A blacksmith fixes iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are $5.243 m$ and $5.231 m$ , respectively at $27^{0} C$ . To what temperature should the ring be heated so as to fit the rim of the wheel?

356^{0} C

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218^{0} C

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562^{0} C

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259^{0}

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Solution

The correct option is A $218^{0} C$
Given,

Initial Temperature, $T_{1} = 27^{0} C$

Initial length, $l_{1} = 5.231 m$

Final length, $l_{2} = 5.243 m$

Now,

$α_{l} = \frac{Δ l}{l_{1} Δ T} = \frac{l_{2} - l_{1}}{l_{1} Δ T}$

$⟹ l_{2} = l_{1} [1 + α_{1} (T_{2} - T_{1})]$

That is,

$5.243 m = 5.231 m [1 + 1.20 \times 10^{- 5} K^{- 1} (T_{2} - 27^{0} C)]$

This gives,

$T_{2} = 218^{0} C$

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Similar questions

Q. A black smith fixes ring on the rime of the wooden wheel of a bullock cart the diameter of the rime and iron ring arc

5.243

and

5.231

resp. at

27^{\circ} C

to what temp, should the ring be heated. So as to fit the rime of the wheel

α = 1.20 \times 10^{- 5}

Q. Iron rim is made slightly smaller than the wooden wheel. The rim is usually heated before fixing into the wooden wheel because on heating the iron rim:

Fixing of iron rim to the wooden wheel of a cart involve
s:

Question 7

Iron rim is made slightly smaller than the wooden wheel. The rim is usually heated before fixing into the wooden wheel because on heating the iron rim
(a) expands and fits onto the wooden wheel
(b) contracts and fits onto the wooden wheel
(c) no change in the size take place
(d) expands first, then on cooling contracts and fits onto the wooden wheel

In Change carts, the iron rim is usually heated before fixing to the wooden wheels, because on heating, the iron:-

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A blacksmith fixes iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are 5.243m and 5.231m, respectively at 270C. To what temperature should the ring be heated so as to fit the rim of the wheel?

The correct option is A 2180CGiven,Initial Temperature, T1=270CInitial length, l1=5.231mFinal length, l2=5.243mNow,αl=Δll1ΔT=l2−l1l1ΔT⟹l2=l1[1+α1(T2−T1)]That is,5.243m=5.231m[1+1.20×10−5K−1(T2−270C)]This gives,T2=2180C

A blacksmith fixes iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are $5.243 m$ and $5.231 m$ , respectively at $27^{0} C$ . To what temperature should the ring be heated so as to fit the rim of the wheel?