Solve the following mathematical problems:

প্রশ্নে বলা হচ্ছে যে, 20ms^-1 সেকেন্ডে ঊর্ধ্বমূখী কোন বেলুন হতে একটি পাথর 10 সেকেন্ডে ভূমিকে স্পর্শ করলে তখন ভূমি হতে বেলুনের উচ্চতা কত হবে?

এটি পদার্থবিজ্ঞানের অঙ্ক । অঙ্কটির সমাধান করতে হলে বেশ কয়েকটি বিষয়ের উপর সম্যক ধারণা থাকতে হবে। 

অভিকর্ষজত্বরণ (g) হলো প্রতি সেকেন্ডে পড়ন্ত বস্তুর বেগ যার মান ঊর্ধ্বমুখী হলে হবে g = -9.8ms^-2 এবং বেগ (Velocity) কে u দ্বারা সূচিত করা হবে। আর সময়কে t দ্বারা সূচিত করে, উচ্চতাকে h দ্বারা সূচিত করতে হবে। এখন আসুন অঙ্কের সমাধানে চলে যাইঃ

পাথরের ক্ষেত্রে, $h=utt+\frac{1}{2}g{t}^2$

= $$\begin{gathered} =\left(20\times 10\right)+\frac{1}{2}\times 9.8\times {10}^2 \\ = 200+490 \\ =690m \end{gathered}$$

বেলুনের ক্ষেত্রে, $$\begin{gathered} h=ut-\frac{1}{2}g{t}^2 \\ =\left(20\times 10\right)-\frac{1}{2}\times 9.8\times {10}^2 \\ =200-490 \\ =290m \\ u=20m{s}^{-1}, g=9.8m{s}^{-2}; t=10 second \end{gathered}$$

The total hight of ballon is = (690 +290)m = 980m (answer)

We get the following conditions,$x\ge 0,y\ge 0,x+y=5,x\le 2,y\le 4$
As the maximum value of x is 2, 
So the values of x may be x = 0, 1, 2.
And as the maximum value of y is 4,
So the values of y may be y = 0, 1, 2, 3, 4.
And according to another condition x + y = 5.
This also approves the first condition because here x+y=2+3 = 5. 

That mean's if we take x = 2 and y = 3,
Then x+y=5 equation will be true.
From here, we get the maximum value of x = 2 and
The maximum value of y = 3.
The maximum value of z= 6x+2y 

$$\begin{gathered} \Rightarrow z=\left(6\times 2\right)\left(2\times 3\right) \\ \Rightarrow z=12+6 \\ \therefore z=18 \end{gathered}$$

Let the number be 10X + Y, (Y is in unit place and X is in the tenth place) 

So the sum of the digits will be X + Y if the sum of the digits subtracted from the number than the result will be 10X + Y - (X + Y) = 9X

So the number resulting will be always the product of 9 so if its need to have 6 in unit place

So the value of X must be 4 as e result it will be $9 \times 4 = 36$ here 6 in the unit place so all the two digit number which has 4 in tenth places will be the number

 So total such number will be 10 (40, 41, 42, 43, 44, 45, 46, 47, 48, 49) (ans.)

50 people will be kept in 3 groups where all group are consisted of the number of people equivalent to prime number.

So, 1^st way is = 2 +5+43 = 50 

2^nd way is = 2 +7+41 = 50

3^rd way is = 2 +11 + 37 = 50 

4^th way is = 2 +17 +31 = 50

5^th way is = 2 +19 +29 = 50 

(ans.) 5 ways.

Radius of semi circular sheet, $r = \frac{28}{2}14m$

Circumference of the sheet = $\mathrm{\pi r}=\left(\frac{22}{7}\times 14\right)=44\mathrm{cm}$

Here, circumference of semi circle will be equal to the circumference of the base of cone.

Now, let the radius of the base of cone = R

According to question = $$\begin{gathered} = 2\mathrm{\pi R}=44 \\ =\mathrm{R}=\frac{44}{2\mathrm{\pi }}=\frac{44}{{\displaystyle \frac{22\times 2}{7}}}=44\times \frac{7}{22\times 2}=7\mathrm{cm} \end{gathered}$$

Here, radius of semi circle equals to the height of cone.

That mean's  l = r = 14 cm 

Now, we know $$\begin{gathered} {}^2={\mathrm{h}}^2+{\mathrm{R}}^2 \\ \Rightarrow {\mathrm{h}}^2={\mathrm{l}}^2-{\mathrm{R}}^2 \\ \Rightarrow {\mathrm{h}}^2={\left(14\right)}^2-{\left(7\right)}^2=196-49=147 \\ \Rightarrow \mathrm{h}=\sqrt{49\times 3}=\sqrt{7^2\times 3}=7\sqrt{3} \\ \therefore \mathrm{h}=7\sqrt{3} \end{gathered}$$

So, the depth is $7\sqrt{3}$ cm and 

The capacity will be = $\left(\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}\right)=\left(\frac{1}{3}\times \frac{22}{7}\times 7^2\times 7\sqrt{3}\right)=622.37c{m}^3$ (ans)