Answer

Verified

61.2k+ views

Hint: Find the original surface area and new surface area of the cube by using given data. Percent increase in the surface area of the cube is found by $x = \dfrac{{{\text{New area - Original area}}}}{{{\text{Original area}}}} \times 100$.

Let us assume the original side to be $a$

Then, the original surface area $S = 6{a^2}$

It is given in the question that the side of the cube is increased by $50\% $

Then, the new side becomes

$a' = a + \dfrac{1}{2}a$

$ \Rightarrow a = \dfrac{{2a + a}}{2}$

$ \Rightarrow a = \dfrac{{3a}}{2}$

Therefore, the new surface area comes out to be

$s = 6{\left( {\dfrac{{3a}}{2}} \right)^2}$

$ \Rightarrow s = 6\left( {\dfrac{{9{a^2}}}{4}} \right)$

$ \Rightarrow s = \dfrac{{(3 \times 9){a^2}}}{2}$

$ \Rightarrow s = \dfrac{{27{a^2}}}{2}$

Hence, let be the $x$ increase of surface area in percentage, which is given as

$x = \dfrac{{{\text{New area - Original area}}}}{{{\text{Original area}}}} \times 100$

$ \Rightarrow x = \dfrac{{\dfrac{{27{a^2}}}{2}{\text{ - }}6{a^2}}}{{6{a^2}}} \times 100$

\[ \Rightarrow x = \dfrac{{\dfrac{{27{a^2} - 12{a^2}}}{2}}}{{6{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{{27{a^2} - 12{a^2}}}{{6{a^2} \times 2}} \times 100\]

\[ \Rightarrow x = \dfrac{{15{a^2}}}{{12{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{5}{4} \times 100\]

$\therefore x = 125\% $

So, the required solution is $(b){\text{ 125}}$.

Note: Whenever we face such types of problems the key point is to have a good grasp of the various formulas of surface area, volume, etc. of solids. Here, in this question, we have calculated both the surface areas before and after the increase in the side of the cube. On comparing the two, the required solution can be reached.

Let us assume the original side to be $a$

Then, the original surface area $S = 6{a^2}$

It is given in the question that the side of the cube is increased by $50\% $

Then, the new side becomes

$a' = a + \dfrac{1}{2}a$

$ \Rightarrow a = \dfrac{{2a + a}}{2}$

$ \Rightarrow a = \dfrac{{3a}}{2}$

Therefore, the new surface area comes out to be

$s = 6{\left( {\dfrac{{3a}}{2}} \right)^2}$

$ \Rightarrow s = 6\left( {\dfrac{{9{a^2}}}{4}} \right)$

$ \Rightarrow s = \dfrac{{(3 \times 9){a^2}}}{2}$

$ \Rightarrow s = \dfrac{{27{a^2}}}{2}$

Hence, let be the $x$ increase of surface area in percentage, which is given as

$x = \dfrac{{{\text{New area - Original area}}}}{{{\text{Original area}}}} \times 100$

$ \Rightarrow x = \dfrac{{\dfrac{{27{a^2}}}{2}{\text{ - }}6{a^2}}}{{6{a^2}}} \times 100$

\[ \Rightarrow x = \dfrac{{\dfrac{{27{a^2} - 12{a^2}}}{2}}}{{6{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{{27{a^2} - 12{a^2}}}{{6{a^2} \times 2}} \times 100\]

\[ \Rightarrow x = \dfrac{{15{a^2}}}{{12{a^2}}} \times 100\]

\[ \Rightarrow x = \dfrac{5}{4} \times 100\]

$\therefore x = 125\% $

So, the required solution is $(b){\text{ 125}}$.

Note: Whenever we face such types of problems the key point is to have a good grasp of the various formulas of surface area, volume, etc. of solids. Here, in this question, we have calculated both the surface areas before and after the increase in the side of the cube. On comparing the two, the required solution can be reached.

Recently Updated Pages

Write a composition in approximately 450 500 words class 10 english JEE_Main

Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main

What is the common property of the oxides CONO and class 10 chemistry JEE_Main

What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main

If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main

The area of square inscribed in a circle of diameter class 10 maths JEE_Main

Other Pages

Nitrene is an intermediate in one of the following class 11 chemistry JEE_Main

Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The resultant of vec A and vec B is perpendicular to class 11 physics JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main