JEE Exam  >  JEE Tests  >  Chapter-wise Tests for JEE Main & Advanced  >  JEE Advanced (Single Correct MCQs): Units and Measurements - JEE MCQ

JEE Advanced (Single Correct MCQs): Units and Measurements - JEE MCQ


Test Description

15 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced (Single Correct MCQs): Units and Measurements

JEE Advanced (Single Correct MCQs): Units and Measurements for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced (Single Correct MCQs): Units and Measurements questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced (Single Correct MCQs): Units and Measurements MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced (Single Correct MCQs): Units and Measurements below.
Solutions of JEE Advanced (Single Correct MCQs): Units and Measurements questions in English are available as part of our Chapter-wise Tests for JEE Main & Advanced for JEE & JEE Advanced (Single Correct MCQs): Units and Measurements solutions in Hindi for Chapter-wise Tests for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Advanced (Single Correct MCQs): Units and Measurements | 15 questions in 30 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chapter-wise Tests for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
JEE Advanced (Single Correct MCQs): Units and Measurements - Question 1

The dimension of   (ε0 : permittivity of free space, Eelectric field)

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 1

Note : Here  represents energy per unit volume

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 2

A quantity X is given by  where ∈0 is the permittivity of the free space, L is a length, ΔV is a potential difference and Δt is a time interval. The dimensional formula for X is the same as that of

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 2

Dimensionally ε0L= Capacitance (c)

1 Crore+ students have signed up on EduRev. Have you? Download the App
JEE Advanced (Single Correct MCQs): Units and Measurements - Question 3

A cube has a side of length 1.2 × 10–2m. Calculate its volume.

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 3

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 4

Pressure depends on distance as,  , where α, β are constants, z is distance, k is Boltzman’s constant and θ is temperature. The dimension of β are

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 4

Unit of k is joules per kelvin or dimensional formula of k is [ML2T–2 θ -1] Note : The power of an exponent is a number.

Therefore, dimensionally 

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 5

A wire of length ℓ = 6 ± 0.06 cm and radius r = 0.5 ± 0.005 cm and mass m = 0.3 ± 0.003 gm. Maximum percentage error in density is

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 5

   

Putting the values Δℓ = 0.06 cm, ℓ = 6 cm; Δr = 0.005 cm; r = 0.5 cm,
m = 0.3 gm; Δm = 0.003 gm

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 6

Which of the following set have different dimensions?

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 6

Electric flux 

∴ Dimensionally

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 7

In a screw gauge, the zero of mainscale coincides with fifth division of circular scale in figure (i). The circular division of screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball in figure (ii) is

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 7

Least count = 0.5/50 = 0.01mm

Zero error = 5 × L.C = 5 × 0.01 mm = 0.05 mm
Diameter of ball = [Reading on main scale] + [Reading on circular scale × L . C] – Zero error = 0.5 × 2 + 25 × 0.01 – 0.05  = 1.20 mm

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 8

A student performs an exper iment for determination of  The error in length ℓ is Δℓ and in time T is ΔT and n is number of times the reading is taken. The measurement of g is most accurate for

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 8

Δℓ and ΔT are least and number of readings are maximum in option (d), therefore the measurement of g is most accurate with data used in this option.

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 9

A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 9


= 0.05 + 0.0625 = 0.1125
⇒ ΔY = 2 × 1011 × 0.1125 = 0.225 × 1011
= 0.2 × 1011 N/m2

Note: We can also take value of y from options given without calculating it as it is same in all options.

∴ Y = ( 2 ± 0.2) × 1011 N/m2

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 10

Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum.
They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm

Least count for time = 0.1 s

If EI, EII and EIII are the percentage errors in g, i.e.,  for students I, II and III, respectively, then

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 10

The time period of a simple pendulum is given by


JEE Advanced (Single Correct MCQs): Units and Measurements - Question 11

A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 11

20 divisions on the vernier scale = 16 divisions of main scale

∴ 1 division on the vernier scale

= 16/20 divisions of main scale 

We know that least count = 1MSD – 1VSD
= 1 mm – 0.8 mm  = 0.2 mm

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 12

Th e den sity of a solid ball is to be deter mined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2 %, the relative percentage error in the density is

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 12

Diameter D = M.S.R. + (C.S.R) × L.C.

 

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 13

In the determination of Young’s modulus   using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 13

The maximum possible error in Y due to l and d are


JEE Advanced (Single Correct MCQs): Units and Measurements - Question 14

The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 14

Reading = M.S.R + No of division of V.S matching the main scale division (1MSD – 1VSD)

= 5.124 cm    Option (b) is correct.

JEE Advanced (Single Correct MCQs): Units and Measurements - Question 15

There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2, respectively, are

Detailed Solution for JEE Advanced (Single Correct MCQs): Units and Measurements - Question 15

446 docs|930 tests
Information about JEE Advanced (Single Correct MCQs): Units and Measurements Page
In this test you can find the Exam questions for JEE Advanced (Single Correct MCQs): Units and Measurements solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Advanced (Single Correct MCQs): Units and Measurements, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE