In a cylindrical container, circular discs are stacked. The first disc is 1 cm in thickness, the second is 2 cm thick, and consecutive discs increase by 1 cm. If the cylinder is 2 m tall, what is the maximum number of discs that can be completely contained in it?
Answers
Step-by-step explanation:
Circular discs 5 mm thickness, are placed one above the other to form a cylinder of curved surface area 462 sq cm. if the radius is 3.5 cm.
As per given the thickness of disc =5mm=
10
5
cm=0.5cm
Radius of disc =3.5 cm
Cuvered surface area of cylinder =462 sq cm
∴2πr=462cm
2
.......................................(1)
Let number of discs are x
The height of cylinder =h= thicknessofdisc×numberofdiscs=0.5x
∴2πr=2×
7
22
×3.5×05x............................(2)
From equation (1) and (2) we get
∴2πr=2×
7
22
×3.5×05x=462
→x=
2×22×3.5×0.5
462×7
=42 discs
Answer: answer is 19
the given sequence of the discs was in AP(arithmetic progression)
the first term t1= 1 ,t2 =2 ,t3 = 3.... therefore d= tn - tn-1 = 3-2 =1
when we add all the numbers from 1 , we can take the last term where the sum of previous numbers is less than 200
1+2+3+4+5+6+7+8+9+10= 55
55+11 =66 ,66+12= 78 ,153+18=171 ,171 +19 =190 ,190+20= 210> 200
therefore the largest number to be added with the sum of perivious numbers to get the answer less than 200 is 19 ,because when 20 is added to the sum of 1-19 is 210 which is more than 200.
- therefore the number of discs in the cylinder is 19