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

Question:

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

(A) $\mathrm{A}_{1}=24.36, \mathrm{~B}_{1}=0.0724, \mathrm{C}_{1}=256.2$

(B) $A_{2}=24.44, B_{2}=16.082, C_{2}=240.2$

(C) $\mathrm{A}_{3}=25.2, \mathrm{~B}_{3}=19.2812, \mathrm{C}_{3}=236.183$

(D) $\mathrm{A}_{4}=25, \mathrm{~B}_{4}=236.191, \mathrm{C}_{4}=19.5$

  1. (1) $\begin{aligned} A_{4}+B_{4}+C_{4}

  2. (2) $\mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1}=\mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2}=\mathrm{A}_{3}+\mathrm{B}_{3}+\mathrm{C}_{3}=\mathrm{A}_{4}+\mathrm{B}_{4}$

    $+C_{4}$

  3. (3) $\mathrm{A}_{4}+\mathrm{B}_{4}+\mathrm{C}_{4}<\mathrm{A}_{1}+\mathrm{B}_{1}+\mathrm{C}_{1}=\mathrm{A}_{2}+\mathrm{B}_{2}+\mathrm{C}_{2}=\mathrm{A}_{3}+\mathrm{B}_{3}$

    $+\mathrm{C}_{3}$

  4. (4) None of these


Correct Option: , 4

Solution:

(4)

$D_{1}=A_{1}+B_{1}+C_{1}=24.36+0.0724+256.2=280.6$

$D_{2}=A_{2}+B_{2}+C_{2}=24.44+16.082+240.2=280.7$

$D_{3}=A_{3}+B_{3}+C_{3}=25.2+19.2812+236.183=280.7$

$D_{4}=A_{4}+B_{4}+C_{4}=25+236.191+19.5=281$

None of the option matches.

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