Question

A cube shaped casting "X" solidifies in 5 min. The solidification time in min for a cube shaped casting "Y" of the same material, which is 8 times heavier than the original casting "X", will be

Answer 1

Answer:

time is 20 min

Explanation:

Given :

As we know that

t=c [v/A] ²

t1=5= c[$\frac {v1}{A1} ]^{2}$

now v2=8v1

which implies each ride is getting doubled  

a2= 4a1

t2=c[ $\frac{8v1}{4a1} ]^{2}$

=4xt1

=4x5

= 20 min

Answer 2

Answer:

20 minutes.

Explanation:

We know that the mass is = volume * density so for the cube we have from the question that the mass of the second case is 8 times mass of the first case and the volume of the second case is also 8 times the volume of the first case.

So, m1 = 8 m2 and v1=8v2 so, a1^3 = 8 a2^3 or a1=2a2. Since, we know that for the cube we have t is directly proportional to (V/A)^2. So, we will have that the ratios of the time t1/t2 = (a1/a2)^2 or 5/t2 = (a1/2a1)^2 which on solving we will get that t2 = 20 minutes.