Class 11 Physics

Units & Measurement

NCERT Solution

Part 1

Question 1: Fill in the blanks

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

Answer: Side = 1 cm = 10-2 m

So, volume = (10-2)3 = 10-6 m

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

Answer: Radius = 2 cm = 20 mm

Height = 10 cm = 100 mm

Surface area of cylinder = 2πr + 2πr2

= 2πr(h + r)

= 2 × 3.14 × 20 (100 + 20)

= 125.6 × 120 = 1.5072

= 1.5 × 104 mm

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

Answer: Speed = 18 km h-1

It can be converted into meter per second as follows:

= 18 × 5/18 = 5 m per second

So, answer is 5 m

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

Answer: Density = 11.3 g cm-3

= 11.3 kg3 m-3

Question 2: Fill in the blanks by suitable conversion of units

(a) 1 kg m2 s–2 = ……....g cm2 s–2

Answer: 1 kg m2 s-2 = (1 kg m2) ÷ s2

`=(1×1000×10^2)/(s^2)` g cm2

= 107 g cm2 s-2

(b) 1 m = ……....... ly

Answer: 1 m `=1/(9.46×10^5)` ly

= 10-16 ly (approx)

(c) 3.0 m s–2 = ……...... km h–2

Answer: 3 ms-2 = `(3×10^(-3)km)/((1/(3600))^2h^2)`

`=3×3600×3600×10^(-3)` km h-2

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

Answer: G = 6.67 × 10-11 Nm2 kg-2

= 6.67 × 10-11 (kg m) ÷ s2 m2 kg-2

= 6.67 × 10-11 kg-1 m3 s-2

`=(6.67×10^(-11)×(10^2)^3)/((10^3)^2)`

= 6.67 × 10-8 cm-3 s-2 g-1

Question 3: A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1J = 1 kg m2 s–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 γ 2 in terms of the new units.

Answer: 1 cal = 4.2 kg m2 s-2

SINew System
n1 = 4.2n2 = ?
M1 = 1 kgM2 = α kg
L1 = 1 mL2 = β m
T1 = 1 sT2 = γ s

Dimensional formula of energy = [M1L2T-2]

Comparing with [MaLbTc], we get

a = 1, b = 2, c = -2

Now, `n_2=n_1[(M_1)/(M_2)]^a[(L_1)/(L_2)]^b[(T_1)/(T_2)]^c`

`=4.2[(1kg)/(αkg)]^1[(1m)/(βm)]^2[(1s)/(γs)]^(-2)`

Or, `n_2=4.2α^(-1)β^(-2)γ^2`

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:

  1. atoms are very small objects
  2. a jet plane moves with great speed
  3. the mass of Jupiter is very large
  4. the air inside this room contains a large number of molecules
  5. a proton is much more massive than an electron
  6. the speed of sound is much smaller than the speed of light.

Answer:

  1. Atoms are much smaller than tip of a needle.
  2. A jet plane moves at greater speed than a car.
  3. The mass of Jupiter is much larger than that of earth.
  4. The air inside this room contains a large number of molecules than air inside a balloon.
  5. No correction required
  6. No correction required