**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 load 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 }= 18×1000/(60×60)= 5 ms^{-1}**

**Distance covered in 1s = 5 m**

**(d) Relative density = 11.3**

**density = 11.3 g/cc = = 11.2 × 10 ^{3} kgm^{-3}**

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

**(a)1 kgn ^{-2} s^{-2} = —–g cm2 s^{-2}**

**(b) 1 m = —– light year****(c) 3 ms ^{-2} = —- kmh^{-2}**

**(d) G = 6.67 × 10**^{-11}Nm^{2}kg^{-2}= —-cm^{3}s^{-2}g^{-1}**Solution.**

**(a)1 kgm ^{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**

**1 m = 1/ 9.46 × 10 ^{15} = 1.053 × 10^{-16} light year**

**(c)3 ms ^{-2} = 3 × 10^{-3} km = 3 × 10^{-3} × 3600 × 3600 kmh^{-2} = 3.888 × 10^{4} kmh^{-2}.**

**(d)G = 6.67 × 10 ^{-11} Nm^{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^{-2} 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. I 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. A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observation and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. what is his estimate on the thickness of hair?**

** Solution: Magnification m= Observed Width(y)/Real Width(x)**

**Hence , x =y/m =3.5/100 = 0.035 mm.**

**Question:5. 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)Lest count of vernier calipers = 1 MSD – 1 VSD**

**= 1 MSD –( 19/20)MSD =( 1/20)MSD = (1/20)mm = 0.005 cm**

**(b)Lest count of screw gauge = Pitch/no. of division on circular scale =(1/100) mm = 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:6. 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:7. 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) 6.320J**

**(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:8. 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)**