JEE Advanced Previous Year Questions (2018 - 2024): Units & Measurements | Physics for JEE Main & Advanced PDF Download

2024

Q1: The dimensions of a cone are measured using a scale with a least count of 2mm. The diameter of the base and the height are both measured to be 20.0cm. The maximum percentage error in the determination of the volume is _______.    [JEE Advanced 2024 Paper 2]
Ans: 3
To determine the maximum percentage error in the volume of the cone, we first need to understand the dependence of the cone's volume on its dimensions: the diameter of the base and its height.

The volume V of a cone is given by the formula:

V = 13 πr²h

where r is the radius of the base and h is the height.

Given that both the diameter (D) of the base and the height (h) are measured to be 20.0 cm, we can find the radius r as follows:

r = D2 = 20.0 cm2 = 10.0 cm

Now, let's denote the errors in measuring the diameter and height as ΔD and Δh respectively.

Since the measurements are taken with a scale having a least count of 2 mm, we have:

ΔD = 2 mm = 0.2 cm
Δh = 2 mm = 0.2 cm

To find the maximum percentage error in the volume, we need to use the formula for the propagation of relative errors.

Considering the volume formula V = 13 πr²h, the relative errors in r and h will propagate into the volume as follows:

The relative error in the radius Δr/r is:

Δrr = ΔDD = 0.2 cm20.0 cm = 0.01

The relative error in the height Δh/h is:

Δhh = 0.2 cm20.0 cm = 0.01

Since V is proportional to r² and h, the total relative error in the volume is given by:

ΔVV = 2 Δrr + Δhh

Substituting the relative errors, we get:

ΔVV = 2(0.01) + 0.01 = 0.02 + 0.01 = 0.03

Hence, the maximum percentage error in the volume is:

0.03 × 100% = 3%

Therefore, the maximum percentage error in the determination of the volume is 3%.

Q2: A dimensionless quantity is constructed in terms of electronic charge e, permittivity of free space ε₀, Planck’s constant h, and speed of light c. If the dimensionless quantity is written as eᵅ ε₀ᵝ hᵞ cᵟ and n is a non-zero integer, then (α, β, γ, δ) is given by:
(a) (2n, −n, −n, −n)
(b) (n, −n, −2n, −n)
(c) (n, −n, −n, −2n)
(d) (2n, −n, −2n, −2n)   [JEE Advanced 2024 Paper 1]
Ans: (a)
To find the values of (α, β, γ, δ) for which the expression e^αε₀^βh^γc^δ is dimensionless, we need to ensure that the product of the quantities has no overall physical dimensions. This means the dimensions of charge (e), permittivity (ε₀), Planck's constant (h), and speed of light (c) must cancel each other out.
Electronic charge (e): The dimension of charge in terms of fundamental dimensions is:
[e] = [A.T]
where A represents the dimension of current and T represents time.
Permittivity of free space (ε₀): The dimension is given by:
[ε₀] = [M^-1L^-3T⁴A²]
Planck’s constant (h): The dimension is:
[h] = [M L²T^-1]
Speed of light (c): The dimension is:
[c] = [L T^-1]
We now write the expression for the dimensionless quantity and set its dimensions to zero:
[e^αε₀^βh^γc^δ] = [A^αT^α] [M^-βL^-3βT^4βA^2β] [M^γL^2γT^-γ] [L^δT^-δ]
Combining the dimensions, we get:

