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Exercise 12.2 (Revised) - Chapter 12 Surface Areas And Volumes class 10 ncert solutions Maths - SaraNextGen [2024-2025]


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On April 24, 2024, 11:35 AM

Exercise 12.2 (Revised) : Chapter 12 - Surface Areas And Volumes - Ncert Solutions class 10 - Maths

Question 1:

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2385/Chapter%2013_html_m238f9bd7.jpg

Given that,

Height (h) of conical part = Radius(r) of conical part = 1 cm

Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm

Volume of solid = Volume of conical part + Volume of hemispherical part

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2385/Chapter%2013_html_m394c3ace.gif

 

Question 2:

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. if each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2392/Chapter%2013_html_m26eb7a3c.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2392/Chapter%2013_html_11caa0f9.jpg

From the figure, it can be observed that

Height (h1) of each conical part = 2 cm

Height (h2) of cylindrical part = 12 − 2 × Height of conical part

= 12 − 2 ×2 = 8 cm

Radius (r) of cylindrical part = Radius of conical part = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2392/Chapter%2013_html_1cf0473f.gif

Volume of air present in the model = Volume of cylinder + 2 × Volume of cones

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2392/Chapter%2013_html_m1d1c33e3.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2392/Chapter%2013_html_mc26851.gif https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2392/Chapter%2013_html_36ca26cb.gif

 

Question 3:

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see the given figure).

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2398/Chapter%2013_html_m26eb7a3c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2398/Chapter%2013_html_m1f158c32.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2398/Chapter%2013_html_m65839d23.jpg

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2398/Chapter%2013_html_2cf420e.gif

Length of each hemispherical part = Radius of hemispherical part = 1.4 cm

Length (h) of cylindrical part = 5 − 2 × Length of hemispherical part

= 5 − 2 × 1.4 = 2.2 cm

Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2398/Chapter%2013_html_m7053252.gif

Volume of 45 gulab jamuns = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2398/Chapter%2013_html_32e84575.gif = 1,127.25 cm3

Volume of sugar syrup = 30% of volume

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2398/Chapter%2013_html_m4ccb7c98.gif

Question 4:

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboids are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the following figure). https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2401/13.3.4_html_m26eb7a3c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2401/13.3.4_html_1000b396.jpg

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2401/13.3.4_html_12df006e.jpg

Depth (h) of each conical depression = 1.4 cm

Radius (r) of each conical depression = 0.5 cm

Volume of wood = Volume of cuboid − 4 × Volume of cones

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2401/13.3.4_html_41df1d9a.gif

 

Question 5:

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2405/Chapter%2013_html_344e94b4.jpg

Height (h) of conical vessel = 8 cm

Radius (r1) of conical vessel = 5 cm

Radius (r2) of lead shots = 0.5 cm

Let n number of lead shots were dropped in the vessel.

Volume of water spilled = Volume of dropped lead shots

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2405/Chapter%2013_html_m2f747c5d.gif

Hence, the number of lead shots dropped in the vessel is 100.

Question 6:

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. [Use π = 3.14]

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2407/Chapter%2013_html_m39d7c35f.jpg

From the figure, it can be observed that

Height (h1) of larger cylinder = 220 cm

Radius (r1) of larger cylinder = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2407/Chapter%2013_html_2e3f287f.gif = 12 cm

Height (h2) of smaller cylinder = 60 cm

Radius (r2) of smaller cylinder = 8 cm

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2407/Chapter%2013_html_m65d8eba1.gif

Mass of 1https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2407/Chapter%2013_html_m77e20f4e.gif iron = 8 g

Mass of 111532.8https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2407/Chapter%2013_html_m77e20f4e.gif iron = 111532.8 × 8 = 892262.4 g = 892.262 kg

Question 7:

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2408/Chapter%2013_html_m26eb7a3c.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2408/Chapter%2013_html_3a11acc.jpg

Radius (r) of hemispherical part = Radius (r) of conical part = 60 cm

Height (h2) of conical part of solid = 120 cm

Height (h1) of cylinder = 180 cm

Radius (r) of cylinder = 60 cm

Volume of water left = Volume of cylinder − Volume of solid

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2408/Chapter%2013_html_m2b7d99a3.gif

Question 8:

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter o the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2420/Chapter%2013_html_mb06ff9d.jpg

Height (h) of cylindrical part = 8 cm

Radius (r2) of cylindrical part = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2420/Chapter%2013_html_m4101345d.gif  cm

Radius (r1) spherical part = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2420/Chapter%2013_html_m586d361c.gif

Volume of vessel = Volume of sphere + Volume of cylinder

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/140/2420/Chapter%2013_html_m375f3d94.gif

Hence, she is wrong.

Also Read : Exercise-12.1-(Revised)-Chapter-12-Surface-Areas-And-Volumes-class-10-ncert-solutions-Maths

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