NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.2 Solutions

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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volume Exercise 13.2 Solutions are based on Volume of Combination of Solids.

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Read More: NCERT Solutions For Class 10 Maths Chapter 13 Surface Areas and Volume

Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

Also check other Exercise Solutions of Class 10 Maths Chapter 13 Surface Areas and Volume

Also Check:

Chapter Related Topics
Surface Areas And Volumes Revision Notes	Area Of Prism	Volume Of A Rectangular Prism Formula
Volume Of A Square Pyramid Formula	Volume Of A Triangular Prism Formula	Surface Area Of A Prism Formula
Surface Area Of A Pyramid Formula	Surface Area Of A Rectangular Prism Formula	Surface Area Of A Square Pyramid Formula
Surface Area of a Cylinder	Area of Hollow Cylinder	Surface Area of a Prism Formula

Also check:

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Class 10 Maths Notes	Permutations and Combinations	Mensuration
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Math Formula	NCERT Solutions for Class 10 Maths	Algebra

CBSE X Related Questions

1.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
View Solution
2.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
3.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :
- $q - 9p$
- $p - 9q$
- $p + 9q$
- $2p + 9q$
View Solution
4.
If the pair of linear equations : \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) is consistent and dependent, then
- \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
View Solution
5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals
- \(\frac{21}{4} \text{ cm}\)
- \(\frac{28}{3} \text{ cm}\)
- \(\frac{12}{7} \text{ cm}\)
- \(5.5 \text{ cm}\)
View Solution
6.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals
- $x^2 + 5x - 4$
- $(x + 3) (-x + 8)$
- $a(x^2 + 5x - 24)$
- $x^2 - 24$
View Solution

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