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In new temperature scale W, freezing point of water is 39^(@)W and boiling point is 239^(@)W. What would be the temperature on new temperature scale for 39^(@)C temperature ? |
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Answer» `200^(@)W` `(39)/(100)=(W-39)/(200)` `2xx39+39=W` `:.W=39xx(200)/(100)+39` `=39xx2+39` `=78+39` `=117^(@)W` |
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