UPSC CDS 1 2022 Math Paper

Given:
Diameter of the spherical ball of lead =6cm
The diameters of the balls are 2cm,4cm and xcm
Concept used:
The volume of a sphere =(4πr3)∕3
Calculation:
Radius of the ball =6∕2=3cm
So, volume of the ball =4∕3×(22∕7)×33
According to the question,
4∕3×(22∕7)×33=4∕3×(22∕7)×13+4∕3×(22∕7)×23+4∕3×(22∕7)×(x∕2)3
⇒4∕3×(22∕7)×33=4∕3×(22∕7)[13+23+(x∕2)3]
⇒27=[1+8+x3∕8]
⇒27=[(72+x3)∕8]
⇒216=(72+x3)
⇒144=x3
So, 53<144<63
Here we will take 5 as the base value.
Now from the options we will check cube of 5.2
So, 5.23=140.608 which is less than 144
Now 5.43=157.464 which is more than 144
So, from this we can conclude that x is in between 5.2 and 5.4
∴ Required answer is 5.2cm<xcm<5.4cm