Bright Tutorials Bright Tutorials
ICSE Class 10 Mathematics Question 13 of 21

Solved 2024 Specimen Paper ICSE Class 10 Mathematics — Question 9

Back to all questions
9
Question

Question 6(iii)

A tent is in the shape of a cylinder surmounted by a conical top. If height and radius of the cylindrical part are 7 m each and the total height of the tent is 14 m. Find the :

(a) quantity of air contained inside the tent.

(b) radius of a sphere whose volume is equal to the quantity of air inside the tent.

Use π=227\pi = \dfrac{22}{7}

Answer

(a) From figure,

Radius of cylindrical part = Radius of conical part = r = 7 m

Height of cylindrical part (H) = 7 m

Height of conical part (h) = 14 - 7 = 7 m.

A tent is in the shape of a cylinder surmounted by a conical top. If height and radius of the cylindrical part are 7 m each and the total height of the tent is 14 m. Find the : ICSE 2024 Maths Specimen Solved Question Paper.

Quantity of air inside the tent = Volume of cylindrical part + Volume of conical part

=πr2H+13πr2h=227×72×7+13×227×72×7=22×49+13×22×49=1078+359.33=1437.33 m3.= πr^2H + \dfrac{1}{3}πr^2h \\[1em] = \dfrac{22}{7} \times 7^2 \times 7 + \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 7 \\[1em] = 22 \times 49 + \dfrac{1}{3} \times 22 \times 49 \\[1em] = 1078 + 359.33 \\[1em] = 1437.33 \text{ m}^3.

Hence, the quantity of air inside the tent = 1437.33 m3.

(b) Let radius of required sphere be R m.

According to question,

Volume of sphere = Quantity of air inside the tent

∴43πR3=πr2H+13πr2h⇒43R3=r2H+13r2h⇒43R3=3r2H+r2h3⇒4R3=3r2H+r2h⇒R3=3×72×7+72×74⇒R3=3×49×7+49×74⇒R3=1029+3434⇒R3=13724⇒R3=343⇒R3=73⇒R=7 m.\therefore \dfrac{4}{3}πR^3 = πr^2H + \dfrac{1}{3}πr^2h \\[1em] \Rightarrow \dfrac{4}{3}R^3 = r^2H + \dfrac{1}{3}r^2h \\[1em] \Rightarrow \dfrac{4}{3}R^3 = \dfrac{3r^2H + r^2h}{3} \\[1em] \Rightarrow 4R^3 = 3r^2H + r^2h \\[1em] \Rightarrow R^3 = \dfrac{3 \times 7^2 \times 7 + 7^2 \times 7}{4} \\[1em] \Rightarrow R^3 = \dfrac{3 \times 49 \times 7 + 49 \times 7}{4} \\[1em] \Rightarrow R^3 = \dfrac{1029 + 343}{4} \\[1em] \Rightarrow R^3 = \dfrac{1372}{4} \\[1em] \Rightarrow R^3 = 343 \\[1em] \Rightarrow R^3 = 7^3 \\[1em] \Rightarrow R = 7 \text{ m}.

Hence, radius of sphere = 7 m.