a weight of 10 kg . If Young's

modulus for its material be 12.5 x 1010 N/m2 and σ for it equal to 0.26, calculate the lateral

compression produced. (g=9.8 m/s2).

Solution : Given:

Diameter D=1 mm=1×10−3𝑚, length L= 3 m. Applied force F= 10 kg weight= 10 ×10=100 N.

Young’s modulus Y=12.5 ×1010N/m2 , Poisson’s ratio σ=0.26 .

Y= 𝐹/𝐴

𝑙/𝐿. Or, l/L=

𝐹/𝐴

𝑌= longitudinal strain., where A= π𝑟2 = 𝜋(0.5 × 10−3)2m2=0.785 ×10-6m2.

σ=𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛

𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =

𝑑/𝐷

𝑙/𝐿 .

Or, d/D= σ longitudinal strain =0.26×𝐹/𝐴

𝑌=

0.26×100

12.5 ×1010× 0.785 ×10−6 = 2.64 × 10−4

Or, d=2.64 × 10−4D= 2.64 × 10−4 × 1 × 10−3 = 2.64 × 10−7𝑚

7. A gold wire 0.32 mm in diameter , elongates by 1 mm when stretched by a force of 330 gm wt

and twists through 1 radian when equal and opposite torques of 1.45×10−5 Nm are applied at

its ends. Find the value of Poisson's ratio for gold. (g=9.8 m/s2 ).

Solution: Given, Twisting couple per unit twist , C=1.45×10−5Nm, diameter d =0.32 mm , radius=

0.16 ×10-3m, Length of the wire =L. Force applied F=330 gm wt= 0.33kg wt=0.33×9.8 N.

We know the relation , torsional rigidity η=𝑌

2(1+𝜎) , Or, 1+σ=

𝑌

2𝜂 . Or, σ=

𝑌

2𝜂− 1…………… (i)

where Young’s modulus Y=𝐹/𝐴

𝑙/𝐿=

𝐹𝐿

𝐴𝑙=

0.33×9.8×𝐿

𝜋(0.16 ×10−3)2×10−3 = 4.0 × 1010𝐿.

Also we have relation, C= 𝜋𝜂𝑅4

2𝐿= 1.45 × 10−5 . Or, η=2LC/𝜋𝑅4 =

2𝐿× 1.45×10−5

𝜋(0.16 ×10−3)4=1.41×1010 L

Putting the values of Y and η in equation (i) we get, σ=4×1010−𝐿

2×1.41×1010 L− 1=1.42−1 = 0.428

8. One end of a wire, 2 mm in diameter and 50 cm in length is twisted through 0.8 radian . Calculate

the shearing strain at the surface of the sphere.

Solution: Given diameter = 2 mm , radius r = 1 mm = 1×10−3𝑚., Length L=0.5m, angle ɵ= 0.8

radian ,

Then, Strain at the surface = Rɵ/L=1×10−3×0.8

0.5= 1.6 × 10−3 𝑟𝑎𝑑𝑖𝑎𝑛.

9. A rectangular bar , 2 cm in breadth and 1 cm in depth and 100 cm in length, is supported at the

ends and a load of 2 kg is applied at its middle. Calculate the depression if the Young’s modulus

of the material of the bar is 2×1011N/𝑚2.

Solution : The depression of the bar at middle is given by y=𝑊𝑙3

48𝑌𝐼 , where weight W= mg= 2 ×9.8

N, l=100 cm=1 m, Young’s modulus Y= 2×1011N/𝑚2 , breadth b= 2 cm=2 ×10−2𝑚, depth d=1 cm

=10−2𝑚, and geometrical moment of inertia I= 𝑏𝑑3

12=

2 ×10−2×10−6

12= 1.66 × 10−9. Hence y=

2×9.8×13

48×2×1011×1.66×10−9 = 1.25 × 10−3𝑚.

10. In an experiment , the diameter of the road was 1.26 cm and the distance between the two knife

edges 70 cm . On putting a load of 900 gm at the middle point , the depression was 0.025 cm.

Calculate the Young’s modulus for the substance.

Solution : Given, y=0.025 cm= 0.025 ×10−2𝑚, W= mg= 0.9 ×9.8 , length l= 0.7 m, diameter d= 1.26

cm, radius r= 0.63 ×10−2𝑚. I= π𝑟4/4 =3.14×(1.26×10−2)4

4

y=𝑊𝑙3

48𝑌𝐼. Or, Y=

𝑊𝑙3

48𝑦I=

0.9××9.8 ×0.73×4

48× 0.025 ×10−2×3.14×(0.63×10−2)4 = 2.0387 × 1011N/𝑚2.

11. Compare the loads required to produce equal depressions for two beams of same material ,

length and weight when one has a circular cross-section and the other has a square cross-section.

Solution: y=𝑊𝑙3

48𝑌𝐼=

𝑊′𝑙3

48𝑌𝐼′ Or, 𝑊

𝑊′ =𝐼

𝐼′ =𝜋𝑟4/4

𝑏𝑑3/12 ….. (1).

Since the weights are equal , volumes are also equal because density is same for both. So, 𝜋𝑟2𝐿 =

𝑏𝑑𝐿. For square cross-section b=d. So, 𝜋𝑟2𝐿 = 𝑏2𝐿. Hence 𝜋𝑟2 = 𝑏2. Hence equation (1)

becomes 𝑊

𝑊′ =𝜋𝑟4/4

𝑏4/12=

3𝜋𝑟4

𝜋2𝑟4 =3

𝜋 .

12. A Steel wire of 1 mm radius is bent in the form of a circular arc of radius 50 cm. Calculate the

bending moment. Given Young’s modulus for steel= 2×1011N/𝑚2

Solution : Cross-section radius r= 1 ×10−3𝑚. , radius of circular arc R=50 cm= 0.5 m., Young’s

modulus Y=2×1011N/𝑚2.

we have the relation bending moment M=YI/R , where I be the geometrical moment of inertia

=𝜋𝑟4

4 , for circular cross-section. Hence M=

2×1011×𝜋×(1 ×10−3)4

4×0.5= 0.314 𝑁 − 𝑚.

13. What couple must be applied to a wire one meter long, 1 mm in diameter in order to twist one

end of it, through 900, the other end remaining fixed. Rigidity of material of the wire is 2.8 ×

1010N-𝑚−2.

Hint: Couple for the angular twist θ, 𝐶 =𝜋𝜂𝑅4

2𝐿𝜃, where 𝜃 = 900 =

𝜋

2 radian, η = 2.8 × 1010N-

m−2, Radius, R =1

2 mm = 0.5 × 10−3m and length of the wire 𝐿 = 1𝑚.

Ans. 43.19 × 10−4 N-m.