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251.	The volume of a solid hemispherical object is 19404 cm3. Its total surface area is –1). 4158 cm22). 4923 cm23). 4518 cm24). 4815 cm2
Answer» Total surface area of solid hemisphere = 3πr² And VOLUME of solid hemisphere $(= \frac{2}{3}{\rm{\pi }}{r^3})$ Where, r = RADIUS of the solid hemisphere Given, Volume of solid hemisphere = 19404 ⇒ (2/3)πr³= 19404 ⇒ (2/3) × (22/7) × r³= 19404 ⇒ r³= (19404 × 3 ×7)/(2 × 22) = 9261 ⇒ r = 21 cm ∴ Total surface area = 3πr² = 3 × (22/7) × (21)² cm² = 4158 cm²

251.

The volume of a solid hemispherical object is 19404 cm3. Its total surface area is –1). 4158 cm22). 4923 cm23). 4518 cm24). 4815 cm2

Answer»

Total surface area of solid hemisphere = 3πr²

And VOLUME of solid hemisphere $(= \frac{2}{3}{\rm{\pi }}{r^3})$

Where, r = RADIUS of the solid hemisphere

Given,

Volume of solid hemisphere = 19404

⇒ (2/3)πr³= 19404

⇒ (2/3) × (22/7) × r³= 19404

⇒ r³= (19404 × 3 ×7)/(2 × 22) = 9261

⇒ r = 21 cm

∴ Total surface area = 3πr²

= 3 × (22/7) × (21)² cm²

= 4158 cm²

Discussion

252.	There are 3 sphere having r cm radius which are completely fit (vertical manner) in a cylinder. Those sphere cut out from the cylinder. What the ratio of total surface area of all three sphere to curved surface area of cylinder?1). 1 ∶ 12). 2 ∶ 33). 3 ∶ 14). 4 ∶ 5
Answer» Solution : It is given that all the three SPHERES FIT correctly into the cylinder so, the height of the cylinder will be equal to the sum of DIAMETER of the three spheres, ===> $h = 3 d$ ( 'h' is the height of the cylinder and 'd' is the diameter of the sphere.) We know that $d = 2r$ so, $h = 3×2r = 6r$ $h = 6r$ Total Surface Area of a Sphere $= 4 \PI r^2$ Total Surface Area of 3 Spheres $= 3× 4 \pi r^2 = 12 \pi r^2$ Curved Surface Area of Cylinder $= 2 \pi r h$ Total Surface Area of $3$ Spheres : Curved Surface Area of Cylinder $= 12 \pi r^2 ÷2 \pi r h$ $= 6 r ÷ h$ $= 6 r ÷ 6 r( h = 6r )$ $= 1 : 1$ *So, the CORRECT option is 1).1 : 1*

252.

There are 3 sphere having r cm radius which are completely fit (vertical manner) in a cylinder. Those sphere cut out from the cylinder. What the ratio of total surface area of all three sphere to curved surface area of cylinder?1). 1 ∶ 12). 2 ∶ 33). 3 ∶ 14). 4 ∶ 5

Answer»

Solution :

It is given that all the three SPHERES FIT correctly into the cylinder so, the height of the cylinder will be equal to the sum of DIAMETER of the three spheres,

===> $h = 3 d$ ( 'h' is the height of the cylinder and 'd' is the diameter of the sphere.)

We know that $d = 2r$ so,

$h = 3×2r = 6r$

$h = 6r$

Total Surface Area of a Sphere $= 4 \PI r^2$

Total Surface Area of 3 Spheres $= 3× 4 \pi r^2 = 12 \pi r^2$

Curved Surface Area of Cylinder $= 2 \pi r h$

Total Surface Area of $3$ Spheres : Curved Surface Area of Cylinder

$= 12 \pi r^2 ÷2 \pi r h$

$= 6 r ÷ h$

$= 6 r ÷ 6 r( h = 6r )$

$= 1 : 1$

So, the CORRECT option is 1).1 : 1

Discussion

253.	The perimeter and the breadth of a rectangle are 52 cm and 12 cm respectively. Find its area (in cm2). 1). 842). 3363). 1684). 252
Answer» We know, PERIMETER of the rectangle = 2 × (Length + BREADTH) Area of the rectangle = Length × Breadth Given, perimeter of the rectangle = 52 & Breadth = 12 ⇒ 2 × (Length + Breadth) = 52 ⇒ Length + 12 = 26 ⇒ Length = 14 ∴ Area of the rectangle = 12 × 14 = 168 cm²

253.

The perimeter and the breadth of a rectangle are 52 cm and 12 cm respectively. Find its area (in cm2). 1). 842). 3363). 1684). 252

Answer»

We know, PERIMETER of the rectangle = 2 × (Length + BREADTH)

Area of the rectangle = Length × Breadth

Given, perimeter of the rectangle = 52 & Breadth = 12

⇒ 2 × (Length + Breadth) = 52

⇒ Length + 12 = 26

⇒ Length = 14

∴ Area of the rectangle = 12 × 14 = 168 cm²

Discussion

254.	A trapezium of area 55 cm2 can be divided into a rectangle and two equal triangles. If the parallel sides of the trapezium are of 15 cm and 7 cm, find the area of the triangle.1). 10 cm22). 20 cm23). 25 cm24). 35 cm2
Answer» Area of trapezium = (height of trapezium × sum of parallel sides)/2 ⇒ Height of trapezium = (55 × 2)/(15 + 7) = 110/22 = 5 cm Now, Area of the TRIANGLES = area of trapezium – area of RECTANGLE = 55 – (5 × 7) = 55 – 35 = 20 cm² As the two triangles are equal, area of one triangle = 20/2 = 10 cm²

254.

A trapezium of area 55 cm2 can be divided into a rectangle and two equal triangles. If the parallel sides of the trapezium are of 15 cm and 7 cm, find the area of the triangle.1). 10 cm22). 20 cm23). 25 cm24). 35 cm2

Answer»

Area of trapezium = (height of trapezium × sum of parallel sides)/2

⇒ Height of trapezium = (55 × 2)/(15 + 7) = 110/22 = 5 cm

Now,

Area of the TRIANGLES = area of trapezium – area of RECTANGLE = 55 – (5 × 7) = 55 – 35 = 20 cm²

As the two triangles are equal, area of one triangle = 20/2 = 10 cm²

Discussion

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