Question.1. Fill in the blanks:

(a)The volume of a cube of side 1 cm is equal to —— m^{3}

(b)The surface area of a solid cylinder of radius 2 cm and height 10 cm is equal to —— (mm)^{2}

(c)A vehicle moving with a speed of 18 km h^{-1} cover —– m in 1 s

(d)The relative density of lead is 11.3 Its density is —– g cm^{-3} or —– kg m^{-3}

Solution. (a)Length L = 1 cm = 10^{-2} m,

Volume, L^{3} = (10^{-2})^{3} = 10^{-6} m^{3}

(b) Given, r = 2 cm = 20 mm, h = 10 cm = 100 mm

Surface area of solid cylinder = (2πr)h

= 2 ×3.14 × 20 × 100 = 1.26 × 10^{4} mm^{2}

(c) Speed, v = 18 kmh^{-1 }

$ = \frac{18 \times 1000}{60 \times 60} $

v = 5 ms^{-1}

Distance covered in 1 sec = 5 m

(d) Relative density = 11.3

density = 11.3 g/cc

$\large = \frac{11.3 \times 10^{-3} kg}{(10^{-2}m)^3} $

= 11.2 × 10^{3} kg m^{-3}

Question:2. Fill in the blank by suitable conversion of units:

(a) 1 kgm^{-2} s^{-2} = …..g cm^{2} s^{-2}

(b) 1 m = ….. light year

(c) 3 m s^{-2} = ….. km h^{-2}

(d) G = 6.67 × 10^{-11} N m^{2} kg^{-2} = ….cm^{3} s^{-2} g^{-1}

Solution: (a)1 kg m^{2} s^{-2}

= 1 × 10^{3} g (10^{2 }cm)^{2} s^{-2}

= 10^{7} g cm^{-2} s^{-2}

(b) 1 light year = 9.46 × 10^{15} m

$\large 1 m = \frac{1}{9.46 \times 10^{15}} l y $

= 1.053 × 10^{-16} light year

(c) 3 ms^{-2} = 3 × 10^{-3} km s^{-2}

$\large = 3 \times 10^{-3} km (\frac{1}{60 \times 60} h)^{-2}$

= 3 × 10^{-3} × 3600 × 3600 km h^{-2}

= 3.888 × 10^{4} km h^{-2}.

(d) G = 6.67 × 10^{-11} N m^{2} kg^{-2}

= 6.67 × 10^{-11} (kg m s^{-1}) m^{2} kg^{-2}

= 6.67 × 10^{-11} m^{3} s^{-1} kg^{-1}

= 6.67 × 10^{-8} cm^{3} s^{-1} g^{-1}

Question:3. A calorie is unit of heat energy and it equal about 4.2 j, where 1 J = kgm^{2} s^{-2}. Suppose, the employ a system of units in which the unit of mass equals α kg, the unit of length equal β m and the unit of time is γ s . show that a calorie has a magnitude of 4.2 α^{-1} β^{-2} γ^{2} in terms of new units.

Solution: 1 calorie = 4.2 J = 4.2 kg m^{2} s^{-2}

If α kg = new units of mass

Then, 1 kg = 1/α new unit of mass = α^{-1} new unit of mass

Similarly, 1 m = β^{-1} new unit of length

1 s = γ^{-1 }new unit of time

Now, 1 calorie = 4.2 (α^{-1} new unit of mass) (β^{-1} new unit of length)^{2} (γ^{-1} new unit of time)^{-2 }

= 4.2 α^{-1} β^{-2} γ^{2} unit of energy. (Proved)

Question: 4. Explain this statement clearly :

(i) To call a dimensionless quantity ‘ large ‘ or ‘small ‘ is meaningless without specifying a standard for comparison .

(ii) In view of this , reword 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 .

Solution: (i) This statement is true , because a dimensionless quantity can be large or small only in comparison to some standard . e.g. angle is dimensionless θ = 90° is larger than 60° but smaller than 120° .

(ii) (a) The size of an atom is smaller than the sharp of tip of a pin .

(b) A Jet plane moves faster than a superfast train.

(c) The mass of jupiter is very large as compares to mass of earth .

(d) The air inside a room contains more number of molecules than one mole of air .

(e) This statement is already correct .

(f) This statement is already correct .

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 is terms of the new unit, if light takes 8 min and 20 sec to cover the distance.

Solution: Velocity of light = c = 1 new unit of length s^{-1}

Time taken by light of sun to reach the earth = t = 8 min 20 s = 8 × 60 + 20 = 500 s

distance between the sun to reach the earth,

x = c × t = 1 new unit of length s^{-1} × 500 s = 500 new units of length.

Question:6. Which of the following is the most precise device for measuring length ?

(a) a Vernier calipers with 20 divisions on the sliding scale coinciding with 19 main scale division

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

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

Soution: (a) Since , 20 VSD = 19 MSD

$ 1 VSD = \frac{19}{20} MSD $

Least count of vernier calipers = 1 MSD – 1 VSD

$\large LC = 1 MSD – \frac{19}{20} MSD $

$ = \frac{1}{20} MSD = \frac{1}{20} mm $

= 0.005 cm

(b) Lest count of screw gauge = Pitch/no. of division on circular scale

$LC = \frac{1}{100} mm $

LC = 0.001 cm

(c) Wavelength of light , λ = 10^{-5} cm = 0.00001 cm

Since most precise devise should have minimum least count, optical instrument is the most precise one.

