JEE Main 2020 — Physics Mechanics

2020mcqeasy

For the four sets of three measured physical quantities as given below. Which of the following options is correct?

$(i)$ ${A}_{1}=24.36,{B}_{1}=0.0724,{C}_{1}=256.2$

$(ii)$ ${A}_{2}=24.44,{B}_{2}=16.082,{C}_{2}=240.2$

$(iii)$ ${A}_{3}=25.2,{B}_{3}=19.2812,{C}_{3}=236.183$

$(iv)$ ${A}_{4}=25,{B}_{4}=236.191,{C}_{4}=19.5$

Official previous-year question

Held on 9 Jan 2020 · Verified 6 Jul 2026.

Options

A
${A}_{4}+{B}_{4}+{C}_{4}<{A}_{1}+{B}_{1}+{C}_{1}<{A}_{3}+{B}_{3}+{C}_{3}<{A}_{2}+{B}_{2}+{C}_{2}$
B
${A}_{1}+{B}_{1}+{C}_{1}={A}_{2}+{B}_{2}+{C}_{2}={A}_{3}+{B}_{3}+{C}_{3}={A}_{4}+{B}_{4}+{C}_{4}$
C
${A}_{1}+{B}_{1}+{C}_{1}<{A}_{2}+{B}_{2}+{C}_{2}={A}_{3}+{B}_{3}+{C}_{3}<{A}_{4}+{B}_{4}+{C}_{4}$
D
${A}_{1}+{B}_{1}+{C}_{1}<{A}_{3}+{B}_{3}+{C}_{3}<{A}_{2}+{B}_{2}+{C}_{2}<{A}_{4}+{B}_{4}+{C}_{4}$

Did you get this right?

Solution

In addition or subtractions, final answer must contain the equal numbers of digits after decimal as the adding or subtracting quantities among which is having the lowest numbers of digits after decimal.

${A}_{1}+{B}_{1}+{C}_{1}=24.36+0.0724+256.2=280.6324=280.6$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal.

${A}_{2}+{B}_{2}+{C}_{2}=24.44+16.082+240.2=280.722=280.7$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal

${A}_{3}+{B}_{3}+{C}_{3}=25.2+19.2812+236.183=280.6642=280.7$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal ${A}_{4}+{B}_{4}+{C}_{4}=25+236.191+19.5=280.691=281$

The Least number of significant digit after decimal in the above expression is 0 So, Sum should conatian 0 significant digit after decimal

Answer should be ${A}_{1}+{B}_{1}+{C}_{1}<{A}_{2}+{B}_{2}+{C}_{2}={A}_{3}+{B}_{3}+{C}_{3}<{A}_{4}+{B}_{4}+{C}_{4}$

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Options

${A}_{4}+{B}_{4}+{C}_{4}<{A}_{1}+{B}_{1}+{C}_{1}<{A}_{3}+{B}_{3}+{C}_{3}<{A}_{2}+{B}_{2}+{C}_{2}$

${A}_{1}+{B}_{1}+{C}_{1}={A}_{2}+{B}_{2}+{C}_{2}={A}_{3}+{B}_{3}+{C}_{3}={A}_{4}+{B}_{4}+{C}_{4}$

${A}_{1}+{B}_{1}+{C}_{1}<{A}_{2}+{B}_{2}+{C}_{2}={A}_{3}+{B}_{3}+{C}_{3}<{A}_{4}+{B}_{4}+{C}_{4}$

${A}_{1}+{B}_{1}+{C}_{1}<{A}_{3}+{B}_{3}+{C}_{3}<{A}_{2}+{B}_{2}+{C}_{2}<{A}_{4}+{B}_{4}+{C}_{4}$

Solution

${A}_{1}+{B}_{1}+{C}_{1}=24.36+0.0724+256.2=280.6324=280.6$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal.

${A}_{2}+{B}_{2}+{C}_{2}=24.44+16.082+240.2=280.722=280.7$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal

${A}_{3}+{B}_{3}+{C}_{3}=25.2+19.2812+236.183=280.6642=280.7$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal ${A}_{4}+{B}_{4}+{C}_{4}=25+236.191+19.5=280.691=281$

The Least number of significant digit after decimal in the above expression is 0 So, Sum should conatian 0 significant digit after decimal

Answer should be ${A}_{1}+{B}_{1}+{C}_{1}<{A}_{2}+{B}_{2}+{C}_{2}={A}_{3}+{B}_{3}+{C}_{3}<{A}_{4}+{B}_{4}+{C}_{4}$

JEE Main 2020 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

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JEE Main 2020 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics