solved problem of first three chapters by asif rasheed

7/28/2019 Solved Problem of First Three Chapters by asif rasheed

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Ideal Science Academy (156-D Gulfishan colony)

Asif Rasheed BS (HONS) Physics 0344 78 46 394 Page 1

CHAPTER: 1Physical quantities and measurement (Problems)

P1.1) Express the following quantities using prefixes.

Solution:(a) 5000gAs 1000g = 1kg so 5000/1000

=5kg Ans

(b)2000000w

As 106

= mega

So 2000000 w

2 x 106

=2MW

(c) 52 x 10

-10

x kg since 1kg =1000g or10

3

g=52 x 10

-10x 10

3g

= 52 x 10-10+3

g

= 52 x 10-7

g

= 5.2 x 101

x 10-7

g

= 5.2 x 10-7+1

g

= 5.2 x 10-6

g =5.2ug Ans

(d) 225 x 10-8

s

=2.25 x 102

x 10-8

s

= 2.25 x 10

2-8

s=2.25 x 10

-6s

= 2.25 us ANS

P1.2) How do prefixes micr

o, nano and pico related to each other.

As we know that,

Micro =10-6

Nano = 10-9

Pico = 10-12

1 p= 1/1000n

1 p= 1/1000000u

1 p= 1/1000u

1 p= 1000n

1 p= 1000000p

P1.3) Your hair grow at the rate of 1mm per day find their growth rate in nms-1

.

As milli = 10-3

Nano = 10-9

1m = 10-6

n OR 1m = 1000000n

By multiplying m on both sides

1mm = 106

nm OR 1mm = 1000000nm

As we know that

One day = 24 hours

One hour = 60 minutes

One minutes = 60s

So

One day = 24 x60 x6

= 86400 s


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So the growth rate in nms-1

is

= 1000000nm/86400s

= 11.57nms-1

Ans

P1.4) Rewrite the followings in standard form:

Solution:

(a)1168 x Solution:

11.68 x 1.168 x x 1.168 x 10

3-27

1.168 x ANS

(b)32 x Solution:

3.2 x x 3.2 x 10

1+5

3.2 x

(c) 725 x

7.25 x x g7.25 x x x 10 -5+3

7.25 x 102-2

g

7.25 x 7.25g ANS

(d) 0.02 x Solution:

0.02 x 10-2

x 10-8

2 x 10-2-8

(as we know that p

owers are added up)

2 x 10-10

ANS

P1.5) Write the following quantities in scientific notation:

Solution:

(a) 6400km

6.4x103km ANS

(b) 380000km

3.8x105

ANS

(c) 300000000ms-1

3.108ms

-1

(d) Seconds in a day:

24x60x60s=86400s

8.64x104s ANS

P1.6) Question on book:

As the zero of Vernier scale is on right so zero error will be positive and if its 4th

division is

conceding with the main scale then the zero error=0.01x4=0.04

Zero error= +0.04cm

And zero correction= -0.04cm

P1.7) A screw gauge has 50 divisions on its circular scale. The pitch of the screw gauge is

0.5mm. What is its least count?

Solution:

Least count= pitch of screw gauge/no. of divisions in circular


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0.5/50=0.01mm

0.01x10-3

m

1x10-5

m

1x10-5

x100cm

1x10-3

cm

0.001cm ANS

P1.8) Which of the following quantities have three significant figures?

Solution:

(a) 3.006m(b)5.05x10-21kg(c) 0.00309kg(d)301.0s

P1.9) What are the significant figures in the following measurements?

(a) 1.009m (It carry all of them 4)(b)0.00450kg

0.00450

It has 3 significant figures

(c) 1.66x10-27kg1.66x10

-27kg

It has 3 significant figures.

(d)2001sIt has 4 significant figures.

P.10) A chocolate wrapper is 6.7cm lon

g, 5.4cm wide. Calculate its area up to reasonable

number of significant figures.

Solution:

Area= length x width

=6.7cm x 5.4cm

= 36.18cm2

Area in significant figure= 36cm2


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CHAPTER: 2

KINEMATICSP2.1) A train moves with a uniform Velocity of 36 for 10S. Find the distance travelled by it.Solution:

(Velocity) V= 36

= 36x1000/60x60= 36000/3600= 10

(Time) t= 10(Distance) S= ?

