PLS ANSWER QUICKLY 2MARROW XAM
64 solid iron spheres,each of radius r and surface area s are melted to form a sphere with surface area s` .find
(i)radius R` of the new sphere
(ii) the ratio of s and s`
Answers
Answered by
0
1. take ^ as pie
64^rcube*4/3=Rcube*4^/3
64rcube=Rcube
4r=R
2.ratio=4^rsquare/4^Rsquare
= rsquare/(4r)square............(from answer of 1)
=1/16
hope it helped
64^rcube*4/3=Rcube*4^/3
64rcube=Rcube
4r=R
2.ratio=4^rsquare/4^Rsquare
= rsquare/(4r)square............(from answer of 1)
=1/16
hope it helped
Answered by
0
volume of 64 spheres = volume of the big sphere
64 × 4/3 ×22/7 × r³ = 4/3 × 22/7 × R³
⇒BY CANCELING WE GET
64 × r³ = R³
4³ × r³ = R³
4r = R
RATIO OF THEIR SURFACE AREAS = 4πr² : 4πR²
= r² : R²
we found that R=4r
r² : (4r)²
r² : 16r²
1: 16
so the ratio of their surface areas id 1 : 16
Best of luck !!!!
64 × 4/3 ×22/7 × r³ = 4/3 × 22/7 × R³
⇒BY CANCELING WE GET
64 × r³ = R³
4³ × r³ = R³
4r = R
RATIO OF THEIR SURFACE AREAS = 4πr² : 4πR²
= r² : R²
we found that R=4r
r² : (4r)²
r² : 16r²
1: 16
so the ratio of their surface areas id 1 : 16
Best of luck !!!!
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