NCERT Grade 11-Units and Measurement-Answers

NCERT Solutions for Class 11 Physics

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1.     (a) 1 cm = m

Volume of the cube = 1 cm3

∴ 1 cm = 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 = 2πr (r + h).

Given that,

r = 2 cm = 2 × 1cm = 2 × 10mm = 20mm

h = 10cm = 10 × 10mm = 100 mm

∴ S = 2 × 3.14 × 20 × (20 + 100) = 15072 = 1.5 × 104 mm2

(c) Using the conversion,

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,

Density of water = 1 g/cm3

Density of lead = Relative density of lead × Density of water = 11.3 × 1 = 11.3 g / cm3

2.    (a) 1 kg = 103 g

1m2 = 104 cm2

1kg m2 s-2 = 1 kg × 1 m2 × 1 s-2

= 103 g × 104 cm2 × 1 s-1 = 107 g cm2 s-2

(b) Light year is the total distance travelled by light in one year.

1 ly = Speed of light × One year

= (3 × 108 m/s) × (365 × 24 × 60 × 60 s)

= 9.64 × 1015 m

(c) 1 m = 10-3 km

Again, 1 s = 1/3600h

1 s = 3600

1 s = (3600)2 h

∴ 3 m s-2 = (3 × 10-3 km) ×((3600)2 h-2) = 3.88 × 104 km h-2

(d) 1 N = 1 kg m s-2

1 kg = 10-3 g-1

1 m3 = 106 cm3

∴ 6.67 × 10-11 N m2 kg-2 = 6.67 × 10-11 × (1 kg m s-2) (1 m2) (1 s-2)

= 6.67 × 10-11 × (1kg × 1 m3 × 1s-2)

= 6.67 × 10-11 × (10-3 g-1) × (106 m3) × (1 s-2)

= 6.67 × 10-11 cm3 s-2 g-1

3.    Given that,

1 calorie = 4.2 (1kg) (1m2) (s-2)

New unit of mass = α kg

Hence, in terms of the new unit, 1 kg = 1/a a = a-1

In terms of the new unit of length,

And, in terms of the new unit of time,

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.

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

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)

1SD – 19/20SD = 1/20mm = 0.005 cm

(b) Least count of screw gauge 

(c) Least count of an optical device = Wavelength of light (λ) ∼ 10-5 m = 0.00001 cm

Hence, it can be inferred that an optical instrument is the most suitable device to measure length.

7.    Magnification(M) = observed width(y)/ real width(x)

Magnification of the microscope(m) = 100

Average width of the hair in the field of view of the microscope(y) = 3.5 mm

real width (x) = y/m

∴ Actual thickness of the hair is 3.5/100 = 0.035 mm.

8.    (a) Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread winding using a metre scale. The diameter of the thread is given by the relation,

Diameter = length of thread winding ⁄ no. of turns

(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. Due to low resolution of human eye, it may not be possible to read very closed divisions pricely.

(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.

9.    Areal magnifications, ml = A-1

10.   (a) Answer: 1

The given quantity is 0.007 m2.

If the number is less than one, then all zeros on the right of the decimal point (but left to the first non-zero) are insignificant. This means that here, two zeros after the decimal are not significant. Hence, only 7 is a significant figure in this quantity.

(b) Answer: 3

The given quantity is 2.64 × 1024 kg.

Here, the power of 10 is irrelevant for the determination of significant figures. Hence, all digits i.e., 2, 6 and 4 are significant figures.

(c) Answer: 4

The given quantity is 0.2370 g cm-3.

For a number with decimals, the trailing zeroes are significant. Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point is also a significant figure.

(d) Answer: 4

The given quantity is 6.320 J.

For  a  number  with  decimals,  the  trailing  zeroes  are  significant. Hence, all  four  digits appearing in the given quantity are significant figures.

(e) Answer: 4

The given quantity is 6.032 Nm-2.

All zeroes between two non-zero digits are always significant.

(f) Answer: 4

The given quantity is 0.0006032 m2.

If the number is less than one, then the zeroes on the right of the decimal point (but left to the first non-zero) are insignificant. Hence, all three zeroes appearing before 6 are not significant figures. All zeros between two non-zero digits are always significant. Hence, the remaining four digits are significant figures.

