(a) The first term of an Arithmetic Progression (A.P) is 8. The ratio of the 7th term to the 9th term is 5 : 8. Calculate the common difference of the progression.
(b) A sphere of radius 2 cm is of mass 11.2g. Find (i) the volume of the sphere ; (ii) the density of the sphere ; (iii) the mass of a sphere of the same material but with radius 3cm. [Take \(\pi = \frac{22}{7}\)].
(b) A sphere of radius 2 cm is of mass 11.2g. Find (i) the volume of the sphere ; (ii) the density of the sphere ; (iii) the mass of a sphere of the same material but with radius 3cm. [Take \(\pi = \frac{22}{7}\)].
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Correct Answer: Option n
Explanation:
(a) \(T_{n} = a + (n - 1)d\) (terms of an AP)
Given a = -8;
\(T_{7} = a + 6d = -8 + 6d\)
\(T_{9} = a + 8d = -8 + 8d\)
\(\frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}\)
\(5(-8 + 8d) = 8(-8 + 6d)\)
\(-40 + 40d = -64 + 48d\)
\(-40 + 64 = 48d - 40d \times 24 = 8d\)
\( d = 3\)
(b) Given r = 2 cm, m = 11.2g
(i) \(V = \frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 2^{3}\)
= \(\frac{704}{21}\)
= \(33.52 cm^{3} = 33.52 \times 10^{-6} m^{3}\)
= \(3.352 \times 10^{-5} m^{3}\)
(ii) \(Density = \frac{mass}{volume}\)
= \(\frac{11.2}{33.52}\)
= \(0.334 g/cm^{3}\)
(iii) \(V = \frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 3^{3}\)
= \(\frac{2376}{21}\)
= \(113.14 cm^{3}\)
\(mass = density \times volume\)
\(0.334 \times 113.14 = 37.789g\)
(a) \(T_{n} = a + (n - 1)d\) (terms of an AP)
Given a = -8;
\(T_{7} = a + 6d = -8 + 6d\)
\(T_{9} = a + 8d = -8 + 8d\)
\(\frac{-8 + 6d}{-8 + 8d} = \frac{5}{8}\)
\(5(-8 + 8d) = 8(-8 + 6d)\)
\(-40 + 40d = -64 + 48d\)
\(-40 + 64 = 48d - 40d \times 24 = 8d\)
\( d = 3\)
(b) Given r = 2 cm, m = 11.2g
(i) \(V = \frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 2^{3}\)
= \(\frac{704}{21}\)
= \(33.52 cm^{3} = 33.52 \times 10^{-6} m^{3}\)
= \(3.352 \times 10^{-5} m^{3}\)
(ii) \(Density = \frac{mass}{volume}\)
= \(\frac{11.2}{33.52}\)
= \(0.334 g/cm^{3}\)
(iii) \(V = \frac{4}{3} \pi r^{3}\)
= \(\frac{4}{3} \times \frac{22}{7} \times 3^{3}\)
= \(\frac{2376}{21}\)
= \(113.14 cm^{3}\)
\(mass = density \times volume\)
\(0.334 \times 113.14 = 37.789g\)