RD Sharma 2022 Mcqs Solutions for Class 10 Maths Chapter 14 - Surface Areas And Volumes

Answer:

The diameter of a sphere = 6 cm

Then radius of a sphere

The diameter of a wire = 2 mm

Then radius of wire

Now,

Volume of sphere = volume of wire

Here,

r = radius

l = length of wire

To remove the decimal from base we should multiply both numerator and denumerator by 100,

We get,

Hence, the correct answer is choice (c).

Answer:

Radius of metallic sphere = 10.5 cm

Therefore,

Volume of the sphere

Now,

Radius of the cone = 3.5 cm

and Height of the cone = 3 cm

Therefore,

Volume of the cone

Number of cone

Dividing eq. (i) and (ii) we get

Number of cone

Number of cone = 126

Hence, the correct answer is choice (b).

Answer:

Let r be the radius of the base and h be the height of conical part.

Since,

Surface area of both part of solid is equal.

i.e.,

But,

Squaring on both side,

Then we get,

From equation (i) putting the value of l in above equation

Hence, the correct answer is choice (b).

Answer:

Volume of sphere = volume of the cone

Hence, the correct answer is choice (a).

Answer:

Given: Height of the cylinder (H)= 6 cm
Let the radius and height of the cone be r cm and h cm respectively.
Since a metalic solid cone is melted to form a solid cylinder of equal radius.
So, volume of cone = volume of the cylinder
$$\begin{gathered} \Rightarrow \frac{1}{3}\mathrm{\pi }{r}^{2}h=\mathrm{\pi }{r}^{2}H \\ \Rightarrow \frac{1}{3}h=H \\ \Rightarrow \frac{1}{3}h=6 \\ \Rightarrow h=18 \mathrm{cm} \end{gathered}$$

Thus,  the height of the cone was 18 cm.
Hence, the correct answer is option (c).

 

Answer:

Let the radius of the cylinder be r cm.
Curved surface area of cylinder = Area of rectangular sheet
$$\begin{gathered} \Rightarrow 2\mathrm{\pi }r\left(40\right)=40\times 22 \\ \Rightarrow 2\times \frac{22}{7}\times r\times 40=40\times 22 \\ \Rightarrow r=3.5 \mathrm{cm} \end{gathered}$$

Hence, the correct answer is option A.

Answer:

Given: Height of solid metal cylinder (h) = 45 cm
Diameter of â€‹solid metal cylinder (d) = 4 cm
Radius of â€‹solid metal cylinder (r) = $\frac{d}{2}=\frac{4}{2}=2 \mathrm{cm}$

Diameter of â€‹solid spheres (D) = 6 cm
Radius of solid spheres (R) = $\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}$

Let the number of solid spheres be n.

Volume of cylinder = volume of n spheres
$$\begin{gathered} \mathrm{\pi }{r}^{2}h=n\times \frac{4}{3}\mathrm{\pi }{R}^3 \\ \Rightarrow {\left(2\right)}^2\times 45=n\times \frac{4}{3}{\left(3\right)}^3 \\ \Rightarrow 4\times 45=n\times 4\times 9 \\ \Rightarrow n=5 \end{gathered}$$

The number of solid spheres is 5.

Hence, the correct answer is option is (b).




 

Answer:

Volume of the spheres are in the ratio 64 : 27. 
$$\begin{gathered} \Rightarrow V_1:V_2=64:27 \\ \Rightarrow \frac{4}{3}{\mathrm{\pi r}}_1^3:\frac{4}{3}{\mathrm{\pi r}}_2^3=64:27 \\ \Rightarrow {\mathrm{r}}_1^3:{\mathrm{r}}_2^3=64:27 \\ \Rightarrow {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^3=\frac{64}{27} \\ \Rightarrow {\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^3={\left(\frac{4}{3}\right)}^3 \end{gathered}$$
Thus, ratio of the radii = 4 : 3
Ratio of the surface area of the spheres will be
$$\begin{gathered} \frac{4{\mathrm{\pi r}}_1^2}{4{\mathrm{\pi r}}_2^2} \\ ={\left(\frac{{\displaystyle r_1}}{{\displaystyle r_2}}\right)}^2={\left(\frac{4}{3}\right)}^2=\frac{16}{9} \end{gathered}$$
So, the ratio is 16 : 9.
Hence the correct answer is option (d).

Answer:

Since h > 2r where h is the height of the cylinder and r is the radius 
So, when a sphere is enclosed in it, the radius of the sphere will be r. 
Thus, the diameter of the sphere will be 2r.
Hence, the correct answer is option (b)

Answer:

When a plane parallel to the base of a cone cuts it, then a frustum and a smaller cone is formed.
The cross-section thus formed will be a circle.
Hence, the correct answer is option (a).

Answer:

Base radius of the hemisphere = r
Since the two hemispheres are joined end to end, it becomes a complete sphere. 
Curved surface area of the new solid = total surface area of the sphere.
$\mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{new} \mathrm{solid}=4{\mathrm{\pi r}}^2$

Hence, the correct answer is option (a).

Answer:

The bucket is in the form of a frustum.
The diameters are respectively, $d_1=44 cm and d_2=24 cm$
Radii of the circular ends = $r_1=22 cm and r_2=12 cm$
$$\begin{gathered} Volume, V=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_1^2+{\mathrm{r}}_1{\mathrm{r}}_2+{\mathrm{r}}_2^2\right) \\ \mathrm{V}=\frac{1}{3}\mathrm{\pi }\times 35\left({22}^2+22\times 12+{12}^2\right) \end{gathered}$$
$$\begin{gathered} V=32706.6 c{m}^3 \\ =32.7 liters \end{gathered}$$
Hence, the correct answer is option (a)

Answer:

Radius of the bigger sphere = r cm
Radius of smaller spheres = r₁ cm
$$\begin{gathered} \frac{Volume of bigger sphere}{Volume of small spheres}=\frac{{\displaystyle \frac{4}{3}{\mathrm{\pi r}}^3}}{\frac{4}{3}\pi r_1^3}=\frac{{\displaystyle r^3}}{r_1^3}=8 \\ \Rightarrow {\left(\frac{r}{r_1}\right)}^3={\left(\frac{2}{1}\right)}^3 \\ \Rightarrow \left(\frac{{\displaystyle r}}{{\displaystyle r_1}}\right)=\left(\frac{{\displaystyle 2}}{{\displaystyle 1}}\right) \end{gathered}$$
Hence, r : r₁ = 2 : 1.
Hence, the correct answer is option (a).

Answer:

The radius of cylindrical pipe

The volume per minute of water flow from the pipe

The radius of cone

Depth of cone = 24 cm

The volume of cone

The time it will take to fill up a conical vessel

Hence, the correct answer is choice (b).

