AP ICET 2013

Question 108

11 cubic meters of steel is cast in to cylindrical bars of diameters 10 cm and the length 1.4 metres. Then the number of such bars that can be cast with the given metal is (Take $$\pi = \frac{22}{7})$$

Solution

Solution 

Volume of given steel = 11$$m^3$$

Volume of the bar = $$\pi\ \left(\frac{0.1}{2}\right)^2\ 1.4\ m^3$$ = $$\frac{22}{7}\frac{0.01}{4}\ 1.4\ m^3\ =\ 11\cdot\frac{0.01}{2}\cdot0.2$$ = $$11\cdot0.01\cdot0.1\ =\ 0.011$$

No. of bars, n =  $$\left(\frac{\left(Volume\ of\ steel\right)}{Volume\ of\ a\ bar}\right)\ =\ $$ 

$$\left(\frac{11}{11\ \times\ 10^{-3}}\right)\ =\ 10^{3\ }\ =\ 1000\ Answer$$


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