The length of a cylinder is measured with a metre rod having least count 0.1 cm. Its diameter is measured with vernier calipers having least count 0.01 cm. If the length and diameter of the cylinder are 5.0 cm and 2.00 cm, respectively, then the percentage error in the calculated value of volume will be

Question

The length of a cylinder is measured with a metre rod having least count 0.1 cm. Its diameter is measured with vernier calipers having least count 0.01 cm. If the length and diameter of the cylinder are 5.0 cm and 2.00 cm, respectively, then the percentage error in the calculated value of volume will be (A) 2 % (B) 1 % (C) 3 % (D) 4 %

Tardigrade Admin · Accepted Answer

Volume of cylinder (V)=π r2l (Δ V/V)×100=2×(Δ r/r)× 100+(Δ l/l)× 100 =2×(0.01/2)×100+(0.1/5)×100 =1 %+2 %=3 %

$2%$

$1%$

$3%$

$4%$

Volume of cylinder $(V) = π r^{2} l$
$\frac{Δ V}{V} \times 100 = 2 \times \frac{Δ r}{r} \times 100 + \frac{Δ l}{l} \times 100$
$= 2 \times \frac{0.01}{2} \times 100 + \frac{0.1}{5} \times 100$
$= 1% + 2% = 3%$

2%

1%

3%

4%

Volume of cylinder (V)=πr2l VΔV​×100=2×rΔr​×100+lΔl​×100 =2×20.01​×100+50.1​×100 =1%+2%=3%

$2%$

$1%$

$3%$

$4%$

Volume of cylinder $(V) = π r^{2} l$
$\frac{Δ V}{V} \times 100 = 2 \times \frac{Δ r}{r} \times 100 + \frac{Δ l}{l} \times 100$
$= 2 \times \frac{0.01}{2} \times 100 + \frac{0.1}{5} \times 100$
$= 1% + 2% = 3%$