Class 11 - Physics - Units and Measurements

Question1.

Fill in the blanks

(a) The volume of a cube of side 1 cm is equal to   .....m3

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2

(c) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s

(d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3.

Answer:

  • Volume of cube = (a)3    (equation 1)

Length of edge 1cm= (1/100) m

Volume of cube = (1/100) m3

Using equation (1); 1cm x 1cm x 1cm

= (1/100) m x (1/100) m x (1/100) m

Therefore 1cm3 = 10-6m3

Hence, the volume of a cube of side 1cm is equal to =10-6m3.

  • The total surface area of a cylinder of radius r and height h is:

Given that:-

r=2cm = (2x1) cm = (2x10) mm r=20mm

h=10cm= (10 x 10) mm =100mm

Surface area of the cylinder S = 2 πr (h+r)

= (2x3.14x20x (20+100))

=15072

=1.5 x 104(mm) 2

Therefore, the surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to =1.5 x 104mm2

  • Given Time, t = 1sec.

Speed s = 18 kmh-1

Using, 1km = 1000m

1hour = 3600sec

Speed s = 18 x (1000/3600) = 5m/sec.

Therefore speed = (distance/time)

Distance = (speed x time) = 5 x 1 =5m.

Therefore, a vehicle moving with speed of 18kmh-1 covers 5m in 1s.

  • Density of lead = (Relative density of lead) x (Density of water)

Density of water = 1g/cm3

     = (11.3 x 1) = 11.3g/cm3

             Again, 1g = (1/1000) kg = (10-3) kg

            1cm3 = (1/100) = 10-2m3

 Therefore, 11.3g/cm3 = (11.3 x 10-3) kg/ (10-2) m-3 = (11.3 x 10-3x106) kgm-3

= (1.13 x 103) kgm-3

Therefore; the relative density of lead is 11.3. Its density is 11.3 g cm–3 or

1.13 x 103 kg m–3

Question2.

Fill in the blanks by suitable conversion of units

(a) 1 kg m2 s–2 = ....g cm2 s–2

(b) 1 m =..... ly

(c) 3.0 m s–2 =.... km h–2

(d) G = 6.67 × 10–11 N m2 (kg)–2 =.... (cm) 3 s–2 g–1

Answer:

  • 1 kg m2 s–2 = 1 x 103g(102 cm)2s-2 = 107 g cm2 s-2
  • 1 light year = 9.46 x 1015m

Therefore 1m = (1/9.46 x 1015) = (1.053 x 10-16) light year.

  • 3ms-2 = 3 x 10-3 km( 1 h/60x60)-2

= 3 x10-3 x 3600 x 3600 kmh-2

=3.0 m s–2 = 3.888 x 10-4 kmh-2

  • Given that

G= 6.67 × 10–11 N m2 (kg)–2

We know that; 1 N = 1 kg m s–2

1 kg = 103 g

1 m = 100 cm = 102 cm

Therefore, we get 

=> 6.67 × 10–11 N m2 kg–2 = 6.67 × 10–11 × (1 kg m s–2) (1 m2) (1kg–2)

=> 6.67 × 10–11 × (1 kg–1 × 1 m3 × 1 s–2)

Converting Kg to g and m to cm;

=> 6.67 × 10–11 × (103 g)-1 × (102 cm) 3 × (1 s–2)

=>  6.67 × 10–11 × 10-3 g-1 × 106 cm3 × (1 s–2)

=> 6.67 × 10–8 cm3 s–2 g–1

Therefore G = 6.67 × 10–11 N m2 (kg)–2 =6.67 × 10–8 (cm) 3 s–2 g–1.

Question 3.

A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kgm2s–2 .

Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s.

Show that a calorie has a magnitude 4.2 α–1 β–2 γ2 in terms of the new units.

