HC Verma Concepts Of Physics Exercise Solutions Of Chapter 2 (Physics and Mathematics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 2 - Physics And Mathematics:

EXERCISE

1. A vector A makes an angle of 20° and B makes an angle of 110° with the X-axis. The magnitudes of these vectors are 3 m and 4 m respectively. Find the resultant.

Sol: 

So, resultant makes an angle with X-axis is 53⁰ + 20⁰ = 73⁰.

2. Let A and B be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angles 30° and 60° respectively, find the resultant.

Sol:

So, resultant makes an angle with X-axis is 15⁰ + 30⁰ = 45⁰.

Alternative method,

X component of OA = 10 cos 30° = 5√3

X component of BC = 10 cos 60° = 5 

Y component of OA = 10 sin 30° = 5

Y component of BC = 1.5 sin 60° = 5√3

R_x = x component of resultant = 5 + 5√3 = 13.66 m

R_y = y component of resultant= 5 + 5√3 = 13.66 m

So, R = Resultant = 19.32 m                                    

If it makes an angle a with positive x-axis

Tan a = x component / y component = 1 
⇒ a = tan^–1 1 = 45⁰.

3. Add vectors A, B and C each having magnitude of 100 unit and inclined to the X-axis at angles 45°, 135° and 315° respectively.

Sol: 

Vector B and C are equal in magnitude but opposite in direction. Therefore, resultant of these two vectors is zero vector. If we add zero vector with another vector, resultant will be that vector.

So, resultant of vectors A, B, and C is same as vector A.

R = 100 unit at 45⁰ with X-axis.

4. Let a = 4 i + 3 j and b = 3 i + 4 j. (a) Find the magnitudes of (a) a, (b) b, (c) a + b and (d) a - b.

Sol:

(a) | a|= √ (4² + 3²) 
Or, | a| = √ (16 + 9) = √25 = 5.

(b) | b|= √ (3² + 4²) 
Or, | b| = √ (9 + 16) = √25 = 5.

(c)  a + b = (4 i + 3 j) + (3 i + 4 j) = 7 i + 7 j

Or, | a + b |= √ (7² + 7²) = 7√2.

(d) a + b = (4 i + 3 j) – (3 i + 4 j) = i – j

Or, | a – b |= √ {1² + (– 1)²} = √2.

5. Refer to figure (2-E1). Find (a) the magnitude, (b) x and y components and (c) the angle with the X-axis of the resultant of OA, BC and DE.

Sol:

Resultant vector diagram 

Angle between vector OA and BC is 90⁰.

Therefore, resultant of OA and BC is

Angle between vector OA and R₁₂ is given by

So, angle between vector DE and R₁₂ is 90⁰ + 36.9⁰ + 30⁰ = 156.9⁰. 

(a) The magnitude of the resultant of OA, BC and DE is 

(c) angle between vector DE and R is given by

angle between vector DE and R is = 180⁰ – 37⁰ = 143⁰

Therefore, angle made by resultant with X-axis is

= (143⁰ – 90⁰) = 53⁰.

(b) X component of resultant is = 1.62*cos 53⁰ = 0.98 m.

And Y component of resultant is = 1.62*sin 53⁰ = 1.3 m.

Alternative method,

X component of OA = 2 cos 30° = √3

X component of BC = 1.5 cos 120° = – 0.75

X component of DE = 1 cos 270° = 0

Y component of OA = 2 sin 30° = 1

Y component of BC = 1.5 sin 120° = 1.3

Y component of DE = 1 sin 270° = – 1

R_x = x component of resultant = √3 – 0.75 + 0 = 0.98 m

R_y = resultant y component = 1 + 1.3 – 1 = 1.3 m

So, R = Resultant = 1.6 m

If it makes an angle a with positive x-axis  

Tan a = x component / y component = 1.32

Or, a = tan^–1 1.32.

6. Two vectors have magnitudes 3 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit?

Sol:

We know, the magnitude of the resultant is given by

(a) (1)² = (3)² + (4)² + 2 * 3 * 4 * cos θ

Or, cos θ = – (24/24) = – 1

Or, θ = 180⁰.

