JEE Advanced Previous Year Questions (2018 - 2023): Units & Measurements | Physics for JEE Main & Advanced PDF Download

2023

Q1: In an experiment for determination of the focal length of a thin convex lens, the distance of the object from the lens is  10 ± 0.1 cm and 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 n is ______.    
[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 v propagate 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 Y is expressed in terms of three derived quantities, namely, the gravitational constant G, Planck's constant ℎ and 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  β from the first equation into the second and third equations, we get : 
For the second equation, substituting  β we get : α + 2(γ + 1) + 3γ = −1, which simplifies to
α = −5γ − 3.
For the third equation, substituting  β we get : −α − (γ + 1)−2γ = −2, which simplifies to α = −3γ − 1.
Setting the two equations for α equal to each other gives : −5γ − 3 = −3γ − 1, which simplifies to γ = −2.
Substituting γ = −2 back into the equations for α and β we 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 ε₀ is 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