Answer:

Volume of sand filled in cylindrical vessel

Clearly,

The volume of conical heap = volume of sand

Hence, the correct answer is choice (c).

Answer:

Height,

The C.S.A. of cone

Hence, the correct answer is choice (d).

Answer:

Radius of cone VAOB

r = 4 cm

Height of cone VAOB

h = 3 cm

The volume of cone VAOB

Hence, the correct answer is choice (a).

Answer:

The C.S.A. of cylinder

S = 264 m²

The volume of cylinder

V = 924 m³

From eq. (i) and (ii),

We get

Putting the value in (i)

Hence, the correct answer is choice (b).

Answer:

Volume of cylinder

Let r be the radius of cone

But,

The volume of cone = volume of cylinder

Hence, Radius of cone = 8 cm.

Hence, the correct answer is choice (d).

Answer:

Ist sphere

…… (i)

IInd sphere

…… (ii)

Divide (i) by (ii) we get,

Now, the ratio of their C.S.A

Hence,

Hence, the correct answer is choice (d).

Answer:

Let r be the radius of single sphere.

Now,

The volume of single sphere = sum of volume of three spheres

Hence, the diameter = 20 × r = 24 cm

Hence, the correct answer is choice (b).

Answer:

Let r be the radius of sphere

But,

Surface area of sphere = C.S.A. of cylinder

Hence, the correct answer is choice (c).

Answer:

The radius of greatest sphere cut off from cylindrical log of wood should be radius of cylindrical log.

i.e., r = 1 cm

The volume of sphere

Hence, the correct answer is choice (a).

Answer:

The radius of sphere, r = 3 cm

The volume of sphere

$$\begin{gathered} =\frac{4}{3}{\mathrm{\pi r}}^3 \\ =\frac{4}{3}\mathrm{\pi }{\left(3\right)}^3 \\ =36\mathrm{\pi } {\mathrm{cm}}^3 \end{gathered}$$

Since,

The sphere fully immersed into the vessel, the level of water be raised by x cm.

Then,

The volume of raised water = volume of sphere

Hence, the correct answer is choice (c).

Answer:

The volume of solid cylinder = 12 × volume of one sphere

The required diameter d = 2 × 2 = 4 cm

Hence, the correct answer is choice (d).

Answer:

Clearly,

The volume of recasted cone = volume of sphere

Hence, the correct answer is choice (b).

Answer:

External radius 

Internal radius 

The volume of hollow sphere

$$\begin{gathered} V=\frac{4}{3}\mathrm{\pi }\left({\mathrm{R}}^3-{\mathrm{r}}^3\right) \\ =\frac{4}{3}\mathrm{\pi }\left(4^3-2^3\right) \end{gathered}$$

Let h be the height of cone.

Clearly,

The volume of recasted cone = volume of hollow sphere

$$\begin{gathered} \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{4}{3}\mathrm{\pi }\left(4^3-2^3\right) \\ \Rightarrow 4^2\mathrm{h}=4\left(4^3-2^3\right) \\ \Rightarrow \mathrm{h}=14 \mathrm{cm} \end{gathered}$$

Hence, the height of cone = 14 cm

Hence, the correct answer is choice (b).

Answer:

The volume of iron piece = 49 × 33 × 24 cm³

Let, r is the radius sphere.

Clearly,

The volume of sphere = volume of iron piece

Hence, the correct answer is choice (a).

Answer:

The ratio of lateral surface to the total surface area of cylinder

Hence, the correct answer is choice (b).

Answer:

Disclaimer: In the the question, the statement given is incorrect. Instead of total height of solid being equal to 3 times the volume 
of cone, the volume of the total solid should be equal to 3 times the volume of the cone.

Let x be the height of cylinder.

Since, volume of the total solid should be equal to 3 times the volume of the cone, 
So,
 $$\begin{gathered} \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}+{\mathrm{\pi r}}^2\mathrm{x}=3\left(\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}\right) \\ \Rightarrow \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}-{\mathrm{\pi r}}^2\mathrm{h}+{\mathrm{\pi r}}^2\mathrm{x}=0 \\ \Rightarrow {\mathrm{\pi r}}^2\mathrm{x}=\frac{2}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ \Rightarrow \mathrm{x}=\frac{2}{3}\mathrm{h} \end{gathered}$$
 

Hence, the height of cylindrical part

Hence, the correct answer is choice (d).

Answer:

Radius of hemisphere = r

Therefore,

The radius of cone = r

and height h = r

Then,

Volume of cone

Hence, the correct answer is choice (b).

Answer:

Given that

and

Then,

The ratio of C.S.A. of cylinders

Hence, the correct answer is choice (a).

Answer:

Radius of cylinder = r

Height = h

= 2r

Since, the sphere fitted the cylinder.

i.e., diameter of sphere = height of cylinder.

Hence, the correct answer is choice (c).

Answer:

Radii of circular ends of frustum

Slant height

{squaring on both sides}

Hence, the correct answer is choice (a).

Answer:

Since,

are similar triangles,

i.e., In

$$\begin{gathered} \frac{OA}{O\text{'}L}=\frac{OV}{O\text{'}V} \\ \Rightarrow \frac{r_1}{r_2}=\frac{h_1}{h_1-h_2} \\ \Rightarrow \left(h_1-h_2\right)r_1=h_1{r}_2 \end{gathered}$$
$$\begin{gathered} \Rightarrow r_1{h}_1-r_1{h}_2=h_1{r}_2 \\ \Rightarrow r_1{h}_1-h_1{r}_2=r_1{h}_2 \\ \Rightarrow h_1\left(r_1-r_2\right)=r_1{h}_2 \\ \Rightarrow \frac{\left(r_1-r_2\right)}{r_1}=\frac{h_2}{h_1} \\ \Rightarrow \frac{\left(r_1-r_2\right)}{r_1}=\frac{1}{2} \\ \Rightarrow 1-\frac{r_2}{r_1}=\frac{1}{2} \\ \Rightarrow \frac{r_2}{r_1}=1-\frac{1}{2}=\frac{1}{2} \end{gathered}$$

Thus, $r_2:r_1=1:2$

Hence, the correct answer is choice (b).

Answer:

Slant height = 10 cm

Total lateral surface area

= 260 $\mathrm{\pi } {\mathrm{cm}}^2$

Hence, the correct answer is choice (c).

Answer:

Hence, the correct answer is choice (c).

Answer:

Radius of top of bucket r₁ = 20 cm

Radius of bottom of bucket r₂ = 8 cm

Height of bucket = 16 cm

The curved surface area of bucket

C.S.A. of bucket

= $1760 c{m}^2$

Hence, the correct answer is choice (a).

