Find the volume of iron required to make an open box whose external dimensions are 36 cmx 25 cm x 16.5 cm, the box being 1.5 cm thick throughout. If 1cm of iron weights 8.5 grams, find the weight of the empty box in kilograms.
Answers
Question:
- Find the volume of iron required to make an open box whose external dimensions are 36 cm x 25 cm x 16.5 cm, the box being 1.5 cm thick throughout. If 1cm³ of iron weights 8.5 grams, find the weight of the empty box in kilograms.
Answer:
- Weight of the empty box is 33.660kg
Step-by-step Explanation
Given:
- Outer length of open box =36 cm
- Outer breadth =25 cm
- Outer height =16.5 cm
- Thickness of iron =1.5 cm
To find:
- The weight of the empty box in kilograms
Solution:
∴ Inner length =36−(2×1.5)=36−3=33 cm
Inner breadth =25−(2×1.5)=25−3=22 cm
Inner height =16.5−1.5=15 cm
∴ Volume of iron used in it = Outer volume − Inner volume
=36×25×16.5−33×22×15
=14850−10890
=3960 cm³
Now, weight of 1 cm³=8.5 g
∴ Total weight =3960×8.5 g
=33660 g
=33.660 kg [1 kg=1000 g]
- The weight of the empty box is 33.660 kg.
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Answer:
Step-by-step explanation:
the external dimensions of the rectangular box =36cm*25cm*16.5cm
external volume of the box = 14850 cubic cm
since the thickness of the box is 1.5 cm and box is open.
internal length = (36-1.5*2)cm= 36-3=33 cm
internal breadth = 25-1.5*2=25-3=22 cm
internal height = 16.5-1.5 = 15 cm [since box is open]
therefore the internal dimensions of the rectangular box= 33cm*22cm*15cm
internal volume of the box = 10890 cubic cm
therefore the volume of the iron = external volume of the box - internal volume of the box
= 14850-10890
=3960 cubic cm
since 1 cubic cm of iron weighs 8.5 grams.
therefore weight of 3960 cubic cm = 3960*8.5=33660 gm
i.e. the weight of the empty box is 33.66 kg.