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NCERT Exemplar · Class 10 Mathematics Surface Areas and Volumes

62 questions · 62 still being checked

EXERCISE 12.4 1–10 (part 6 of 7)

  1. Exercise 1

    A solid metallic hemisphere of radius 8\displaystyle 8 cm is melted and recasted into a right circular cone of base radius 6\displaystyle 6 cm. Determine the height of the cone.

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    NCERT’s answer
    28.$\displaystyle 44$ cm
    Melting conserves volume, so the hemisphere's volume equals the cone's. \[V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(8)^3 = \frac{1024\pi}{3} \text{ cm}^3 \] \[V_{\text{cone}} = \frac{1}{3}\pi R^2h = \frac{1}{3}\pi(6)^2h = 12\pi h \] \[12\pi h = \frac{1024\pi}{3} \implies h = \frac{1024}{36} = \frac{256}{9} \]Answer: \(\displaystyle h = \dfrac{256}{9} \approx 28.44\) cm

  2. Exercise 2

    A rectangular water tank of base 11 m×6 m\displaystyle 11 \mathrm{~m} \times 6 \mathrm{~m} contains water upto a height of 5\displaystyle 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5\displaystyle 3.5 m, find the height of the water level in the tank.

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    NCERT’s answer
    8.$\displaystyle 6$ m
    The water's volume is unchanged by the transfer. \[V_{\text{water}} = 11\times6\times5 = 330 \text{ m}^3 \] \[V_{\text{cylinder}} = \pi r^2h = \frac{22}{7}(3.5)^2h = 38.5h \] \[38.5h = 330 \implies h = \frac{330}{38.5} = \frac{60}{7} \]Answer: \(\displaystyle h = \dfrac{60}{7} \approx 8.57\) m

  3. Exercise 3

    How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36\displaystyle 36 cm, 25\displaystyle 25 cm and 16.5\displaystyle 16.5 cm provided the thickness of the iron is 1.5\displaystyle 1.5 cm. If one cubic cm of iron weighs 7.5\displaystyle 7.5 g, find the weight of the box.

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    NCERT’s answer
    $\displaystyle 3960$ \(\displaystyle cm^{3}\), $\displaystyle 29.7$ kg
    Iron fills the walls and base; the top is open, so only the base subtracts a thickness from the height. \[V_{\text{ext}} = 36\times25\times16.5 = 14850 \text{ cm}^3 \] \[l' = 36-2(1.5)=33,\quad b' = 25-2(1.5)=22,\quad h' = 16.5-1.5=15 \] \[V_{\text{int}} = 33\times22\times15 = 10890 \text{ cm}^3 \] \[V_{\text{iron}} = 14850-10890 = 3960 \text{ cm}^3 \] \[\text{Weight} = 3960\times7.5 = 29700 \text{ g} = 29.7 \text{ kg} \]Answer: \(\displaystyle 3960\) cm\(\displaystyle ^3\) of iron; weight \(\displaystyle 29.7\) kg

  4. Exercise 4

    The barrel of a fountain pen, cylindrical in shape, is 7\displaystyle 7 cm long and 5\displaystyle 5 mm in diameter. A full barrel of ink in the pen 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?

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    NCERT’s answer
    480000 words
    A full barrel writes 3300 words; scale that rate to the bottle's volume. \[V_{\text{barrel}} = \pi r^2h = \frac{22}{7}(0.25)^2(7) = 1.375 \text{ cm}^3 \] \[V_{\text{bottle}} = \frac{1}{5}\text{ L} = 200 \text{ cm}^3 \] \[\text{Words} = \frac{200}{1.375}\times3300 = \frac{1600}{11}\times3300 = 480000 \]Answer: \(\displaystyle 480000\) words

  5. Exercise 5

    Water flows at the rate of 10m/minute through a cylindrical pipe 5\displaystyle 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40\displaystyle 40 cm and depth 24\displaystyle 24 cm?

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    NCERT’s answer
    $\displaystyle 51$ minutes $\displaystyle 12$ sec
    Flow rate is the pipe's cross-section times speed; time is the vessel's volume divided by that rate. \[\text{Rate} = \pi r^2v = \frac{22}{7}(0.25)^2(1000) = \frac{1375}{7} \text{ cm}^3/\text{min} \] \[V_{\text{cone}} = \frac{1}{3}\pi R^2H = \frac{1}{3}\cdot\frac{22}{7}(20)^2(24) = \frac{70400}{7} \text{ cm}^3 \] \[t = \frac{70400/7}{1375/7} = \frac{70400}{1375} = 51.2 \text{ min} \]Answer: \(\displaystyle 51.2\) min \(\displaystyle = 51\) min \(\displaystyle 12\) s

