Home | | Physics 11th std | Solved Example Problems

# Solved Example Problems

Physics : Nature of Physical World and Measurement : Book Back Exercise, Example Numerical Question with Answers, Solution : Solved Example Problems

## Solved Example Problems for Measurement of length

Example 1.1

From a point on the ground, the top of a tree is seen to have an angle of elevation 60°. The distance between the tree and a point is 50 m. Calculate the height of the tree?

Solution

Angle θ = 60°

The distance between the tree and a point = 50 m

Height of the tree (h) = ?

For triangulation method tan

h = x tan θ

= 50 × tan 60°

= 50 × 1.732

h = 86.6 m The height of the tree is 86.6 m.

Example 1.2

The Moon subtends an angle of 1° 55 at the base line equal to the diameter of the Earth. What is the distance of the Moon from the Earth? (Radius of the Earth is 6.4 × 106 m)

Solution Radius of the Earth = 6.4 × 106 m

From the Figure 1.5 AB is the diameter of the Earth (b)= 2 × 6.4 × 106 m Distance of the Moon from the Earth x = ? Example 1.3

A RADAR signal is beamed towards a planet and its echo is received 7 minutes later. If the distance between the planet and the Earth is 6.3 × 1010 m. Calculate the speed of the signal?

Solution

The distance of the planet from the Earth d = 6.3 × 1010 m The speed of signal Solved Example Problems for Error Analysis

Example 1.4

In a series of successive measurements in an experiment, the readings of the period of oscillation of a simple pendulum were found to be 2.63s, 2.56 s, 2.42s, 2.71s and 2.80s. Calculate (i) the mean value of the period of oscillation (ii) the absolute error in each measurement (iii) the mean absolute error (iv) the relative error (v) the percentage error. Express the result in proper form.

Solution  Solved Example Problems for Propagation of errors

## (i) Solved Example Problem for Error in the sum of two quantities

Example 1.5

Two resistances R1  = (100 ± 3) Ω, R2  = (150 ± 2) Ω, are connected in series. What is their equivalent resistance?

Solution

Equivalent resistance R = ?

Equivalent resistance R = R1 + R2 ## (ii) Solved Example Problem for Error in the difference of two quantities

Example 1.6

The temperatures of two bodies measured by a thermometer are t1 = (20 + 0.5)°C, t2 = (50 ± 0.5)°C. Calculate the temperature difference and the error therein.

Solution ## (iii) Example Problem for Error in the product of two quantities

Example 1.7

The length and breadth of a rectangle are (5.7 ± 0.1) cm and (3.4 ± 0.2) cm respectively. Calculate the area of the rectangle with error limits.

Solution ## (iv) Example Problem for Error in the division or quotient of two quantities

Example 1.8

The voltage across a wire is (100 ± 5)V and the current passing through it is (10±0.2) A. Find the resistance of the wire.

Solution ## (v) Example Problem for Error in the power of a quantity

Example 1.9

A physical quantity x is given by x If the percentage errors of measurement in a, b, c and d are 4%, 2%, 3% and 1% respectively then calculate the percentage error in the calculation of x.

Solution The percentage error in x is given by The percentage error is x = 17.5%

Solved Example Problems for Significant Figures

## Rules of Significant Figures

Example 1.10

State the number of significant figures in the following

i.        600800

ii.       400

iii.      0.007

iv.      5213.0x

v.       2.65 × 1024 m

vi.      0.0006032

Solution

Solution: i) four ii) one iii) one iv) five v) three vi) four

## Rounding Off

Example 1.11

Round off the following numbers as indicated

i) 18.35 up to 3 digits

ii) 19.45 up to 3 digits

iii) 101.55 × 106 up to 4 digits

iv) 248337 up to digits 3 digits

v) 12.653 up to 3 digits.

Solution

i) 18.4 ii) 19.4 iii) 101.6 × 106 iv) 248000 v) 12.7

### Example:

1) 3.1 + 1.780 + 2.046 = 6.926

Here the least number of significant digits after the decimal is one. Hence the result will be 6.9.