11.   Length of sheet, l = 4.234 m

Breadth of sheet, b = 1.005 m

Thickness of sheet, h = 2.01 cm = 0.0201 m

The given table lists the respective significant figures:

Hence, area and volume both must have least significant figures i.e., 3. Surface area of the sheet = 2 (l × b + b × h + h × l)

= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)

= 2(4.25517 + 0.0202005 + 0.0851034)

= 2 × 4.36

= 8.72 m2

Volume of the sheet = l × b × h

= 4.234 × 1.005 × 0.0201

= 0.0855 m3

This number has only 3 significant figures i.e., 8, 5, and 5.

12.   Mass of grocer’s box = 300 kg

Mass of gold piece I = 20.15 g = 0.02015 kg

Mass of gold piece II = 20.17 g = 0.02017 kg

(a) Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg

In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg.

(b) Difference in masses = 20.17 – 20.15 = 0.02 g

In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

13.   

Percentage error in P = 13 %

Value of P is given as 3.763.

By rounding off the given value to the first decimal place, we get P = 3.8.

14.   (a) Answer: Correct

Dimension of y = M0L1T0

Dimension of a = M0L1T0

Dimension of =  = M0L0T0

∵ Dimension of L.H.S = Dimension of R.H.S

Hence, the given formula is dimensionally correct.

(b) Answer: Incorrect

y = a sin vt

Dimension of y = M0L1T0

Dimension of a = M0L1T0

Dimension of vt = M0L1T× M0L0T1 = M0L1T0

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.

(c) Answer: Incorrect

Dimension of y = M0L1T0

Dimension of  = M0L1T-1

Dimension of  = M0L1T1

But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.

(d) Answer: Correct

Dimension of y = M0L1T0

Dimension of a = M0L1T0

Dimension of  = M0L0T0

Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of y and a are the same. Hence, the given formula is dimensionally correct.

15.   Given the relation,

Dimension of m = M1L1T0

Dimension of m0 = M1L0T0

Dimension of v = M0L1T-1

Dimension of v2 = M0L2T-2

Dimension of c = M0L1T-1

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that of R.H.S. This is only possible when the factor,  is dimensionless i.e., is dimensionless. This is only possible if is divided by.  Hence,  the  correct relation is  

16.   Radius of hydrogen atom, r = 0.5 A = 0.5 × 10-10 m

1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms = 6.023 × 1023 × 0.524 × 10-30

= 3.16 × 10-7 m3

17.   Radius of hydrogen atom, r = 0.5 A = 0.5 × 10-10 m

Now, 1 mole of hydrogen contains 6.023 × 1023 hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms, Va = 6.023 × 1023 × 0.524 × 10-30

= 3.16 × 10-7 m3

Molar volume of 1 mode of hydrogen atoms at STP.

Hence, the molar volume is 7.08 × 104 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.

18.   Line of sight is defined as an imaginary line joining an object and an observer’s eye.

When we observe nearby stationary objects such as trees, houses, etc. while sitting in a moving train, they appear to move rapidly in the opposite direction because the line of sight changes very rapidly.

On the other hand, distant objects such as trees, stars, etc. appear stationary because of the large distance. As a result, the line of sight does not change its direction rapidly.

19.   Diameter of Earth’s orbit = 3 × 1011  m

Radius of Earth’s orbit, r = 1.5 × 1011 m

Let the distance parallax angle be 1″ = 4.847 × 10-6 rad.

Let the distance of the star be D.

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1″.

20.   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

Using the relation,

Where,

Diameter of Earth’s orbit, d = 3 × 1011 m

Distance of the star from the Earth, D = 405868 × 1011 m

21.   It is indeed very true that precise measurements of physical quantities are essential for the development of science. For example, ultra-shot laser pulses (time interval ~ 10-15 s) are used to measure time intervals in several physical and chemical processes.

X-ray spectroscopy is used to determine the inter-atomic separation or inter-planer spacing.

The development of mass spectrometer makes it possible to measure the mass of atoms precisely.

22.   (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 =  3.3 × 1012 m2

Hence, volume of rain water, V = A × h = 7.09 × 1012 m3

Density of water, p = 1 × 103 kg m-3

Hence, mass of rain water = × V = 7.09 × 1015 kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 1015 kg.