  6. Exercise 6

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

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    NCERT’s answer
    \(\displaystyle 74.25 \mathrm{~m}^3, 80.61 \mathrm{~m}^2\)
    Canvas covers only the cone's curved surface. \[V = \frac{1}{3}\pi r^2h = \frac{1}{3}\cdot\frac{22}{7}(4.5)^2(3.5) = 74.25 \text{ m}^3 \] \[l = \sqrt{r^2+h^2} = \sqrt{20.25+12.25} = \sqrt{32.5} \approx 5.7 \text{ m} \] \[\text{CSA} = \pi rl = \frac{22}{7}(4.5)(5.7) \approx 80.61 \text{ m}^2 \]Answer: Volume \(\displaystyle \approx74.25\) m\(\displaystyle ^3\); canvas \(\displaystyle \approx80.61\) m\(\displaystyle ^2\)

  7. Exercise 7

    A factory manufactures 120000\displaystyle 120000 pencils daily. The pencils are cylindrical in shape each of length 25\displaystyle 25 cm and circumference of base as 1.5\displaystyle 1.5 cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs 0.05\displaystyle 0.05 per dm2\displaystyle \mathrm{dm}^2.

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    NCERT’s answer
    Rs $\displaystyle 2250$
    Curved surface equals base circumference times length, so the radius is never needed. \[\text{CSA}_{\text{one}} = 1.5\times25 = 37.5 \text{ cm}^2 \] \[\text{CSA}_{\text{total}} = 37.5\times120000 = 4500000 \text{ cm}^2 = 45000 \text{ dm}^2 \] \[\text{Cost} = 45000\times0.05 = 2250 \]Answer: Rs \(\displaystyle 2250\)

  8. Exercise 8

    Water is flowing at the rate of 15\displaystyle 15 km/h through a pipe of diameter 14\displaystyle 14 cm into a cuboidal pond which is 50\displaystyle 50 m long and 44\displaystyle 44 m wide. In what time will the level of water in pond rise by 21\displaystyle 21 cm?

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    NCERT’s answer
    $\displaystyle 2$ hours
    Speed of flow \(\displaystyle v = 15\ \text{km/h} = 15000\ \text{m/h}\); pipe radius \(\displaystyle r = 7\ \text{cm} = 0.07\ \text{m}\). \[\text{Flow rate} = \pi r^2 v = \frac{22}{7} \times (0.07)^2 \times 15000 = 231\ \text{m}^3/\text{h} \] Volume needed to raise the pond by \(\displaystyle 0.21\ \text{m}\): \[V = 50 \times 44 \times 0.21 = 462\ \text{m}^3 \] \[t = \frac{462}{231} = 2\ \text{h} \] Answer: $\displaystyle 2$ hours.

  9. Exercise 9

    A solid iron cuboidal block of dimensions 4.4 m×2.6 m×1 m\displaystyle 4.4 \mathrm{~m} \times 2.6 \mathrm{~m} \times 1 \mathrm{~m} is recast into a hollow cylindrical pipe of internal radius 30\displaystyle 30 cm and thickness 5\displaystyle 5 cm. Find the length of the pipe.

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    NCERT’s answer
    $\displaystyle 112$ m
    Volume is conserved when the block is recast into the pipe. \[V_{\text{cuboid}} = 440 \times 260 \times 100 = 11{,}440{,}000\ \text{cm}^3 \] Internal radius \(\displaystyle r = 30\ \text{cm}\), thickness \(\displaystyle 5\ \text{cm}\), so external radius \(\displaystyle R = 35\ \text{cm}\). \[\text{Cross-section} = \pi(R^2 - r^2) = \frac{22}{7}(35^2 - 30^2) = \frac{22}{7}\times 325 = \frac{7150}{7}\ \text{cm}^2 \] \[\text{Length} = \frac{V_{\text{cuboid}}}{\text{Cross-section}} = 11{,}440{,}000 \div \frac{7150}{7} = 11200\ \text{cm} = 112\ \text{m} \] Answer: $\displaystyle 112$ m.

  10. Exercise 10

    500\displaystyle 500 persons are taking a dip into a cuboidal pond which is 80\displaystyle 80 m long and 50\displaystyle 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m3\displaystyle 0.04 \mathrm{~m}^3?

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    NCERT’s answer
    0.$\displaystyle 5$ cm
    Displaced volume for $\displaystyle 500$ persons: \[V = 500 \times 0.04 = 20\ \text{m}^3 \] Rise in level over the pond's base area: \[h = \frac{V}{l \times b} = \frac{20}{80 \times 50} = \frac{20}{4000} = 0.005\ \text{m} \] \[h = 0.5\ \text{cm} \] Answer: $\displaystyle 0.5$ cm.