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

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 *x *=* *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__

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 + *R*2

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*

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

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*

Example 1.9

A physical quantity *x i*s 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__

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

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

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

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

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 *L*3*T* −2

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.

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 *l*1/2 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

__Numerical Problems__

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-1

Time taken by the pulse for to and fro :

t = T/2 = 80s/2 = 40s

*v = d/t*

*d* = 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 m

Area of the circle A = ?

A = πr2 = 3.14 × 3.12 × 3.12 = 30.566016

According 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 γ*a ** *using dimensional analysis.

Answers:

γ ∝ *l*m Fb mc

γ = K *l*m Fb mc

K - dimensionless constant of proportionality

a,b,c - powers of *l*, F, *m*

Dimensional Formula of F = [MLT-2]

Dimensional Formula of linear density

Applying the principle of homogeneity of dimension

b + 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

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.