Math, asked by shanthraju19, 7 months ago

the circumference of base of a cylinderical vessel is 132 cm and its outer diameter is 28 cm the length of the pipe is 35 cm find the mass of the pipe if 1cm of woods has a mass of o.6 g answer this follow me​

Answers

Answered by TrishaNikhilJaiswal
2

Answer:

NCERT TEXTBOOK QUESTIONS SOLVED

EXERCISE 13.1 (Page 213)

Q1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required .for snaking the box.

(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs. 20.

Sol: (i) Here, l = 1.5 m, b = 1.25 m

∵ It is open from the top.

∵ Its surface area = [Lateral surface area] + [Base area]

= [2(l + b)h] + [l × b]

= [2(1.50 + 1.25)0.65 m2] + [1.50 × 1.25 m2]

= [2 × 2.75 × 0.65 m2] + [1.875 m2]

= 3.575 m2 + 1.875 m2 = 5.45 m2

∵ The total surface area of the box = 5.45 m2

∴ Area of the sheet required for making the box = 5.45 m2

(ii) Rate of sheet = Rs. 20 per m2

Cost of 5.45 m2 = Rs. 20 × 5.45

⇒ Cost of the required sheet NCERT TEXTBOOK QUESTIONS SOLVED

EXERCISE 13.1 (Page 213)

Q1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required .for snaking the box.

(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs. 20.

Sol: (i) Here, l = 1.5 m, b = 1.25 m

∵ It is open from the top.

∵ Its surface area = [Lateral surface area] + [Base area]

= [2(l + b)h] + [l × b]

= [2(1.50 + 1.25)0.65 m2] + [1.50 × 1.25 m2]

= [2 × 2.75 × 0.65 m2] + [1.875 m2]

= 3.575 m2 + 1.875 m2 = 5.45 m2

∵ The total surface area of the box = 5.45 m2

∴ Area of the sheet required for making the box = 5.45 m2

(ii) Rate of sheet = Rs. 20 per m2

Cost of 5.45 m2 = Rs. 20 × 5.45

⇒ Cost of the required sheet = Rs.109

Answered by Anonymous
0

......

......

....

....

..

...

.

.

.

.

.

.

....

.. ....

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Similar questions