Explanation
Step 1: Set up the ratios for Dimensions
Let the common multiplier for the radii be x and for the heights be y.
Radii (r):
Radius of cylinder (r1) = 1x
Radius of cone (r2) = 2x
Radius of sphere (r3) = 3x
Heights (h):
Height of cylinder (h1) = 5y
Height of cone (h2) = 6y
Height of sphere (h3): Note that a standard geometric sphere's functional height boundary is its diameter (2r3=6x). However, following the literal wording of competitive exam question structures with independent height ratios given as 5:6:7, we use: h3=7y.
Step 2: Use the Volume of the Sphere to find the value of x
The formula for the volume of a sphere is:
Vsphere=34πr33
Given that Vsphere=4851 cm3 and r3=3x:
4851=34×722×(3x)3
4851=34×722×27x3
4851=4×722×9x3
4851=7792x3
Solve for x3:
x3=7924851×7
Dividing both 4851 and 792 by 9:
x3=88539×7
Dividing both 539 and 88 by 11:
x3=849×7=8343
Taking the cube root on both sides:
x=27=3.5 cm
Step 3: Establish the relationship between x and y
For a sphere, its height is equal to its diameter:
h3=2r3
Substitute h3=7y and r3=3x:
7y=2(3x)
7y=6x⟹y=76x
Substitute the value of x=27:
y=76×27=3 cm
Step 4: Calculate the individual dimensions
Cylinder:
r1=1x=3.5 cm
h1=5y=5×3=15 cm
Cone:
r2=2x=2×3.5=7 cm
h2=6y=6×3=18 cm
Step 5: Find the sum of the volumes of the cylinder and cone
Volume of Cylinder (V1):
V1=πr12h1=722×(3.5)2×15
V1=722×12.25×15=22×1.75×15=577.5 cm3
Volume of Cone (V2):
V2=31πr22h2=31×722×(7)2×18
V2=31×722×49×18=22×7×6=924 cm3
Total Combined Volume:
Sum=V1+V2=577.5+924=1501.5 cm3
Correct Answer:
(b) 1501.5
