
When freezing, water increases its volume by $\dfrac{1}{11}$. By what part of its volume will ice decrease when it melts and turns back into the water?\[\]
A. $\dfrac{1}{11}$\[\]
B. $\dfrac{1}{10}$\[\]
C.$\dfrac{1}{12}$\[\]
D.$\dfrac{1}{13}$\[\]
Answer
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Hint: We denoted the volume before freezing as $V$ and after freezing as ${{V}_{n}}$. We use the given information on the increase in volume after freezing to express ${{V}_{n}}$ in terms of $V.$ We find the decrease in volume ${{V}_{n}}-V$ and the part of the volume of decrease of the volume after freezing as $\dfrac{{{V}_{n}}-V}{V}$\[\]
Complete step-by-step solution
Let us assume that the volume of the water before freezing is $V$. We are given the question that during freezing the volume of the water increases by part$\dfrac{1}{11}$. The increase in the volume of the water is $\dfrac{1}{11}$ part of $V$ which is $\dfrac{1}{11}\times V$. Let us denote the new volume of the water after freezing is ${{V}_{n}}$. The new volume is the sum of the volume before freezing $V$ and the increase in volume during freezing $\dfrac{1}{11}\times V$. We have
\[{{V}_{n}}=V+\dfrac{1}{11}V=V\left( 1+\dfrac{1}{11} \right)=V\left( \dfrac{11+1}{11} \right)=V\times \dfrac{12}{11}=\dfrac{12V}{11}....\left( 1 \right)\]
We know that after freezing some part of the water will turn into ice. The volume of ice and water at this point is ${{V}_{n}}$. When it will melt and if melts completely back to only water the volume ${{V}_{n}}$will decrease to the volume water before freezing which is $V.$ So the decrease in volume is ${{V}_{n}}-V$. We put the value of ${{V}_{n}}$ from equation (1) and have the decrease in volume as
\[{{V}_{n}}-V=\dfrac{12V}{11}-V=V\left( \dfrac{12}{11}-1 \right)=V\left( \dfrac{12-11}{11} \right)=\dfrac{V}{12}\]
So the fractional part of new volume which has decreased is
\[\dfrac{{{V}_{n}}-V}{{{V}_{n}}}=\dfrac{\dfrac{V}{11}}{\dfrac{12V}{11}}=\dfrac{1}{12}\]
So the correct choice is A.
Note: We need to be careful that we are asked to find the decreased amount of volume ${{V}_{n}}-V$ as a part of volume after freezing ${{V}_{n}}$ not volume before freezing $V.$ We note that when water freezes to solid its volume increases and when vaporizes to gas its volume decreases.
Complete step-by-step solution
Let us assume that the volume of the water before freezing is $V$. We are given the question that during freezing the volume of the water increases by part$\dfrac{1}{11}$. The increase in the volume of the water is $\dfrac{1}{11}$ part of $V$ which is $\dfrac{1}{11}\times V$. Let us denote the new volume of the water after freezing is ${{V}_{n}}$. The new volume is the sum of the volume before freezing $V$ and the increase in volume during freezing $\dfrac{1}{11}\times V$. We have
\[{{V}_{n}}=V+\dfrac{1}{11}V=V\left( 1+\dfrac{1}{11} \right)=V\left( \dfrac{11+1}{11} \right)=V\times \dfrac{12}{11}=\dfrac{12V}{11}....\left( 1 \right)\]
We know that after freezing some part of the water will turn into ice. The volume of ice and water at this point is ${{V}_{n}}$. When it will melt and if melts completely back to only water the volume ${{V}_{n}}$will decrease to the volume water before freezing which is $V.$ So the decrease in volume is ${{V}_{n}}-V$. We put the value of ${{V}_{n}}$ from equation (1) and have the decrease in volume as
\[{{V}_{n}}-V=\dfrac{12V}{11}-V=V\left( \dfrac{12}{11}-1 \right)=V\left( \dfrac{12-11}{11} \right)=\dfrac{V}{12}\]
So the fractional part of new volume which has decreased is
\[\dfrac{{{V}_{n}}-V}{{{V}_{n}}}=\dfrac{\dfrac{V}{11}}{\dfrac{12V}{11}}=\dfrac{1}{12}\]
So the correct choice is A.
Note: We need to be careful that we are asked to find the decreased amount of volume ${{V}_{n}}-V$ as a part of volume after freezing ${{V}_{n}}$ not volume before freezing $V.$ We note that when water freezes to solid its volume increases and when vaporizes to gas its volume decreases.
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