Q.1) You have a circular pizza cut into 3 equal parts, and one piece is taken out. Estimate the degree measure of a single piece of pizza and convert it into radians. Also, find the radian measure of the angle of the remaining part of the pizza.
Answer:
When a circular pizza is cut into 3 equal parts, each piece’s degree measure is 360°/3 = 120°.
To convert this into radians, we use the conversion factor 1° = 0.01745 rad.
Thus, the single piece of pizza measures 120° × 0.01745 rad/° = 2.09 rad.
Since one piece is taken out, two pieces of pizza remain, and their total degree measure is 2 × 120° = 240°.
Converting this to radians, we get 240° × 0.01745 rad/° = 4.19 rad.
Q.2) You have the length of a pendulum (1.5 ± 0.01) m and the acceleration due to gravity (9.8 ± 0.1) m/s^2. Calculate the time period of the pendulum with uncertainty.
Answer:
Given the length of the pendulum (l ± Δl) = (1.5 ± 0.01) m and acceleration due to gravity (g ± Δg) = (9.8 ± 0.1) m/s^2.
To find the time period (T ± ΔT), we use the formula T = 2π √(ℓ / g).
Substituting the given values with uncertainties, we get:
T ± ΔT = 2π [(1.5 m ± 0.7%) / (9.8 m/s^2 ± 1.0%)]^1/2.
In division, the percentage uncertainties are added, resulting in:
T ± ΔT = 2π (0.15 ± 1.7%)^1/2.
Considering the square root operation, the percentage uncertainty is halved:
(T ± ΔT) = 2π (0.15 ± 0.85%).
Therefore, the time period of the pendulum is (T ± ΔT) = 2.4 ± 0.9%.
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Q.3) Find the area of a rectangular sheet with length (l ± Δl) = (1.50 ± 0.02) m and width (w ± Δw) = (0.20 ± 0.01) m. Calculate the area (A ± ΔA).
Answer:
Given the length of the rectangular sheet (l ± Δl) = (1.50 ± 0.02) m and width (w ± Δw) = (0.20 ± 0.01) m.
To find the area (A ± ΔA), we use the formula A = l × w.
Substituting the given values with uncertainties, we get:
(A ± ΔA) = (1.50 ± 0.02) m × (0.20 ± 0.01) m.
First, we calculate the percentage uncertainties in length and width:
Percentage uncertainty in length = (0.02 / 1.50) × 100% = 1.33%.
Percentage uncertainty in width = (0.01 / 0.20) × 100% = 5%.
Now, we write the percentage uncertainties in the formula:
(A ± ΔA) = (1.50 m ± 1.33%) × (0.20 m ± 5%).
In multiplication, the percentage uncertainties are added:
(A ± ΔA) = (1.50 m × 0.20 m) ± (1.33% + 5%),
(A ± ΔA) = 0.30 m^2 ± 6.33%.
Converting back to fractional uncertainty, we get:
(A ± ΔA) = (0.30 ± 0.02) m^2.
Q.4) Calculate the answers with appropriate significant figures:
- (a) 246.24 + 238.278 + 98.3 = 582.818, round off to 582.8.
- (b) 1.4 × 2.639 + 117.25 = 120.9446, round off to 120.9.
- (c) (2.66 × 10^4) – (1.03 × 10^3) = 2.5579 × 10^4, round off to 2.56 × 10^4.
- (d) (112 × 0.456) / (3.2 × 120) = 0.133, round off to 0.13.
- (e) 168.99 × 9 = 1520.91, round off to 2 × 10^3 = 2000.
- (f) 1023 + 8.5489 = 1031.55, round off to 1032.
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Q.5) Calculate the answers with appropriate significant figures:
A) mp / me = 1.67 × 10-27 kg / 9.1096 × 10-31 kg
(b) The ratio of charge on electron ‘qe‘ to mass of electron ‘me‘
qe / me = 1.6 × 1019 C / 9.1096 × 10-31 kg
Q.6) Find the dimensions of:
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Q.7) Show that:
- (a) KE = (1/2)mv^2 and
- (b) PEg = mgh are dimensionally correct.
(a) To show that KE = (1/2)mv^2 is dimensionally correct:
KE represents the kinetic energy of an object in motion. When work is done on the object, the kinetic energy stored is equal to the work done. The formula for work is W = F * d,
where F is the force applied and d is the displacement. Substituting KE = F * d into the equation:
KE = (1/2)mv^2 F * d = (1/2)mv^2
Now, let’s analyze the dimensions: Force (F) has dimensions [MLT^-2] Displacement (d) has dimensions [L] Mass (m) has dimensions [M] Velocity (v) has dimensions [LT^-1]
So, the dimensions of the left-hand side (LHS) and the right-hand side (RHS) are: LHS = [MLT^-2] * [L] = [ML^2T^-2] RHS = [M] * [LT^-1]^2 = [ML^2T^-2]
Since both sides have the same dimensions ([ML^2T^-2]), the equation is dimensionally correct.
(b) To show that PEg = mgh is dimensionally correct:
PEg represents the gravitational potential energy of an object at a height h above the ground. When work is done on the object, the potential energy stored is equal to the work done. The formula for work is W = F * d,
where F is the force applied and d is the displacement. Substituting PEg = F * d into the equation:
PEg = mgh F * d = mgh
Now, let’s analyze the dimensions: Force (F) has dimensions [MLT^-2] Displacement (d) has dimensions [L] Mass (m) has dimensions [M] Gravitational acceleration (g) has dimensions [LT^-2] Height (h) has dimensions [L]
So, the dimensions of the left-hand side (LHS) and the right-hand side (RHS) are: LHS = [MLT^-2] * [L] = [ML^2T^-2] RHS = [M] * [LT^-2] * [L] = [ML^2T^-2]
Since both sides have the same dimensions ([ML^2T^-2]), the equation is dimensionally correct.
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