(b) (5)² = (3)² + (4)² + 2 * 3 * 4 * cos θ

Or, cos θ = (0/24) = 0  

Or, θ = 90⁰.

(c) (7)² = (3)² + (4)² + 2 * 3 * 4 * cos θ

Or, cos θ = (24/24) = 1

Or, θ = 0⁰.

7. A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.

Sol:

Vector diagram

The magnitude of displacement vector (R) is

= √ {(2+4)² + (0.5)²}

= √ {36 + 0.25} = 6.02 km.

Its angle with WE axis is α

→ a = tan^-1 (0.5/6)

Or, α = tan^-1 (1/12) = 4.8⁰.

8. A carom board (4 ft x 4 ft square) has the queen at the centre. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the centre to the front edge, (b) from the front edge to the hole and (c) from the centre to the hole.

Sol:

Let, CD be ‘x’

Then, AD = (2 – x); FB = AC = 2 ft; BD = 4 ft.

From ∆ BFE, tan θ = (FE/FB) = x/2 and

From ∆ ABD, tan θ = (AD/BD) = (2 – x)/4

Therefore, x/2 = (2 – x)/4   

Or, x = 2/3 = CD

(a) The magnitude of displacement of the queen from the centre to the front edge is

→ BE = √ (FB² + FE²)

Or, BE = √ {2² + (2/3)²}

Or, BE = √ (40/9) = (2/3) √ 10 ft.

(b) The magnitude of displacement of the queen from the front edge to the hole is

→ AB = √ (BD² + AD²)

Or, AB = √ [4² + {2 – (2/3)}²]

Or, AB = √ (160/9) = (4/3) √ 10 ft.

(c) The magnitude of displacement of the queen from the centre to the hole is

→ AE = √ (AC² + CE²)

Or, AE = √ (2² + 2²) = 2√2 ft.

9. A mosquito net over a 7 ft * 4 ft bed is 3 ft high. The net has a hole at one corner of the bed through which a mosquito enters the net. It flies and sits at the diagonally opposite upper corner of the net. (a) Find the magnitude of the displacement of the mosquito, (b) Taking the hole as the origin, the length of the bed as the X-axis, its width as the Y-axis, and vertically up as the Z-axis, write the components of the displacement vector.

Sol:

Mosquito enters the net through ‘A’. It flies and sits at the diagonally opposite upper corner of the net (that is at ‘G’).

Therefore, the displacement of the mosquito is given by

→ R = 7 i + 4 j + 3 k

(a) The magnitude of the displacement of the mosquito is

= √ (7² + 4² + 3²)          

= √ (49 + 16 + 9) = √ 74 ft.

(b) X component of the displacement vector is 7 ft.

Y component of the displacement vector is 4 ft.

Z component of the displacement vector is 3 ft.

10. Suppose a is a vector of magnitude 4.5 unit due north.

What is the vector (a) 3 a, (b) – 4 a?

Sol: 

(a) 3 a = (3 * 4.5) unit due north = 13.5 unit due north.

(b) – 4 a = (– 4 * 4.5) unit due north = 18.0 unit due south.

11. Two vectors have magnitudes 2 m and 3 m. The angle between them is 60°. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.

Sol: 

(a) We know,

The scalar product of the two vectors is given by

→ A.B = | A|| B| cos θ                 

Or, A.B = 2 * 3 cos 60° = 3 m².

(b) We know,

The vector product of the two vectors is given by

→ A*B = | A|| B| sin θ n 

Or, | A*B| = 2 * 3 cos 60° = 3√3 m².

12. Let A₁, A₂, A₃, A₄, A₅, A₆, A₇ be a regular hexagon. Write the x-components of the vectors represented by the six sides taken in order. Use the fact that the resultant of these six vectors is zero, to prove that cos 0 + cos π/3 + cos 2π/3 + cos 3π/3 + cos 4π/3 + cos 5π/3 = 0. Use the known cosine values to verify the result.