Answer:

Radius of top of bucket

Radius of bottom of bucket

Height of bucket, h = 24 cm.

C.S.A. of the bucket

= 2750 cm²

Area of bottom

The cost of painting its C.S. ,

Hence, the correct answer is choice (c).

Answer:

Let r be the radius of cylinder.

Area of circular base of cylinder

The height of cylinder h = 8 cm

The C.S.A. of cylinder

Clearly,

The diameter of cylinder

Hence, the correct answer is choice (b).

Answer:

Let the radius and height of the original cylinder be R and h, respectively.
Now, the radius of the new cylinder = $\frac{R}{2}$
Then, the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is given by
$$\begin{gathered} \mathrm{\pi }{\left(\frac{R}{2}\right)}^{2}h :\mathrm{\pi } R^{2}h \\ =\frac{1}{4} : 1 \\ =1 : 4 \end{gathered}$$

Hence, the correct answer is option C.

Answer:

A cone is converted into a cone.

So,

Volume of cone = Volume of cylinder

Hence, the correct answer is choice (b).

Answer:

For conical portion

Curved surface area of the conical portion

For cylindrical portion we have

Then,

Curved surface area of cylindrical portion

Area of canvas used for making the tent

Hence, the correct answer is choice (d).

Answer:

Here,

Diameter of sphere = 6 cm

Radius of sphere

Volume of the sphere

Now,

Diameter of cylinder = 4 cm

^{Radius of cylinder}

Height of the cylinder = 45 cm

Then,

Volume of the cylinder

The number of solid sphere

The number of solid sphere is 5.

Hence, the correct answer is choice (c).

Answer:

Radius of the sphere = 6 cm.

Volume of the sphere

and

Radius of the cylinder = 8 cm

Volume of the cylinder

Therefore,

Volume of the sphere = volume of the cylinder

or

Hence, the correct answer is choice (a).

Answer:

Height of the bucket = 40 cm

Radius of the upper part of bucket = 35 cm

R₁ = 35 cm and

R₂ = 14 cm

The volume of the bucket

Hence, the correct answer is choice (b).

Answer:

Since,

Therefore,

In and

$\frac{O\text{'}V}{OV}=\frac{O\text{'}C}{OA}$

The ratio of the volume of upper part and the cone,

From eq. (i) and (ii),

We get,

Hence, the correct answer is choice (d).

Answer:

Let VAB be cone of height 30 cm and base radius r₁ cm.

Suppose it is cut off by a plane parallel to the base at a height h₂ from the base of the cone.

Clearly $∆VOD~∆VO\text{'}B$

Therefore,

$$\begin{gathered} \frac{OV}{O\text{'}V}=\frac{OD}{O\text{'}B} \\ \Rightarrow \frac{h_1}{30}=\frac{r_2}{r_1} \end{gathered}$$

But,

$$\begin{gathered} Volume of cone VCD=\frac{1}{27}Volume of cone VAB \\ \Rightarrow \frac{1}{3}\mathrm{\pi }{\left({\mathrm{r}}_2\right)}^2{\mathrm{h}}_1=\frac{1}{27}\left(\frac{1}{3}\mathrm{\pi }{\left({\mathrm{r}}_1\right)}^{2}30\right) \\ \Rightarrow {\left(\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\right)}^2{\mathrm{h}}_1=\frac{10}{9} \\ \Rightarrow {\left(\frac{{\mathrm{h}}_1}{30}\right)}^2{\mathrm{h}}_1=\frac{10}{9} \\ \Rightarrow {\mathrm{h}}_1=10 \end{gathered}$$

Hence,

Required height 

Hence, the correct answer is choice (c).

Answer:

Let r be the radius of the base of solid.

Clearly,

The volume of solid = 3 × volume of cone

Vol. of cone + Vol. of cylinder = 3 Volume of cone

Vol. of cylinder = 2 Vol. of cone

Thus,

The height of cylinder

Hence, the correct answer is choice (b).

Answer:

The volume of reservoir

The volume of reservoir = 176 m²

Hence, the correct answer is choice (a).

Answer:

(i) The surface area of the front face of the victory stand = ar (ABCM) + ar (DMNE) + ar (NFGH) 


The surface area of the front face of the victory stand = 50 × 12 +​ 50 × 40 + â€‹ 50 × 24
= 50 × (12 +​ 40 + â€‹24)
​= 50 × (76)
= 3800 cm² 
Hence, the correct answer is option (d).


(ii) The total surface area of the victory stand = TSA of  position 3+ TSA of  position 1 + TSA of  position 2 − 2 × ( overlapping area of position 3 and 1) − 2 × ( overlapping area of position 2 and 1) 
= 2 × (50 × 12 + 12 × 40 + 50 × 40) â€‹+ 2 × (50 × 40 + 40 × 40 + 50 × 40) + 2 × (50 × 40 + 40 × 24 + 50 × 24) − 2 × (40 × 12 ) − 2 × (40 × 24)
= 2 × [(600 + 480 + 2000) â€‹+ (2000 + 1600 + 2000) + (2000 + 960 + 1200)] − (2 × 40) × 36
= 2 × (12840) − 2880
​= 25680 − 2880
=22880 cm² 
Hence, the correct answer is option (b).


(iii) The volume of the box designated for the player ranking third in the competition = volume of cuboid of position 3
= (50 × 40 × 12)                              (∵ Volume of cuboid = l × b × h)
= 24000 cm³
Thus, the volume of the box designated for the player ranking third in the competition  is 24000 cm³.

Hence, the correct answer is option (b).


(iv) The volume of the box designated for the player ranking second in the competition = volume of cuboid of position 2
= (50 × 40 × 24)                              (∵ Volume of cuboid = l × b × h)
= 48000 cm³

Thus, the volume of the box designated for the player ranking second in the competition  is 48000 cm³.
Hence, the correct answer is option (a).


(v) The volume of the box designated for the winner = volume of cuboid of position 1
= 50 × 40 × (16 + 24)                               (∵ Volume of cuboid = l × b × h)
= 50 × 40 × 40
= 80000 cm³

Thus, the volume of the box designated for the winner is 80000 cm³.
Hence, the correct answer is option (c).

