मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

A Brass Rod of Length 50 Cm and Diameter 3.0 Mm is Joined to a Steel Rod of the Same Length and Diameter. What is the Change in Length of the Combined Rod at 250 °C, If the Original Lengths Are at 40.0 °C? is There a ‘Thermal Stress’ Developed at the Junction? the Ends of the Rod Are Free to Expand (Co-efficient of Linear Expansion of Brass = 2.0 × 10–5 K–1, Steel = 1.2 × 10–5 K–1).

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प्रश्न

A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).

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उत्तर १

Initial temperature, T1 = 40°C

Final temperature, T2 = 250°C

Change in temperature, ΔT = T2 – T= 210°C

Length of the brass rod at T1l1 = 50 cm

Diameter of the brass rod at T1d1 = 3.0 mm

Length of the steel rod at T2l2 = 50 cm

Diameter of the steel rod at T2d2 = 3.0 mm

Coefficient of linear expansion of brass, α1 = 2.0 × 10–5K–1

Coefficient of linear expansion of steel, α2 = 1.2 × 10–5K–1

For the expansion in the brass rod, we have:

`("Change in length"(trianglel_1))/("Original length"(l_1))" = alpha_1triangleT`

`:. trianglel_1 = 50 xx (2.1xx 10^(-5))xx210`

= 0.2205 cm

For the expansion in the steel rod, we have:

`("Change in length"(trianglel_2))/("Original length"(l_2))" = alpha_2triangleT`

`:.trianglel_2 = 50xx (1.2xx10^(-5))xx210`

= 0.126 cm

Total change in the lengths of brass and steel,

Δl = Δl1 + Δl2

= 0.2205 + 0.126

= 0.346 cm

Total change in the length of the combined rod = 0.346 cm

Since the rod expands freely from both ends, no thermal stress is developed at the junction.

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उत्तर २

Here `l_"brass" = l_"steel" = 50 cm, d_"brass" = d_"steel" = 3mm = 0.3 cm, trianglel_"brass" = ?, trianglel_"steel" = ?`

`triangleT = 250 - 40 = 210 ""^@C`

`alpha_"brass" = 2xx10^(-5) ""^@C^(-1) and alpha_"steel" = 1.2 xx 10^(-5) ""^@C^(-1)`

Now `trianglel_"brass" = alpha_"brass" xx l_"brass" xx triangleT`

`= 2xx10^(-5) xx 50 xx 210 = 0.21 cm`

Now `trianglel_"steel" = alpha_"steel" xx l_"steel" xx triangleT`

`= 1.2 xx 10^(-5) xx 50 xx 210`

`=0.126 cm = 0.13 cm`

`:. "Total change in length," trianglel = trianglel_"brass" + trianglel_"steel" = 0.21 + 0.13 = 0.34 cm`

Since the rod is not clamped at its ends, no thermal stress developed at the junction.

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पाठ 10: Thermal Properties of Matter - Exercises [पृष्ठ २९५]

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एनसीईआरटी Physics Part 1 and 2 [English] Class 11
पाठ 10 Thermal Properties of Matter
Exercises | Q 10 | पृष्ठ २९५

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