NIMCET 2027 Area And Volume Previous Year Questions (PYQs)
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Showing 5 questions
If r is the radius of the circle given below, what is the area of the shaded region?
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Solution
The square is inscribed in the circle.
So the diagonal of the square = diameter of circle = $2r$.
Let side of square be $a$.
Using diagonal formula:
$ a\sqrt{2} = 2r $
$ a = \dfrac{2r}{\sqrt{2}} = r\sqrt{2} $
Area of square:
$ a^2 = (r\sqrt{2})^2 = 2r^2 $
The shaded region is exactly half of the square (a triangular half).
So shaded area =
$ \dfrac{1}{2} \times 2r^2 = r^2 $
The number of solid spheres ,each of diameter 3 cm that could be moulded to form a solid metal cylinder of height 54cm and diameter 4 cm is?
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Solution
Solution:
Diameter of sphere = 3 cm → Radius = 1.5 cm
Diameter of cylinder = 4 cm → Radius = 2 cm
Height of cylinder = 54 cm
Volume of one sphere:
$$ V_s = \frac{4}{3} \pi (1.5)^3 = 4.5\pi\ \text{cm}^3 $$
Volume of cylinder:
$$ V_c = \pi (2)^2 (54) = 216\pi\ \text{cm}^3 $$
Number of spheres:
$$ \text{Number} = \frac{V_c}{V_s} = \frac{216\pi}{4.5\pi} = 48 $$
∴ The required number of spheres = 48.
Diameter of sphere = 3 cm → Radius = 1.5 cm
Diameter of cylinder = 4 cm → Radius = 2 cm
Height of cylinder = 54 cm
Volume of one sphere:
$$ V_s = \frac{4}{3} \pi (1.5)^3 = 4.5\pi\ \text{cm}^3 $$
Volume of cylinder:
$$ V_c = \pi (2)^2 (54) = 216\pi\ \text{cm}^3 $$
Number of spheres:
$$ \text{Number} = \frac{V_c}{V_s} = \frac{216\pi}{4.5\pi} = 48 $$
∴ The required number of spheres = 48.
What is the diameter of the largest circle that can be drawn on a chessboard so that its entire circumference gets covered by the black squares and no part of the circumference falls on any white space, given that the chessboard has black and white squares of size one inch?
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Solution
The largest such circle fits diagonally across black squares only.
The maximum possible diameter comes out to be √10 inches.
From a height of $16$ meters a ball fell down and each time it bounces half the distance back.
What is the total distance traveled?
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Solution
Distance = $16 + 2(8 + 4 + 2 + 1 + \cdots)$
The infinite sum inside equals $8/(1-\frac12)=16$.
Total distance = $16 + 2\times16 = 48$ m.
NIMCET 2027
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and More.
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