Answer:

(i) The width of the tunnel = AB
Radius of the circle (r) = $5\sqrt{2} \mathrm{m}$

$$\begin{gathered} \mathrm{In} △\mathrm{AOB}, \mathrm{by} \mathrm{using} \mathrm{pythagoras} \mathrm{Theorem} \\ A{B}^2=A{O}^2+B{O}^2 \\ \Rightarrow A{B}^2={\left(5\sqrt{2}\right)}^2+{\left(5\sqrt{2}\right)}^2 \\ \Rightarrow A{B}^2=2{\left(5\sqrt{2}\right)}^2 \\ \Rightarrow A{B}^2=100 \\ \Rightarrow AB=10 \mathrm{m} \end{gathered}$$

Thus, the width of the tunnel is 10 m.
Hence, the correct answer is option (b).

(ii) Drawing a perpendicular OM on AB from the centre of the circle O.
Since, the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AM = MB and ∠OMB =90º



$$\begin{gathered} \mathrm{In} △\mathrm{OMB}, \mathrm{by} \mathrm{using} \mathrm{pythagoras} \mathrm{Theorem} \\ O{B}^2=M{O}^2+B{M}^2 \\ \Rightarrow {\left(5\sqrt{2}\right)}^2=M{O}^2+{\left(5\right)}^2 \\ \Rightarrow 50=M{O}^2+25 \\ \Rightarrow M{O}^2=50-25 \\ \Rightarrow M{O}^2=25 \\ \Rightarrow MO=5 \end{gathered}$$

Thus, the height of the tunnel is 5 m.
Hence, the correct answer is option (c).

(iii) Radius of the circle (r) = $5\sqrt{2} \mathrm{m}$
$\begin{array}{rcl}\mathrm{\theta } & =& 360°-90°\\ & =& 270°\end{array}$
The perimeter of cross-section of the tunnel = $2\pi r\times \frac{\theta }{360°}+AB$
$$\begin{gathered} =2\times \mathrm{\pi }\times 5\sqrt{2}\times \frac{270°}{360°}+10 \\ =2\times \mathrm{\pi }\times 5\sqrt{2}\times \frac{3}{4}+10 \\ =\frac{15\sqrt{2}}{2}\mathrm{\pi }+10 \\ \mathit{=}\left(\frac{15}{\sqrt{2}}\mathrm{\pi }+10\right) \mathrm{m} \end{gathered}$$

Thus, the perimeter of cross-section of the tunnel is $\left(\frac{15\mathrm{\pi }}{\sqrt{2}}+10\right) \mathrm{m}$.
Hence, the correct answer is option (b).

(iv)  Radius of the circle (r) = $5\sqrt{2} \mathrm{m}$​
$\begin{array}{rcl}\mathrm{\theta } & =& 360°-90°\\ & =& 270°\end{array}$
The area of cross-section of the tunnel = $\frac{1}{2}\times r^2\times \left[\left(\frac{\mathrm{\pi }}{180°}\right)\theta -\sin \theta \right]$
$$\begin{gathered} =\frac{1}{2}\times {\left(5\sqrt{2}\right)}^2\times \left[\left(\frac{\mathrm{\pi }}{180°}\right)270°-\sin \left(270°\right)\right] \\ =\frac{1}{2}\times 50\times \left[\left(\frac{\mathrm{\pi }}{2}\right)3-\left(-1\right)\right] \left[\because \sin (270°) = \sin (90° + 180°) = - \sin 90° = - 1\right] \\ =25\times \left[\frac{3\mathrm{\pi }}{2}+1\right] \\ =\left(\frac{75\mathrm{\pi }}{2}+25\right) {\mathrm{m}}^2 \end{gathered}$$

Thus, the perimeter of cross-section of the tunnel is $\left(\frac{75\mathrm{\pi }}{2}+25\right){\mathrm{m}}^2$.
Hence, the correct answer is option (a).

(v) The length of the tunnel = 9 km = 9000 m
The perimeter of circular face of the tunnel = $\left(\frac{15\mathrm{\pi }}{\sqrt{2}}\right) \mathrm{m}$

The surface area of the inner circular face = length of the tunnel × perimeter of cross-section of the tunnel 
$$\begin{gathered} =\left(\frac{15\mathrm{\pi }}{\sqrt{2}}\right)\mathrm{m}\times 9000 \mathrm{m} \\ =\frac{15\mathrm{\pi }}{2}\mathrm{m}\times \sqrt{2}\times 9000 \mathrm{m} \\ =67500\sqrt{2}\mathrm{\pi } {\mathrm{m}}^2 \end{gathered}$$

Thus, the surface area of the inner circular face is $67500\sqrt{2}\mathrm{\pi } {\mathrm{m}}^2$.
Hence, the correct answer is option (d).

Answer:

(i) Diameter of cylinder (d) = 42 cm
Radius of cylinder (r) = $\frac{r}{2}=\frac{42}{2}$ = 21 cm

The outer surface area of the hemispherical part = $2\pi r^2$
$$\begin{gathered} =2\times \frac{22}{7}\times 21\times 21 \\ =2772 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the outer surface area of the hemispherical part is 2772 cm².
Hence, the correct answer is option (d).

(ii) Height of the cylinder (h) = 63 − 21 = 42 cm 

The outer surface area of the cylindrical part of the vessel = $2\pi rh$
$$\begin{gathered} =2\times \frac{22}{7}\times 21\times 42 \\ =5544 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the outer surface area of the cylindrical part of the vessel is 5544 cm².
Hence, the correct answer is option (a).

(iii) The volume of the hemi-spherical part of the vessel = $\frac{2}{3}{\mathrm{\pi r}}^3$
$$\begin{gathered} =\frac{2}{3}\times \frac{22}{7}\times {\left(21\right)}^3 \\ =19404 {\mathrm{cm}}^3 \end{gathered}$$

Thus, the volume of the hemi-spherical part of the vessel is 19404 cm³.
Hence, the correct answer is option (b).

(iv) The volume of the cylindrical portion of the vessel = $\pi r^{2}h$
$$\begin{gathered} =\frac{22}{7}\times {\left(21\right)}^2\times 42 \\ =58212 {\mathrm{cm}}^3 \end{gathered}$$

Thus, the volume of the cylindrical portion of the vessel is 58212 cm³.
Hence, the correct answer is option (d).

(v) The total surface area of the vessel = CSA of hemisphere + CSA of cylinder
$$\begin{gathered} =2{\mathrm{\pi r}}^2+2\mathrm{\pi rh} \\ =2772 +5544 \\ =8316 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the  total surface area of the vessel is 8316 cm².
Hence, the correct answer is option (a).

Answer:

(i) Height of frustum (h) = OO' = 24 cm
Radius of top of frustum (R) = A'O' = 22.5 cm
Radius of bottom of frustum (r) = AO = 12.5 cm

Let l be the slant height of the frustum.
$$\begin{gathered} l=\sqrt{h^2+{\left(R-r\right)}^2} \\ =\sqrt{{\left(24\right)}^2+{\left(22.5-12.5\right)}^2} \\ =\sqrt{576+{\left(10\right)}^2} \\ =\sqrt{576+100} \\ =\sqrt{676} \\ =26 \mathrm{cm} \end{gathered}$$

Thus, the slant height of the bucket is 26 cm.
Hence, the correct answer is option (b).

