Answer: Option C. -> 50 < t < 75 mins
:
C
Area=$\pi$r${}^2$=6.25$\pi$
Rate of flow= ($\pi$r${}^2$x1000) mm${}^3$/min
=6250$\pi$ mm${}^2$/min
Volume of cone=$\frac{1}{3}$ $\pi$r${}^2$h = $\frac{1}{3}$ $\pi$ (200)${}^2$x 240 mm${}^3$
=80$\pi$x40000
=3200000$\pi$ mm${}^3$
Thus, total time taken is:
$\frac{\left(Volumeofconeinm{m}^3\right)}{\left(Rateofwaterflowinm{m}^3/min\right)}$ = 51.2min
Thus answer option (c) is correct.

Answer: Option A. -> 128 cm2
:
A

Required Area = Area of triangle AFG + Area of triangle DEF + Area of triangle BCD =
$(\frac{1}{2}\times 6\times 8)+(\frac{1}{2}\times 8\times 6)+(\frac{1}{2}\times 10\times 16)$
= 128 cm${}^2$

Answer: Option A. -> 128 cm2
:
C
Ans: c.

Answer: Option A. -> abc
:
A
Assume sides to be x,y,z. Then xy=a, yz=b, xz=c.

Multiplying: (xyz)${}^2$= abc; =N${}^2$. (Vol=xyz)
Hence Ans: (a)

Answer: Option D. -> (π2) - 1
:
D
Ans: d)
Soln: Consider the double shaded area.

Area: (Area of quadrant) – (Area of triangle).
($\frac{\pi }{4}$)-($\frac{1}{2}$ x 1 x 1). Total shaded area is twice of this
= 2 ($\frac{\pi }{4}$) - ($\frac{1}{2}$ x 1 x 1); = $\frac{\pi }{2}$ -1,

Answer: Option D. -> None of these
:
D
Answer- (d) None of these
Solution:- Calculate the altitude of the equilateral triangle. Then according to the conditions given,
Diameter of the circle circumscribing the equilateral triangle= Altitude of the equilateral triangle + Diameter of the smaller cirlce.
Solve the above equation for the required answer.
Alternate Soln: The triangle’s dimension is used to calculate the radii of the outermost and the second circle.


Radius of the bigger circle: R=$\frac{\left(1.5\right)}{cos30}$; =$\sqrt{3}$ units.
Radius of smaller circle:
r = 1.5 tan 30. = $\frac{\frac{3}{2}}{\sqrt{3}}$ r= $\frac{\sqrt{3}}{2}$.
Radius of smallest circle:
$\frac{\left(\right(radiusofbiggest)-(radiusof{2}^{nd}\left)\right)}{2}$;
=$\frac{\sqrt{3}}{4}$
Hence ans: d (none of these)

Answer: Option A. -> equilateral
:
A
Ans: a) equilateral.
Substitute values and check. 2,2,2 for equilateral; 1,1,2 for isosceles, 3,4,5 for right angled. The equation holds only for equilateral triangles

Answer: Option B. -> √3(r+2r√3)24
:
B
Soln:

The angle shown is 30 degrees. Thus AB=r$\sqrt{3}$ (30-60-90 triangle).
Thus side of a triangle: 2r + 2r$\sqrt{3}$.
Area of the equilateral triangle: $\frac{\sqrt{3}(r+2r\sqrt{3}{)}^2}{4}$Hence Ans: b

Answer: Option B. -> 16AB
:
B
Ans: b) $\frac{1}{6}$AB.

Soln:

Let PB=r=PQ. QR=x. In triangle PRB, RB=2r-x. PR= r+x.
Since PRB is right angled, $(r+x{)}^2=(2r-x{)}^2+r^2$ .
Solving we get x=$\frac{r}{3}$=$\frac{AB}{6}$

Answer: Option C. -> 24π
:
C

From the fig:X = $\sqrt{{13}^2-5^2}$ = 12x Thus, circumference= 24 $\pi$
Hence Ans: c