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