Board |
SCERT, Kerala |

Text Book |
NCERT Based |

Class |
Plus One |

Subject |
Physics Textbook Solution |

Chapter |
Chapter 2 |

Chapter Name |
Units and Measurement |

Category |
Plus One Kerala |

## Kerala Syllabus Plus One Physics Textbook Solution Chapter 2 Units and Measurement

- Chapter 1 : Physical World
- Chapter 2 : Units and Measurement
- Chapter 3 : Motion in a Straight Line
- Chapter 4: Motion in a Plane
- Chapter 5: Law of Motion
- Chapter 6: Work, Energy, and Power
- Chapter 7: Systems of Particles and Rotational Motion
- Chapter 8: Gravitation
- Chapter 9: Mechanical Properties of Solids
- Chapter 10: Mechanical Properties of Fluids
- Chapter 11: Thermal Properties of Matter
- Chapter 12: Thermodynamics
- Chapter 13: Kinetic Theory
- Chapter 14: Oscillations
- Chapter 15: Waves

##### Plus One Physics Part I

##### Plus One Physics Part II

### Chapter 2 Units and Measurement Textbook Solution

Fill in the blanks (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2 (c) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s (d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3.

(a) 1 cm =

Volume of the cube = 1 cm^{3}

But, 1 cm^{3} = 1 cm × 1 cm × 1 cm =

∴1 cm^{3} = 10^{–6} m^{3}

Hence, the volume of a cube of side 1 cm is equal to 10^{–6} m^{3}.

(b) The total surface area of a cylinder of radius *r* and height *h* is

*S* = 2π*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 × 10^{4} mm^{2}

(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/cm^{3}

Again, 1g =

1 cm^{3} = 10^{–6} m^{3}

1 g/cm^{3} =

∴ 11.3 g/cm^{3} = 11.3 × 10^{3} kg/m^{3}

Fill in the blanks by suitable conversion of units (a) 1 kg m2 s–2 = ....g cm2 s–2 (b) 1 m = ..... ly (c) 3.0 m s–2 = .... km h–2 (d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1.

^{}

(a) 1 kg = 10^{3} g

1 m^{2} = 10^{4} cm^{2}

1 kg m^{2} s^{â€“2} = 1 kg Ã— 1 m^{2} Ã— 1 s^{â€“2}

=10^{3} g Ã— 10^{4} cm^{2 }Ã— 1 s^{â€“2} = 10^{7} g cm^{2} s^{â€“2}

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

1 ly = Speed of light Ã— One year

= (3 Ã— 10^{8} m/s) Ã— (365 Ã— 24 Ã— 60 Ã— 60 s)

= 9.46 Ã— 10^{15} m

(c) 1 m = 10^{â€“3} km

Again, 1 s =

1 s^{â€“1} = 3600 h^{â€“1}

1 s^{â€“2} = (3600)^{2} h^{â€“2}

âˆ´3 m s^{â€“2} = (3 Ã— 10^{â€“3} km) Ã— ((3600)^{2} h^{â€“2})= 3.88 Ã— 10^{4} km h^{â€“2}

(d) 1 N = 1 kg m s^{â€“2}

1 kg = 10^{â€“3} g^{â€“1}

1 m^{3} = 10^{6} cm^{3}

âˆ´ 6.67 Ã— 10^{â€“11} N m^{2} kg^{â€“2} = 6.67 Ã— 10^{â€“11} Ã— (1 kg m s^{â€“2}) (1 m^{2}) (1 s^{â€“2})

= 6.67 Ã— 10^{â€“11} Ã— (1 kg Ã— 1 m^{3} Ã— 1 s^{â€“2})

= 6.67 Ã— 10^{â€“11} Ã— (10^{â€“3} g^{â€“1}) Ã— (10^{6} cm^{3}) Ã— (1 s^{â€“2})

= 6.67 Ã— 10^{â€“8} cm^{3} s^{â€“2} g^{â€“1}

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.

^{}

Given that,

1 calorie = 4.2 (1 kg) (1 m^{2}) (1 s^{–2})

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,

∴1 calorie = 4.2 (1 α^{–1}) (1 β^{–2}) (1 γ^{2}) = 4.2 α^{–1} β^{–2} γ^{2}

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:

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

^{}

The given statement is true because a dimensionless quantity may be large or small in comparision 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.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:

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

The given statement is true because a dimensionless quantity may be large or small in comparision 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.

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?

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

Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light ?

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.

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

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.

Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ? (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?

(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 using a metre scale. The diameter of the thread is given by the relation,

(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.

The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement.