Formula: S= Vavx t

= (10) x (10)

S= 100 m ANS

P2.2) A train starts from nest. It moves through 1km in 100S with uniform acceleration. What

will be its speed at the end of 100S?

Solution:

(Distance) S= 1km =1000 m

(Time) t= 100 S

(Velocity)Vi= 0 m/s

Vf=?

By using formula:

S= vit+ 1000= 0(t)+ a 1000= a (10000)

2x1000/10000=a

A= 0.2 ANSNow by using first equation of motion:

Vf= v ;+at

Vf= 0=(0.2)(100)

Vf= 20

ANS

P2.3) A car has a Velocity of 10m/s. At accelerate at 0.2 m for half minute. Find thedistance travelled during this time and the final Velocity of the car.

Solution:

(Initial Velocity) Vi= 10m/s

(Acceleration) a= 0.2m/(time) t= minutes= 30s

(final velocity) Vf= ?

S=?

By using 1st

equation of motion:

Vf= Vi+at

Vf= (10)+(0.2)(30)

Vf= 10+6

Vf= 16mBy using 3

rdequation of motion to find s:

2aS= V-V2aS= V-V2(0.2)= ()-()0.4s= 256-100

S= 156/.4

S= 390m ANS


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P2.4) A tennis ball is hit vertically upward with a Velocity of 30m/s. it takes 3s to reach the

highest point. Calculate the maximum highest reached by the ball. How long it will take to

return to the ground?

Solution:

(Initial Velocity) Vi= 30m/s

(Time) t1= 3s

(Height) S=?

Time required returning to the ground t2=?

g = -10m/The value of g will be negative because the ball will be decelerating.

Now by using the 2nd

equation of the motion:

S= vit+ (10)()= 90+(-5)(9)

= 90-45

Height S= 45m ANS

P 2.6) A train starts from the nest with an acceleration of 0.5m. Find i

ts speed in when it has moved through 100m.

Solution:

Initial Velocity Vi= 0

Acceleration a= 0.5Distance s= 100m

Final Velocity Vf=?

To find the final Velocity we have to find the time. By using 2nd

equation of motion:

S= vit+1/2By putting the values:

100= (0)t+1/2(0.5)()100= (0.5)100= 0.25

=

=400

= 400

Taking square root on both sides:=T= 20s

Now for Vf, we have formula:

Vf= Vi+at

Vf= 0=(0.5)(20)

Vf= 10mNow to convert 10m/s into km/h, we will multiply it with 3600 nad divide it by 1000.

So, Vf= 10 x

Vf= 36 ANS

P2.8) A cricket ball is hit vertically upward and returns to the ground 6s later. Calculate:

(i) Maximum height reached by the ball(ii) Initial Velocity of the ball

Solution:

Acceleration g = -10m/Time t= 6s

Time for upward= t1= 6/2= 3s

Height= s =?


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Initial Velocity= Vi?

Final Velocity= Vf= 0

By using 1st equation of motion:

Vf= vi+gt

O= vi+(-10)(3)

O= vi-30

30=vi

Vi=30mBy using 3

rdequation of motion:

2aS= -2x(-10) x s= 0 (30)

-20s= -900

20s= 900

S= 900/20

S= 45m ANS

When brakes are applied the speed of train decreases from the 96km/h to 48km/h. In)P2.9

800m how much distance will it cover before coming to rest? (Assume the retardation is

constant)

Solution:

The situation can be divided into two parts. The parts 1 data is as follows:

Initial Velocity Vi= 96 km/h

= 96x1000/3600= 26.66m/s

Final Velocity Vf= 48km/h

= 48x1000/3600= 13.33m/s

Distance s= 800m

Acceleration a=?

By using 3rd

equation of motion:

2aS= V- V2a(800)= ()-()1600s= -533.35

a= -533.35/1600

a= -0.33m/vi= 48km/h= 13.33m/s

vf= 0 m/s

s=?

Again by using 3rd

equation of motion:

2aS= v v2(-0.3)s= ()- ()-0.6s= -177.688

S= -177.688/-0.6

S= 266.53m

P2.10) In problem 2.9 find the time taken by the train to stop after the application of brakes.

Solution:

Initial Velocity Vi= 96km/h= 26.667m/s

Final Velocity Vf= 0m/s

Acceleration a= -0.33m/Time t=?

