#everydayquiz #10Questions
.If the ratio of diagonals of two cubes is 4 : 3, then the ratio of the surface areas of the two cubes respectively is :
(a) 7 : 9
(b) 16 : 9
(c) 64 : 27
2.If a hemispherical dome has an inner radius 14m, then its volume (in m^3) is :
(a) 5749. 33m3
(b) 5749m3
(c) 5740m3
(d) None of these
3.A circus tent is cylindrical up to a height of 3 m and conical above it. If its diameter is 105m and the slant height of the conical part is 63m, then the total area of the canvas required to make the tent is
(a) 11385m^2
(b) 10395m^2
(c) 9900m^2
(d) 990m^2
4.A spherical lead ball of radius 10cm is melted and small lead balls of radius 5 mm are made. The total numbers of possible small lead balls is :
(a) 8000
(b) 400
(c) 800
(d) 125
5.A water tank is 30m long, 20m wide and 12m deep. It is made of iron sheet which is 3m wide. The tank is open at the top. If the cost of iron sheet is Rs. 10per meter. Find the total cost of iron required to build the tank?
(a)Rs.6000
(b)Rs.5000
(c)Rs. 5500
(d)Rs. 5800
6.A right angled triangle with its sides 5cm, 12cm and 13cm is revolved about the side 12cm. Find the volume of the solid formed?
(a)942cm^3
(b)298cm^3
(c)314cm^3
(d)302cm^3
7.A cube of edge 6 cm is painted on all sides and then cut into unit cubes. The number of unit cubes with no sides painted is :
(a) 0
(b) 64
(c) 186
(d) 108
8.Three spherical balls or radii 1 cm, 2 cm and 3 cm are melted to form a single spherical ball. In the process, the loss of material is 25%. The radius of the new ball is:
(a) 6 cm
(b) 5 cm
(c) 3cm
(d) 2cm
9.If a sphere radius r is divided into four identical parts, then the total surface area of the four parts is :
(a) 4πr^2 square units
(b) 2πr^2 square units
(c) 8πr^2 square units
(d) 3πr^2 square units
10.The radius of a cylinder is 10 cm and height is 4 cm. The number of centimetres that may be added either to the radius or to the height to get the same increase in the volume of the cylinder is :
(a) 5
(b) 4
(c) 25
(d) 16
ANSWERS AND SOLUTION :
1.(b) Ratio of Surface area = (ratio of diagonls)^2
= (4/3)^2 = 16/9
2.(a) V = 2/3 πr^3 = 2/3*22/7 (14)^3=5749.33m^3
3.(a) Total area of the canvas = 2πrh+ πrl
= πr (2h+l)
= 22/7 * 105/2 (2 * 3 + 63)
= 22/7 * 105/2 * 69 = 11385 sq. metre
4.(a) Volume of bigger ball 4/3 πr^3
= 4/3 π ×10×10 ×10 cu.cm
Volume of smaller ball = 4/3 π(0.5)^3
= 4/3 π(0.5)^3
so Probable number of smaller balls
= (4/3 π ×10 ×10 ×10)/(4/3 π ×0.5 × 0.5 ×0.5)= 8000
5.(a)
6.(c)
7.(b) Volume of bigger cube = 6 * 6 * 6
= 216 cu. Cm
Volume of unit cube = 1 * 1 * 1
= 1 cu.cm
Number of uncoloured cubes
= 4 * 4 * 4
= 64, because edge of uncoloured cube = 4 cm
8.(c)Volume of the new ball
= 3/4 * 4/3 π(r1^3+ r2^3+ r3^3 )
= π(1^3+ 2^3+ 3^3 )
= π(1+ 8+ 27) = 36πcubic cm
so 4/3 πr^3 = 36π ⇒ r^3 = (36 ×3)/4 =27
so r = 3 cm
9.(c) Required total surface area
= 4πr^2+4 × πr^2=8πr^2 sq.units
10.(a) Let radius be increased by x cm. Volume of cylinder
= π × 10^2 (4+x)
∴ π(10+x)^2 × 4 = π(10)^2 (4+x)
=> (10+x)^2 = 25 (4+x)
=> 100 + 20x + x^2 = 100 + 25x
=>x^2 – 5x = 0
=> x(x -5) = 0
x = 5cm 1.(b) Ratio of Surface area = (ratio of diagonls)^2
= (4/3)^2 = 16/9
2.(a) V = 2/3 πr^3 = 2/3*22/7 (14)^3=5749.33m^3
3.(a) Total area of the canvas = 2πrh+ πrl
= πr (2h+l)
= 22/7 * 105/2 (2 * 3 + 63)
= 22/7 * 105/2 * 69 = 11385 sq. metre
4.(a) Volume of bigger ball 4/3 πr^3
= 4/3 π ×10×10 ×10 cu.cm
Volume of smaller ball = 4/3 π(0.5)^3
= 4/3 π(0.5)^3
so Probable number of smaller balls
= (4/3 π ×10 ×10 ×10)/(4/3 π ×0.5 × 0.5 ×0.5)= 8000
5.(a)
6.(c)
7.(b) Volume of bigger cube = 6 * 6 * 6
= 216 cu. Cm
Volume of unit cube = 1 * 1 * 1
= 1 cu.cm
Number of uncoloured cubes
= 4 * 4 * 4
= 64, because edge of uncoloured cube = 4 cm
8.(c)Volume of the new ball
= 3/4 * 4/3 π(r1^3+ r2^3+ r3^3 )
= π(1^3+ 2^3+ 3^3 )
= π(1+ 8+ 27) = 36πcubic cm
so 4/3 πr^3 = 36π ⇒ r^3 = (36 ×3)/4 =27
so r = 3 cm
9.(c) Required total surface area
= 4πr^2+4 × πr^2=8πr^2 sq.units
10.(a) Let radius be increased by x cm. Volume of cylinder
= π × 10^2 (4+x)
∴ π(10+x)^2 × 4 = π(10)^2 (4+x)
=> (10+x)^2 = 25 (4+x)
=> 100 + 20x + x^2 = 100 + 25x
=>x^2 – 5x = 0
=> x(x -5) = 0
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