Class 11 - Unit # 01: Physical Quantities and Measurements - Solved Numericals


 

Problem 1

What is the percent uncertainty in the measurement 3.67 ± 0.25 m? (Ans: 6.8%)

Given Data:

  • Measured value (x) = 3.67 m
  • Absolute uncertainty (Δx) = 0.25 m
  • Percent Uncertainty = ?

Solution:

The formula for percent uncertainty is:

Percent Uncertainty =
Absolute Uncertainty (Δx)
Measured Value (x)
× 100%

Substituting the given values:

Percent Uncertainty =
0.25
3.67
× 100%
Percent Uncertainty = 0.0681 × 100%
Percent Uncertainty = 6.81%

Result:

The percentage uncertainty in length is 6.81%.

Problem 2

What is the area, and its approximate uncertainty, of a circle with radius 3.7 × 104 cm? (Ans: 4.3 × 109 cm2 ± 5.4%)

Given Data:

  • Radius (r) = 3.7 × 104 cm
  • Least count/uncertainty in radius (Δr) = 0.1 × 104 cm
  • Area (A) = ?
  • Uncertainty in Area (ΔA) = ?

Solution:

First, let's calculate the area of the circle using the formula:

A = πr2
A = 3.1416 × (3.7 × 104)2
A = 3.1416 × 1.369 × 109
A ≈ 4.30 × 109 cm2

For a power function like r2, the percentage uncertainty in area is twice the percentage uncertainty in the radius:

% Uncertainty in A = 2 × (
Δr
r
) × 100%
% Uncertainty in A = 2 × (
0.1 × 104
3.7 × 104
) × 100%
% Uncertainty in A = 2 × 0.027 × 100% ≈ 5.4%

Now, find the absolute uncertainty in Area (ΔA):

ΔA = 5.4% of A =
5.4
100
× (4.30 × 109)
ΔA ≈ 0.23 × 109 cm2

Result:

The area of the circle is 4.30 × 109 cm2 and its absolute uncertainty is 0.23 × 109 cm2 (or 5.4%).

Problem 3

An aeroplane travels at 850 km/h. How long does it take to travel 1.00 km? (Ans: 4.23 s)

Given Data:

  • Speed of aeroplane (v) = 850 km/h
  • Distance traveled (S) = 1.00 km
  • Time taken (t) = ?

Solution:

According to the standard definition of uniform speed:

S = v × t ⇒ t =
S
v

Substituting the values to get time in hours:

t =
1.00
850
hours

Converting hours into seconds (1 hour = 3600 seconds):

t =
1.00
850
× 3600
t =
3600
850
t ≈ 4.24 seconds

Result:

It will take 4.24 seconds for the aeroplane to travel a distance of 1.00 km.

Problem 4

A rectangular holding tank 25.0 m in length and 15.0 m in width is used to store water for a short period of time in an industrial plant. If 2980 m3 of water is pumped into the tank, what is the depth of the water? (Ans: 7.95 m)

Given Data:

  • Length of the tank (l) = 25.0 m
  • Width of the tank (w) = 15.0 m
  • Volume of water (V) = 2980 m3
  • Depth of water (h) = ?

Solution:

The total volume of a rectangular prism layout is given by:

V = l × w × h

Rearranging the formula to isolate the depth variable (h):

h =
V
l × w

Substituting the given configuration details:

h =
2980
25.0 × 15.0
h =
2980
375
h ≈ 7.95 m

Result:

The depth of the water inside the rectangular holding tank is 7.95 m.

Problem 5

Find the volume of a rectangular underground water tank that has storage facilities dimensions of 1.9 m by 1.2 m by 0.8 m. (Ans: 1.824 m3)

Given Data:

  • Length (l) = 1.9 m
  • Width (w) = 1.2 m
  • Depth / Height (h) = 0.8 m
  • Volume (V) = ?

Solution:

The total volume formula for a rectangular prism container shape is:

V = l × w × h

Substituting the matrix dimensions:

V = 1.9 × 1.2 × 0.8
V = 1.824 m3

Result:

The volume of the rectangular tank is 1.824 m3.

