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Surface Areas and Volumes

Short Answer Type

341. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

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342. Find the volume of a sphere whose surface area is 154 cm².

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343.

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of र 498.96. If the cost of white-washing is र2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

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344.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the

(i) radius r´ of the new sphere,
(ii) ratio of S and S’.

(i) Volume of a solid iron sphere

equals 4 over 3 πr cubed

∴ Volume of 27 solid iron spheres =

27 open parentheses 4 over 3 πr cubed close parentheses equals 36 πr cubed

∴ Volume of the new sphere = 36πr³ Let the radius of the new sphere be r’. Then,

Volume of the new sphere

equals 4 over 3 πr apostrophe cubed

According to the question,

4 over 3 πr apostrophe cubed equals 36 πr cubed

rightwards double arrow space space space space space space space space space space space space space space space space space straight r apostrophe cubed equals fraction numerator left parenthesis 36 πr cubed right parenthesis cubed over denominator 4 straight pi end fraction rightwards double arrow space space space space space space space space space space space space space space space space space straight r apostrophe cubed equals 27 straight r cubed rightwards double arrow space space space space space space space space space space space space space space space space space straight r apostrophe equals left parenthesis 27 straight r cubed right parenthesis to the power of bevelled 1 third end exponent rightwards double arrow space space space space space space space space space space space space space space space space space straight r apostrophe equals left parenthesis 3 cross times 3 cross times 3 straight r cubed right parenthesis to the power of begin inline style bevelled 1 third end style end exponent rightwards double arrow space space space space space space space space space space space space space space space space space space straight r apostrophe equals 3 straight r

Hence, the radius r’ of the new-sphere is 3r.
(ii) S = 4πr² S’ = 4π(3r)²

therefore space space space fraction numerator straight S over denominator straight S apostrophe end fraction equals fraction numerator 4 πr squared over denominator 4 straight pi left parenthesis 3 straight r right parenthesis squared end fraction equals 1 over 9 equals 1 space colon space 9

Hence, the ratio of S and S’ is 1 : 9.

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345. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

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346. If the number of square centimetres on the surface of a sphere is equal to the number of cubic centimetres in its volume, what is the diameter of the sphere?

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347. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

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348. Find the volume of a sphere whose surface area is 55.44 cm².

open parentheses straight pi equals 22 over 7 close parentheses

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349.

A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.

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350. The outer diameter of a spherical shell is 10 cm and the inner diameter is 8 cm. Find the volume of the metal contained in the shell.

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Previous Year Papers

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Test Yourself

Short Answer Type

344. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r´ of the new sphere,(ii) ratio of S and S’.

344.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the

(i) radius r´ of the new sphere,
(ii) ratio of S and S’.