[e^αε₀^βh^γc^δ] = [M^-β+γL^-3β+2γ+δT^α+4β-γ-δA^α+2β]
For the quantity to be dimensionless, the exponents of M, L, T, and A must all be zero:
-β + γ = 0
-3β + 2γ + δ = 0
α + 4β - γ - δ = 0
α + 2β = 0
From the fourth equation:
α + 2β = 0 ⟹ α = -2β
Substituting α = -2β into the third equation:
-2β + 4β - γ - δ = 0 ⟹ 2β - γ - δ = 0 ⟹ γ + δ = 2β
From the first equation:
-β + γ = 0 ⟹ γ = β
Substituting γ = β into γ + δ = 2β:
β + δ = 2β ⟹ δ = β
Thus, we have:
α = -2β
γ = β
δ = β
Let β = -n where n is a non-zero integer. Then, we can write the values as:
α = -2(-n) = 2n
β = -n
γ = -n
δ = -n
Hence, the tuple (α, β, γ, δ) is given by: (2n, -n, -n, -n)
So, the correct option is Option A: (2n, -n, -n, -n)

2023

Q1: In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is10 ± 0.1 cmand the distance of its real image from the lens is  20 ± 0.2 cm. The error in the determination of focal length of the lens is n%. The value of nis ______. 
[JEE Advanced 2023 Paper 1 ]
Ans: 1
Given :
- Object distance u = 10.0 ± 0.1cm
- Image distance v = 20.0 ± 0.2cm
According to the lens formula for a thin lens : 

So, we can calculate the focal length :

=

= = 20/3 cmNext, we need to calculate the error in the determination of the focal length. For that, we find the differential of the lens formula :
Then, the derivative of the equation gives us the change in the focal length (df)in terms of the changes in the object distance (du)and the image distance (dv):

For maximum error, we get :

This equation tells us how errors in (u)and vpropagate to an error in f. Now, when you compute the relative error in the focal length, you get : 

Plugging in your values of  u = 10 cm,  du = 0.1 cm,  v = 20 cm,  dv = 0.2 cm, and  f = 20/3 cm, you indeed get: 

So, the error in the focal length of the lens is indeed 1% (i.e., n = 1). 

Q2: Young's modulus of elasticity Yis expressed in terms of three derived quantities, namely, the gravitational constant G, Planck's constant ℎand the speed of light c, as Y = C^αℎ^βG^γ. Which of the following is the correct option?
(a) α = 7, β = -1, γ = -2
(b) α = -7, β = -1, γ = -2
(c) α = 7, β = -1, γ = 2
(d) α = -7, β = 1, γ = -2 [JEE Advanced 2023 Paper 2 ]
Ans:(a)
Given the equation, Y = C^αℎ^βG^γ
where Y represents Young's modulus, c the speed of light, h is Planck's constant, G is the gravitational constant, and α, β, and γ are powers to be determined.
We start by expressing the quantities involved in terms of their fundamental dimensions: Mass (M), Length (L), and Time (T). The equation in terms of dimensions is :

Next, by equating the powers of the dimensions M, L, and T on both sides of the equation, we form three separate equations:
1. Equating the powers of M, we get : β − γ = 1
2. Equating the powers of L, we get : α + 2β + 3γ= −1
3. Equating the powers of T, we get : −α − β− 2γ = −2
Now we can solve this system of equations.
From the first equation, we get β =  γ + 1.
Substituting βfrom the first equation into the second and third equations, we get :
For the second equation, substituting βwe get : α + 2(γ + 1) + 3γ = −1, which simplifies to
α = −5γ − 3.
For the third equation, substituting βwe get : −α − (γ + 1)−2γ = −2, which simplifies to α = −3γ − 1.
Setting the two equations for αequal to each other gives : −5γ − 3 = −3γ − 1, which simplifies to γ = −2.
Substituting γ = −2back into the equations for αand βwe find γ = 7 and β = −1.
So, the solution to the system is α = 7,β = −1,γ = −2, which corresponds to Option A.

2022

Q1: In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, electron mass m_e, Planck's constant ℎ, and Coulomb's constant , where ε₀is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is. The value of α + β + γ + ∂is _______.

Compare:
On solving: α = 3, β = 2,  γ  = -3,  ∂ = 2So, α + β + γ + ∂ = 4

Q2: Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is   0.5 mm. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

2018