JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 13)
If the surface area of a cube is increasing at a
rate of 3.6 cm2/sec, retaining its shape; then
the rate of change of its volume (in cm3/sec),
when the length of a side of the cube is
10 cm, is :
9
10
18
20
Explanation
For cube of side 'a'
A = 6a2 and V = a3
Given $${{dA} \over {dt}} = 3.6$$
$$ \Rightarrow $$$$ 12a{{da} \over {dt}}$$ = 3.6
$${{dV} \over {dt}} = 3{a^2}.{{da} \over {dt}} = 3{a^2}\left( {{{3.6} \over {12a}}} \right)$$
at a = 10
$${{dV} \over {dt}} = 9$$
A = 6a2 and V = a3
Given $${{dA} \over {dt}} = 3.6$$
$$ \Rightarrow $$$$ 12a{{da} \over {dt}}$$ = 3.6
$${{dV} \over {dt}} = 3{a^2}.{{da} \over {dt}} = 3{a^2}\left( {{{3.6} \over {12a}}} \right)$$
at a = 10
$${{dV} \over {dt}} = 9$$
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