Exercises (20-35)
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 the tangent is tanθ =dy/dx =4x.
x - 1 2 3 4 5 6 7 8 9 10
y - 2 8 18 32 50 72 98 128 162 200
ANSWER: (It is difficult to draw the full graph here, hence up to x =4 has been drawn. Students should try themselves to draw it on a graph paper)
 |
| Graph for Q-20 |
For example, a tangent AB is drawn on the curve at x = 2 {i.e. at point P(2,8)}. The slope of this tangent =tan∠PAQ = 8 divisions/1 division =8.
Similarly, by drawing tangents at x =4, 6, and 8 we can find out that tanθ =16, 24 and 32 respectively.
To verify, let us take any point on the curve, say R. Its coordinates are, x =3.2 (approx), y = 20. So, x² = 3.2² =10.24; And 2x² = 20.48 ≈ y.
By taking other random points on the curve it can be shown that y =2x².
Now draw a tangent at R and calculate the tanθ which is = 20 divisions/1.6 divisions =12.5 ≈4*3.2 =4x. Since R is a random point hence verified.
Similarly, by drawing tangents at x =4, 6, and 8 we can find out that tanθ =16, 24 and 32 respectively.
To verify, let us take any point on the curve, say R. Its coordinates are, x =3.2 (approx), y = 20. So, x² = 3.2² =10.24; And 2x² = 20.48 ≈ y.
By taking other random points on the curve it can be shown that y =2x².
Now draw a tangent at R and calculate the tanθ which is = 20 divisions/1.6 divisions =12.5 ≈4*3.2 =4x. Since R is a random point hence verified.
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.
ANSWER: y = sin x
→dy/dx = cos x
→dy = cos x* dx
So for a very small change in x = dx, the change in y is dy. Here, x =π/3 and dx = π/100. Hence,
dy = cos π/3 * π/100
= ½ * π/100 =π/200 =0.0157
→dy/dx = cos x
→dy = cos x* dx
So for a very small change in x = dx, the change in y is dy. Here, x =π/3 and dx = π/100. Hence,
dy = cos π/3 * π/100
= ½ * π/100 =π/200 =0.0157
22. The electric current in a charging R-C circuit is given by i = i₀.e⁻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 = 10RC.
ANSWER: To find out the rate of change of current, we should differentiate i with respect to the time t.
di/dt = i₀*d(e⁻t/RC)/d(-t/RC) * d(-t/RC)/dt
= e⁻t/RC *i₀ *(-1/RC)
= (-i₀/RC)*e⁻t/RC
(a) At t= 0, di/dt = (-i₀/RC)*e⁰ =(-i₀/RC)*1
→di/dt = -i₀/RC
(b) At t = RC,
di/dt = (-i₀/RC)*e⁻t/RC
= (-i₀/RC)*e⁻¹
= -i₀/RCe
(c) At t = 10RC
di/dt = (-i₀/RC)*e⁻¹⁰
= -i₀/RCe¹⁰
di/dt = i₀*d(e⁻t/RC)/d(-t/RC) * d(-t/RC)/dt
= e⁻t/RC *i₀ *(-1/RC)
= (-i₀/RC)*e⁻t/RC
(a) At t= 0, di/dt = (-i₀/RC)*e⁰ =(-i₀/RC)*1
→di/dt = -i₀/RC
(b) At t = RC,
di/dt = (-i₀/RC)*e⁻t/RC
= (-i₀/RC)*e⁻¹
= -i₀/RCe
(c) At t = 10RC
di/dt = (-i₀/RC)*e⁻¹⁰
= -i₀/RCe¹⁰
23. The electric current in a discharging R-C circuit is given by i = i₀.e⁻t/RC where i₀, R and C are constant parameters of the circuit and t is time. Let i₀ = 2.00 A, R = 6.00x10⁵ 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.