Problem 6

Two students derive following equations in which x refers to distance traveled, v the speed, a the acceleration, and t the time (the subscript 'o' means a quantity at time t = 0):
(a) x = vt2 + 2at
(b) x = vot + 2at2
Which of these could possibly be correct according to a dimensional check? (Ans: Equation b is dimensional correct)

Standard Dimensions:

  • Distance [x] = [L]
  • Velocity [v] = [LT-1]
  • Acceleration [a] = [LT-2]
  • Time [t] = [T]

Solution Analysis:

Checking Equation (a): x = vt2 + 2at

Dimension of Left Hand Side (L.H.S):

[L.H.S] = [x] = [L] —— (Equation i)

Dimension of Right Hand Side (R.H.S):

[R.H.S] = [vt2] + [2at]
[R.H.S] = ([LT-1][T2]) + ([LT-2][T])
[R.H.S] = [LT] + [LT-1] —— (Equation ii)

Comparing Equation (i) and (ii), [L.H.S] ≠ [R.H.S]. Thus, equation (a) is incorrect.

Checking Equation (b): x = vot + 2at2

Dimension of Left Hand Side (L.H.S):

[L.H.S] = [x] = [L] —— (Equation iii)

Dimension of Right Hand Side (R.H.S):

[R.H.S] = [vot] + [2at2]
[R.H.S] = ([LT-1][T]) + ([LT-2][T2])
[R.H.S] = [L] + [L] = [L] —— (Equation iv)

Comparing Equation (iii) and (iv), [L.H.S] = [R.H.S]. Thus, equation (b) is correct.

Result:

Student (a)'s derived equation is dimensionally incorrect and Student (b)'s derived equation is dimensionally correct.

Problem 7

One hectare is defined as 104 m2. One acre is 4 × 104 ft2. How many acres are in one hectare? (Hint: 1 m = 3.28 ft) (Ans: 2.69 acres)

Given Data:

  • 1 Hectare = 104 m2
  • 1 Acre = 4 × 104 ft2
  • Conversion factor: 1 m = 3.28 ft ⇒ 1 m2 = (3.28)2 ft2 = 10.7584 ft2

Solution:

Step 1: Convert the value of 1 Hectare from m2 into ft2:

1 Hectare = 104 × 10.7584 ft2 = 107,584 ft2

Step 2: Find total number of acres inside one hectare by dividing their area layouts:

Number of Acres =
Area of Hectare (in ft2)
Area of Acre (in ft2)
Number of Acres =
107,584
4 × 104
=
107,584
40,000
Number of Acres ≈ 2.69 acres

Result:

There are approximately 2.69 acres in one hectare.

Problem 8

A watch factory claims that its watches gain or lose not more than 10 seconds in a year. How accurate is this watch, expressed as a percentage? (Ans: 3.17 × 10-5 %)

Given Data:

  • Error / Uncertainty (Δt) = 10 s
  • Total Time period (t) = 1 year = 365 days × 24 hrs × 60 mins × 60 s = 31,536,000 s
  • Percentage Error / Accuracy = ?

Solution:

Percentage Error =
Δt
t
× 100%
Percentage Error =
10
31,536,000
× 100%
Percentage Error = 3.17 × 10-7 × 100%
Percentage Error = 3.17 × 10-5%

Result:

The variance accuracy error percentage in this watch is 3.17 × 10-5%.

Problem 9

The diameter of the Moon is 3480 km. What is the volume of the Moon? How many Moons would be needed to create a volume equal to that of Earth? (Hint: Radius of Earth = 6380 km) (Ans: Vmoon = 2.21 × 1010 km3, 49 Moons)

Given Data:

  • Diameter of Moon (Dm) = 3480 km
  • Radius of Moon (Rm) = Dm / 2 = 1740 km
  • Radius of Earth (Re) = 6380 km
  • Volume of Moon (Vm) = ?
  • Number of Moons needed (N) = ?

Solution:

Both celestial bodies are spherical. The volume of a sphere is given by:

V =
4
3
π R3
1. Volume of the Moon (Vm):
Vm =
4
3
× 3.1416 × (1740)3
Vm = 1.3333 × 3.1416 × 5.268 × 109
Vm ≈ 2.21 × 1010 km3
2. Volume of the Earth (Ve):
Ve =
4
3
× 3.1416 × (6380)3
Ve = 1.3333 × 3.1416 × 2.597 × 1011
Ve ≈ 1.09 × 1012 km3
3. Number of Moons in Earth (N):
N =
Ve
Vm
=
1.09 × 1012
2.21 × 1010
N ≈ 49.3 Moons

Result:

The volume of the Moon is 2.21 × 1010 km3 and approximately 49 Moons would be needed to match the volume of the Earth.

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