Quantitative Aptitude
VOLUME AND SURFACE AREA MCQs
1 hectare = 10,000 m2
So, Area = (1.5 x 10000) m2 = 15000 m2.
Depth = \(\frac{5}{100}m = \frac{1}{20}m\)
So, Volume = (Area x Depth) = \(\left(15000\times\frac{1}{20}\right)m^{3}=750m^{3.}\)
2(15 + 12) x h = 2(15 x 12)
h = \(\frac{180}{27}m=\frac{20}{3}m.\)
So, volume = \(\left(15\times12\times\frac{20}{3}\right)m^{3} = 1200m^{3}\)
Let the length of the wire be h.
Radius = \(\frac{1}{2}mm=\frac{1}{20}cm. Then,\)
\(\frac{22}{7}\times\frac{1}{20}\times\frac{1}{20}\times h= 66 \)
h = \(\left(\frac{66\times20\times20\times7}{22}\right)=8400cm=84 m.\)
External radius = 4 cm,
Internal radius = 3 cm.
Volume of iron = \(\left(\frac{22}{7}\times\left[(4)^{2}-(3)^{2}\right]\times21
\right)cm^{3}\)
= \(\left(\frac{22}{7}\times7\times1\times21\right)cm^{3}\)
= 462 cm3.
So, Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.
Volume of water displaced = (3 x 2 x 0.01) m3
= (0.06 x 1000) kg
= 60 kg.
Total volume of water displaced = (4 x 50) m3 = 200 m3.
So, Rise in water level = \(\left(\frac{200}{40\times20}\right)m0.25m=25cm.\)
l = 10 m,
h = 8 m.
So, r = l2 - h2 = (10)2 - 82 = 6 m.
So, Curved surface area = \(\pi rl = \left(\pi\times6\times10\right)m^{2}=60\pi m^{2}\)
Area of the wet surface = [2(lb + bh + lh) - lb]
= 2(bh + lh) + lb
= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m2
= 49 m2