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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Thermal Properties of Matter

NTA Abhyas 2020 · Physics · Thermal Properties of Matter

The graph of elongation of a rod of a substance A with temperature...

The graph of elongation of a rod of a substance AA with temperature rise is shown in the first graph. A& liquid BB contained in a cylindrical vessel made out of substance AA (graduated in millilitres)millilitres) at 0°C0 ° C is heated gradually. The readings of the liquid level in the vessel corresponding to different temperatures are shown in the second graph. The real coefficient of volume expansion of the liquid is

The graph of elongation of a rod of a substance A with temperature rise is shown in the first graph. A & liquid B& The graph of elongation of a rod of a substance A with temperature rise is shown in the first graph. A & liquid B

Options

  1. A2.7×10−5∘C−12.7 \times 1{0}^{- 5}{}^{∘}C^{-1}
  2. B15.4×10−5∘C−115.4 \times 1{0}^{- 5}{}^{∘}C^{-1}
  3. C16.2×10−5∘C−116.2 \times 1{0}^{- 5}{}^{∘}C^{-1}
  4. D151.2×10−5∘C−1151.2 \times 1{0}^{- 5}{}^{∘}C^{-1}

Correct answer

C. 16.2×10−5∘C−116.2 \times 1{0}^{- 5}{}^{∘}C^{-1}

Step-by-step solution

From the graph (a),

The graph of elongation of a rod of a substance A with temperature rise is shown in the first graph. A & liquid B

the slope of line is&

m=1001.2300m = \dfrac{\text{1001.2}}{\text{300}}

ΔLΔT=(1001.2−1000.0)mm(300−0)∘C\dfrac{\Delta \text{L}}{\Delta \text{T}} = \dfrac{\left(\text{1001.2} - \text{1000.0}\right) \text{mm}}{{\left(3 0 0 - 0\right)}^{∘} \text{C}}

=4×10−6m∘C−1= 4 \times 1{0}^{- 6}\text{m}{}^{∘}C^{-1}

But ΔLΔT=Lα\dfrac{\Delta \text{L}}{\Delta \text{T}} = \text{L} \alpha

∴α=4×10−61m∴ \alpha = \dfrac{4 \times 1{0}^{- 6}}{1 m}

=4×10−6∘C−1= 4 \times 1{0}^{- 6}{ }^{∘}C^{-1}

As shown in graph (b),
The graph of elongation of a rod of a substance A with temperature rise is shown in the first graph. A & liquid B
the slop of line is

ΔVaΔT=(1003−1000)mL(20−0)∘C=320mL∘C−1\dfrac{\Delta {V}_{a}}{\Delta \text{T}} = \dfrac{\left(1003 - 1000\right)\text{mL}}{\left(20 - 0\right) ∘\text{C}} = \dfrac{3}{20}\text{mL}{}^{∘}C^{-1}

But, ΔVaΔT=Vγa\dfrac{\Delta {V}_{a}}{\Delta \text{T}} = \text{V} {γ}_{a}

∴γa=320×103=150×10−6∘C−1∴ {γ}_{a} = \dfrac{3}{20 \times 1{0}^{3}} = 150 \times 1{0}^{- 6}{}^{∘}C^{-1}

Since, γa=γr−γc{γ}_{a} = {γ}_{r} - {γ}_{c}

∴γr=γa+γc=γa+3α∴ {γ}_{r} = {γ}_{a} + {γ}_{c} = {γ}_{a} + 3 \alpha

=(150+12)×10−6∘C−1= \left(150 + 12\right) \times 1{0}^{- 6}{}^{∘}C^{-1}

=16.2×10−5∘C−1= \text{16.2} \times 1{0}^{- 5}{}^{∘}C^{-1}
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