Class 11-science NCERT Solutions Physics Chapter 1 - Units and Measurements
Units and Measurements Exercise 10
Solution 1
(a) 1 cm =
Volume of the cube = 1 cm3
But, 1 cm3 = 1 cm 
1 cm 1 cm = 
1 cm3 = 10-6 m3
Hence, the volume of a cube of side 1 cm is equal to 10-6 m3.
(b) The total surface area of a cylinder of radius r and height h is
S = 2 
r (r + h).
Given that,
r = 2 cm = 2 1 cm = 2 10 mm = 20 mm
h = 10 cm = 10 10 mm = 100 mm
= 15072 = 1.5 104 mm2
(c) Using the conversion,
1 km/h = 
Therefore, distance can be obtained using the relation:
Distance = Speed Time = 5 1 = 5 m
Hence, the vehicle covers 5 m in 1 s.
(d) Relative density of a substance is given by the relation,
Relative density = 
Density of water = 1 g/cm3
Again, 1g = 
kg
1 cm3 = 10-6 m3
1 g/cm3 = 
11.3 g/cm3 = 11.3 103 kg/m3
Solution 2
Units and Measurements Exercise 11
Solution 3
Solution 4
The given statement is true because a dimensionless quantity may be large or small in comparison to some standard reference. For example, the coefficient of friction is dimensionless. The coefficient of sliding friction is greater than the coefficient of rolling friction, but less than static friction.
(a) An atom is a very small object in comparison to a soccer ball.
(b) A jet plane moves with a speed greater than that of a bicycle.
(c) Mass of Jupiter is very large as compared to the mass of a cricket ball.
(d) The air inside this room contains a large number of molecules as compared to that present in a geometry box.
(e) A proton is more massive than an electron.
(f) Speed of sound is less than the speed of light.
Solution 5
Distance between the Sun and the Earth:
= Speed of light 
Time taken by light to cover the distance
Given that in the new unit, speed of light = 1 unit
Time taken, t = 8 min 20 s = 500 s
Distance between the Sun and the Earth = 1 500 = 500 units
Solution 6
A device with minimum count is the most suitable to measure length.
(a) Least count of vernier callipers
= 1 standard division (SD) - 1 vernier division (VD)
(b) Least count of screw gauge = 
(c) Least count of an optical device = Wavelength of light 
10-5 cm
= 0.00001 cm
Hence, it can be inferred that an optical instrument is the most suitable device to measure length.
Solution 7
Magnification of the microscope = 100
Average width of the hair in the field of view of the microscope = 3.5 mm
Actual thickness of the hair is 
= 0.035 mm.
Solution 8
(b) It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.
(c) A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.
Solution 9
Area of the image of the house formed on the screen = 1.55 m2
= 1.55
104 cm2Arial magnification, ma =
Linear magnifications, ml = 
Solution 10
Solution 11
Solution 12
Solution 13
Units and Measurements Exercise 12
Solution 14
Solution 15
Solution 16
Solution 17