ANSWER: Given i₀ = 2 A,
R =6 x 10⁵ Ω
C =0.500 µF = 5 x10⁻⁷ F
Hence -t/RC = -t/(6x10⁵ * 5x10⁻⁷) = -t/(30/100)
=-t/0.30
So the electric current in a discharging R-C circuit is,
i = 2*e⁻t/0.30
R =6 x 10⁵ Ω
C =0.500 µF = 5 x10⁻⁷ F
Hence -t/RC = -t/(6x10⁵ * 5x10⁻⁷) = -t/(30/100)
=-t/0.30
So the electric current in a discharging R-C circuit is,
i = 2*e⁻t/0.30
(a) At t = 0.3 s, the current, i = 2*e⁻¹ =2/e A.
(b) For the rate of change of current, we find out di/dt.
di/dt = 2*d(e-t/0.30)/d(-t/0.30)*d(-t/0.30)/dt
= 2*e-t/0.30*(-1/0.30)
= -2*(10/3)*e-t/0.30
= -20e-t/0.30/3
at t = 0.30 s, di/dt = -20e⁻¹/3 =-20/3e A/s.
(c) Since at t = 0.30, di/dt = -20/3e A/s,
increase in current in 0.01 s =0.01*(-20/3e) =-0.20/3e A (approx).
Hence the current at 0.31 s = Current at t = 0.30 s + increase in 0.01 s
=2/e - 0.2/3e A
=2/e - 2/30e A
=(60 - 2)/30e A
=58/30e A
=5.8/3e A
24. Find the area bounded under the curve y = 3x²+6x+7 and the X-axis between x = 5 and x = 10.
ANSWER: Let us consider a point P (x,y) on the curve between x=5 and x=10. Imagine a very very small increment dx, next to x. Now the area of the small strip is
dA = y.dx
Dividing the whole area into strips of thickness dx, total area will be the integration of all areas between x=5 and x=10. Hence,
A= ∫dA = ∫y.dx = ∫(3x²+6x+7)dx
=[x³+3x²+7x+C]
{between limits x=5 and x=10, C is a constant of integration}
→A=[(10³+3*10²+70+C)-(5³+3*5²+35+C)]
=[1370-235]
= 1135.
dA = y.dx
 |
| Sketch for Q-24 |
Dividing the whole area into strips of thickness dx, total area will be the integration of all areas between x=5 and x=10. Hence,
A= ∫dA = ∫y.dx = ∫(3x²+6x+7)dx
=[x³+3x²+7x+C]
{between limits x=5 and x=10, C is a constant of integration}
→A=[(10³+3*10²+70+C)-(5³+3*5²+35+C)]
=[1370-235]
= 1135.
25. Find the area enclosed by the curve y = sin x and X-axis between x = 0 and x = π.
ANSWER: Same as in the previous problem,
Area A = ∫dA =∫y.dx =∫sin x.dx
→A = [-cos x+C] [Between limits x = 0 and x = π; Cis a constant of integration]
= [(-cos π+C) -(-cos 0+C)]
= [1-(-1)]
= 2.
Area A = ∫dA =∫y.dx =∫sin x.dx
→A = [-cos x+C] [Between limits x = 0 and x = π; Cis a constant of integration]
= [(-cos π+C) -(-cos 0+C)]
= [1-(-1)]
= 2.
26. Find the area bounded by the curve y = e⁻x, the X-axis, and the Y-axis.
ANSWER: Here one limit of the x-coordinate is Y-axis, i.e. x= 0. For the other limit let us see what is the maximum value of x. y = e⁻x = 1/ex . The maximum value of x = ∞ because then y = 1/∞ =0.
So we have to integrate between x = 0 to x =∞.
Thus A = ∫dA =∫y.dx =∫e⁻x.dx
→A = [-e⁻x + +C] {between x=0 to x=∞}
= [-1/ex+C]
= [-1/∞ -(-1/e⁰)]
= [0 + 1]
= 1.
So we have to integrate between x = 0 to x =∞.
Thus A = ∫dA =∫y.dx =∫e⁻x.dx
→A = [-e⁻x + +C] {between x=0 to x=∞}
= [-1/ex+C]
= [-1/∞ -(-1/e⁰)]
= [0 + 1]
= 1.
