Question

Mark the correct alternative of the following.
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?

Answer 1

Correct option is C. $$a_2v_1=a_1v_2$$
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 $$a_1$$

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

$$\Rightarrow$$ $$a_1=2\pi r,$$ $$a_2=2\pi R$$

$$\Rightarrow$$ $$r=\dfrac{a_1}{2\pi},\,R=\dfrac{a_2}{2\pi}$$

Also,

$$v_1=\pi r^2a_2$$

$$v_1=\pi\left(\dfrac{a_1}{2\pi}\right)^2a_2$$

$$\therefore$$ $$v_1=\dfrac{a_1^2a_2}{4\pi}$$ ---- ( 1 )

Now,

$$v_2=\pi R^2a_1$$

$$v_2=\pi\left(\dfrac{a_2}{2\pi}\right)^2a_1$$

$$\therefore$$ $$v_2=\dfrac{a_2^2a_1}{4\pi}$$ ---- ( 2 )

Dividing ( 1 ) by ( 2 ) we get,

$$\Rightarrow$$ $$\dfrac{v_1}{v_2}=\dfrac{\dfrac{a_1^2a_2}{4\pi}}{\dfrac{a_2^2a_1}{4\pi}}$$

$$\Rightarrow$$ $$\dfrac{v_1}{v_2}=\dfrac{a_1}{a_2}$$

$$\therefore$$ $$a_2v_1=a_1v_2$$