   

Sol:

It is given that, the resultant of six vectors is zero. So, components of it resultant vector also zero.

So, R_x = A₁ cos 0⁰ + A₂ cos π/3 + A₃ cos 2π/3 + A₄ cos 3π/3 + A₅ cos 4π/3 + A₆ cos 5π/3 = 0

Magnitude of A₁, A₂, A₃, A₄, A₅, A₆, and A₇ all are same.

Therefore,

cos 0⁰ + cos π/3 + cos 2π/3 + cos 3π/3 + cos 4π/3 + cos 5π/3 = 0.

Similarly,

R_y = A₁ sin 0⁰ + A₂ sin π/3 + A₃ sin 2π/3 + A₄ sin 3π/3 + A₅ sin 4π/3 + A₆ sin 5π/3 = 0

Or, sin 0⁰ + sin π/3 + sin 2π/3 + sin 3π/3 + sin 4π/3 + sin 5π/3 = 0.

13. Let a = 2 i + 3 j + 4 k and b = 3 i + 4 j + 5 k. Find the angle between them.

Sol:

We know, a. b = | a|| b| Cos θ

Or, Cos θ = a. b /| a|| b|                    

→ a. b = (2 i + 3 j + 4 k ).(3 i + 4 j + 5 k)

Or, a. b = 6 + 12 + 20 = 38

→ | a|| b|= {√ (2² + 3² + 4²)} {√ (3² + 4² + 5²)}

Or, | a|| b| = √ (29) * √ (50) = √ (1450)

So, cos θ = 38/√ (1450)

Or, θ = cos^-1 {38/√ (1450)} = 3.7⁰.

14. Prove that A • (A * B) = 0.

Sol:

As, A * B = A B sin θ n

AB sin θ n is a vector which is perpendicular to the plane containing A and B, this implies that it is also perpendicular to A

As dot product of two perpendicular vector is zero.

Thus,

A • (A * B) = 0.

15. If A = 2 i+ 3 j+ 4 k and B = 4 i+ 3 j+ 2 k, find A x B.

Sol:

A x B =          

           

= i (6 – 12) – j (4 – 16) + k (6 – 12)

= – 6 i + 12 j – 6 k.

16. If A, B, C are mutually perpendicular, show that C * (A * B) = 0. Is the converse true?

Sol:

   

Given: that A, B and C are mutually perpendicular.

A * B is a vector which direction is perpendicular to the plane containing A and B.

Also C is perpendicular to A and B

∴ Angle between C and A * B is 0⁰ or 180⁰.

So,    C * (A * B) = 0 

The converse is not true.

For example, if two of the vector are parallel, then also C * (A * B) = 0

So, they need not be mutually perpendicular.

17. A particle moves on a given straight line with a constant speed v. At a certain time it is at a point P on its straight line path. O is a fixed point. Show that OP * v is independent of the position P.

Sol:

  

OP * v = (OP) v sin θ n = (AO) v n

It can be seen from the figure, OQ = OP sin θ = OP’ sin θ’.

So, whatever may be the position of the particle, the magnitude and direction of OP * v remain constant.

∴ OP * v is independent of the position P.

18. The force on a charged particle due to electric and magnetic fields is given by F = qE + qv * B. Suppose E is along the X-axis and B along the Y-axis. In what direction and with what minimum speed v should a positively charged particle be sent so that the net force on it is zero?

Sol:

  

Given: net force on charge is equal to zero.

∴ F = qE + qv * B = 0

Or, E + v * B = 0

Or, E = – (v * B)

So, the direction of v * B should be opposite to the direction of E. Hence, v should be in the positive yz-plane and magnitude should be equal.

→ E = v B sin θ

Or, v = E/B sin θ

For minimum velocity (v_min) sin θ should be maximum. [Sin θ = 1, θ = 90⁰]

V_min = E/B along positive z-axis.

19. Give an example for which A • B = C • B but A ≠ C.

Sol:                                     

A = 2 i; B = 3 j; C = 9 k.