(ii) Curved surface area of the frustum = $\mathrm{\pi l}\left(\mathrm{R}+\mathrm{r}\right)$
$$\begin{gathered} =\frac{22}{7}\times 26\times \left(22.5+12.5\right) \\ =\frac{22}{7}\times 26\times \left(35\right) \\ =2860 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the curved surface area of the frustum is 2860 cm².
Hence, the correct answer is option (a).

(iii)  Height of cylinderical base (h₁) = 6 cm
Radius of cylinderical base (r) = OA = 12.5 cm
Curved surface area of the cylindrical base = $2\mathrm{\pi }r{h}_1$
$$\begin{gathered} =2\times \frac{22}{7}\times 12.5\times 6 \\ =\frac{3300}{7} \\ =471.42 {\mathrm{cm}}^2 \left(\mathrm{Approximately}\right) \end{gathered}$$

Thus, the curved surface area of the cylindrical base is 471.42 cm².
Hence, the correct answer is option (c).

(iv) Total surface area of the metallic sheet used to make the bucket = CSA of cylinderical base + CSA of frustrum + CSA of circular bottom
$$\begin{gathered} =2{\mathrm{\pi rh}}_1+\mathrm{\pi l}\left(\mathrm{R}+\mathrm{r}\right)+{\mathrm{\pi r}}^2 \\ =471.42+2860+\frac{22}{7}\times {\left(12.5\right)}^2 \left[\mathrm{From} \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right)\right] \\ =471.42+2860+\frac{3437.5}{7} \\ =471.42+2860+491.07 \\ =3822.49 {\mathrm{cm}}^2 \\ \approx 3822.50 {\mathrm{cm}}^2 \left(\mathrm{Approximately}\right) \end{gathered}$$

Thus, the total surface area of the metallic sheet used to make the bucket is 3822.5 cm².
Hence, the correct answer is option (d).

(v) The volume of the water that the bucket can hold = Volume of the frutrum

$$\begin{gathered} =\frac{\mathrm{\pi h}}{3}\left(R^2+Rr+r^2\right) \\ =\frac{1}{3}\times \frac{22}{7}\times 24\times \left[{\left(22.5\right)}^2+\left(22.5\right)\left(12.5\right)+{\left(12.5\right)}^2\right] \\ =\frac{22}{7}\times 8\times \left[506.25+281.25+156.25\right] \\ =\frac{22}{7}\times 8\times 943.75 \\ =\frac{166100}{7} \\ =23728.57 {\mathrm{cm}}^3 \\ =\frac{23728.57}{1000} \mathrm{litres} \left(\because 1 {\mathrm{cm}}^3=1 \mathrm{litre}\right) \\ =23.72857 \mathrm{litres} \\ \approx 23.728 \mathrm{litres} \left(\mathrm{Approximately}\right) \end{gathered}$$

Thus, the volume of water that the bucket can hold is 23.728 litres.

Hence, the correct answer is option (b).

Answer:

(i) Length of the cuboid (l) = 15 cm
 Breadth of the cuboid (b) = 10 cm
Height of the cuboid (h) = 3.5 cm
The volume of the cuboid = l × b × h 
= 15 × 10 × 3.5
= 525 cm³

Thus, the volume of the cuboid is 525 cm³.
Hence, the correct answer is option (c).

(ii) Diameter of each conical depression (d) = 1 cm
∴ Radius of each conical depression (r) = $\frac{d}{2}=\frac{1}{2}=0.5 \mathrm{cm}$
Height of each conical depression (h₁) = 1.4 cm
The volume of the conical depression = $\frac{1}{3}\mathrm{\pi }{r}^2{h}_1$
$$\begin{gathered} =\frac{1}{3}\times \frac{22}{7}\times {\left(\frac{1}{2}\right)}^2\times 1.4 \\ =\frac{1}{3}\times \frac{22}{7}\times \frac{1}{4}\times 1.4 \\ =\frac{1.1}{3} \\ =\frac{11}{30}{\mathrm{cm}}^3 \end{gathered}$$

Thus, the volume of the conical depression is $\frac{11}{30}$cm³.
Hence, the correct answer is option (a).

(iii) The volume of the wood in the stand = Volume of the cuboid − 4 times the volume of the conical depression 
$$\begin{gathered} =\left(l\times b\times h\right)-4\times \left(\frac{1}{3}\pi r^{\mathit{2}}h\right) \\ =525-4\times \left(\frac{11}{30}\right) \left[\mathrm{From} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right)\right] \\ =525-1.467 \\ =523.5326 \\ \approx 523.53 {\mathrm{cm}}^3 \end{gathered}$$

Thus, the volume of the wood in the stand is 523.53 cm³.
Hence, the correct answer is option (b).

(iv)  The surface area of the cuboid = TSA of cuboid
$$\begin{gathered} =2\left(lb+bh+hl\right) \\ =2\times \left[\left(15\times 10\right)+\left(10\times 3.5\right)+\left(3.5\times 15\right)\right] \\ =2\times \left(150+35+52.5\right) \\ =2\times 237.5 {\mathrm{cm}}^2 \\ =475 {\mathrm{cm}}^2 \end{gathered}$$

Thus, the surface area of the cuboid is 475 cm².
Hence, the correct answer is option (d).

(v) Let l be the slant height of each conical depression.
Radius of each conical depression (r) = $\frac{d}{2}=\frac{1}{2}=0.5 \mathrm{cm}$
Height of each conical depression (h₁) = 1.4 cm
$$\begin{gathered} l=\sqrt{{\left(r\right)}^2+{\left(h_1\right)}^2} \\ \Rightarrow l=\sqrt{{\left(0.5\right)}^2+{\left(1.4\right)}^2} \\ \Rightarrow l=\sqrt{0.25+1.96} \\ \Rightarrow l=\sqrt{2.21} \\ \Rightarrow l=1.48 \mathrm{cm} \end{gathered}$$

The surface area of four conical cavities = $4\times \pi rl$
$$\begin{gathered} =4\times \frac{22}{7}\times \frac{1}{2}\times 1.48 \\ =\frac{130.24}{14} \\ =9.30 {\mathrm{cm}}^2 \\ \approx 9.28 {\mathrm{cm}}^2 \left(\mathrm{Approximately}\right) \end{gathered}$$

Thus, the surface area of four conical cavities is 9.28 cm².
Hence, the correct answer is option (b).