Answer:

Given that,

1 calorie = 4.2 J   (1J = 1 kg m2s–2)

Therefore, 1 calorie =4.2 kg m2s–2 = 4.2 × (1 kg) × (1 m2) × (1 s–2)     … (1)

Given that;

Unit of mass equals α kg => 1 kg mass = (1/α) mass

Unit of length equals β m => 1 m length = (1/ β) length

Unit of time is γ s        => 1 s time = (1/ γ) time

Putting the values in equation (1), we get

 New unit of mass = α kg

=4.2 × (1 kg) × (1 m2) × (1 s–2)

= 4.2 × (1/ α) × (1/ β) 2 × (1/ γ) –2

= 4.2 × (α–1) × (β–1)2 × (γ–1) –2

=4.2 α–1 β–2 γ2 unit of energy.

Question 4.

Explain this statement clearly:

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”.

In view of this, reframe the following statements wherever necessary:

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

Answer:

The statement is true, because a dimensionless quantity can be large or small when compared with the unit of measurement.

  • The size of atom is much smaller than the sharp tip of a pin.
  • A jet plane moves faster than a superfast train.
  • The mass of Jupiter is very large compared to the mass of earth.
  • The air inside this room contains a large number of molecules than in one mole of air.
  • A proton is heavier than an electron.

Mass of a proton is 1.67 × 10-24 g while mass of electron 9.11 × 10-28 g, therefore statement is true.

  • The speed of sound is less than the speed of light.

Speed of sound 346 m/ sec while speed of light is 3 x 108 m / sec in air at 25 0C, therefore statement is true.

Question 5.

A new unit of length is chosen such that the speed of light in vacuum is unity.

What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer:

Given;

Speed of light in vacuum = unity

Distance between sun and earth = (speed of light in vacuum) x (time)

= (3 x108ms-1) x 500s

Therefore new unit of length = speed of light in vacuum

= (3 x 108) ms-1

Therefore distance between sun and earth in terms of new unit

= (3 x 108 x 500)/ (3 x 108)

=500 units according to new unit.

Question 6.

Which of the following is the most precise device for measuring length?

(a) a Vernier callipers with 20 divisions on the sliding scale

(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale

(c) an optical instrument that can measure length to within a wavelength of light

Answer:

  • Least count of Vernier callipers =(1/20)= 0.05cm
  • Least count of screw gauge = (1/100) = 0.01cm
  • Least count of an optical wave =(1/105) = 0.00001cm

Optical instrument is the most precise device for measuring length.

Question7.

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100.

He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm.

What is the estimate on the thickness of hair?

Answer:

Given;

Average width of the hair in the field of view of the microscope = 3.5mm

Magnification produced by microscope = 100

Using (magnification m) = (size of image) (v)/ (size of object) (O)

=> Size of object (O) = (size of image)/ (magnification m)

Therefore thickness of hair = (Thickness of image)/ (magnification)

= (3.5mm)/ (100)

=0.035mm.

                                                                                                                                                                                

Question 8.

Answer the following:

(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread?

(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale.

Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

(c)The mean diameter of a thin brass rod is to be measured by Vernier callipers.

Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Answer:

  • As the diameter of a thread is so small that it cannot be measured using a metre scale.
  • Number of turns of the thread to be wound to get turns closely one another.

Let measure of length (l) of the windings on the scale.

Number of turns = ‘n’.

Therefore diameter of thread = (1/n)

  • Least count = (pitch) /(number of divisions in circular scale)

i.e. least count decreases when number of divisions on the circular scale increases. Thereby accuracy would increase,

but it is impossible to take precise reading due to low resolution of human eye.

(c)    By taking a large number of measurements of the same quantity, it is quite possible that the majority of

these measurements will have small errors which might be positive or negative. When average values are taken positive and negative errors are likely to cancel each other

Question 9.

The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen,

and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement?

Answer:

Given:

Area of photograph of house =     1.75 cm2

Area of image of house projected on screen = 1.55m2 = (1.55 x 104) cm2

Therefore, areal of magnification of the projector screen, mareal;

mareal = (area of image)/(area of object)

= (1.55 x 104) cm2/ (1.75cm2)

The linear magnification will be mlinear = √ ( mareal) = √ (8857) = 94.1

Question 10.

State the number of significant figures in the following:

(a) 0.007 m2

(b) 2.64 × 1024 kg

(c) 0.2370 g cm–3

 (d) 6.320 J

(e) 6.032 N m–2

(f) 0.0006032 m2

Answer:

  • The given quantity is 0.007 m2.