Area of the house on the slide = 1.75 cm^{2}

Area of the image of the house formed on the screen = 1.55 m^{2}

= 1.55 × 10^{4} cm^{2}

Arial magnification, *m*_{a} =

∴Linear magnifications, *m*_{l} =

State the number of significant figures in the following : (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm–3 2019-20 36 PHYSICS (d) 6.320 J (e) 6.032 N m–2 (f) 0.0006032 m2

(a) **Answer: **1

The given quantity is 0.007 m^{2}.

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, One 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 × 10^{24} 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 One non-zero digits are always significant.

(f) **Answer: **4

The given quantity is 0.0006032 m^{2}.

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 One non-zero digits are always significant. Hence, the remaining four digits are significant figures.

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

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:

Quantity

Number

Significant Figure

*l*

4.234

4

*b*

1.005

4

*h*

0.0201

3

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 m^{2}

Volume of the sheet = *l *× *b* × *h*

= 4.234 × 1.005 × 0.0201

= 0.0855 m^{3}

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

The mass of a box measured by a grocer’s balance is 2.30 kg. One gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ?

Mass of grocer’s box = 2.300 kg

Mass of gold piece **I** = 20.15g = 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.

A physical quantity P is related to four observables a, b, c and d as follows : = ( ) 3 2 P ab/ c d The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

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

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion : (a) y = a sin 2π t/T (b) y = a sin vt (c) y = (a/T) sin t/a (d) y a = ( )2 (sin 2 / + cos 2 / ) πt T t T π (a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

(a) **Answer: **Correct

Dimension of *y* = M^{0} L^{1} T^{0}

Dimension of *a* = M^{0} L^{1} T^{0}

Dimension of = M^{0} L^{0} T^{0}

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* = M^{0} L^{1} T^{0}

Dimension of *a* = M^{0} L^{1} T^{0}

Dimension of *vt* = M^{0} L^{1} T^{–1} × M^{0} L^{0} T^{1 }= M^{0} L^{1} T^{0}

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* = M^{0}L^{1}T^{0}

Dimension of = M^{0}L^{1}T^{–1}

Dimension of= M^{0} L^{–1} T^{1}

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* = M^{0} L^{1} T^{0}

Dimension of *a* = M^{0} L^{1} T^{0}

Dimension of = M^{0} L^{0} T^{0}

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.

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :m m 1 v 0 = − 2 1/2 . Guess where to put the missing c.

Given the relation,

Dimension of *m* = M^{1} L^{0} T^{0}

Dimension of = M^{1} L^{0} T^{0}

Dimension of *v* = M^{0} L^{1} T^{–1}

Dimension of *v*^{2} = M^{0} L^{2} T^{–2}

Dimension of *c* = M^{0} L^{1} T^{–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., (1 – *v*^{2}) is dimensionless. This is only possible if *v*^{2} is divided by c^{2}. Hence, the correct relation is

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ?

Radius of hydrogen atom, *r* = 0.5 = 0.5 × 10^{–10} m

Volume of hydrogen atom =

1 mole of hydrogen contains 6.023 × 10^{23} hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms = 6.023 × 10^{23} × 0.524 × 10^{–30}

= 3.16 × 10^{–7} m^{3}

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 Å). Why is this ratio so large ?

^{}

Radius of hydrogen atom, *r* = 0.5 = 0.5 × 10^{–10} m

Volume of hydrogen atom =

Now, 1 mole of hydrogen contains 6.023 × 10^{23} hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms, *V*_{a} = 6.023 × 10^{23} × 0.524 × 10^{–30}

= 3.16 × 10^{–7} m^{3}

Molar volume of 1 mole of hydrogen atoms at STP,

*V*_{m} = 22.4 L = 22.4 × 10^{–3} m^{3}

Hence, the molar volume is 7.08 × 10^{4} 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.

Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

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.

The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s One locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?

Diameter of Earth’s orbit = 3 × 10^{11} m

Radius of Earth’s orbit, *r* = 1.5 × 10^{11} m

Let the distance parallax angle be= 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.

∴ We have

Hence, 1 parsec ≈ 3.09 × 10^{16} m.

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 One locations of the Earth six months apart in its orbit around the Sun ?

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

= 3 × 10^{8} × 365 × 24 × 60 × 60 = 94608 × 10^{11} m

∴4.29 ly = 405868.32 × 10^{11} m

1 parsec = 3.08 × 10^{16} m

∴4.29 ly = = 1.32 parsec

Using the relation,

But, 1 sec = 4.85 × 10^{–6} rad

∴

Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

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.

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.

