Q10 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Chemistry

A charcoal cube having side length \(10 \mathrm{~cm}\) is chopped lengthwise into 5 equal pieces. Then the effective adsorption power

  • Correct answer
    A
    increases by 2.33 times
  • B
    decreases by 2.33 times
  • C
    increases by 2.14 times
  • D
    decreases by 2.14 times

Solution

Initial volume of cube \(=(10)^3=1000 \mathrm{~cm}^3\) \(V\) when cube is chopped along length then it will form cuboid in which length \(=6 \mathrm{~cm}\) \(\begin{aligned} \text {breadth } & =10 \mathrm{~cm} \\ \text {height } & =10 \mathrm{~cm} \end{aligned}\) Adsorption power \(\propto\) more surface area \(\propto 6 \mathrm{~cm}\) \(\propto 6 \times(10)^2 \propto 600 \mathrm{~cm}^2\) Final surface area of the cuboid \(\begin{aligned} & =2(l \times b+b \times h+h \times l) \text { Area of cuboid } \\ & =2(2 \times 10+100+20) \\ & =280 \mathrm{~cm}^2 \end{aligned}\) For final cuboid area \(=280 \times 5=1400 \mathrm{~cm}^2\) So, final adsorption power \(\propto 1400 \mathrm{~cm}^2\) \(\begin{aligned} \frac{\text { Final adsorption power }}{\text { Initial adsorption power }} & =\frac{1400}{600}=\frac{14}{6}=\frac{7}{3} \\ & =2.33 \text { times } \end{aligned}\)

Hence, the correct answer is option: 'A'.

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