h =ut+$\frac{1}{2}gt^{2}$
= (20×20)+$\frac{1}{2}$x9.8×10$^{2}$
= 200+490
= 690 m
বেলুনর ক্ষেত্রে,
h = ut+$\frac{1}{2}gt^{2}$
= (20×20)+$\frac{1}{2}$x9.8×10$^{2}$
=200-490
=-290m
= 290m
u=20ms$^{-1}$;g=9.8ms$^{-2}$; t=10 second
so, The total higth of ballon is =(690+290)m = 980m
Question 2:
z = 18.
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 ofx may be x = 0,1,2.
And as the maximum value of y is 4,
So the values ofy may be y = O, 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
=> (6×2)+(2×3)
=> z = l2 + 6
so, z = 18.
so, total such number will be 10(40- 41, 42, 43, 44, 45,46,47, 48,49)
Let the number be 1OX + Y. (Y is in unit place and X is i.” 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 1OX + Y – (X + Y) = 9X
so, the number resulting will be always the product of 9 so if its 5 need to have 6 in unit place
So the value of X must be 4 as e result it will be 9 X 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)
5 ways
50 people will be kept in 3 groups where all group are consisted of the number 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
5 ways
$622.37cm^{3}$
Radius of semi Circular sheet. r = $\frac{28}{2}$= 14 cm
so, Circumference of the sheet = $\pi$r= ( $\frac{22}{7}$x 14) = 44cm
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 = 2$\pi$R = 44
=> R = $\frac{44}{2\pi}=\frac{44}{\frac{22\times }{7}}=44\times \frac{7}{22\times 2}$ = 7cm
Here. radius of semi circle equals to the height of cone.
that mean’s I = r = 14cm
Now. we know I$^{2}$ = h$^{2}$+R$^{2}$
= h$^{2}$ = I$^{2}$ – R$^{2}$
=> h$^{2}$=(14)$^{2}$-(7)$^{2}$=196-49 = 147
=> h = $\sqrt{49\times 3}=\sqrt{7^{2}\times 3}= 7\sqrt{3}$
so, h = $7\sqrt{3}$
So. the depth is $7\sqrt{3}$ cm and
The capacity will be = $(\frac{1}{3}\pi^{2h})=(\frac{1}{3}\times\frac{22}{7}\times 7^{2}\times 7\sqrt{3})=622.37cm^{3}$
