The diameter of a roller is 250 cm and length is 140 cm.
If it takes 500 complete revolutions to level a play ground,
determine the cost of levelling at the rate of 50 paise per square metre.
rekhagautam109:
you are in 8th class
yess
am i right
ok
so you don't understand 9th class question
physics mein Hamne Focus Pole Concave convex bhi padh liya h
abtk khud nhi smajh payi is question ko
i understand Cause I'm in ICSE... which is One year Faster than CBSE
kuch nhi Practice navigates to Perfection
very good
Answers
Answered by
3
Answer:
ANSWER
Given that r=42cm and h=120cm
Area covered by the roller in one revolution=Curved Surface Area of the road roller.
=2πrh
=2×
7
22
×42×120
31680cm
2
Area covered by the roller in 500 revolutions =31680×500
=15840000cm
2
=
10000
15840000
=1584m
2
[10000cm
2
=1sq.m]
Cost of levelling per 1sq.m = Rs.
100
75
Thus, cost of levelling the play ground =
100
1584×75
= Rs.1188.
thank you
Answered by
4
D= 250 , r= 125cm
length =140cm
circumference= 2πr
500*140cm= 70000cm
so....
area
= 2πrh
= 2*22/7*1.25*1.40
= 44*.20*1.25
= 11 metre *500 revolution
*
50paise /100= $ 0.5 /m²
= 5500 m²*0.5 m²
= $2750 ans...
hlo
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