If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant.

This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity.

  • The given quantity is (2.64 × 1024) kg.

Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures.

  • The given quantity is 0.2370 g cm–3.

For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure.

  • The given quantity is 6.320 J.

For a number with decimals, the trailing zeroes are significant. Hence, all four digits appearing in the given quantity are significant figures.

  • The given quantity is 6.032 Nm–2

The given quantity is 0.0006032 m2

All zeroes between two non-zero digits are always significant.

  • If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant.
  • Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant.
  • Hence, the remaining four digits are significant figures.

Question 11.

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively.

Give the area and volume of the sheet to correct significant figures.

Answer:

Given:

Length of sheet, l = = 4.234 m

Breadth of sheet, b = 1.005 m

Thickness of sheet, h = 2.01 cm = 0.0201 m

The given table lists the respective significant figures:

Physics Class 11 Unit & Measurement Significant Figures

Hence, area and volume both must have least significant figures i.e., 3.

Surface area of the sheet = 2 (l × b + b × h + h × l)

= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)

= 2 (4.25517 + 0.02620 + 0.08510)

= (2 × 4.360)

= 8.72 m2

Volume of the sheet = (l × b × h)

= (4.234 × 1.005 × 0.0201)

= 0.0855 m3

This number has only 3 significant figures i.e., 8, 5, and 5.

Question 12.

The mass of a box measured by a grocer’s balance is 2.300 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box.

What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Answer:

Given:

Mass of grocer’s box m= 2.300 kg

Mass of gold piece I m1= 20.15g = 0.02015 kg

Mass of gold piece II m2= 20.17 g = 0.02017 kg

Total mass of the box = (m+m1+m2) = (2.3 + 0.02015 + 0.02017) = 2.34032 kg

As the least number of significant figures in the mass of box is 2.So, the maximum number of significant figures should be = 2.

Therefore total mass =2.3kg

Difference in masses = (m2 – m1) = (20.17 – 20.15) = 0.02 g

There are two significant digits. So, the difference in the masses to the correct significant figures is 0.02g.

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

Question 13.

A physical quantity P is related to four observables a, b, c and d as follows:

P = (a3b2) / (√c d)

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively.

What is the percentage error in the quantity P? If the value of P calculated using the above relation turns out to be 3.763,

to what value should you round off the result?

Answer:

Given:

P = (a3b2) / ((Vc) d)

(∆P/P) = 3(∆a/a) + 2(∆b)/b) + (1/2) (∆c/c) + (∆d/d)

=(∆P/P) x 100)%  = (((3x(∆a/a) x 100) + ((2x(∆b/b) x 100)+((1/2) (∆c/c) x 100)+( (∆d/d x 100)%

 = (3 x 1% + 2 x 3% + (1/2) x 4% +2%)

Where ((∆a/a) x 100 = 1%, (∆b/b) x 100 =3%, (∆c/c) x 100 =4% and (∆d/d x 100=2%

=3 % +6% +2%+2% =13%

Percentage error in P = 13 %

Value of P is given as 3.763.

By rounding off the given value to the first decimal place, we get P = 3.8.

Question 14.

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:

(a) y = a sin 2π t/T

(b) y = a sin vt

(c) y = (a/T) sin t/a

(d) y = ( a√2) (sin 2 π /T/ +cos 2 π/T  )

(a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

Answer:

(a) Answer: Correct

y = a sin (2πt/T)

Dimension of y = M0 L1 T0

Dimension of a = M0 L1 T0

Dimension of sin (2πt/T) = M0 L0 T0

As Dimension of L.H.S = Dimension of R.H.S

Therefore, the given formula is dimensionally correct.

 

(b) Answer: Incorrect

y = a sin (vt)

Dimension of y = M0 L1 T0

Dimension of a = M0 L1 T0

Dimension of vt =( M0 L1 T–1) × (M0 L0 T1) = M0 L1 T0

But the argument of the trigonometric function must be dimensionless, which is not so in the given case.

