NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas And Volumes

Ans : Given that , 
Volume of cubes= \(  (\text { Edge })^{3}=64\)
\(  (\text { Edge })^{3}=64\)
Edge \(  =4 \mathrm{cm} \)

If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm. 
Surface area of cuboids 
\(  \begin{array}{l}{=2(l b+b h+l h)} \\ {=2(4 \times 4+4 \times 8+4 \times 8)} \\ {=2(16+32+32)} \\ {=2(16+64)} \\ {=2 \times 80=160 \mathrm{cm}^{2}}\end{array} \) 

Ans : 
It can be observed that radius (r) of the cylindrical pa t and the hemispherical part is the same (i.e., 7 cm). 
Height of hemispherical part = Radius = 7 cm 
Height of cylindrical part (h) = 13-7 = 6 cm 
Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part \( =2 \pi r h+2 \pi r^{2}  \)
Inner Surface area of vessel \( \begin{array}{l}{=2 \times \frac{22}{7} \times 7 \times 6+2 \times \frac{22}{7} \times 7 \times 7} \\ {=44(6+7)=44 \times 13} \\ {=572 \mathrm{cm}^{2}}\end{array}  \) 

Ans : 
It can be observed that the radius of the conical part and the hemispherica part is same (i.e., 3.5cm). 
Height of hemispherical part=Radius(r) =3.5 =7/2cm
Height Of conical part (h) = 15.5 -3.5 = 12 cm
Slant Height \(  \begin{array}{l}{=\sqrt{r^{2}+h^{2}}} \\ {=\sqrt{\left(\frac{7}{2}\right)^{2}+(12)^{2}}=\sqrt{\frac{49}{4}+144}=\sqrt{\frac{49+576}{4}}} \\ {=\sqrt{\frac{625}{4}}=\frac{25}{2}}\end{array} \)  of conical part + CSA of hemispherical part 
Total surface area of toy CSA of conica part + CSA of hemispherical part
\( \begin{array}{l}{=\pi r l+2 \pi r^{2}} \\ {=\frac{22}{7} \times \frac{7}{2} \times \frac{25}{2}+2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}} \\ {=137.5+77=214.5 \mathrm{cm}^{2}}\end{array}  \) 

Ans : 
From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube's edge, i.e., 7cm. 
Radius (r) of hemispherical part=7/2=3.5cm
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part -Area of base of hemispherical part
\( =6(\mathrm{Edge})^{2}+2 \pi r^{2}-\pi r^{2}=6(\mathrm{Edge})^{2}+\pi v^{2}  \)
Total surface area of solid \( \begin{array}{l}{=6(7)^{2}+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}} \\ {=294+38.5=332.5 \mathrm{cm}^{2}}\end{array} \) 

Ans : 
Diameter of hemisphere =Edge Of cube = l
Radius of hemisphere = l/2 
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of base of hemispherical part 
=\( 6(\text { Edge })^{2}+2 \pi r^{2}-\pi r^{2}=6(\text { Edge })^{2}+\pi r^{2}  \)
Total surface area of solid \( \begin{array}{l}{=6 l^{2}+\pi \times\left(\frac{l}{2}\right)^{2}} \\ {=6 l^{2}+\frac{\pi I^{2}}{4}} \\ {=\frac{1}{4}(24+\pi) l^{2} \text { unit }^{2}}\end{array}  \)