The volume of a box with sides as L - 3 - 0-02 m- B - 4 - 0-01 m -H -3 - 0-02 m will be

Volume = L × B × H Let most probable error in volume is e_H ∴ e^2_v = ( δ (LBH)δ Le_L )^2+ ( δ (LBH)δ Be_B )^2+ ( δ (LBH)δ He_H )^2 ⇒ e^2_v = (BHe_L)^2+ ...

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Standard VII

Mathematics

Area of Circle

The volume of...

Question

The volume of a box with sides as $L = 3 \pm 0.02 m, B = 4 \pm 0.01 m, H = 3 \pm 0.02 m$ will be

36 \pm 0.302 m^{3}

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36 \pm 0.351 m^{3}

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36 \pm 0.428 m^{3}

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36 \pm 0.271 m^{3}

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Solution

The correct option is B $36 \pm 0.351 m^{3}$
Volume $= L \times B \times H$
Let most probable error in volume is $e_{H}$
$∴ e_{v}^{2} = {(\frac{δ (L B H)}{δ L} e_{L})}_{L}^{2} + {(\frac{δ (L B H)}{δ B} e_{B})}_{B}^{2} + {(\frac{δ (L B H)}{δ H} e_{H})}_{H}^{2}$

$\Rightarrow e_{v}^{2} = (B H e_{L})^{2} + (L H e_{B})^{2} + (L B e_{H})^{2}$

$\Rightarrow e_{v} = \sqrt{(12)^{2} \times (0.02)^{2} + (9)^{2} \times (0.01)^{2} + (12)^{2} \times (0.02)^{2}}$

$\Rightarrow e_{v} = 0.351 m^{3}$

Hence volume of box $= 36 \pm 0.351 m^{3}$

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The volume of a box with sides as L=3±0.02 m,B=4±0.01 m,H=3± 0.02 m will be

The correct option is B 36±0.351 m3Volume =L×B×H Let most probable error in volume is eH ∴e2v=(δ(LBH)δLeL)2+(δ(LBH)δBeB)2+(δ(LBH)δHeH)2 ⇒e2v=(BHeL)2+(LHeB)2+(LBeH)2 ⇒ev=√(12)2×(0.02)2+(9)2×(0.01)2+(12)2×(0.02)2 ⇒ev=0.351m3 Hence volume of box =36±0.351 m3

The volume of a box with sides as $L = 3 \pm 0.02 m, B = 4 \pm 0.01 m, H = 3 \pm 0.02 m$ will be