Question

33. If mass of a cube increases by 1% and its side
increases by 0.5%, then percentage change in it
density will be
(1) - 0.5%
(2) 0.5%
(3) 1%
(4) 2%​

Answer 1

answer is 2% it also writen also in mass

Answer 2

Answer:

The percentage change in the density of cube will be equal to - 0.5% when mass of a cube increases by $1\%$ and side  of cube increases by $0.5\%$.

Therefore, option (2) is correct.

Explanation:

Given, the percentage increase in the mass of the cube = 1%

The % increase in the side of cube = 0.5%

We know that the volume of the cube, $V= l^3$

The density of the cube can be calculated from the ratio of the mass per unit volume of the cube.

$Density \;(d) =\frac{m}{V}$

$d =\frac{m}{l^3}$

Percentage change in the density of cube can be calculated as:

$\frac{\triangle d}{d}\times 100=\frac{\triangle m}{m}\times 100 - 3 \frac{\triangle l}{l}\times 100$

$\frac{\triangle d}{d}\times 100 = 1\% -3(0.5\%)$

$\frac{\triangle d}{d}\times 100 =-0.5\%$

Therefore, the percentage change in the density of cube will be -0.5%.

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