Answer:

Statement-2 (Reason): If n cubes each of volume a³ cubic units are joined end to end to form a cuboid. Then the surface area of the resulting cuboid is 2(2n + 1)a² square units.

Volume of each cube = a³ cubic units
⇒ Side of each cube = a units

Since, there are n such cubes joined end to end to form a cuboid.
So, the length of the cuboid (l) = na
Breadth of the cuboid (b) = a
Height of the cuboid (h) = a

Now,
$$\begin{gathered} \mathrm{TSA} \mathrm{of} \mathrm{cuboid} = 2\times \left[lb+bh+hl\right] \\ =2\times \left[\left\{\left(na\right)\times a\right\}+\left(a\times a\right)+\left\{a\times \left(na\right)\right\}\right] \\ =2\times \left[n{a}^2+a^2+n{a}^2\right] \\ =2\times \left[2n{a}^2+a^2\right] \\ =2\times \left[a^2\left(2n+1\right)\right] \\ =2\left(2n+1\right)a^2 \mathrm{square} \mathrm{units} \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): Three cubes each of volume 8 cubic centimeters are joined end to end to form a cuboid. The surface area of the resulting cuboid is 28 cm².
Number of cubes (n) = 3
Volume of cube = 8 cubic centimeters             .....(1)
Also, volume of cube = a³                               .....(2)
From (1) and (2), we get
a³  = 8 cubic centimeters
⇒ a  = 2 cm
Now, the surface area of the resulting cuboid = 2(2n + 1)a² square units.                 (From Statement-2)
= 2(2×3 + 1) × (2)²
= 2(6 + 1) × 4
= 2(7) × 4
= 14 × 4
= 56 square units

Thus, Statement-1 is false.

So,  Statement-1 is false, Statement-2 is true.
Hence, the correct answer is option (d).

Answer:

Statement-2 (Reason): If n cubes each of surface area S are joined end to end to form a cuboid, then the volume of the resulting cuboid is $n{\left(\frac{S}{6}\right)}^{\frac{3}{2}}.$

Let the side of each cube be a units and surface area be S square units.
Surface area of each cube (S)= 6a² square units
$$\begin{gathered} \Rightarrow S = 6a^2 \\ \Rightarrow \frac{S}{6}=a^2 \\ \Rightarrow {\left(\frac{S}{6}\right)}^{\frac{1}{2}}=a .....\left(1\right) \end{gathered}$$

Since, there are n such cubes joined end to end to form a cuboid.
So, the length of the cuboid (l) = na
Breadth of the cuboid (b) = a
Height of the cuboid (h) = a

Now,
$$\begin{gathered} \text{Volume} \mathrm{of} \mathrm{cuboid} = l\times b\times h \\ =na\times a\times a \\ =n{a}^3 \\ =n{\left({\left(\frac{S}{6}\right)}^{\frac{1}{2}}\right)}^3 \left[\mathrm{From} \left(1\right)\right] \\ =n{\left(\frac{S}{6}\right)}^{\frac{3}{2}} \mathrm{cubic} \mathrm{units} \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): Two cubes each of surface area 96 cm² are joined end to end to form a cuboid. The volume of the resulting cuboid is 128 cm³.
Number of cubes (n) = 2
Surface area of cube (S)= 96 cm²       
Now, the volume of the resulting cuboid = $n{\left(\frac{S}{6}\right)}^{\frac{3}{2}}$ cubic units                (From Statement-2)
$$\begin{gathered} =2\times {\left(\frac{S}{6}\right)}^{\frac{3}{2}} \\ =2\times {\left(\frac{96}{6}\right)}^{\frac{3}{2}} \\ =2\times {\left(16\right)}^{\frac{3}{2}} \\ =2\times {\left(4\right)}^3 \\ =2\times 64 \\ =128 {\mathrm{cm}}^3 \end{gathered}$$
Thus, Statement-1 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Hence, the correct answer is option (a).

 

Answer:

Statement-2 (Reason): If two solid right cylinders of the same height and base radii r₁, r₂ are melted and recast into a cylinder of the same height, then the radius of the base of the cylinder is $\frac{\sqrt{r_1^2+r_2^2}}{2}.$

Radius of the base​ of the first cylinder =  r₁
Radius of the base​ of the second cylinder = r₂ 
Height of the base​ of both the cylinders = h

Radius of the base​ of the the resulting cylinder = R
Height of the base​ of the resulting cylinder = h                     (∵ Height of the resulting cylinder is same as the initial two cylinders)
Volume of the resulting cylinder = Sum of the volume of the two cylinders
$$\begin{gathered} \Rightarrow \pi R^{2}h=\pi {r_1}^{2}h+\pi {r_2}^{2}h \\ \Rightarrow \pi h{R}^2=\pi h\left({r_1}^2+{r_2}^2\right) \\ \Rightarrow R^2=\left({r_1}^2+{r_2}^2\right) \\ \Rightarrow R=\sqrt{{r_1}^2+{r_2}^2} \end{gathered}$$

Thus, Statement-2 is false.

Statement-1 (Assertion): If two solid right cylinders of the same height and base radii 3 cm and 4 cm are melted and recast into a cylinder of the same height, then the radius of the base of the new cylinder is 5 cm.

Radius of the base​ of the first cylinder = 3
Radius of the base​ of the second cylinder = 4 
Let the height of the base​ of both the cylinders = h

Let the radius of the base​ of the the resulting cylinder = R
Height of the base​ of the resulting cylinder = h                                 (∵ Height of the resulting cylinder is same as the initial two cylinders)
Volume of the resulting cylinder = Sum of the volume of the two cylinders
$$\begin{gathered} \Rightarrow \pi R^{2}h=\pi {r_1}^{2}h+\pi {r_2}^{2}h \\ \Rightarrow \pi h{R}^2=\pi h\left(3^2+4^2\right) \\ \Rightarrow R^2=\left(3^2+4^2\right) \\ \Rightarrow R=\sqrt{9+16} \\ \Rightarrow R=\sqrt{25} \\ \Rightarrow R=5 \mathrm{cm} \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).