Hence, the given formula is dimensionally incorrect.

(c) Answer: Incorrect

y = (a/T) sin (t/a)

Dimension of y = M0L1T0

Dimension of (a/T) = M0L1T–1

Dimension of (t/a) = M0 L–1 T1

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

 (d) Answer: Correct

y = (a√2) (sin (2πt/T) + cos (2πt/T))

Dimension of y = M0 L1 T0

Dimension of a = M0 L1 T0

Dimension of (t/T) = M0 L0 T0

Since the argument of the trigonometric function must be dimensionless (which is true in the given case),

the dimensions of y and a are the same. Hence, the given formula is dimensionally correct. 

Question 15.

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c.

(This relation first arose as a consequence of special relativity due to Albert Einstein).

A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

m = (mo)/ (1-v2)1/2

Guess where to put the missing c.

Answer:

Given the relation:-

m = (mo)/ (1-v2)1/2

Dimension of m = M1L0T0

Dimension of = M1 L0 T0

Dimension of v = M0 L1 T–1

Dimension of v2 = M0 L2 T–2

Dimension of c = M0 L1 T–1

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S.

This is only possible when the factor, (1 − υ2)1/2 is dimensionless i.e.

 (1 – v2) is dimensionless.

This is only possible if v2 is divided by c2

Therefore, the correct relation is

m = (mo)/ (1-v2/c2)1/2

Question 16.

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m.

The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms?

Answer:

Given:

Radius of hydrogen atom = 0.5 Å = (0.5 x 10-10) m 

Volume of 1 hydrogen atom = (4/3) π r3

= (4/3) x (22/7) x (0.5 x 10-10) = (0.5237 x 10-30) m3

Now one mole of hydrogen contains 6.023 x 1023 hydrogen atoms

 Therefore Volume of 1 mole of hydrogen atoms

= volume of one atom x NA

= (0.5237 x 10-30) x 6.02 x 1023 = 3.154 x 10-7m3

Question 17.

One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume).

What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Answer:

Given:

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10–10 m

Volume of hydrogen atom = (4/3) π r3

= (4/3) x (22/7) x (0.5 x 10-10)3

=0.524 x 10-30 m3

Now I mole of hydrogen contains 6.023 x 1023 hydrogen atoms.

Therefore, Volume of hydrogen atom, VA = (6.023 x 1023 x 0.524 x 10-30)

=3.16 x 10-7 m3

Molar volume of 1 mole of hydrogen atoms at STP,

Vm = 22.4 L = 22.4 × 10–3 m3

Therefore (Vm/VA) = (22.4 x 10-3)/ (3.16 x 10-7) = 7.08 x 104

Hence, the molar volume is 7.08 × 104 times higher than the atomic volume. For this reason,

the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

Question 18.

Explain this common observation clearly: If you look out of the window of a fast moving train, the nearby trees,

houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary.

(In fact, since you are aware that you are moving, these distant objects seem to move with you).

Answer:

Line of sight is an imaginary line joining an object and an observer’s eye. When a train moves, the line of sight of a nearby tree changes its direction of motion rapidly. As a result tree appears to run in opposite direction.

On the contrary, the line of sight of far off objects does not change its direction. As a result moon, stars etc. all appear stationary.

Question 19.

The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars.

The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun.

That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011m. However, even the nearest stars are so distant that with such a long baseline,

they show parallax only of the order of 1” (second) of arc or so.

A parsec is a convenient unit of length on the astronomical scale.

It is the distance of an object that will show a parallax of 1” (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun.

How much is a parsec in terms of metres?

Answer:

Diameter of Earth’s orbit = 3 × 1011 m

Radius of Earth’s orbit, r = 1.5 × 1011 m

Let the distance parallax angle be = 4.847 × 10–6 rad.

Let the distance of the star be D.

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of.

Therefore, 

θ = (r/D)

D = (r/ θ) = (1.5 x 1011)/ (4.847 x 10-6)

= (0.309 x 10 -6) ≈ 3.09 x 1016 m

Hence, 1 parsec ≈ 3.09 × 1016 m

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