Q)  Adventure camps are the perfect place for the children to practice decision making for themselves without parents and teachers guiding their every move. Some students of a school reached for adventure at Sakleshpur. At the camp, the waiters served some students with a welcome drink in a cylindrical glass and some students in a hemispherical cup whose dimensions are shown below.

Adventure camps are the perfect place for the children to practice decision making for themselves without parents and teachers guiding their every move.

After that they went for a jungle trek. The jungle trek was enjoyable but tiring. As dusk fell, it was time to take shelter. Each group of four students was given a canvas of area 551m2 . Each group had to make a conical tent to accommodate all the four students. Assuming that all the stitching and wasting incurred while cutting, would amount to 1m2 , the students put the tents. The radius of the tent is 7m.

1. The volume of cylindrical cup is:

a) 295.75 cm3.      b) 7415.5 cm3          c) 384.88 cm3.           d) 404.25 cm3

2. The volume of hemispherical cup is:

a) 179.67 cm3.       b) 89.83 cm3         c) 172.25 cm3.        d) 210.60 cm3

3. Which container had more juice and by how much?

a) Hemispherical cup, 195 cm3                b) Cylindrical glass, 207 cm3

c) Hemispherical cup, 280.85 cm3         d) Cylindrical glass, 314.42 cm3

4. The height of the conical tent prepared to accommodate four students is

a) 18m     b) 10m      c) 24m    d) 14m

5. How much space on the ground is occupied by each student in the conical tent

a) 54 m2   b) 38.5 m2      c) 86 m2   d) 24 m2

Ans:

1. Cylindrical cup volume:

The volume of cylindrical cup is = π r 2 h

D_c_y_l = 7 cm, R_c_y_l = \frac{7}{2} cm

H_c_y_l = 10.5 cm

Volume of cylindrical cup =

= \frac {22}{7} \times (\frac{7}{2})^2 \times (10.5)

= 11 \times \frac{7}{2} \times 10.5

= 404.25 cm3

Hence, option d) is correct option.

2. Hemispherical cup volume:

The volume of hemispherical cup is = \frac{2}{3} \pi r^3

D_h_e_m_ = 7 cm, R_h_e_m_ = \frac{7}{2} cm

Volume of hemispherical cup =

= \frac{2}{3} \times \frac {22}{7} \times (\frac{7}{2})^3

= \frac{2}{3} \times 11 \times \frac{7}{2}  \times \frac{7}{2}

= \frac{11 \times 49}{3 \times 2}

= 89.83 cm3

Hence, option b) is correct option.

3. Larger container:

As we can see that the volume of cylindrical cup is more than the hemispherical cup, hence cylindrical cup will hold more juice.

Since the difference in the capacity = capacity of the cylindrical cup – capacity of the hemispherical  cup

= 404.25 – 89.83

= 314.42 cm3

Hence, option d) is correct option.

4. Height of the tent:

Given that the canvas area = 551 m2

Wastage of canvas = 1 m2

Net canvas used for making the tent = original area of canvas – wastage

= 551 – 1 = 550 m2

Since this canvas is used to make the conical tent,

The curved surface area of a conical tent is = \pi r l

550 = \pi \times R_c_o_n \times L_c_o_n

Given that the Radius of the conical tent, R_c_o_n = 7 m

550 = (\frac{22}{7}) \times 7 \times L_c_o_n

L_c_o_n = (\frac{550}{22}) = 25 m

Now, from the adjoining figure: 

(R_c_o_n)^2 + (H_c_o_n)^2 = (L_c_o_n)^2

(H_c_o_n)^2 = (L_c_o_n)^2 - (R_c_o_n)^2

Here we have, R_c_o_n = 7 m, L_c_o_n= 25 m

(H_c_o_n)^2 = 25^2 - 7 ^2 = 625 – 49 = 576

H_c_o_n = 24 m

Therefore, height of the conical tent is 24 m

Hence, option c) is correct option.

5. Ground space by each student in conical tent:

Ground space of the conical tent = area of the base of tent

The base area of a conical tent is : π r2

R_c_o_n = 7 m

The base area of a conical tent =  (\frac{22}{7}) \times (7)^2

= 22 x 7 = 154 cm2

Since the tent can contain 4 students, hence the ground space will be equally shared by the 4 students.

Therefore, space covered by each student= \frac{154}{4}

= 38.5 cm2

Hence, option b) is correct option. 

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