
If the edges of the cube are halved, then its volume become is:
$
{\text{a}}{\text{. 4 times}} \\
{\text{b}}{\text{. 8 times}} \\
{\text{c}}{\text{. }}\dfrac{1}{8}{\text{times}} \\
{\text{d}}{\text{. }}\dfrac{1}{2}{\text{times}} \\
$
Answer
555.3k+ views
Hint: - Volume of cube $ = {\left( {{\text{Side}}} \right)^3}$
Let, the side of the cube is $a$
As we know the volume$\left( V \right)$ of the cube is $ = {\left( {{\text{Side}}} \right)^3}$
$ \Rightarrow V = {a^3}..........\left( 1 \right)$
Now, if the edges of the cube is halved therefore new edge becomes,
${a_1} = \dfrac{a}{2}$
So, the volume$\left( {{V_1}} \right)$of the new cube is $ = {\left( {\dfrac{a}{2}} \right)^3} = \dfrac{{{a^3}}}{8}$
$\left( {{V_1}} \right) = \dfrac{{{a^3}}}{8}..........\left( 2 \right)$
So, from equation (1) and (2)
${V_1} = \dfrac{V}{8}$
So, the new volume becomes $\dfrac{1}{8}$times the previous volume.
Hence option (c) is correct.
Note: - In such types of questions the key concept we have to remember is that always remember the formula of volume of the cube, then calculate the new volume of the cube according to given condition, then we will get the required answer.
Let, the side of the cube is $a$
As we know the volume$\left( V \right)$ of the cube is $ = {\left( {{\text{Side}}} \right)^3}$
$ \Rightarrow V = {a^3}..........\left( 1 \right)$
Now, if the edges of the cube is halved therefore new edge becomes,
${a_1} = \dfrac{a}{2}$
So, the volume$\left( {{V_1}} \right)$of the new cube is $ = {\left( {\dfrac{a}{2}} \right)^3} = \dfrac{{{a^3}}}{8}$
$\left( {{V_1}} \right) = \dfrac{{{a^3}}}{8}..........\left( 2 \right)$
So, from equation (1) and (2)
${V_1} = \dfrac{V}{8}$
So, the new volume becomes $\dfrac{1}{8}$times the previous volume.
Hence option (c) is correct.
Note: - In such types of questions the key concept we have to remember is that always remember the formula of volume of the cube, then calculate the new volume of the cube according to given condition, then we will get the required answer.
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