Question

. The length, breadth and thickness of a block
of metal were measured with the help of
Vernier Calipers. The measurements are
L = (5.250 + 0.001)cm, b = (3.450 + 0.01)cm
t = (1.740 + 0.001)cm
Find the percentage error in volume of the
block.​

Answer 1

Given

Measured values of length , breadth and thickness of a metal block are

• length , l = (5.250 + 0.001) cm
• breadth , b = (3.450 + 0.01) cm
• thickness , t = (1.740 + 0.001) cm

To find

• Percentage error in the volume of block

Knowledge required

• Error of a product

Suppose Z is a quantity such that

Z = A×B×C

and measured values of A , B and C are A ± Δ A , B ± Δ B , C ± Δ C respectively.

then,

Relative error in the quantity Z will be given by

$\boxed{\sf{\dfrac{\triangle Z}{Z}=\dfrac{\triangle A}{A}+\dfrac{\triangle B}{B}+\dfrac{\triangle C}{C}}}$

Hence, the rule is : When three quantities are multiplied or divided , the relative error in the result is the sum of the relative errors in the multipliers.

Solution

→  Let, volume of cuboid be V

the formula for calculating volume of cuboid block is

length × breadth × thickness

so, using rule for error of a product

$\implies\sf{\dfrac{\triangle V}{V}=\dfrac{\triangle l}{l}+\dfrac{\triangle b}{b}+\dfrac{\triangle t}{t}}\\\\\\\implies\sf{\dfrac{\triangle V}{V} =\dfrac{0.001}{5.250}+\dfrac{0.01}{3.450}+\dfrac{0.001}{1.740}}\\\\\\\implies\sf{\dfrac{\triangle V}{V}=0.0002 + 0.003 + 0.0006}\\\\\\\implies\red{\sf{\dfrac{\triangle V}{V}=0.0038\:\:\:\:\;(approx.)}}$

Converting , into percentage error by multiplying by 100% both sides

$\implies\sf{\dfrac{\triangle V}{V}\times 100\% = 0.0038\times 100\%}\\\\\\\implies\boxed{\boxed{\red{\sf{percentage\:error\:in\:volume=0.38 \;\%}}}\orange{\bigstar}}$

Answer .