A • B = (2 i). (3 j) = 0

C • B = (9 k). (3 j) = 0 and A ≠ C.

20. Draw a graph from the following data. Draw tangents at x = 2, 4, 6 and 8. Find the slopes of these tangents. Verify that the curve drawn is y = 2x² and the slope of tangent is tan θ = dy/dx = – 4x.

Sol:

From the graph, we can say curve is a parabola.

Let the equation of curve be

 y = ax² + bx + c

Putting (1, 2), (2, 8) and (3, 18) we get following three equation.

2 = a + b + c ---------------- 1

8 = 4a + 2b + c ------------- 2

18 = 9a + 3b + c ------------ 3

By solving above three equation, we get

a = 2; b = 0; c = 0

So, equation of the curve is y = 2x²

The slope of tangent is tan θ = dy/dx =  = 4x

Slope of tangent @ 2 = 4 * 2 = 8

Slope of tangent @ 4 = 4 * 4 = 16

Slope of tangent @ 6 = 4 * 6 = 24 and

Slope of tangent @ 8 = 4 * 8 = 32.

21. A curve is represented by y = sin x. If x is changed from π/3 to π/3 + π/100, find approximately the change in y.

We have y = sin x

For small changes ∆x, the approximately the change in y is given by

→ ∆y = (dy/dx)∆x

Or, ∆y = [d(sin x)/dx]∆x

Or, ∆y = cos x (∆x) = [cos (π/3)]*(π/3)

Or, ∆y = 0.0157.

22. The electric current in a charging R-C circuit is given by i = i_oe^{- t /RC} where i₀, R and C are constant parameters of the circuit and t is time. Find the rate of change of current at (a) t = 0, (b) t = RC, (c) t = 10 RC.

Sol:

Given: that i = i_oe^{- t /RC}.

Rate of change of current = di/dt = – (1/RC) i_oe^{- t /RC}.

(a) The rate of change of current at t = 0

(b) The rate of change of current at t = RC

(c) The rate of change of current at t = 10 RC

23. The electric current in a discharging R-C circuit is given by i = i_oe^-t/RC where i₀, R and C are constant parameters and t is time. Let i_o = 2.00 A, R = 6.00 x 10⁵ Ω and C = 0.500 µF. (a) Find the current at t = 0.3 s. (b) Find the rate of change of current at t = 0.3 s. (c) Find approximately the current at t = 0.31 s.

Sol:    

Given: R = 6.00 x 10⁵ Ω; C = 0.500 µF; i_o = 2.00 A; RC = (6 *10⁵) * (0.5*10^-6) = 0.3 s.

(a) The current at t = 0.3 s is

→ i = i_oe^-t/RC = 2*e^-(0.3/0.3)

Or, i = (2/e) A = 0.736 A.

(b) Rate of change of current = di/dt = – (1/RC) i_oe^{- t /RC}

→ di/dt|_{t = 0.3s} = – 2/0.3 e^-1 = – 20/3e  A/s.

(c) The current at t = 0.31 s is

→ i = i_oe^-t/RC = 2 * e^-(0.31/0.3) 

Or, i = (5.8/3e) A = 0.712 A. 

24. Find the area bounded under the curve y = 3x² + 6x + 7 and the X-axis with the ordinates at x = 5 and x = 10.

Sol:

  

The area bounded under the curve y = 3x² + 6x + 7 and the X-axis is given by 

25. Find the area enclosed by the curve y = sin x and the X-axis between x = 0 and x = π.

Sol:

The area enclosed by the curve y = sin x and the X-axis between x = 0 and x = π is given by

26. Find the area bounded by the curve y = e^-x, the X-axis and the Y-axis.

Sol:

 

The area bounded by the curve y = e^-x, the X-axis and the Y-axis is given by   

27. A rod of length L is placed along the X-axis between x = 0 and x = L. The linear density (mass/Length) ρ of the rod varies with the distance x from the orgin as ρ = a + bx. (a) Find the SI units of a and b. (b) Find the mass of the rod in terms of a, b and L.

