Question

a bucket is the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. find the capacity and surface area of the bucket. also, find cost of milk which can completely fill the container at the rate of Rs. 25 per litre.

Answer 1

firstly u have to find volume of frustum and after that u have to find slant height .now u have to find out TSA of frustum. At last u have to multiply the the TSA with the cost as given above. 2. 2
volume of frustum=1/3π{R +r +Rr) h

2. 2
TSA of frustum =πl(R+r)+πR+πr

Answer 2

Step-by-step explanation:

AnswEr:-

Volume of the Bucket = $\sf\dfrac{\pi rh}{3}$

$\sf\; r^2_1 + r^2_2 + r_1 r_2$

Given :-

Height = 30 cm

$\sf\; r_1 = 20 \;cm$

$\sf\; r_2 = 10 \; cm$

So, the capacity of the bucket = $\sf\; 3.14 = \dfrac{30}{3}$

$:\implies\sf\; 20^2 + 10^2 + 20 \times\; 10\; cm^3$

$:\implies\sf\; 21980\; cm = 21.980 \; litres$

Now, According to Question :-

Cost of 1 litre of milk = 40 Rs.

Cost of 21980 litres of milk = Rs 21.980 × 40

Rs = 879.20

$\rule{150}2$

Surface area of the bucket :-

Curved surface area of the bucket + Surface area of the bottom

$:\implies\sf\; \pi(r_1 + r_2 ) + \pi \; r_2^2$

Where,

$:\implies\sf\; l = \sqrt{h^2 + (r_1 + r_2)}$

$:\implies\sf\; l = \sqrt{900 + 100\; cm}$

$:\implies\sf\; 31.62 cm$

$\rule{150}2$

Therefore, Surface area of the bucket =

$:\implies\sf\dfrac{22}{7}\times\; 31.62 (20 + 10) + \dfrac{22}{7} \times \; 10^2$

$:\implies\sf\; \dfrac{22}{7} \times 1048.6$

$:\implies\large\boxed{\sf{\red{3295.6 \;cm\;(approx)}}}$

$\rule{150}2$