A steel rod at 25° C is observed to be 1 m long when measured by another metal scale which is correct at 0 ° C. The exact length of steel rod at 0 ° C is (α text steel =12 × 10-6 / ° C , α text metal =20 × 10-6 / ° C )

Question

A steel rod at 25° C is observed to be 1 m long when measured by another metal scale which is correct at 0 ° C. The exact length of steel rod at 0 ° C is (α text steel =12 × 10-6 / ° C , α text metal =20 × 10-6 / ° C ) (A) 1.00002 m (B) 1.0002 m (C) 0.998 m (D) 0.9998 m

Tardigrade Admin · Accepted Answer

A s both steel rod and metal scale expands at

2 5^{\circ} C

,
we use True value of length of rod at

2 5^{\circ} C

=

Scale reading

\times (1 + α (θ^{'} - θ))

where,

α =

coefficient of linear expansion of scale
(

= 20 \times 1 0^{- 6} /^{\circ} C

)

θ^{'} =

temperature at which observation is taken

(= 2 5^{\circ} C_{r})

and

θ =

temperature at which metre scalc reads correctly

(= 0^{\circ} C)

Substituting values in Eq. (i), we get
Length of rod at

2 5^{\circ} C

= 1 \times [1 + 20 \times 1 0^{- 6} (25 - 0)]

= 1 + 5 \times 1 0^{- 4} = 1.0005 m

Now, if

l_{2} =

length of steel rod at

2 5^{\circ} C

and

l_{1} =

length of steel rod at

0^{\circ} C

, then we have

l_{2} = l_{1} (1 + α Δ θ)

Substituting values, we have

1.0005 = l_{1} [1 + 12 \times 1 0^{- 6} \times (25 - 0)]

\Rightarrow l_{1} = \frac{1.0005}{1.0003} = 1.000199 = 1.0002 m

Q. A steel rod at 25∘C is observed to be 1m long when measured by another metal scale which is correct at 0∘C. The exact length of steel rod at 0∘C is (αsteel ​=12×10−6/∘C,αmetal ​=20×10−6/∘C)

1.00002 m

1.0002 m

0.998 m

0.9998 m

Q. A steel rod at $2 5^{\circ} C$ is observed to be $1 m$ long when measured by another metal scale which is correct at $0^{\circ} C$ . The exact length of steel rod at $0^{\circ} C$ is
$(α_{steel} = 12 \times 1 0^{- 6} /^{\circ} C, α_{metal} = 20 \times 1 0^{- 6} /^{\circ} C)$