Formula Vf= vi+at

O= 26.677+(-0.33)t


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-26.66= -033t

T= -26.667/-0.33

T= 80.80

Vf-vi=at

0-26.677/-0.3344=t

T=80s ANS

P2.11) A car moves with uniform Velocity of 5s it comes to rest in the next 10s/ Find

deceleration and total distance covered by the car?

Solution:

Initial Velocity Vi= 40mTime t = 10s

Final Velocity Vf= 0

Retardation a=?

Total distance S=?

As we know thwt:

a= Vf-Vi/t

a= 0-40/40

a= -4 m ANSDistance travelled in 1

stfive seconds.

= vxt= 40x5

S1= 200m

Average Velocity for next 10 seconds.

Vav= 40+0/2= 20mS2= Vav*t

= 20x10

= 200mTotal distance S= S1 + S2

= 200+200 = 400m ANS


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Chapter No 3Dynamics

P 3.1) A force of 20 N moves a body with an acceleration of 20ms-2

what is its mass?

Solution:Force F= 20 N Acceleration a= 20ms

-2Mass m=?

Formula F= ma

M=F/a

20/2 = 10 kg Ans

P 3.2) Weight is 147 N what is its mass?

Solution:

Weight w=147 N

Acceleration g = 10ms-2

Mass m=?

Formula W = mg m = W / g

147/10 = 14.7 kg Ans

P3.3) How much force is needed to prevent a b

ody of mass 10kg from falling?

Solution:

Mass m=10kg

Force F=a

The force needed to prevent the body from falling is equal to the weight of the body

F = W

W=mg

F=mg

F= 10 x 10= 100 N Ans

P3.4) Find the acceleration produce by a force of 100 N in a mass of 50 kg?

Solution:

Acceleration a =?

Force F = 100 N

Mass m = 50kg

Formula:

F= ma a= F/m a=100/50 = 2ms-2

P3.5) A body has weight 20N how much force is required to move it vertically upward with an

acceleration of 20ms-2

?

Solution:

Weight W = 20N

Acceleration a = 20ms-2


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Force F=?

To find out force we have to first calculate the Mass of the body

To find out the mass to use W= mg m = w/g m= 20/10 m = 2kg

So he net force will take the body upward will be

Net force F = W

F= ma

W = mg

The g will be negative because body is moving upward so

W = -mg

Net force F = F-W

ma m(-g )

ma + mg

m(a +g)

2 (2+10)

Force F= 24N Ans

P3.6) Two masses 52kg and 48kg are attached to the end of the string that passes over a

frictionless pulley. Find the tension in the string and acceleration in the body? When the

masses are moving vertically.

Solution:

m1 = 52kg

m2 = 58kg

T =?

a =?

First we find tension in the string

Formula:

T = 2m1m2 g / m1+m2

T = 2(52) (48) (10)/ 52+48

T = 499.2N

T = 500N approximately Ans

Now we will find the acceleration

Formula:a = (m1-m2) g / m1+m2

a = (52-48) (10)/ 52+48

a= 4x10/100

a= 40/100

a= 0.4ms-2

Ans


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P3.7) Two masses 26kg and 24kg are attached to the end of a string which passes over a

frictionless pulley. 26kg is lying over a smooth horigalal table .24kg mass is moving vertically

downward. Find the tension in string and acceleration in bodies.

Solution:

m1 = 24kg m2 = 26kg T =? a =?

Formula:

T = m1 m2g / m1 +m2

T = (24) (26) (10)/24+26

T = 124.8N

T = 125N

Formula: a = m1g / m1+m2

= 24x10 / 24+ 26

a = 408ms-2

Ans

P3.8) How much time is required to change 22 Ns momentum b

y a force of 20 N?

Solution:

(Initial momentum) Pi = 22Ns

Pf= 0Ns F = 20N t =?

Formula:

Pf Pi/ t = F

F = Pf- Pi /t

= 0-22/ 20

t = -1.1s

As time cannot be negative to

t = 1.1s ANS

P3.9) How much is the force of friction between a wooden block of mass 5 kg and the

horizontal marble floor? The coefficient of friction between the wood and marble is 0.6

Solution:

Fr =? m= 5 kg u= 0.6 Formula Fr = UF

F = mg

5 x 10 = 50 NFr =UF

Fr = 0.6 x 50 = 30 N ANS

solved problem of first three chapters by asif rasheed

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