104
Playing in a ball pool is good entertainment for kids. Suhana bought $600$ new balls of diameter $7$ cm to fill in the pool for her kids. The cuboidal box containing $600$ balls has dimensions $42$ cm $\times 91$ cm $\times 50$ cm ($l \times b \times h$).
Based on above information, answer the following questions :
(i) Find the volume of one ball.
(ii) $10$ balls are painted with neon colours. Determine the area of painted surface.
(iii) (a) Find the volume of empty space in the box.
OR
(iii) (b) The lowermost layer of the balls covers the base of the box edge to edge when balls are placed evenly adjacent to each other. (A) How much area is covered by one ball? (B) How many balls are there in lowermost layer?
Based on above information, answer the following questions :
(i) Find the volume of one ball.
(ii) $10$ balls are painted with neon colours. Determine the area of painted surface.
(iii) (a) Find the volume of empty space in the box.
OR
(iii) (b) The lowermost layer of the balls covers the base of the box edge to edge when balls are placed evenly adjacent to each other. (A) How much area is covered by one ball? (B) How many balls are there in lowermost layer?
Show SolutionHide Solution↓
Solution: (i) Volume of one ball $= \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} = \frac{539}{3}$ or $179.67$ cu. cm [1 mark]
(ii) Area of painted surface $= 10 \times 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 1540$ sq. cm [1 mark]
(iii) (a) Volume of box $= 42 \times 91 \times 50 = 191100$ cu. cm [1/2 mark]
Volume of $600$ balls $= 600 \times \frac{539}{3} = 107800$ cu. cm [1/2 mark]
Volume of empty space $= 191100 - 107800 = 83300$ cu. cm [1 mark]
OR
(iii) (b) (A) Area covered by one ball $= 7 \times 7 = 49$ sq. cm [1 mark]
(B) Number of balls in lowermost layer $= 78$ [1 mark]
(ii) Area of painted surface $= 10 \times 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 1540$ sq. cm [1 mark]
(iii) (a) Volume of box $= 42 \times 91 \times 50 = 191100$ cu. cm [1/2 mark]
Volume of $600$ balls $= 600 \times \frac{539}{3} = 107800$ cu. cm [1/2 mark]
Volume of empty space $= 191100 - 107800 = 83300$ cu. cm [1 mark]
OR
(iii) (b) (A) Area covered by one ball $= 7 \times 7 = 49$ sq. cm [1 mark]
(B) Number of balls in lowermost layer $= 78$ [1 mark]