27. A rod of length L is placed along the X-axis between x = 0 and x = L. The linear density (mass/length) ⍴ of the rod varies with the distance x from the origin as ⍴ = 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.
ANSWER: (a) Given ρ = a+bx
Since ρ is linear density i.e. mass per unit length, the SI unit of ρ will be kg/m. Now both the terms 'a' and 'bx' will have the same unit.
Hence the unit of a = kg/m
and the unit of b = kg/m² because x is multiplied to b.
(b) Let us consider an infinitesimally small length dx next to any length x from the origin. Mass of this small length dm =ρ*dx. Integrating such small masses between x = 0 to x = L will give the total mass of the rod. So,
m = ∫dm =∫ρ.dx = ∫(a+bx)dx =[ax +bx²/2 +C]
{Between the limits x = 0 to x = L}
→m = [aL + bL²/2+C -0-0-C] = aL + bL²/2.
Since ρ is linear density i.e. mass per unit length, the SI unit of ρ will be kg/m. Now both the terms 'a' and 'bx' will have the same unit.
Hence the unit of a = kg/m
and the unit of b = kg/m² because x is multiplied to b.
(b) Let us consider an infinitesimally small length dx next to any length x from the origin. Mass of this small length dm =ρ*dx. Integrating such small masses between x = 0 to x = L will give the total mass of the rod. So,
m = ∫dm =∫ρ.dx = ∫(a+bx)dx =[ax +bx²/2 +C]
{Between the limits x = 0 to x = L}
→m = [aL + bL²/2+C -0-0-C] = aL + bL²/2.
28. The momentum p of a particle changes with time t according to the relation dp/dt = (10 N) + (2 N/s)t. If the momentum is zero at t = 0, what will be the momentum at t = 10 s?
ANSWER: The momentum at time t is given by
p = ∫dp = ∫{(10 N)+(2 N/s)t}dt
= (10 N)t +(2 N/s)t²/2 +C
= (10t +t²) + C N-s
Where C is a constant of integration. Given at t = 0, p=0. Putting it in this,
0 = 10*0 + 0² +C
→C = 0. So the value of the integration constant is zero and,
p = (10t + t²) N-s
Now the momentum at t = 10 s is
p = (10*10 +10²) N-s =200 N-s =200 kg-m/s.
{Since N-s = (kg-m/s²)*s =kg-m/s}
p = ∫dp = ∫{(10 N)+(2 N/s)t}dt
= (10 N)t +(2 N/s)t²/2 +C
= (10t +t²) + C N-s
Where C is a constant of integration. Given at t = 0, p=0. Putting it in this,
0 = 10*0 + 0² +C
→C = 0. So the value of the integration constant is zero and,
p = (10t + t²) N-s
Now the momentum at t = 10 s is
p = (10*10 +10²) N-s =200 N-s =200 kg-m/s.
{Since N-s = (kg-m/s²)*s =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.
ANSWER: Given dy/dx = x²
→dy = x²dx
Integrating both sides,
→∫dy = ∫x²dx
→y = x³/3 +C
Where C is a constant of integration.
→dy = x²dx
Integrating both sides,
→∫dy = ∫x²dx
→y = x³/3 +C
Where C is a constant of integration.
30. Write the number of significant digits in (a) 1001, (b) 100.1, (c) 100.10, (d) 0.001001.
ANSWER: Zeros between digits are significant. Hence the number of significant digits in (a) 1001 is 4(four).
in (b) 100.1 is 4(four).
(c) The zero next to a non-zero digit after the decimal is also a significant digit. Hence the number of significant digits in 100.10 is 5(five).
(d) If the integer part is zero, any number of continuous zeros just after the decimal part is insignificant. Hence in 0.001001, the two zeros after the decimal are insignificant. So the total number of significant digits here are 4(four).
in (b) 100.1 is 4(four).
(c) The zero next to a non-zero digit after the decimal is also a significant digit. Hence the number of significant digits in 100.10 is 5(five).
(d) If the integer part is zero, any number of continuous zeros just after the decimal part is insignificant. Hence in 0.001001, the two zeros after the decimal are insignificant. So the total number of significant digits here are 4(four).
