A length of 2m is measured using a metal tape at 10-C- It is calibrated at a temperature 30-C- The actual length is - - 1- 10-4 - 1C-

The correct option is B 1.996 m Actual length at 30^∘C is L L = L_0( + 20T) 2 = L_0 [1 + 10^ 4× (20) L_0 = 2(1 + 0.002) = 1.996 ...

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Standard XII

Physics

Error and Uncertainty

A length of ...

Question

A length of $2 m$ is measured using a metal tape at $10^{\circ} C$ . It is calibrated at a temperature $30^{\circ} C$ . The actual length is $(α = 1 \times 10^{- 4} / 1 C^{\circ})$ .

2.004 m

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1.996 m

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1.004 m

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0.004 m

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Solution

The correct option is B $1.996 m$
Actual length at $30^{\circ} C$ is $L$
$L = L_{0} (+ 20 T)$
$2 = L_{0} [1 + 10^{- 4} \times (20)$
$L_{0} = \frac{2}{(1 + 0.002)} = 1.996 m$ .

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A length of 2m is measured using a metal tape at 10∘C. It is calibrated at a temperature 30∘C. The actual length is (α=1×10−4/1C∘).

The correct option is B 1.996mActual length at 30∘C is LL=L0(+20T)2=L0[1+10−4×(20)L0=2(1+0.002)=1.996 m.

A length of $2 m$ is measured using a metal tape at $10^{\circ} C$ . It is calibrated at a temperature $30^{\circ} C$ . The actual length is $(α = 1 \times 10^{- 4} / 1 C^{\circ})$ .

The correct option is B $1.996 m$
Actual length at $30^{\circ} C$ is $L$
$L = L_{0} (+ 20 T)$
$2 = L_{0} [1 + 10^{- 4} \times (20)$
$L_{0} = \frac{2}{(1 + 0.002)} = 1.996 m$ .