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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

As shown in diagram, AB is a rod of length 30 cm and area of cross...

As shown in diagram, ABAB is a rod of length 3030 cm and area of cross section 1.0cm21. 0{cm}^{2} and thermal conductivity 336SIunits336SIunits. The ends AA and BB are maintained at temperatures 20°C20 ° Cand 40°C40 ° C, respectively. A point CC of this rod is connected to a box DD, containing ice at 0°C0 ° C, through a highly conducting wire of negligible heat capacity. The rate at which ice melts in the box is (assume latent heat of fusion for ice Lf=80calg−1{L}_{f} = 80calg^{- 1})

As shown in diagram, AB is a rod of length 30 cm and area of cross section 1. 0 cm ^ 2 and thermal conductivity

Options

  1. A84mgs−184mgs^{- 1}
  2. B84gs−184gs^{- 1}
  3. C20mgs−120mgs^{- 1}
  4. D40mgs−140mgs^{- 1}

Correct answer

D. 40mgs−140mgs^{- 1}

Step-by-step solution

Thermal resistance of AC=LKA=0.1336×10−4=103336=R(let)<br/><br/><imgalt=""src="/api/proxy−image?url=https\text{AC} = \dfrac{\text{L}}{\text{KA}} = \dfrac{\text{0.1}}{3 3 6 \times 1 {0}^{- 4}} = \dfrac{1 {0}^{3}}{3 3 6} = \text{R} \left(\text{let}\right)<br /><br /><img alt="" src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fnta_abhyas%252Fjee_main%252Fimages%252FEM0029825_Ans.jpg%3Falt%3Dmedia \text{ and } clean=1 \text{ and } v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fnta_abhyas%2Fjee_main%2Fimages%2FEM0029825_Ans.jpg?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fnta_abhyas%2Fjee_main%2Fimages%2FEM0029825_Ans.jpg?alt=media" style="width: 243px; height: 130px;" /><br /><br />Thermal resistance of \text{BC} = \dfrac{\text{0.2}}{3 3 6 \times 1 {0}^{- 4}} =2 \text{R}<br /><br />Heat flow rates are<br /><br />{H}_{1} = 20R;H2=402R=20R<br/><br/>undefined<br/><br/>=13440103=13.44 W<br/><br/>Rateofmeltingofice<br/><br/>=HLf=13.44/4.280gs−1=40mgs−1\dfrac{2 0}{\text{R}} \text{;} {H}_{2} = \dfrac{4 0}{2 \text{R}} = \dfrac{2 0}{\text{R}}<br /><br />undefined<br /><br /> =\dfrac{1 3 4 4 0}{1 {0}^{3}} = \text{13.44 W}<br /><br />Rate of melting of ice<br /><br />=\dfrac{H}{\text{L}\text{f}} = \dfrac{13.44/4.2}{80}{g s}^{-1} =40mgs^{- 1}
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