A slab of ice of 11 inches in length, 8 inches in breadth and 2 inches in thickness was melted and resolidified into the form of a rod of 8 inches diameter. The length of such a rod is nearest to
Answer: B Since volume remains constant We can say Volume =pie*r2*h 8*11*2 =L*pie*42 i.e L=3.5 inch
Q. No. 14:
The length of the largest possible rod that can be placed in cubical room is 35√3 m. What is the surface area of the largest possible sphere that fit with in the cubical room?
Answer: B The length of the largest possible rod in the cubical room =a√3. So, a√3 = 35√3 => a=35, where a is the side of cube. So, diameter of the sphere that can fit in the cube of side 35 is also 35m. Radius =35/2 So surface area = 4*pie*r2= 4*22/7*35/2*35/2 = 3850 sq m
Q. No. 15:
A large solid sphere is melted and moulded to form identical right circular cones with base radius and height same as the radius of the sphere. One of these cones is melted and moulded to form a smaller solid sphere. What is the ratio of the surface area of the smaller sphere to the surface area of the larger sphere?
Answer: D Let the radius of the original sphere =R and let the number of identical cones =n So, 4/3*pie*R3=n *(pie/3*R2*R
) =>n=4 Volume of smaller sphere = pie/3* R3 Radius of smaller sphere = (R3/4)1/3 = R/22/3 =>Surface area of smaller sphere/ surface area of larger sphere = =>1:24/3
Q. No. 16:
The ratio of volume of a cylinder to that of cone is 25:4. If the heights are interchanged, then the ratio is 100:9. What is the ratio, if only the radius are interchanged?
Answer: B Let the radius of cone = a Height of the cone = b Radius of cylinder = A Height of cylinder = B Given, (1/3 * pie * a*a*b) / (pie * A*A*B) = 4/25 => (a2b)/(3A2B) = 4/25.........(i) If the heights are interchanged, the ratio is, (1/3 * pie* a2B)/(pie*A2b) = 9/100 => (a2B)/(3A2b) = 9/100...........(ii) We need the ratio of the volumes, when radius are interchanged = (A*A*b)/(3*a*a*B). Thus by given two equation, (A*A*b) / (a*a*B) = 100/27. Thus required ratio is 1/3 * 100/27 = 100:81. Hence, the ratio of cylinder : cone = 81:100
Q. No. 17:
Pankaj has finished fencing 2/3 of the area of his rectangular garden that is 10ft wide. When he finishes fencing another 60 sq ft of the garden, he would have finished 3/4 of the area. What is the length of the garden?
Answer: B If the area of the garden is "X", then then difference between 3/4 and 2/3 of the area is 60 sq ft. i.e (3/4 - 2/3) *X = 60 => X = 720 sq ft. Length = Area/ Breadth = 720/10 = 72 ft.
Q. No. 18:
A cylinder inscribed inside a cone having base radius 21 cm and height 21 cm. Find the maximum possible volume of the cylinder?
Answer: A Let the height of the cylinder =h => height of small cone just above the top circular surface of cylinder = (21-h) Since, that small cone and the original complete cone are similar, Height of small cone = Radius of small cone = (21-h) But, radius of small cone = radius of cylinder Thus, Volume of the cylinder = pie * (21-h)*(21-h)* h V= pie/2 * (21-h)*(21-h)*(2h) Sum of all the variables terms in the above expression = 21-h+21-h+2h = constant Thus , product of these terms would be maximum when all the variables terms are equal, since sum is constant V= pie/2 * 42/3 * 42/3 * 42/3 = 4312 sq cm.