Question

Find the curved surface area of cylindrical Container of curd with one end open whose height and radius 30 cm and 14 cm respectively



Answer
given ,
diameter =84cm
length =120 cm
area covered by roller in one.
Revolution = curved surface area of the road roller.
= 2πrh
​=
$2 \times \frac{22}{7} \times 42 \times 120$
= 2×22×6×120
= 31680 sq cm.
= area covered by roller in 500 revolution.
= 31680 × 500
= 15840000 sq cm.
=
$\frac{15840000}{10000}$
( 10000 sq cm = 1sq m )

cost of levelling per 1 sq m = Rs.
$\frac{75}{100}$
= 1584×75/100 = 118800/100
= Rs₹ :- 1188/-.​

Answer 1

Answer:

Given ,

• diameter =84cm
• length =120 cm

area covered by roller in one.

Revolution = curved surface area of the road roller.

= 2πrh

= $2 \times \dfrac{22}{7} \times 42 \times 120$

= 2×22×6×120

= 31680 sq cm.

= area covered by roller in 500 revolution.

= 31680 × 500

= 15840000 sq cm.

= $\dfrac{15840000}{10000}$

( 10000 sq cm = 1sq m )

cost of levelling per 1 sq m = Rs. $\dfrac{75}{100}$

= 1584×75/100 = 118800/100

= Rs₹ :- 1188/-.

Answer 2

Answer:

Given ,

diameter =84cm

length =120 cm

area covered by roller in one.

Revolution = curved surface area of the road roller.

= 2πrh

= 2 \times \dfrac{22}{7} \times 42 \times 1202×

7

22

×42×120

= 2×22×6×120

= 31680 sq cm.

= area covered by roller in 500 revolution.

= 31680 × 500

= 15840000 sq cm.

= \dfrac{15840000}{10000}

10000

15840000

( 10000 sq cm = 1sq m )

cost of levelling per 1 sq m = Rs. \dfrac{75}{100}

100

75

= 1584×75/100 = 118800/100

= Rs₹ :- 1188/-.

Step-by-step explanation:

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