2) 12.637 - 2.42 = 10.217

Here the least number of significant digits after the decimal is two. Hence the result will be 10.22

## Multiplication and Division

### Example:

1) 1.21 × 36.72 = 44.4312 = 44.4

Here the least number of significant digits in the measured values is three. Hence the result when rounded off to three significant digits is 44.4

2) 36.72 ÷ 1.2 = 30.6 = 31

Here the least number of significant digits in the measured values is two. Hence the result when rounded off to significant digit becomes 31.

Solved Example Problems for Application of the Method of Dimensional Analysis

Solved Example Problems

## (i) To convert a physical quantity from one system of units to another

Example 1.12

Convert 76 cm of mercury pressure into Nm−2 using the method of dimensions.

Solution

In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm−2

The dimensional formula of pressure P is [ML−1T−2] Example 1.13

If the value of universal gravitational constant in SI is 6.6x10−11 Nm2 kg−2, then find its value in CGS System?

Solution

Let GSI be the gravitational constant in the SI system and Gcgs in the cgs system. Then The dimensional formula for G is M−1 L3T −2 ## (ii) To check the dimensional correctness of a given physical equation

Example 1.14

Check the correctness of the equation using dimensional analysis method

Solution Both sides are dimensionally the same, hence the equations is dimensionally correct.

## (iii) To establish the relation among various physical quantities

Example 1.15

Obtain an expression for the time period T of a simple pendulum. The time period T depends on (i) mass ‘m’ of the bob (ii) length ‘l’ of the pendulum and (iii) acceleration due to gravity g at the place where the pendulum is suspended. (Constant k = 2π) i.e

Solution Here k is the dimensionless constant. Rewriting the above equation with dimensions Comparing the powers of M, L and T on both sides, a=0, b+c=0, -2c=1

Solving for a,b and c a = 0, b = 1/2, and c = −1/2

From the above equation T = k. m0  l1/2 g−1/2 Example 1.16

The force F acting on a body moving in a circular path depends on mass of the body (m), velocity (v) and radius (r) of the circular path. Obtain the expression for the force by dimensional analysis method. (Take the value of k=1) where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k = 1, we have Comparing the powers of M, L and T on both sides From the above equation we get ## 1. In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 sec. If the speed of sound in water is 1460 ms-1. What is the distance of enemy submarine?Answers:The speed of sound in water v = 1460 ms-1Time taken by the pulse for to and fro :t = T/2 = 80s/2 = 40sv = d/td = v × T/2 = 1460 × 40= 58400 m or 58.40 km.Ans: (58.40 km)2. The radius of the circle is 3.12 m. Calculate the area of the circle with regard to significant figures.Answers:Radius of the circle r = 3.12 mArea of the circle A = ?A = πr2 = 3.14 × 3.12 × 3.12 = 30.566016According to the rule of significant fig,A = 30.6 m2 [Given data has three sig. fig.]Ans: (30.6 m 2)3. Assuming that the frequency γ of a vibrating string may depend upon i) applied force (F) ii) length (l) iii) mass per unit length (m), prove that γa using dimensional analysis.Answers:γ ∝ lm Fb mcγ = K lm Fb mcK - dimensionless constant of proportionalitya,b,c - powers of l, F, mDimensional Formula of F = [MLT-2]Dimensional Formula of linear density Applying the principle of homogeneity of dimensionb + c = 0     ....(1)a + b -c = 0 ….. (2) 4. Jupiter is at a distance of 824.7 million km from the Earth. Its angular diameter is measured to be 35.72˝. Calculate the diameter of Jupiter. Ans: (1.428 × 105 km)5. The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement.Answers:The errors in both Z & T are least count errors. Ans: (6%)

Tags : Nature of Physical World and Measurement | Physics , 11th Physics : UNIT 1 : Nature of Physical World and Measurement
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th Physics : UNIT 1 : Nature of Physical World and Measurement : Solved Example Problems | Nature of Physical World and Measurement | Physics