 

Answer:

Statement-2 (Reason): If S₁, S₂ are surface areas of two spheres and V₁, V₂ are their volumes, then $\frac{V_1}{V_2}={\left(\frac{S_1}{S_2}\right)}^{\frac{3}{2}}.$ 
Surface areas of the first and second spheres are S₁ and S₂ respectively.
Let the radii of the first and second spheres be r₁ â€‹and r₂ â€‹respectively.
Now,
$$\begin{gathered} \frac{S_1}{S_2}=\frac{4\pi {\left(r_1\right)}^2}{4\pi {\left(r_2\right)}^2} \left(\because \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{sphere}=4\pi r^2\right) \\ \Rightarrow \frac{S_1}{S_2}=\frac{{\left(r_1\right)}^2}{{\left(r_2\right)}^2} \\ \Rightarrow \frac{S_1}{S_2}={\left(\frac{r_1}{r_2}\right)}^2 \\ \Rightarrow {\left(\frac{S_1}{S_2}\right)}^{\frac{1}{2}}=\left(\frac{r_1}{r_2}\right) .....\left(1\right) \end{gathered}$$
Since, volumes of the first and second spheres are V₁ â€‹and V₂ â€‹respectively.
$$\begin{gathered} \frac{V_1}{V_2}=\frac{{\displaystyle \frac{4}{3}}\pi {\left(r_1\right)}^3}{{\displaystyle \frac{4}{3}}\pi {\left(r_2\right)}^3} \left(\because \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{sphere}=\frac{4}{3}\pi r^3\right) \\ \Rightarrow \frac{V_1}{V_2}=\frac{{\displaystyle {\left(r_1\right)}^3}}{{\displaystyle {\left(r_2\right)}^3}} \\ \Rightarrow \frac{V_1}{V_2}={\left(\frac{r_1}{r_2}\right)}^3 \\ \Rightarrow \frac{V_1}{V_2}={\left({\left(\frac{S_1}{S_2}\right)}^{\frac{1}{2}} \right)}^3 \left[\mathrm{Using} \left(1\right)\right] \\ \Rightarrow \frac{V_1}{V_2}={\left(\frac{S_1}{S_2}\right)}^{\frac{3}{2}} \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): If surface areas of two spheres are in the ratio 16 : 9, then their volumes are in the ratio 64: 27.

Let the surface areas of the first and second spheres are S₁ and S₂ respectively.
⇒ S₁ : S₂ = 16 : 9
$\Rightarrow \frac{S_1}{S_2}=\frac{16}{9} .....\left(2\right)$
Now, let the volumes of the first and second spheres are V₁ â€‹and V₂ â€‹respectively.
⇒ V₁ â€‹: V₂ = 64 : 27
Then,
$$\begin{gathered} \frac{V_1}{V_2}={\left(\frac{S_1}{S_2}\right)}^{\frac{3}{2}} \left[\mathrm{Using} \left(2\right)\right] \\ \Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}={\left(\frac{16}{9}\right)}^{\frac{3}{2}} \\ \Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}={\left({\left(\frac{4}{3}\right)}^2\right)}^{\frac{3}{2}} \\ \Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}={{\left(\frac{4}{3}\right)}^2}^{\times \frac{3}{2}} \\ \Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}={\left(\frac{4}{3}\right)}^3 \\ \Rightarrow \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{64}{27} \end{gathered}$$
Thus, Statement-2 is true.

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).






 

Answer:

Statement-2 (Reason): If radii of two spheres are in the ratio 2 : 3, their surface areas are in the ratio 4 : 9.

Let the radii of the first and second spheres be r₁ â€‹and r₂ â€‹respectively and the surface areas of the first and second spheres are S₁ and S₂ respectively.
⇒ r₁ â€‹: r₂ = 2 : 3
Now,
$$\begin{gathered} \frac{S_1}{S_2}=\frac{4\pi {\left(r_1\right)}^2}{4\pi {\left(r_2\right)}^2} \left(\because \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{sphere}=4\pi r^2\right) \\ \Rightarrow \frac{S_1}{S_2}=\frac{{\left(r_1\right)}^2}{{\left(r_2\right)}^2} \\ \Rightarrow \frac{S_1}{S_2}={\left(\frac{r_1}{r_2}\right)}^2 \\ \Rightarrow \frac{S_1}{S_2}={\left(\frac{2}{3}\right)}^2 \\ \Rightarrow \frac{S_1}{S_2}=\frac{4}{9} \end{gathered}$$
⇒ S₁ â€‹: S₂ = 4 : 9

Thus, Statement-2 is true.

Statement-1 (Assertion): If volumes of two spheres are in the ratio 343 : 125, then their radii are in the ratio 7 : 5.

Let the volumes of the first and second spheres are V₁ â€‹and V₂ â€‹respectively and the radii of the first and second spheres be r₁ â€‹and r₂ â€‹respectively.
⇒ V₁ â€‹: V₂ = 343 : 125

$$\begin{gathered} \frac{V_1}{V_2}=\frac{{\displaystyle \frac{4}{3}}\pi {\left(r_1\right)}^3}{{\displaystyle \frac{4}{3}}\pi {\left(r_2\right)}^3} \left(\because \mathrm{Volume} \mathrm{of} \mathrm{a} \mathrm{sphere}=\frac{4}{3}\pi r^3\right) \\ \Rightarrow \frac{343}{125}=\frac{{\displaystyle {\left(r_1\right)}^3}}{{\displaystyle {\left(r_2\right)}^3}} \\ \Rightarrow \frac{343}{125}={\left(\frac{r_1}{r_2}\right)}^3 \\ \Rightarrow {\left(\frac{343}{125}\right)}^{\frac{1}{3}}=\left(\frac{r_1}{r_2}\right) \\ \Rightarrow {\left({\left(\frac{7}{5}\right)}^3\right)}^{\frac{1}{3}}=\left(\frac{r_1}{r_2}\right) \\ \Rightarrow {{\left(\frac{7}{5}\right)}^3}^{\times \frac{1}{3}}=\left(\frac{r_1}{r_2}\right) \\ \Rightarrow \left(\frac{7}{5}\right)=\left(\frac{r_1}{r_2}\right) \end{gathered}$$
⇒ r₁ â€‹: r₂ = 7 : 5

Thus, Statement-1 is true.
But Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).
 
 

Answer:

Statement-2 (Reason): The surface area of the sphere is 2πr(h + r).
Sphere does not have any height.
Also, the surface area of the sphere of radius r units is $4\pi r^2$ square units.

Thus, Statement-2 is false.

Statement-1 (Assertion): If a right circular cylinder of radius r and height h(h > 2r) just encloses a sphere, then the diameter of sphere is 2r.
A right circular cylinder of radius r and height h(h > 2r) just encloses a sphere, then the maximum diameter possible for sphere is the width of the cylinder, i.e., the diameter of the cylinder.