Sol:

(a) [ρ] = [a] = [bx] = mass/length

Therefore, the SI units of a is kg m^-1 and

The SI units of b is kg m^-2.

(b) 

                       

→ dm = ρ dx

Or, m = ∫ dm

Or, m = ∫ ρ dx

Or, m = ∫ (a + bx) dx

Or, m = [ax + bx²/2  

Or, m = aL + bL²/2.

28. The momentum p of a particle changes with time t according to the relation dp/dt = (10 N) + (2N/s) t. If the momentum is zero at t = 0, what will the momentum be at t = 10 s?

Sol:  

Given: dp/dt = (10 N) + (2N/s) t

Or, dp = [(10 N) + (2N/s) t] dt

Or, p = ∫dp = ∫ [(10 N) + (2N/s) t] dt

Or, p = 10t + t² + c

At t = 0; p = 0 [given] ⇒ c = 0

∴the momentum at t is p_t = 10t + t²

So, the momentum at t = 10 is

→ P_t=10 = 10 * 10 + 10² 

Or, P_t=10 = 200 N-s = 200 kg-m/s.

29. The changes in a function y and the independent variable x are related as dy/dx = x². Find y as a function of x.

Sol:

The change in a function of y and the independent variable x are related as dx/dy = x².

⇒ dy = x² dx

Taking integration of both sides,

→ ∫dy = ∫ x²dx

Or, y = x³/3 + c

∴ y as a function of x is represented by y = x³/3 + c.

30. Write the number of significant digits in (a) 100.1 (b) 100.10, (c) 100.10, (d) 0.001001.

Sol:

The number significant digits

(a) 1001 No. of significant digits = 4.

(b) 100.1 No. of significant digits = 4.

(c) 100.10 No. of significant digits = 5.

(d) 0.001001 No. of significant digits = 4.

31. A metre scale is graduated at every millimetre. How many significant digits will be there in a length measurement with this scale?

Sol:

The metre scale is graduated at every millimeter.

1 m = 1000 mm

The minimum no. of significant digit may be 1(e.g.3 mm) and the maximum no. of significant digits may be 4 (e.g.1000 mm)

So, the no. of significant digits may be 1, 2, 3 or 4.

32. Round the following numbers to 2 significant digits (a) 3472, (b) 84.16, (c) 2.55 and (d) 28.5.

Sol: 

2 significant digits of numbers are following

(a) Value = 3500

(b) Value = 84

(c) Value = 2.6 

(d) Value = 28.

33. The length and the radius of a cylinder measured with a slide callipers are found to be 4.54 cm and 1.75 cm respectively. Calculate the volume of the cylinder.

Sol: 

Given: for the cylinder

Length = l = 4.54 cm; radius = r = 1.75 cm.

Volume = πr²l = π * (4.54) * (1.75)²

So, volume V = πr²l

Or, V = (3.14) * (4.54) * (1.75)²

Or, V = 43.6577 cm³

Since, the minimum no. of significant digits on a particular term is 3, the result should have 3 significant digits.

So, it is to be rounded off to 3 significant digits, V = 43.7 cm³.

34. The thickness of a glass plate is measured to 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the average thickness of the plate from this data.

Sol: 

We know that,                                 

Average thickness = (2.17 + 2.17 + 2.18)/3

= 2.1733 mm

Rounding off to 3 significant digits, average thickness = 2.17 mm.

35. The length of the string of a simple pendulum i measured with a metre scale to be 90.0 cm. The radius of the bob plus the length of the hook is calculated be 2.13 cm using measurements with a slide callipers. What is the effective length of the pendulum? (The effective length is defined as the distance between the point of suspension and the centre of the bob.)

Sol:

Actual effective length = (90.0 + 2.13) cm.

But, in the measurement 90.0 cm, the no. of significant digits is only 2.

So, the addition must be done by considering only 2 significant digits of each measurement.

So, effective length = 90.0 + 2.1 = 92.1 cm.

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