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Electric Current Physics Current Electricity Physics Class 12th

Due to cold weather a 1m water pipe of cross-sectional area 1 cm² is filled with ice at -10°C. Resistivity heating is used to melt the ice. Current of 0.5A is passed through 4 k $Ω$ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required?

(Given latent heat of fusion for water/ice = 3.33 × 10⁵ J kg^-¹, specific heat of ice = 2 × 10³ J kg^-1 and density of ice = 10³ kg/m³)

Option 1 -

3.53 s

Option 2 -

0.353 s

Option 3 -

35.3 s

Option 4 -

70.6 s

5 Views | Posted 2 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Energy required to melt

Q = $M S Δ T + M L$

$\Rightarrow 1 0^{- 1} \times 2 \times 1 0^{3} \times 10 + 1 0^{- 1} \times 3.33 \times 1 0^{5}$

->3.53 × 10⁴ J

Heat produce in wire

H = l²RT

$Q = 3.53 \times 1 0^{4} = {(\frac{1}{2})}^{2} \times (4 \times 1 0^{3}) \times t$

$t = \frac{3.53 \times 1 0^{4} \times 4}{4 \times 1 0^{3}} = 35.3 s e c$

Energy required to meltQ = <math> <mrow> <mi>M</mi> <mi>S</mi> <mi>Δ</mi> <mi>T</mi> <mo>+</mo> <mi>M</mi> <mi>L</mi> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815998485"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mo>⇒</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mo>×</mo> <mn>2</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mo>+</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mo>×</mo> <mn>3</mn> <mo>.</mo> <mn>3</mn> <mn>3</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>5</mn> </mrow> </msup> </mrow> </math>      <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815998486"> </o:OLEObject></xml><! [endif]->->3.53 × 104 JHeat produce in wireH = l2RT  <math> <mrow> <mi>Q</mi> <mo>=</mo> <mn>3</mn> <mo>.</mo> <mn>5</mn> <mn>3</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>×</mo> <mrow> <mo> (</mo> <mrow> <mn>4</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mo>)</mo> </mrow> <mo>×</mo> <mi>t</mi> </mrow> </math> <math> <mrow> <mi>t</mi> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> <mo>.</mo> <mn>5</mn> <mn>3</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>×</mo> <mn>4</mn> </mrow> <mrow> <mn>4</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mn>5</mn> <mo>.</mo> <mn>3</mn> <mi>s</mi> <mi>e</mi> <mi>c</mi> </mrow> </math>             <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815998487"> </o:OLEObject></xml><! [endif]->              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1815998488"> </o:OLEObject></xml><! [endif]->

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