Numerical Problems: Units and Measurements | Physics Class 11 - NEET PDF Download

Q1. The velocity of a freely falling body changes as g ^p h ^q where g is acceleration due to gravity and h is the height. The values of p and q are
Sol: [v] = [g]^p[h]^q
[M⁰L¹T^-1] = [M⁰L¹T^-2]^p [L¹]^q = [M⁰L^p+qT^-2p]
⇒ 1 = p + q and -1 = -2p
⇒ p = ¹/₂, q = ¹/₂

Q2. The dimensions of ^a/_b in the equation P = a - t²/bx, where P is pressure, x is distance and t is time, are
Sol: [a] = [T²] and [b] = [a - t²/[P][x]] = T²/_{[M^-1L^-1T^-2] [L]}
⇒ [b] = [M^-1T⁴]
So [^a/_b] = [T²]/_[M^-1T⁴] = [MT^-2]

Q3. The density of wood is 0.5 g/cm³ in cgs system of units. The corresponding value of SI units is
Sol: Density = M/V
1 cm = 10^-2 m; 1 gm = 10^-3 kg
0.5 g/cm³ = 0.5 (10^-3) kg / (10^-2)³ m
= 0.5 × 10³ × 10⁶ = 500

Q4. A cube has side 1.2 × 10^-2 m. Its volume will be recorded as
Sol: v = l³ = 1.728 × 10⁶
Length has two significant figures v = 1.7 × 10⁶ m³

Q5. In a certain system of units, 1 unit of time is 5 sec, 1 unit of mass is 20 kg and unit of length is 10 m. In this system, one unit of power will correspond to
Sol: [P] = [M L² T^-3]
Unit of P = 20 kg × (10 m)² × (5 sec)^-3 = 20 × 100 × (1/125) = 16 W

Q6. To keep an object moving in a circle at constant speed requires a force F ∝ m ^a v ^b r ^c. According to dimensional analysis the a, b, c are
Sol: F ∝ m ^a v ^b r ^c
i.e. [F] = [m]^a[v]^b[r]^c
or [M L T^-2] = [M]^a[L T^-1]^b[L]^c
or [M¹ L¹ T^-2] = [M]^a[L ^{b + c} T^-b]
Comparing LHS and RHS
a = 1, b + c = 1 & -b = -2
⇒ a = 1, b = 2, c = -1

Q7. A wire is of mass (0.3 ± 0.003) gm, the radius is (0.5 ± 0.005) cm and length is (6 ± 0.1) cm. The maximum percentage error in density is
Sol: Δρ/ρ × 100
= Δm/m × 100 + Δl/l × 100 + 2Δr/r × 100
= 0.003/0.3 × 100 + 0.1/6 × 100 + 2 × 0.005/0.5 × 100
= 1% + 1.67% + 2% = 4%

Q8. The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by relation of the type f = c m ^x k ^y, where c is a dimensionless constant. The values of x and y are
Sol: f = c m^x k^y
[M⁰L⁰T^-1] = [M]^x[M L²T^-2]^y
M⁰L⁰T^-1 = M^x+yL^2yT^-2y
x + y = 0 ...(1)
-2y = -1 ...(2)
From (2), y = 1/2
From (1), x + 1/2 = 0 ⇒ x = -1/2
Thus, x = -1/2, y = 1/2

Q9. The density of a cube is measured by measuring its mass and length of its sides. If the maximum errors in the measurement of its mass and length are 4% and 3% respectively, the maximum error in density is
Sol: Density (d) is defined as:
d = Mass / L³
% error in d = % error in mass + 3 [% error in length]
Calculating the maximum error:
= 4% + 3 [3%] = 13%

Q10. If the units of length and force are increased four times, then the unit of energy will
Sol: Dimensionally,
E = M L² T^-2
E = (M L T^-2) (L)
= (4)(M L T^-2) (4 L) = 16 (M L² T^-2)