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Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - JEE MCQ


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11 Questions MCQ Test - Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced

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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 1

The dimensions of the quantities in one (or more) of the following pairs are the same. Identify the pair (s)

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 1

t = F × r × sin θ; W = F × d × cosqθ
Dimensionally, light year = wavelength = [L]

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 2

The pairs of physical quantities that have the same dimensions is (are) :

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 2

Reynold’s number
= Coefficient of friction = [M0L0T0]
Note :  Curie is the unit of radioactivity (number of atoms decaying per second) and frequency also has the unit per second.

Latent heat = Q/m and Gravitation potential = W/m

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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 3

The SI unit of inductance, the henry can be written as

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 3


*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 4

Let [ε0] denote the dimensional formula of the permittivity of the  vacuum, and [μ0] that of the permeability of the vacuum. If M = mass, L = length,  T = time and I = electric current,

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 4


*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 5

A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 5

As the length of the string of simple pendulum is exactly l m (given), therefore the error in length Δl = 0.
Further the possibility of error in measuring time is 1s in 40s.



*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 6

Using the expression 2d sin θ = λ, one calculates the values of d by measuring the corresponding angles q in the range θ to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0°

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 6


on dividing (i) and (ii) we get

As θ increases from 0° to 90°, cot θ decreases and therefore    decreases option (d) is correct

This value of  decreases as θ increases from 0° to 90°

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 7

Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of length L and a unit of mass M. Then the correct option(s) is(are)

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 7

L ∝ hx cy Gz

Dimensionally


∴ x – z = 0 ⇒ x = z
∴ 2x + y + 3z = 1 and –x – y – 2z = 0
On solving we get


C, D are correct options


∴ x – z = 1
2x + y + 3z = 0
–x – y – 2z = 0
On solving we get

A is the correct option.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 8

Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 8

Vernier callipers

1 MSD = 1cm/8 = 0.125cm

5 VSD = 4MSD

∴ 1 VSD = 0.1cm
L.C = 1MSD – 1VSD
= 0.125cm – 0.1cm
= 0.025cm
Screw gauge One complete revolution = 2M.S.D
If the pitch of screw gauge is twice the L.C of vernier callipers then pitch = 2 × 0.025 = 0.05cm.
L.C of screw Gauge

(b) is a correct option Now if the least count of the linear scale of the screw gauge is twice the least count of venier callipers then.
L.C of linear scale of screw gauge = 2 × 0.025 = 0.05cm.
Then pitch = 2 × 0.05 = 0.1cm.

Then L.C of screw gauge = 0.1/100 cm = 0.001cm = 0.01mm.
(c) is a correct option.

 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 9

In terms of potential difference V, electric current I, permittivity ε0, permeability μ0 and speed of light c, the dimensionally correct equation(s) is(are)

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 9

We know that




⇒ Option (d) is incorrect

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 10

A length–scale (l) depends on the permittivity (ε) of a dielectric material. Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for I is(are) dimensionally correct?

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 10

We know that, dimensionally 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 11

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is   The values of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52s, 0.56s, 0.57s, 0.54s and 0.59s. The least count of the watch used for the measurement of time period is 0.01s. Which of the following statement(s) is (are) true?

Detailed Solution for Test: MCQs (One or More Correct Option): Units and Measurements | JEE Advanced - Question 11

% error in measurement of ‘r’ 

∴ % error in the measurement of ‘T’
% error in the value of g

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