In This Post we are providing Chapter-2 UNITS AND MEASUREMENTS NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.
NCERT MOST IMPORTANT QUESTIONS ON UNITS AND MEASUREMENTS
1.E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively. Determine the dimensions of
Ans:
Thus, it is dimension less.
2.Two resistances R1 = 100 and R2 = 200 are connected in series. Then what is the equivalent resistance?
Ans:
3.If velocity, time and force were chosen the basic quantities, find the dimensions of mass?
Ans:
4.A calorie is a unit of heat or energy and it equals about 4.2 J where . Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals m, the unit of time is y s. Show that a calorie has a magnitude in terms of the new units.
Ans. Given that,
1 calorie = 4.2 (1 kg) 
New unit of mass = α kg
Hence, in terms of the new unit, 1 kg =
In terms of the new unit of length,
And, in terms of the new unit of time,
5. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450).
Ans. Let the distance between the ship and the enemy submarine be
.
Speed of sound in water = 1450 m/s
Time lag between transmission and reception of Sonar waves = 77 s
In this time lag, sound waves travel a distance which is twice the distance between the ship
and the submarine (2S).
Time taken for the sound to reach the submarine
∴ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km
6.One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1). Why is this ratio so large?
Ans. Radius of hydrogen atom, r = 0.5 
=
m
Volume of hydrogen atom =
Now, 1 mole of hydrogen contains 
hydrogen atoms.
∴ Volume of 1 mole of hydrogen atoms,
Molar volume of 1 mole of hydrogen atoms at STP,
Hence, the molar volume is 
times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.
7.The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?
Ans. Distance of the star from the solar system = 4.29 ly
1 light year is the distance travelled by light in one year.
1 light year = Speed of light × 1 year
= 
m
∴4.29 ly = 
m
1 parsec = 
m
∴4.29 ly =
= 1.32 parsec
Using the relation,
Where,
Diameter of Earth’s orbit, d=
8.Estimate the average mass density of a sodium atom assuming its size to be about 2.5. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg. Are the two densities of the same order of magnitude? If so, why?
Ans. Diameter of sodium atom = Size of sodium atom = 2.5 
Radius of sodium atom, r =
= 
m
Volume of sodium atom, V =
According to the Avogadro hypothesis, one mole of sodium contains atoms and has a mass of 23 g or 
kg.
∴ Mass of one atom =
Density of sodium atom, p =
It is given that the density of sodium in crystalline phase is 970 kg
.
Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.
9. It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.
Ans.
The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.
Distance of the Moon from the Earth = 
m
Distance of the Sun from the Earth =
m
Diameter of the Sun = 
m
It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:
Hence, the diameter of the Moon is 
m.
10. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):
(a) the total mass of rain-bearing clouds over India during the Monsoon
(b) the mass of an elephant
(c) the wind speed during a storm
(d) the number of strands of hair on your head
(e) the number of air molecules in your classroom.
Ans. (a) During monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of water column, h = 215 cm = 2.15 m
Area of country, A = 
Hence, volume of rain water, V = A × h =
Density of water, p = 
Hence, mass of rain water = p× V = 
kg
Hence, the total mass of rain-bearing clouds over India is approximately kg.
(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say
).
Volume of water displaced by the ship, 
Now, move an elephant on the ship and measure the depth of the ship (
) in this case.
Volume of water displaced by the ship with the elephant on board, 
Volume of water displaced by the elephant = 
Density of water = D
Mass of elephant = AD 
(c) Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.
(d) Area of the head surface carrying hair = A
With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.
∴Area of one hair = 
Number of strands of hair 
(e) Let the volume of the room be V.
One mole of air at NTP occupies 22.4 l i.e., 
volume.
Number of molecules in one mole = 
∴Number of molecules in room of volume V
=
= 
= 
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