Question : 7. 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 the view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

Sol: $\large Magnification (m) = \frac{observed \; width}{Real \; width} = \frac{y}{x} $

$\large m = \frac{3.5}{100} = 0.035 mm$

Question:8. (a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?

(b)A screw gauge has pitch of 1 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the 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 calipers. Why is set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?

Solution:(a) Meter scale can not measure small diameter of thread. No of turns of the thread to be wound to get turns closely one another.

Let l = measuring length windings of the scale which contains n no. of turns.

Diameter of thread = l/n

(b) Least count =Pitch / no. of division on circular scale i. e. least count decreases when no of division on the circular scale increases. Thereby accuracy would increases. Practically, however, it is impossible to take precise reading due to low resolute of human eye.

(c) Large no. of observation (say 50) have more reliable result, than smaller no. of observations (say 3). Because probability of making random error in positive side of a physical quantity would be same that of in negative side. Therefore, when no of observations are large random errors would be cancelled each other and hence result would reliable.

Question:9. The photograph of a house occupies an area 1.75 cm^{2} on a 35 mm sides. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m^{2}. What is the magnification of the projected screen arrangement?

Solution. Area of object = 1.75 cm^{2}

Area of image = 1.55 m^{2} = 1.55 × 10^{4} cm^{2}

$Area \; magnification =\frac{Area \;of\; image}{Area \;of \;object }$

$\large = \frac{1.55 \times 10^4}{1.75}$

= 8857

$ Liner \;magnification =\sqrt{8857} $

= 94.1

Question:10. State the number of significant figures in the following ;

(i) 0.007 m^{2}

(ii) 2.64 × 10^{24} kg

(iii) 0.2370 g cm^{-3}

(iv) 6.320 J

(v) 6.032 Nm^{-2}

(iv) 0.0006032

Solution: No, of significant figures:

(i) one

(ii) three

(iii) four

(iv) four

(v) four

(vi) four

Question:11. The length breadth the thickness of a metal sheet are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct number of significant figure.

Solution. Given, length, l= 4.234 m, Breadth, b = 1.005 m, thickness, t = 2.01 cm = 2.01 × 10^{-2} m

Area of sheet = 2 (lb + bt + tl)

= 2[4.234 × 1.005 + 0.0201 + 0.0201 × 4.234]

= 8.7209478 m^{2}

= 8.72 m^{2 }(running off to 3 significant digits)

Volume = lbt

= 4.234 × 1.005 × 0.0201

= 0.0855289

= 0.0855 m^{3} (rounding off to 3 significant digits)

Question:12.The mass of a box measured by a grocer’s Balance is 2.3 kg . Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is

(a) Total mass of the box

(b) the difference in masses of gold pieces to correct significant figures.

Solution: Mass of the box, m = 2.3 kg

Mass of one gold piece, m_{1} = 20.17 g = 0.02017 kg

Since the result is correct only up to one place of decimal, therefore after rounding off, total mass = 2.3 kg

Difference in masses = m_{2} – m_{1 }

= 20.17 – 20.15

= 0.02 g (correct up to 2 place of decimal)

Question:13. A physical quantity P is related to four observatives a, b, c, and d as follow :

$\displaystyle P = \frac{a^3 b^2}{\sqrt{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? it is value of P calculate using the above relation turns out to be 3.763, to what value should you round off the result?

Solution: $ \displaystyle P = \frac{a^3 b^2}{\sqrt{c} d } $

percentage error in P,

$ \displaystyle \frac{\Delta P}{P}\times 100 = \pm (3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{1}{2}\frac{\Delta c}{c} + \frac{\Delta d}{d})\times 100 $

= ±(3 × 1% + 2×3% + (1/2)×4% + 1×2% )

= ±13 %

∵ 13% error has two significant figures .

Therefore , if P ≅ 3.763 , P would be rounded off to 3.8

Question:14. A book with many printing errors contains four different formulae for the displacement y of a particle undergoing a certain periodic motion :

(i) $\large y = a sin\frac{2 \pi t}{T} $

(ii) y = a sin vt

(iii) $\large y = \frac{a}{T} sin(\frac{t}{a})$

(iv) $\large y = a \sqrt{2} ( sin\frac{2\pi t}{T} + cos\frac{2\pi t}{T}) $

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

Solution: As angle is dimensionless

(i) $\frac{2\pi t}{T}$ is dimensionless .This equation is correct

(ii) v t is not dimensionless . This equation is not correct

L.H.S = L

R.H.S = L sin LT^{-1} × T = L sin L

(angle is not dimensionless)

(iii) $\frac{t}{a}$ is not dimensionless . This equation is not correct, as angle is not dimensionless

(iv) $\frac{2\pi t}{T}$ is dimensionless .This equation is correct.

Question:15 . A famous relation in physics related ‘moving mass’ m to the ‘rest mass’ m_{0} of a particle in terms of its speed υ and the speed of light c. This relation first arose as a consequence of special relativity by Albert Einstein. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

$\large m = \frac{m_0}{(1-v^2)^{1/2}} $ . Guess where to put the missing c.

Solution: c^{2} : square of velocity of light should be put in the denominator of v^{2} , so that v^{2}/c^{2} becomes dimensionless & dimension of L.H.S becomes dimension of R.H.S . Therefore the correct relation is :

$\large m = \frac{m_0}{(1 – \frac{v^2}{c^2})^{1/2}} $