(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 × 10^{12} m^{2}

Hence, volume of rain water, *V *= *A* × *h* = 7.09 × 10^{12} m^{3}

Density of water, *ρ* = 1 × 10^{3} kg m^{–3}

Hence, mass of rain water = *ρ* × *V* = 7.09 × 10^{15} kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 10^{15} kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say *d*_{1}).

Volume of water displaced by the ship, *V*_{b} = *A* *d*_{1}

Now, move an elephant on the ship and measure the depth of the ship (*d*_{2}) in this case.

Volume of water displaced by the ship with the elephant on board, *V*_{be}= *Ad*_{2}

Volume of water displaced by the elephant = *Ad*_{2} – *Ad*_{1}

Density of water = *D*

Mass of elephant = *AD* (*d*_{2} – *d*_{1})

(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 = π*r*^{2}

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., 22.4 × 10^{–3} m^{3} volume.

Number of molecules in one mole = 6.023 × 10^{23}

∴Number of molecules in room of volume *V*

== 134.915 × 10^{26}*V*

= 1.35 × 10^{28}*V*

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030 kg, radius of the Sun = 7.0 × 108 m

Mass of the Sun, *M* = 2.0 × 10^{30} kg

Radius of the Sun, *R* = 7.0 × 10^{8} m

Volume of the Sun, *V* =

Density of the Sun =

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.

When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72" of arc. Calculate the diameter of Jupiter.Distance of Jupiter from the Earth, *D* = 824.7 × 10^{6} km = 824.7 × 10^{9} m

Angular diameter =

Diameter of Jupiter = *d*

Using the relation,

Fill in the blanks (a) The volume of a cube of side 1 cm is equal to .....m3 (b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ...(mm)2 (c) A vehicle moving with a speed of 18 km h–1 covers....m in 1 s (d) The relative density of lead is 11.3. Its density is ....g cm–3 or ....kg m–3.

(a) 1 cm =

Volume of the cube = 1 cm^{3}

But, 1 cm^{3} = 1 cm × 1 cm × 1 cm =

∴1 cm^{3} = 10^{–6} m^{3}

Hence, the volume of a cube of side 1 cm is equal to 10^{–6} m^{3}.

(b) The total surface area of a cylinder of radius *r* and height *h* is

*S* = 2π*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 × 10^{4} mm^{2}

(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/cm^{3}

Again, 1g =

1 cm^{3} = 10^{–6} m^{3}

1 g/cm^{3} =

∴ 11.3 g/cm^{3} = 11.3 × 10^{3} kg/m^{3}

Fill in the blanks by suitable conversion of units (a) 1 kg m2 s–2 = ....g cm2 s–2 (b) 1 m = ..... ly (c) 3.0 m s–2 = .... km h–2 (d) G = 6.67 × 10–11 N m2 (kg)–2 = .... (cm)3 s–2 g–1.

^{}

(a) 1 kg = 10^{3} g

1 m^{2} = 10^{4} cm^{2}

1 kg m^{2} s^{â€“2} = 1 kg Ã— 1 m^{2} Ã— 1 s^{â€“2}

=10^{3} g Ã— 10^{4} cm^{2 }Ã— 1 s^{â€“2} = 10^{7} g cm^{2} s^{â€“2}

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

1 ly = Speed of light Ã— One year

= (3 Ã— 10^{8} m/s) Ã— (365 Ã— 24 Ã— 60 Ã— 60 s)

= 9.46 Ã— 10^{15} m

(c) 1 m = 10^{â€“3} km

Again, 1 s =

1 s^{â€“1} = 3600 h^{â€“1}

1 s^{â€“2} = (3600)^{2} h^{â€“2}

âˆ´3 m s^{â€“2} = (3 Ã— 10^{â€“3} km) Ã— ((3600)^{2} h^{â€“2})= 3.88 Ã— 10^{4} km h^{â€“2}

(d) 1 N = 1 kg m s^{â€“2}

1 kg = 10^{â€“3} g^{â€“1}

1 m^{3} = 10^{6} cm^{3}

âˆ´ 6.67 Ã— 10^{â€“11} N m^{2} kg^{â€“2} = 6.67 Ã— 10^{â€“11} Ã— (1 kg m s^{â€“2}) (1 m^{2}) (1 s^{â€“2})

= 6.67 Ã— 10^{â€“11} Ã— (1 kg Ã— 1 m^{3} Ã— 1 s^{â€“2})

= 6.67 Ã— 10^{â€“11} Ã— (10^{â€“3} g^{â€“1}) Ã— (10^{6} cm^{3}) Ã— (1 s^{â€“2})

= 6.67 Ã— 10^{â€“8} cm^{3} s^{â€“2} g^{â€“1}

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.

