PYQ
Q1.
The equation of state of some gases can be expressed as:
(P + a / V2) (V − b) = RT
where P is pressure, V is volume, T is absolute temperature and a, b, R are constants.
The dimensions of ‘a’ are:
(a) [ML5T-2]
(b) [M0L3T0]
(c) [ML-1T-2]
(d) [M0L6T0]
Given: (P + a/V²)(V − b) = RT
⇒ a/V² has same dimension as pressure
Pressure = [ML-1T-2]
⇒ a = Pressure × V²
⇒ a = [ML-1T-2] × [L3]2
⇒ a = [ML5T-2]
Answer: (a)
Q2.
The time period of oscillation ‘T’ of a gas bubble under water depends upon P, ρ and E, where ‘P’ is the static pressure, ‘ρ’ is the density of water and ‘E’ is the total energy associated with the formation of the bubble. Using the method of dimensions, show that, for constant values of ‘P’ and ‘E’, the time period is directly proportional to the square root of the density ‘ρ’.
Assume: T ∝ Pa ρb Ec
Dimensions:
T = [T]
P = [ML-1T-2]
ρ = [ML-3]
E = [ML2T-2]
Substitute:
[T] = [ML-1T-2]a × [ML-3]b × [ML2T-2]c
Equating powers:
M: a + b + c = 0
L: -a - 3b + 2c = 0
T: -2a - 2c = 1
Solving → b = 1/2 (for constant P and E)
⇒ T ∝ ρ1/2
Hence proved.
PYQ
Q4.
In the relation, y = r sin(ωt + kx), the dimensional formula for kx or ωt is same as:
(a) r / ω
(b) r / y
(c) ωt / r
(d) yr / ωt
In sin(ωt + kx), the argument must be dimensionless.
⇒ ωt is dimensionless → [ω] = [T-1]
⇒ kx is dimensionless → [k] = [L-1]
Hence, kx and ωt are dimensionless quantities.
Only option (b) r/y is dimensionless (since both have same dimension).
Answer: (b)
Q5.
Assertion: When we change the unit of measurement of a quantity, its numerical value changes.
Reason: Smaller the unit of measurement smaller is its numerical value.
(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) If both Assertion and Reason are true but Reason is not the correct explanation.
(c) If Assertion is true but Reason is false.
(d) If both Assertion and Reason are false.
Assertion: True → Changing unit changes numerical value.
Reason: False → Smaller unit gives larger numerical value (not smaller).
Example: 1 m = 100 cm → numerical value increases.
Answer: (c)
PYQ
Q6.
If P, Q, R are physical quantities having different dimensions, which of the following combinations can never be a meaningful quantity?
(a) (P − Q)/R
(b) PQ − R
(c) PQ/R
(d) (PR − Q²)/R
Physical quantities can be added/subtracted only if they have same dimensions.
In (a): P − Q → not possible (different dimensions)
Hence expression is meaningless.
Answer: (a)
Q7.
Taking into account of the significant figures, what is the value of 9.99 m − 0.0099 m?
(a) 9.9801 m
(b) 9.98 m
(c) 9.980 m
(d) 9.9 m
9.99 − 0.0099 = 9.9801
In subtraction, result is limited to least decimal places (2 decimal places).
⇒ Final answer = 9.98 m
Answer: (b)
Q8.
In the expression P = EL2 m-5 G-2, E, l, m and G denote energy, angular momentum, mass and universal gravitational constant respectively. Show that P is a dimensionless quantity.
Dimensions:
E = [ML²T⁻²]
l = [ML²T⁻¹]
m = [M]
G = [M⁻¹L³T⁻²]
Substitute in P:
P = (ML²T⁻²)1/2 × (ML²T⁻¹)-5 × (M⁻¹L³T⁻²)-2
Simplifying powers → all M, L, T cancel out
⇒ P = [M⁰L⁰T⁰] = dimensionless
Hence proved.
PYQ
Q9.
Use 𝐀 = î − 2ĵ + k̂, 𝐁 = 2î − k̂ and 𝐂 = î − xĵ as reference vectors.
(i) Find the value of x if (𝐀 + 𝐁) is perpendicular to 𝐂.
(ii) Prove that dot product is distributive over addition using the above vectors.
(iii) Find a vector parallel to 𝐁 and having same magnitude as 𝐀.
(iv) Find a vector which is perpendicular to both 𝐀 & 𝐁.
A = (1, -2, 1), B = (2, 0, -1), C = (1, -x, 0)
(i) (A + B) ⟂ C ⇒ (A + B)·C = 0
A + B = (3, -2, 0)
Dot product = 3(1) + (-2)(-x) = 3 + 2x = 0
⇒ x = -3/2
(ii) A·(B + C) = A·B + A·C (verified by expansion)
(iii) Required vector = |A| × (B/|B|)
(iv) Perpendicular to both ⇒ A × B
= determinant gives required vector.
Final answers obtained accordingly.
Q10.
(i) Derive an expression for the magnitude of centripetal acceleration of a body moving with uniform speed v along a circular path of radius r.
(ii) An insect trapped in a circular groove of radius 12 cm moves along the groove steadily and completes seven revolutions in 100 sec. What will be the angular speed and linear speed of motion?
(i) Centripetal acceleration:
a = v² / r
(ii) Number of revolutions = 7 in 100 s
⇒ angular speed ω = 2π × (7/100)
⇒ ω = 14π / 100 rad/s
Radius r = 12 cm = 0.12 m
Linear speed v = ωr
= (14π / 100) × 0.12
Final values can be calculated numerically.
PYQ
Q11.
State Kepler’s law of areas. Show that it is a consequence of conservation of angular momentum.
Kepler’s Second Law: A line joining a planet and the sun sweeps equal areas in equal intervals of time.
Areal velocity = (1/2) r²ω
Angular momentum L = m r²ω
⇒ Areal velocity = L / (2m)
Since angular momentum is conserved, areal velocity remains constant.
Hence proved.
Q12.
Derive the law of conservation of linear momentum from Newton’s third law of motion.
Consider two bodies A and B.
Force on A by B = − Force on B by A
F₁ = −F₂
Using Newton’s second law:
dp₁/dt = − dp₂/dt
⇒ dp₁/dt + dp₂/dt = 0
⇒ d(p₁ + p₂)/dt = 0
⇒ Total momentum remains constant.
Hence, linear momentum is conserved.
Q13.
Derive the following kinematic relations by graphical method:
(a) v = u + at
(b) s = ut + ½at²
From velocity-time graph:
Slope = acceleration ⇒ a = (v − u)/t
⇒ v = u + at
Area under v-t graph gives displacement:
Area = rectangle + triangle
= ut + ½ (v − u)t
Substitute v = u + at:
s = ut + ½ at²
Hence derived.
