# Question #74469

Apr 21, 2016

$2.5 \times {10}^{-} 3 {m}^{-} 2$

#### Explanation:

The cylinder with mass $2 k g \mathmr{and}$ area of cross section $5 c {m}^{2}$ is just able to be attached to the ceiling with the adhesive. Changed to SI units for sake of conformity.

The adhesive is able to balance the weight of the cylinder $m g$, where $g$ is acceleration due to gravity.

It implies that the adhesive has just capacity of holding downwards force per unit area $= \text{Weight"/"Area of cross section}$
$= \frac{2 g}{5 \times {10}^{-} 4} N {m}^{-} 2$
When the mass if the cylinder is changed to $10 k g$.
We see that the mass has increased by $\frac{10}{2} = 5 \text{ times}$

As the holding power of the adhesive does not change, therefore the area of cross section must be increased by the same factor for the new cylinder to stay attached with the ceiling.
Therefore, required area of cross section $= 5 \times {10}^{-} 4 \times 5$
$= 2.5 \times {10}^{-} 3 {m}^{-} 2$