Maharashtra Board Class 11 Physics Chapter 1 Units and Measurements Solution

Maharashtra Board Class 11 Physics Solution Chapter 1 – Units and Measurements

Balbharati Maharashtra Board Class 11 Physics Solution Chapter 1: Units and Measurements. Marathi or English Medium Students of Class 11 get here Units and Measurements full Exercise Solution.

 Std Maharashtra Class 11 Subject Physics Chapter 1 Chapter Name Units and Measurements

1.) Choose the correct option.

i) [L1M1T-2] is the dimensional formula for

(A) Velocity (B) Acceleration

(C) Force (D) Work

Given dimensions are dimensions of Newton hence the force.

ii) The error in the measurement of thesides of a rectangle is 1%. The error inthe measurement of its area is

(A) 1% (B) 1/2%

(C) 2% (D) None of the above.

We know that for rectangle

A= lb

ΔA/A= ΔL/L+ Δb/b

ΔA/A= 1%+1%=2%.

iii) Light year is a unit of

(A) Time (B) Mass

(C) Distance (D) Luminosity

Light year the distance travelled by light in one year hence it is the unit of distance.

iv) Dimensions of kinetic energy are the same as that of

(A) Force (B) Acceleration

(C) Work (D) Pressure

Kinetic energy is nothing but form of work hence having same units.

v) Which of the following is not a fundamental unit?

(A) cm (B) kg

Volt is unit of voltage and voltage is not fundamental quantity.

i) Star A is farther than star B. Which starwill have a large parallax angle?

We have relation between parallax angle(θ) and distance of planet from earth (D) as

Θα1/D

As star A is farther than star B i.e

DA>DB Therefore  θAB . therefore star B is having larger parallax angle than star A.

ii) What are the dimensions of the quantity L √(l/g), l being the length and g the acceleration due to gravity?

iii) Define absolute error, mean absoluteerror, relative error and percentage error.

Absolute error– absolute error in given observations is defined as The magnitude of the difference between mean value and each individual value .

Let a1, a2, a3……. an are readings taken during observations and let Δa1 be absolute error in a1

Then,

Δa1 =│amean-a1

Δa2 =│amean-a2

Δan=│amean-an

Mean absolute error-The arithmetic mean of all the absoluteerrors is called mean absolute error. If Δamean is mean absolute error then

Δamean=(Δa1+Δa2+Δa3+……+Δan)/n

Relative error- it is the ratio of mean absolute error to its arithmetic mean value is called relative error.

Relative error= Δamean/amean.

Percentage error-When relative error is expressedin percentage it is called percentage error and given by

Percentage error = Δamean/amean 100%.

iv) Describe what is meant by significant figures and order of magnitude.

The number of digits in a measurement about which we are certain, plus one additional digit, the first one about which we are not certain is known as significant figures. More the significant figures more the accuracy of measurement. For example if 10.2kg is the measured weight of rice bag then 10 is certain and 2 is uncertain and total that is 3 is the significant figure.

The magnitude of any physical quantity can be expressed as A×10n where ‘A’ is a number such that 0.5 ≤ A<5 and ‘n’ is an integer called the order of magnitude.

If radius of Earth = 6400 km

= 6400×103m

= 0.64×107m

Hence order of magnitude is 7.

v) If the measured values of two quantities are A±ΔA and B ±ΔB, ΔA and ΔBbeing the mean absolute errors. What is the maximum possible error in A± B?

Show that if Z= A/B then

ΔZ/Z= ΔA/A+ ΔB/B.

If the measured values of two quantities are A±ΔA and B ±ΔB, ΔA and ΔB being the mean absolute errors then possible error in A ± B is given by ±ΔA±ΔB hence all possible 4 values of error are

+ΔA+ΔB

-ΔA+ΔB

-ΔA-ΔB

+ΔA-ΔB

Maximum value is +ΔA+ΔB .

vi) Derive the formula for kinetic energy ofa particle having mass m and velocity vusing dimensional analysis.

Let m is the mass and v is the velocity of the object.Kinetic energy of the particle is given by

E= K mavb

The dimensional formula of mass, [m]=[M]

The dimensional formula of velocity, [v]=[M0L1T⁻¹]

Dimension of kinetic energy =[M1L2T⁻2]

By substituting all the dimensions we get

K mavb = [M1L2T⁻2]

= [M1] [L2T⁻2]

K mavb = [M1] [L1T⁻¹]2

K mavb = [M1] [V]2since [v]=[L1T⁻¹]

Comparing both sides we get a= 1 and b=2.

Then formula for kinetic energy will become

E= K m1v2

Experimentally k = 1/2

Final formula for kinetic energy is (½) mv2.

Updated: April 23, 2022 — 11:47 pm