31. A meter scale is graduated at every millimeter. How many significant digits will be there in a length measurement with the scale?
ANSWER: Since the least count of the scale is 1 millimeter, Up to the measurement 9 mm, the number of the significant digits will be 1(one). Up to 99 mm the number of significant digits will be 2(two). Up to 999 mm it is 3(three) and above it, it is 4(four).
32. Round the following numbers to 2 significant digits.
(a) 3472
(b) 84.16
(c) 2.55 and
(d) 28.5
ANSWER: (a) 3472 - The second significant digit is 4. It is to be rounded. The digit next to it is 7 which is greater than 5. Hence the second digit should be increased by 1. The third and fourth digits should be dropped and replaced by zeros because they appear to the left of the decimal. Thus 3472 becomes 3500 on rounding to 2 significant digits.
(b) 84.16 - The second significant digit is 4. Next to it is 1 which is less than five. So the digits 1 and 6 are dropped. And 84.16 becomes 84 on rounding to 2 significant digits.
(c) 2.55 - The second significant digit is 5. Next to it is 5. Since the second significant digit which is to be rounded is odd, it will be increased by 1, and from the third on will be dropped. So 2.55 on rounding become 2.6.
(d) 28.5 - The second significant digit is 8. It is even and is to be rounded. The next digit is 5. So the second significant digit will not be increased by 1, it being even digit. The third digit will be dropped. So 28.5 will become 28 on rounding to two significant digit.
(c) 2.55 - The second significant digit is 5. Next to it is 5. Since the second significant digit which is to be rounded is odd, it will be increased by 1, and from the third on will be dropped. So 2.55 on rounding become 2.6.
(d) 28.5 - The second significant digit is 8. It is even and is to be rounded. The next digit is 5. So the second significant digit will not be increased by 1, it being even digit. The third digit will be dropped. So 28.5 will become 28 on rounding to two significant digit.
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.
ANSWER: L = 4.54 cm, r = 1.75 cm. The volume of the cylinder V = πr²L =π*1.75²*4.54 = 43.679 cm³. Since the minimum number of significant digits in the multiplication is three, the product will also have three significant digits. The third significant digit to be rounded is 6. Next to it is 7 which is greater than 5. So the digit 6 will be increased by 1 and the next digits will be dropped. Now volume V =43.7 cm³.
34. The thickness of a glass plate is measured to be 2.17 mm, 2.17 mm and 2.18 mm at three different places. Find the average thickness of the plate from this data.
ANSWER: The average thickness
=(2.17+2.17+2.18)/3 = 2.1733 mm =2.17 mm.
Since the minimum number of significant digits in the quantities are 3, the result will also have 3 significant digits. The third digit in the result is 7. Next to it is 3 which is less than 5. So the third digit 7 will not be increased and digits next to it will be dropped.
=(2.17+2.17+2.18)/3 = 2.1733 mm =2.17 mm.
Since the minimum number of significant digits in the quantities are 3, the result will also have 3 significant digits. The third digit in the result is 7. Next to it is 3 which is less than 5. So the third digit 7 will not be increased and digits next to it will be dropped.
35. The length of the string of a simple pendulum is measured with a meter scale to be 90.0 cm. The radius of the bob plus the length of the hook is calculated to be 2.13 cm using measurements with a slide-calipers. What is the effective length of the pendulum? (The effective length is defined as the distance between the point of suspension and the center of the bob).
ANSWER: The effective length of the pendulum will be = 90.0
+ 2.13 cm
-------------------
The first column where a doubtful digit occurs is the one next to the decimal point (90.0). All digits right to this column must be dropped after proper rounding. The table will be rewritten as
90.0
+2.1
-------------
= 92.1 cm
So the effective length of the pendulum is 92.1 cm.
+ 2.13 cm
-------------------
The first column where a doubtful digit occurs is the one next to the decimal point (90.0). All digits right to this column must be dropped after proper rounding. The table will be rewritten as
90.0
+2.1
-------------
= 92.1 cm
So the effective length of the pendulum is 92.1 cm.
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Links to the Chapters
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CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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