Consider a ship of known base area floating in the s Measure its depth in sea (say d1). Volume of water displaced by the ship, Vb = Ad1

Now, move an elephant on the ship and measure the depth of the ship (d2) in this case. Volume of water displaced by the ship with the elephant on board, Vbe = Ad2

Volume of water displaced by the elephant = Ad2 – Ad1

Density of water = D

Mass of elephant = AD (d2 – d1)

(c) Wind speed during a storm can be measured by an 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 = πr2

Number of strands of hair = total surface area / area of hair

(e) Let the volume of the room be V.

 One mole of air at NTP occupies 22.4 l i.e., 22.4 × 10-3 m3 volume.

Number of molecules in one mole = 6.023 × 1023

∴ Number of molecules in room of volume V

 = 2.6888 × 1025 V

23.   Mass of the Sun, M = 2.0 × 1030 kg

Radius of the Sun, R = 7.0 × 108 m Volume

The density of the Sun is in the density range of solids and liquids. This high density is attributed to the intense gravitational attraction of the inner layers on the outer layer of the Sun.

24.   Distance of Jupiter from the Earth, D = 824.7 × 106 km = 824.7 × 109 m

Angular diameter = 35.72″ = 35.72 × 4.874 × 10-6 rad

Diameter of Jupiter = d

 Using the relation,

25.   Incorrect; on dimensional ground

The relation is. tan θ = v.

Dimension of R.H.S = M0L1T1

Dimension of L.H.S = M0L0T0

(∵ The trigonometric function is considered to be a dimensionless quantity)

Dimension of R.H.S is not equal to the dimension of L.H.S. Hence, the given relation is not correct dimensionally.

To make the given relation correct, the R.H.S should also be dimensionless. One way to achieve this is by dividing the R.H.S by the speed of rainfall v’.

Therefore, the relation reduces to  This relation is dimensionally correct.

26.   Difference in time of caesium clocks = 0.02 s

Time required for this difference = 100 years

= 100 × 365 × 24 × 60 × 60 = 3.15 × 109 s

In 3.15 × 109 s, the caesium clock shows a time difference of 0.02 s.

In 1s, the clock will show a time difference of  .

Hence, the accuracy of a standard caesium clock in measuring a time interval of 1 s is 

27.   Diameter of sodium atom = Size of sodium atom = 2.5 Å

Radius of sodium atom, r = 

=  1.25 × 10-10 m

Volume of sodium atom, V = 

According to the Avogadro hypothesis, one mole of sodium contains 6.023 × 1023 atoms and has a mass of 23 g or 23 × 10-3 kg.

∴ Mass of one atom = 

Density of sodium atom, p = mass of atom / vollume =  4.67 × 10-3 kg/m3

It is given that the density of sodium in crystalline phase is 970 kg m3.

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.

28.   Radius of nucleus r is given by the relation,

Now,  the mass of a nuclei M is equal to its mass number i.e.,

M = A amu = A × 1.66 × 10-27 kg

Density of nucleus,

This relation shows that nuclear mass depends only on constant . Hence, the nuclear mass densities of all nuclei are nearly the same.

Density of sodium nucleus is given by,

29.   Time taken by the laser beam to return to Earth after reflection from the Moon = 2.56 s

Speed of light = 3 × 108 m/s

Time taken by the laser beam to reach Moon = 

Radius of the lunar orbit = Distance between the Earth and the Moon = 1.28 × 3 × 108 = 3.84 × 108 m = 3.84 × 105 km

30.   Let the distance between the ship and the enemy submarine be ‘S’.

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

31.   Time taken by quasar light to reach Earth = 3 billion years

= 3 × 109 × 365 × 24 × 60 × 60 s

Speed of light = 3 × 10m/s

Distance between the Earth and quasar

= (3 × 108) × (3 × 109 × 365 × 24 × 60 × 60)

= 283824 3 × 1022 m

= 2.8 × 1022 m

32.   

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 = 3.84 × 108 m

Distance of the Sun from the Earth = 1.496 × 1011 m

Diameter of the Sun = 1.39 × 109 m

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:

Hence, the diameter of the Moon is 1.39 × 109 m.

33.   One relation consists of some fundamental constants that give the age of the Universe by:

Where,

t = Age of Universe

e = Charge of electrons = 1.6 × 10-19 C

ε0 = Absolute permittivity

m= Mass of protons = 1.67 × 10-27 kg

me = Mass of electrons = 9.1 × 10-31 m kg

c = Speed of light = 3 × 108 m/s

G = Universal gravitational constant = 6.67 × 1011 Nm2 kg-2

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