MCQ

Two steel sheets each of length a_{1} and breadth a_{2} are used to prepare the surfaces of two right circular cylinders ⇀ one having volume v_{1} and height a_{2} and other having volume v_{2} and height a_{1}. Then,

#### Options

*v*_{1}=*v*_{2}*a*_{1}*v*_{1}=*a*_{2}*v*_{2}*a*_{2}*v*_{1}=*a*_{1}*v*_{2}- \[\frac{v_1}{a_1} = \frac{v_2}{a_2}\]

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#### Solution

Let the radius of the base of the cylinders be r and R.

Now, let sheet with length a_{1} be used to form a cylinder with volume v_{1}

So,

\[2\pi r = a_1 \]

\[ \Rightarrow r = \frac{a_1}{2\pi}\]

Volume

\[v_1 = \pi r^2 h = \pi \left( \frac{a_1}{2\pi} \right)^2 a_2 = \frac{{a_1}^2 a_2}{4\pi}\]

Similarly, let sheet with length a2 be used to form a cylinder with volume v2

So,

\[2\pi r = a_1 \]

\[ \Rightarrow r = \frac{a_1}{2\pi}\]

Volume

\[v_2 = \pi R^2 h = \pi \left( \frac{a_2}{2\pi} \right)^2 a_1 = \frac{{a_2}^2 a_1}{4\pi}\]

Now,

\[\frac{v_1}{v_2} = \frac{{a_1}^2 a_2}{{a_2}^2 a_1} = \frac{a_1}{a_2}\]

\[ \Rightarrow v_1 a_2 = v_2 a_1\]

Concept: Surface Area of Cylinder

Is there an error in this question or solution?

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