^{}

Given that,

1 calorie = 4.2 (1 kg) (1 m^{2}) (1 s^{–2})

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,

∴1 calorie = 4.2 (1 α^{–1}) (1 β^{–2}) (1 γ^{2}) = 4.2 α^{–1} β^{–2} γ^{2}

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:

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

^{}

The given statement is true because a dimensionless quantity may be large or small in comparision 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.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:

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

The given statement is true because a dimensionless quantity may be large or small in comparision 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.

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance ?

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

Which of the following is the most precise device for measuring length : (a) a vernier callipers with 20 divisions on the sliding scale (b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale (c) an optical instrument that can measure length to within a wavelength of light ?

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.

A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair ?

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.

Answer the following : (a)You are given a thread and a metre scale. How will you estimate the diameter of the thread ? (b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ? (c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only ?

(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 using a metre scale. The diameter of the thread is given by the relation,

(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.

The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m2. What is the linear magnification of the projector-screen arrangement.

Area of the house on the slide = 1.75 cm^{2}

Area of the image of the house formed on the screen = 1.55 m^{2}

= 1.55 × 10^{4} cm^{2}

Arial magnification, *m*_{a} =

∴Linear magnifications, *m*_{l} =

State the number of significant figures in the following : (a) 0.007 m2 (b) 2.64 × 1024 kg (c) 0.2370 g cm–3 2019-20 36 PHYSICS (d) 6.320 J (e) 6.032 N m–2 (f) 0.0006032 m2

(a) **Answer: **1

The given quantity is 0.007 m^{2}.

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, One 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 × 10^{24} 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 One non-zero digits are always significant.

(f) **Answer: **4

The given quantity is 0.0006032 m^{2}.

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 One non-zero digits are always significant. Hence, the remaining four digits are significant figures.

The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

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:

Quantity | Number | Significant Figure |

| 4.234 | 4 |

| 1.005 | 4 |

| 0.0201 | 3 |

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 m^{2}

Volume of the sheet = *l *× *b* × *h*

= 4.234 × 1.005 × 0.0201

= 0.0855 m^{3}

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

The mass of a box measured by a grocer’s balance is 2.30 kg. One gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures ?

Mass of grocer’s box = 2.300 kg

Mass of gold piece **I** = 20.15g = 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.

A physical quantity P is related to four observables a, b, c and d as follows : = ( ) 3 2 P ab/ c d The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

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

A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion : (a) y = a sin 2π t/T (b) y = a sin vt (c) y = (a/T) sin t/a (d) y a = ( )2 (sin 2 / + cos 2 / ) πt T t T π (a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds.

(a) **Answer: **Correct

Dimension of *y* = M^{0} L^{1} T^{0}

Dimension of *a* = M^{0} L^{1} T^{0}

Dimension of = M^{0} L^{0} T^{0}

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* = M^{0} L^{1} T^{0}

Dimension of *a* = M^{0} L^{1} T^{0}

Dimension of *vt* = M^{0} L^{1} T^{–1} × M^{0} L^{0} T^{1 }= M^{0} L^{1} T^{0}

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* = M^{0}L^{1}T^{0}

Dimension of = M^{0}L^{1}T^{–1}

Dimension of= M^{0} L^{–1} T^{1}

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* = M^{0} L^{1} T^{0}

Dimension of *a* = M^{0} L^{1} T^{0}

Dimension of = M^{0} L^{0} T^{0}

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.

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ mo of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :m m 1 v 0 = − 2 1/2 . Guess where to put the missing c.

Given the relation,

Dimension of *m* = M^{1} L^{0} T^{0}

Dimension of = M^{1} L^{0} T^{0}

Dimension of *v* = M^{0} L^{1} T^{–1}

Dimension of *v*^{2} = M^{0} L^{2} T^{–2}

Dimension of *c* = M^{0} L^{1} T^{–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., (1 – *v*^{2}) is dimensionless. This is only possible if *v*^{2} is divided by c^{2}. Hence, the correct relation is

The unit of length convenient on the atomic scale is known as an angstrom and is denoted by Å: 1 Å = 10–10 m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atoms ?

Radius of hydrogen atom, *r* = 0.5 = 0.5 × 10^{–10} m

Volume of hydrogen atom =

1 mole of hydrogen contains 6.023 × 10^{23} hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms = 6.023 × 10^{23} × 0.524 × 10^{–30}

= 3.16 × 10^{–7} m^{3}

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 Å). Why is this ratio so large ?

