# NCERT Exemplar Solution for CBSE Class 10 Mathematics: Surface Areas and Volumes Exercise 12.4 (I)

Here you will get CBSE Class 10 Mathematics chapter 12, Surface Areas and Volumes: NCERT Exemplar Problems and Solutions. This part brings solutions to questions from exercise 12.4 that includes only the Long Answer Type Questions. every question has been provided with a detailed solution.

*Class 10 Maths NCERT Exemplar Solutions*

Here you get the CBSE Class 10 Mathematics chapter 12, Surface Areas and Volumes: NCERT Exemplar Problems and Solutions. This part of the chapter includes solutions of Question Number 1 to 10 from Exercise 12.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Surface Areas and Volumes. This exercise comprises only the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

**CBSE Class 10 Mathematics Syllabus 2017-2018**

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Surface Areas and Volumes:**

**Exercise 12.4**

**Long Answer Type Questions (Q. No. 1-10)**

**Question. 1 **A solid metallic hemisphere of radius 8 cm is melted and recasted into a right circular cone of Base radius 6cm. Determine the height of the cone.

**Solution.**

Hence, the required height of cone, *h *= 28.44cm

**Question. 2 **A rectangular water tank of base 11m × 6m contains water upto a height of 5m. If the water in the tank is transferred to a cylindrical tank of radius 3.5m, find the height of the water level in the tank.

**Solution.**

Given, height of water in a length, rectangular tank of base dimensions 11m and 6m is 5m.

Therefore, volume of the water in rectangular tank = 11 ×6 × 5= 330m^{3}

Also, given radius of the cylindrical tank, *r* = 3.5m

Let height of water level in cylindrical tank be *h.*

Hence, the height of water level in cylindrical tank is 8.6m.

**Question. 3 **How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36cm, 25cm and 16.5cm provided the thickness of the iron is 1.5cm. If one cubic centimetre of iron weights 7.5g, then find the weight of the box.

**Solution.**

Given, external dimensions of a box are 36cm × 25cm × 16.5cm.

Therefore, external volume of an open box = 36 × 25 × 16.5 = 14850cm^{3}

**Question. ****4 **The barrel of a fountain pen, cylindrical in shape, is 7cm long and 5mm in diameter. A full barrel of ink in the pin is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth of a litre?

**Solution.**

Given, length of the barrel of a fountain pen = 7cm

**Question. ****5 **Water flows at the rate of 10m min^{-}^{1} through a cylindrical pipe 5m in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40cm and depth 24cm?

**Solution.**

Given, speed of water flow = 10m min^{-}^{1} = 1000 cm/min

**Question. ****6 **A heap of rice is in the form of a cone of diameter 9m and height 3.5m. Find the volume of the rice. How much canvas cloth is required to just cover heap?

**Solution.**

Given, height of the conical heap of rice, *h *= 3.5m

And diameter of the conical heap of rice = 9m

∴ Radius of the conical heap of rice, *r *= 9/2 m

Hence, 80.61 m^{2} canvas cloth is required to just cover the heap of rice.

**Question.**** 7 **A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25cm and circumference of base as 1.5cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs.0.05 per dm^{2}.

**Solution.**

Given, length of one pencil which is cylindrical in shape, *h* = 25cm

And circumference of base of pencil = 1.5 cm

**Question. ****8 **Water is flowing at the rate of 15kmh^{−1 }through a pipe of diameter 14cm into a cuboidal pond which is 50m long and 44m wide. In what time will the level of water in pond rise by 21cm?

**Solution.**

Given, speed of water flowing through the pipe, *h* = 15kmh^{−1 }= (15 × 1000) = 15000 mh^{-}^{1}

Length of the pond, *l*= 50m

And, width of the pond, *b* = 44m

And, level upto which water is to be filled, *h* = 21 cm = 21/100 m

**Question. ****9 **A solid iron cuboidal block of dimensions 4.4m × 2.6m × 1m is recast into a hollow cylindrical pipe of internal radius 30cm and thickness 5cm. Find the length of the pipe.

**Solution.**

Here, volume of a solid iron cuboidal block = 4.4m × 2.6m × 1m = 11.44m^{3}

Also, internal radius of hollow cylindrical pipe, *r*_{1} = 30cm = 0.3m

And thickness of hollow cylindrical pipe = 5cm = 0.05m

So, external radius of hollow cylindrical pipe, *r*_{2 }=* r*_{1}* *+ Thickness = 0.3 + 0.05 = 0.35m

Let length of hollow cylindrical pipe be *h* m.

**Question.**** 10 **500 persons are taking a dip into a cuboidal pond which is 80m Long and 50m broad. What is the rise

of water level in the pond, if the average displacement of the water by a person is 0.04m^{3 }?

**Solution.**

Let the rise of water level in the pond be *h*m*, *when 500 persons are taking a dip into a cuboidal pond.

Given, length of the cuboidal pond = 80m

And breadth of the cuboidal pond = 50m

Now, volume for the rise of water level in the pond = Length × Breadth ´ Height

= 80 × 50 ´ *h *= 4000*h*m^{3}

Given that average displacement of the water by a person = 0.04m^{3}

So, the average displacement of the water by 500 persons = 500 ´ 0.04m^{3}

So, according to the given condition,

Volume for the rise of water level in the pond = Average displacement of the water by 500 persons

Hence, the required rise of water level in the pond is 0.5cm.

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