A covered wooden box has the inner measures as 115 cm, 75 cm and 35 cm and the thickness of wood is 2.5 cm. Find the volume of the wood :
(A) 81000 cu.cm
(B) 81775 cu.cm
(C) 82125 cu.cm
(D) None of these
Solution:
The external measures of the box are (115 + 5) cm, (75 + 5) cm, and (35 + 5) cm i.e., 120 cm, 80 cm and 40 cm Volume of the wood : = External volume - Internal volume = [(120 × 80 × 40) - (115 × 75 × 35)] cm3 = (384000 - 301875) cm3 = 82125 cm3
13.
A metallic sphere of radius 5 cm is melted to make a cone with base of the same radius. What is the height of the cone ?
(A) 5 cm
(B) 10 cm
(C) 15 cm
(D) 20 cm
Solution:
Let the height of the cone be h cm Then, $$\eqalign{ & \frac{4}{3}\pi \times {\left( 5 \right)^3} = \frac{1}{3}\pi \times {\left( 5 \right)^2} \times h \cr & \Rightarrow h = 20\,cm \cr} $$
14.
What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?
V1, V2, V3 and V4 are the volumes of four cubes of side lengths x cm, 2x cm, 3x cm and 4 cm respectively. Some statements regarding these volumes are given below :
(i) V1 + V2 + 2V3 V4
(ii) V1 + 4V2 + V3 V4
(iii) 2(V1 + V3) + V2 = V4
Which of these statements area correct ?
The radius of a sphere is equal to the radius of the base of a right circular cone, and the volume of the sphere is double the volume of the cone. The ratio of the height of the cone to the radius of its base is :
(A) 2 : 1
(B) 1 : 2
(C) 2 : 3
(D) 3 : 2
Solution:
Let the radius of cone and the sphere be R and the height of the cone be H Volume of sphere $$ = \frac{4}{3}\pi {r^3}$$ Volume of cone $$ = \frac{1}{3}\pi {r^2}h$$ According to given information : $$\eqalign{ & \Rightarrow \frac{4}{3}\pi {R^3} = 2 \times \frac{1}{3}\pi {R^2}H \cr & \Rightarrow 4R = 2H \cr & \Rightarrow \frac{H}{R} = \frac{4}{2}\,Or\,2:1 \cr} $$
18.
What length of solid cylinder 2 cm in diameter must be taken to cast into a hollow cylinder of external diameter 12 cm, 0.25 cm thick and 15 cm long ?
(A) 42.3215 cm
(B) 44.0123 cm
(C) 44.0625 cm
(D) 44.6023 cm
Solution:
External radius = 6 cm Internal radius = (6 - 0.25) = 5.75 cm Volume of material in hollow cylinder : $$\eqalign{ & = \left[ {\frac{{22}}{7} \times \left\{ {{{\left( 6 \right)}^2} - {{\left( {5.75} \right)}^2}} \right\} \times 15} \right]{\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times 11.75 \times 0.25 \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{22}}{7} \times \frac{{1175}}{{100}} \times \frac{{25}}{{100}} \times 15} \right){\text{c}}{{\text{m}}^3} \cr & = \left( {\frac{{11 \times 705}}{{56}}} \right){\text{c}}{{\text{m}}^3} \cr} $$ Let the length of solid cylinder be h Then, $$\eqalign{ & \frac{{22}}{7} \times 1 \times 1 \times h = \left( {\frac{{11 \times 705}}{{56}}} \right) \cr & \Rightarrow h = \left( {\frac{{11 \times 705}}{{56}} \times \frac{7}{{22}}} \right) \cr & \Rightarrow h = 44.0625\,{\text{cm}} \cr} $$
19.
The diameter of a spare is 8 cm. It is melted and drawn into a wire of diameter 3 mm. The length of the wire is :
(A) 36.9 m
(B) 37.9 m
(C) 38.9 m
(D) 39.9 m
Solution:
Let the length of the wire be h Then, $$\eqalign{ & \pi \times \frac{3}{{20}} \times \frac{3}{{20}} \times h = \frac{4}{3}\pi \times 4 \times 4 \times 4 \cr & \Rightarrow h = \left( {\frac{{4 \times 4 \times 4 \times 4 \times 20 \times 20}}{{3 \times 3 \times 3}}} \right)cm \cr & \Rightarrow h = \left( {\frac{{102400}}{{27}}} \right)cm \cr & \Rightarrow h = 3792.5\,cm \cr & \Rightarrow h = 37.9\,m \cr} $$
20.
A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is: