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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CBSE X Related Questions

  • 1.
    If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

      • \(\frac{7}{2}\)
      • \(\frac{6}{7}\)
      • \(\frac{-7}{6}\)
      • \(\frac{7}{6}\)

    • 2.
      Case Study - 1 : A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled. The diameter of spherical ball is 21 cm. Cylindrical pool - Outer diameter is 50 cm and inner diameter is 40 cm. Height of solid base is 14 cm. Height of water filled is 7 cm. Observe the figure and answer the following questions : (i) Determine the total height of the fountain. (ii) Find the volume of the ball. (iii) (a) If one-third of the ball is submerged in the water, find the volume of the water filled in the pool. OR (iii) (b) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.


        • 3.
          The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

            • \((x + 2), (x - 5)\)
            • \((x + 5), (x - 2)\)
            • \((x + 1), (x - 5)\)
            • \((x - 1), (x + 2)\)

          • 4.
            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

              • \(4 \text{ cm}\)
              • \(4\sqrt{2} \text{ cm}\)
              • \(8 \text{ cm}\)
              • \(2\sqrt{2} \text{ cm}\)

            • 5.
              The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

                • 2
                • \(-6\)
                • \(-2\)
                • 6

              • 6.
                Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
                Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

                  • 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 the Assertion (A).
                  • Assertion (A) is true, but Reason (R) is false.
                  • Assertion (A) is false, but Reason (R) is true.

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