^{}

Radius of hydrogen atom, *r* = 0.5 = 0.5 × 10^{–10} m

Volume of hydrogen atom =

Now, 1 mole of hydrogen contains 6.023 × 10^{23} hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms, *V*_{a} = 6.023 × 10^{23} × 0.524 × 10^{–30}

= 3.16 × 10^{–7} m^{3}

Molar volume of 1 mole of hydrogen atoms at STP,

*V*_{m} = 22.4 L = 22.4 × 10^{–3} m^{3}

Hence, the molar volume is 7.08 × 10^{4} 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.

Explain this common observation clearly : If you look out of the window of a fast moving train, the nearby trees, houses etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).

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.

The principle of ‘parallax’ in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s One locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈ 3 × 1011m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?

Diameter of Earth’s orbit = 3 × 10^{11} m

Radius of Earth’s orbit, *r* = 1.5 × 10^{11} m

Let the distance parallax angle be= 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.

∴ We have

Hence, 1 parsec ≈ 3.09 × 10^{16} m.

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 One locations of the Earth six months apart in its orbit around the Sun ?

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

= 3 × 10^{8} × 365 × 24 × 60 × 60 = 94608 × 10^{11} m

∴4.29 ly = 405868.32 × 10^{11} m

1 parsec = 3.08 × 10^{16} m

∴4.29 ly = = 1.32 parsec

Using the relation,

But, 1 sec = 4.85 × 10^{–6} rad

∴

Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.

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.

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.

(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 × 10^{12} m^{2}

Hence, volume of rain water, *V *= *A* × *h* = 7.09 × 10^{12} m^{3}

Density of water, *ρ* = 1 × 10^{3} kg m^{–3}

Hence, mass of rain water = *ρ* × *V* = 7.09 × 10^{15} kg

Hence, the total mass of rain-bearing clouds over India is approximately 7.09 × 10^{15} kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say *d*_{1}).

Volume of water displaced by the ship, *V*_{b} = *A* *d*_{1}

Now, move an elephant on the ship and measure the depth of the ship (*d*_{2}) in this case.

Volume of water displaced by the ship with the elephant on board, *V*_{be}= *Ad*_{2}

Volume of water displaced by the elephant = *Ad*_{2} – *Ad*_{1}

Density of water = *D*

Mass of elephant = *AD* (*d*_{2} – *d*_{1})

(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 = π*r*^{2}

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., 22.4 × 10^{–3} m^{3} volume.

Number of molecules in one mole = 6.023 × 10^{23}

∴Number of molecules in room of volume *V*

== 134.915 × 10^{26}*V*

= 1.35 × 10^{28}*V*

The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 × 1030 kg, radius of the Sun = 7.0 × 108 m

Mass of the Sun, *M* = 2.0 × 10^{30} kg

Radius of the Sun, *R* = 7.0 × 10^{8} m

Volume of the Sun, *V* =

Density of the Sun =

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.

Distance of Jupiter from the Earth, *D* = 824.7 × 10^{6} km = 824.7 × 10^{9} m

Angular diameter =

Diameter of Jupiter = *d*

Using the relation,

### PDF Download

#### Chapter 2: Units and Measurement Textbook Solution

#### Chapter 2: Units and Measurement Textbook Solution - Preview

#### Plus One Physics Chapter Wise Textbook Solutions

- Chapter 1: Physical World
- Chapter 2: Units and Measurement
- Chapter 3: Motion in a Straight Line
- Chapter 4: Motion in a Plane
- Chapter 5: Law of Motion
- Chapter 6: Work, Energy, and Power
- Chapter 7: Systems of Particles and Rotational Motion
- Chapter 8: Gravitation
- Chapter 9: Mechanical Properties of Solids
- Chapter 10: Mechanical Properties of Fluids
- Chapter 11: Thermal Properties of Matter
- Chapter 12: Thermodynamics
- Chapter 13: Kinetic Theory
- Chapter 14: Oscillations
- Chapter 15: Waves

##### Plus One Physics Part I

##### Plus One Physics Part II

**comment and share**this article if you found it useful. Give your valuable suggestions in the comment session or contact us for any details regarding the HSE Kerala Plus One syllabus, Previous year question papers, and other study materials.

### Plus One Physics Related Links

Plus One Physics Notes |
Click Here |

Plus One Physics Textbook Solutions |
Click Here |

Plus One Focus Area | Click Here |

Plus One Previous Year Questions with Solution | Click Here |

Plus One Latest Syllabus | Click Here |

### Other Related Links

Plus One Physics | Click Here |

Plus One Chemistry | Click Here |

Plus One Mathematics | Click Here |

Plus One Botany | Click Here |

Plus One Zoology | Click Here |

Plus One Computer Science | Click Here |

Plus One English | Click Here |

**drop a comment below**and we will get back to you at the earliest.