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UPSEE 2019 Solved Paper

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Question : 1 of 150

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A wire has a mass

(0.1 \pm 0.001)\,\mathrm{g}

(0.5 \pm 0.005)\,\mathrm{mm}

(10 \pm 0.1)\,\mathrm{cm}

The maximum percentage error in the measurement of its density is

$1\%$
$2\%$
$3\%$
$4\%$

Solution:

Here, mass of wire,

m=(0.1 \pm 0.001)\,\mathrm{g}

Radius,

r=(0.5 \pm 0.005)\,\mathrm{mm}

and length,

L=(10 \pm 0.1)\,\mathrm{cm}

As density of a wire,

\rho = \frac{\text{ mass }}{\text{ volume }} = \frac{\text{ mass }}{\text{ area } \times \text{ length }}

\Rightarrow \rho = \frac{m}{\pi R^{2} \times L}

So relative error in density,

\pm \frac{\Delta \rho}{\rho} = \pm \left[ \frac{\Delta m}{m} + 2 \frac{\Delta R}{R} + \frac{\Delta L}{L} \right]

\Rightarrow \frac{\Delta \rho}{\rho} = \frac{0.001 \times 10^{-3}}{0.1 \times 10^{-3}} + 2 \times \frac{0.005 \times 10^{-3}}{0.5 \times 10^{-3}}

+ \frac{0.1 \times 10^{-2}}{10 \times 10^{-2}}

\Rightarrow \frac{\Delta \rho}{\rho} = 0.01 + 0.02 + 0.01

\Rightarrow \frac{\Delta \rho}{\rho} = 0.04

Maximum percentage error

= \frac{\Delta \rho}{\rho} \times 100 = 0.04 \times 100 = 4\%

Hence, the maximum percentage error in the measurement of density is

4\%

.

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