Surface area of a solid sphere of radius 5 cm is equal to the difference of outer and inner surface area of a hollow sphere of thickness 1 cm, Product of the inner and outer radius will be
By the given condition
4(pi)*5^2 = 4(pi)*r^2 - 4(pi)*(r-1)^2 (let the radius of hollow sphere be r)
By solving equation
find r and (r-1)
r*(r-1) = 12*13 = 156
In a leap year, there are 366 days. Out of these 364 days make 52 weeks and hence 52 saturday. Remaining 2 days can be: Sun-Mon, Tue-Wed, Wed-Thu, Fri-Sat, Sat-Sun.
Total number of possible outcomes = 7
No of saturday outcomes = 2 (Fri-Sat, Sat-Sun)
Then probability of 53 saturdays = 2/7
let log4(base2)=p
therefore 2^p=4
so p = 2,
Again log8(base2)=p
therefore 2^p=8
so p = 3,
Last Again log16(base2)=p
therefore 2^p=16
so p=4,
By the possibility
let last value be logx(base2)=p
but by the condition, p should be 5
therefore 5=logx(base2)
represent in square form 2^5=x
so x = 32
p=log32(base2)
let the smaller side be x
then diagonal is 3x
by using pythagoras theorem
largest side is 2(sqrt2)x
then the ratio of side
2(sqrt2)x : x
therefore 2 sqrt2 : 1
option no 2
A cooper wire when bent in the form of square enclose an area of 48400 cm^2. If the same wire is bent in the form of circle, then the area enclosed by it will be.
let the side of square be a
then the area of square = 48400 cm^2
therefore a^2 = 48400 cm^2
a = 220 cm
when same copper wire bents in the form of circle then
perimeter of square = perimeter of circle
4a = 2(22/7)r (Where r is the radius of circle)
r = 140
then area of circle = (22/7)140*140 = 61600 cm^2
option no 9