NCERT Solutions Class 11 Physics Chapter 2 Units and Measurements
(a) The volume of a cube of side 1 cm is equal to ________ m3.
Solution:
The side of the cube is 1 cm.
Volume \( V \) of a cube = side3
\
Converting \( \text{cm}^3 \) to \( \text{m}^3 \):
\
Therefore, the volume of a cube of side 1 cm is \( 1 \times 10^{-6} \, \text{m}^3 \).
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ________ (mm)2.
Solution:
Radius \( r \) = 2.0 cm
Height \( h \) = 10.0 cm
Surface area \( A \) of a cylinder = \( 2\pi r(h + r) \)
\
\
Converting \( \text{cm}^2 \) to \( \text{mm}^2 \):
\
Therefore, the surface area of a solid cylinder is \( 4800\pi \, \text{mm}^2 \approx 15079.64 \, \text{mm}^2 \).
(c) A vehicle moving with a speed of 18 km h-1 covers ________ m in 1 s.
Solution:
Speed \( v \) = 18 km/h
Converting speed to m/s:
\
Distance covered in 1 s = speed × time
\
Therefore, a vehicle moving with a speed of 18 km h-1 covers \( 5 \, \text{m} \) in 1 s.
(d) The relative density of lead is 11.3. Its density is ________ g cm-3 or ________ kg m-3.
Solution:
Relative density (R.D.) = 11.3
Density of water = 1 g/cm3
\
Converting \( \text{g/cm}^3 \) to \( \text{kg/m}^3 \):
\
Therefore, the density of lead is \( 11.3 \, \text{g/cm}^3 \) or \( 11300 \, \text{kg/m}^3 \).
Question 2.2: Fill in the blanks by suitable conversion of units
(a) 1 kg m2 s-2 = ________ g cm2 s-2
Solution:
We need to convert kg to g and m to cm:
\
\
Therefore,
\
\
Therefore, 1 kg m2 s-2 = \( 10^7 \) g cm2 s-2.
(b) 1 m = ________ ly
Solution:
1 light year (ly) is the distance light travels in one year:
\
Therefore,
\
Therefore, 1 m = \( 1.057 \times 10^{-16} \) ly.
(c) 3.0 m s-2 = ________ km h-2
Solution:
We need to convert m/s2 to km/h2:
1 m = 0.001 km
\
Therefore,
\
Therefore, 3.0 m s-2 = 38880 km h-2.
(d) G = 6.67 x 10-11 N m2 (kg)-2 = ________ (cm)3 s-2 g-1.
Solution:
We need to convert N to dynes, m to cm, and kg to g:
1 N = 105 dynes
\
\
Therefore,
\
\
\
Therefore, G = \( 6.67 \times 10^{-8} \) (cm)3 s-2 g-1.
Question 2.3: Conversion of units for a calorie
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 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, and 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.
Solution:
Given: \( 1 \, \text{cal} = 4.2 \, \text{J} \)
\
In the new system of units:
1 unit of mass = \( \alpha \, \text{kg} \)
1 unit of length = \( \beta \, \text{m} \)
1 unit of time = \( \gamma \, \text{s} \)
We need to express 1 J in terms of the new units:
\
\
Therefore,
\
Thus, a calorie has a magnitude of \( 4.2 \alpha^{-1} \beta^{-2} \gamma^{2} \) in terms of the new units.