NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.5 Solutions

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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volume Exercise 13.5 Solutions are based on all the topics covered in the chapter, surface areas and volumes.

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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

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Chapter Related Topics
Surface Areas And Volumes Revision Notes	Area Of Prism	Volume Of A Rectangular Prism Formula
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CBSE X Related Questions

1.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution
2.
PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is
- \(a^2 + (a + 2)^2 = (2b)^2\)
- \(b^2 = a + 4\)
- \(2a^2 + 1 = b^2\)
- \(b^2 = a + 1\)
View Solution
3.
A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.
View Solution
4.
Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5.
Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution
5.
Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is
- \(2\pi r^3\)
- \(3\pi r^3\)
- \(5\pi r^3\)
- \(4\pi r^3\)
View Solution
6.
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))
- \(314 \sqrt{2}\) \(\text{cm}^{2}\)
- \(314\) \(\text{cm}^{2}\)
- \(\frac{3140}{3}\) \(\text{cm}^{2}\)
- \(3140 \sqrt{2}\) \(\text{cm}^{2}\)
View Solution

View All

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You can also check

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.5 Solutions

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

CBSE X Related Questions

1.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

2.
PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

3.
A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.

4.
Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5.
Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).

5.
Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is

6.
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.5 Solutions

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

CBSE X Related Questions

1.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

2.PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

3.A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.

4.Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5. Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).

5.Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is

6.A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

2.
PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

3.
A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.

4.
Assertion (A) : If probability of happening of an event is \(0.2p\), \(p>0\), then \(p\) can't be more than 5.
Reason (R) : \(P(\bar{E}) = 1 - P(E)\) for an event \(E\).

5.
Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is

6.
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))