So, the maximum diameter of sphere = the diameter of cylinder 
                                                            = 2r                    (∵Diameter of cylinder = 2 × radius of the cylinder)

Thus, Statement-1 is true.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): If A₁ and A₂ denote the surface areas of circular bases of a frustum of height h, then its volume is $\pi h\left(A_1^2+A_2^2+\sqrt{A_1{A}_2}\right).$

Let the surface areas of circular bases of a frustum of height h be A₁ and A₂ respectively and radius be r₁ and r₂ respectively.
$$\begin{gathered} A_1=\mathrm{\pi }{\left(r_1\right)}^2 \mathrm{and} A_2=\mathrm{\pi }{\left(r_2\right)}^2 \\ \Rightarrow \frac{A_1}{\mathrm{\pi }}={\left(r_1\right)}^2 \mathrm{and} \frac{A_2}{\mathrm{\pi }}={\left(r_2\right)}^2 \\ \Rightarrow \sqrt{\frac{A_1}{\mathrm{\pi }}}=r_1 \mathrm{and} \sqrt{ \frac{A_2}{\mathrm{\pi }}}=r_2 .....\left(1\right) \end{gathered}$$

Now, volume of frustrum = $\frac{\pi h}{3}\times \left[{\left(r_1\right)}^2+r_1{r}_2+{\left(r_2\right)}^2\right]$
$$\begin{gathered} =\frac{\mathrm{\pi }h}{3}\times [{\left(\sqrt{\frac{{\mathrm{A}}_1}{\mathrm{\pi }}}\right)}^2+\sqrt{\frac{{\mathrm{A}}_1}{\mathrm{\pi }}}\sqrt{\frac{{\mathrm{A}}_2}{\mathrm{\pi }}}+{\left(\sqrt{\frac{{\mathrm{A}}_2}{\mathrm{\pi }}}\right)}^2] \\ =\frac{\mathrm{\pi }h}{3}\times [\frac{{\mathrm{A}}_1}{\mathrm{\pi }}+\sqrt{\frac{{\mathrm{A}}_1}{\mathrm{\pi }}\times \frac{{\mathrm{A}}_2}{\mathrm{\pi }}}+\frac{{\mathrm{A}}_2}{\mathrm{\pi }}] \\ =\frac{\mathrm{\pi }h}{3}\times [\frac{{\mathrm{A}}_1}{\mathrm{\pi }}+\frac{1}{\mathrm{\pi }}\sqrt{{\mathrm{A}}_1{\mathrm{A}}_2}+\frac{{\mathrm{A}}_2}{\mathrm{\pi }}] \\ =\frac{\mathrm{\pi }h}{3}\times \frac{1}{\mathrm{\pi }}\times [{\mathrm{A}}_1+\sqrt{{\mathrm{A}}_1{\mathrm{A}}_2}+{\mathrm{A}}_2] \\ =\frac{h}{\text{3}}\times [{\mathrm{A}}_1+\sqrt{{\mathrm{A}}_1{\mathrm{A}}_2}+{\mathrm{A}}_2] \end{gathered}$$

Thus, Statement-2 is false.

Statement-1 (Assertion): The radii of the ends of a frustum of a cone are 3 cm and 4 cm. If the height of the frustum is 6 cm, then its volume is 74π cm³.

Radius of one end of frustum (r₁) = 3 cm
Radius of one end of frustum (r₂) = 4 cm 
Height of frustum (h) = 6 cm 

Now, volume of frustrum = $\frac{\pi h}{3}\times \left[{\left(r_1\right)}^2+r_1{r}_2+{\left(r_2\right)}^2\right]$
 $$\begin{gathered} =\frac{1}{3}\times \pi \times 6\times \left[{\left(3\right)}^2+\left(3\right)·\left(4\right)+{\left(4\right)}^2\right] \\ =\pi \times 2\times \left[9+12+16\right] \\ =\pi \times 2\times \left[37\right] \\ =\pi \times 74 \\ =74\pi {\mathrm{cm}}^3 \end{gathered}$$

Thus, Statement-1 is true.

Hence, the correct answer is option (c).

Answer:

Statement-2 (Reason): If the radius of the base of a right circular cone of volume V is halved and the height is doubled, then the volume of the new cone so formed 2V.

Let the radius and height of a right circular cone of volume V  be r and h respectively.
Then,
The volume of the right circular cone (V) = $\frac{1}{3}\pi r^{2}h$                     .....(1)

Let the radius and height of the new right circular cone of volume V₁  be r₁ and h₁ respectively.
Then, $r_1=\frac{r}{2} \mathrm{and} h_1=2h$                               .....(2)
The volume of the right circular cone (V₁) = $\frac{1}{3}\pi {r_1}^2{h}_1$                .....(3)        

Substituting (2) in (3), we get      
$$\begin{gathered} V_1=\frac{1}{3}\pi {r_1}^2{h}_1 \\ \Rightarrow V_1=\frac{1}{3}\pi \times {\left(\frac{r}{2}\right)}^2\times \left(2h\right) \\ \Rightarrow V_1=\frac{1}{3}\pi \times \frac{r^2}{4}\times 2h \\ \Rightarrow V_1=\frac{1}{3}\pi \times \frac{r^2}{2}\times h \\ \Rightarrow V_1=\frac{1}{2}\times \left(\frac{1}{3}\pi r^{2}h\right) \\ \Rightarrow V_1=\frac{1}{2}\times V \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow V_1=\frac{V}{2} \end{gathered}$$

So, the volume of the new cone formed is half of the volume of the given cone.

Thus, Statement-2 is false.

Statement-1 (Assertion): If the radius of the base of a right circular cylinder is halved and the height is doubled, then the volume of the cylinder so formed is doubled.

Let the radius and height of a right circular cylinder of volume V  be r and h respectively.
Then,
The volume of the right circular cylinder (V) = $\pi r^{2}h$                     .....(1)

Let the radius and height of the new right circular cylinder of volume V₁  be r₁ and h₁ respectively.
Then, $r_1=\frac{r}{2} \mathrm{and} h_1=2h$                               .....(2)
The volume of the right circular cylinder (V₁) = $\pi {r_1}^2{h}_1$                .....(3)        

Substituting (2) in (3), we get      
$$\begin{gathered} V_1=\pi {r_1}^2{h}_1 \\ \Rightarrow V_1=\pi \times {\left(\frac{r}{2}\right)}^2\times \left(2h\right) \\ \Rightarrow V_1=\pi \times \frac{r^2}{4}\times 2h \\ \Rightarrow V_1=\pi \times \frac{r^2}{2}\times h \\ \Rightarrow V_1=\frac{1}{2}\times \left(\pi r^{2}h\right) \\ \Rightarrow V_1=\frac{1}{2}\times V \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow V_1=\frac{V}{2} \end{gathered}$$

So, the volume of the new cylinder so formed half of the volume of the given cylinder.

Thus, Statement-1 is false.

Disclaimer: None of the options is correct.