Question

can you solve only mcqs not the otherones

5c3d0e02c4fcebd7bdf7384ea511160c.pdf

Answer 1

$\sf{Q1 . }$

If a sphere is inscribed in a cube , then the ratio of the volume of the cube tongue volume of sphere is

$\sf{A1}$

$\mathsf{volume \: of \: cube \: = {(side)}^{3} }$

${\huge{V}}1 = ( {a)}^{3}$

Now the length of side of cube is equal to diameter of sphere is that a.

We know that radius is half the diameter.

$Radius = \frac{a}{2}$

We get

${ \huge{V}}2 = \frac{4}{3} \pi \: \bigg ( \dfrac{a}{2} \bigg) ^{3}$

${ \huge{V}}2 = \frac{4}{3} \pi \: \bigg ( \dfrac{a {}^{3} }{8} \bigg)$

$\mathfrak{ratio} = \frac{v2}{v1} = \dfrac{ \frac{4}{3} \times \pi \times ( \frac{a {}^{3} }{8} )}{ {a}^{3} }$

$\longrightarrow \: \dfrac{ {a}^{3} \times 3 \times 8 }{4 \times \pi \times {a}^{3} } = \frac{6}{11}$

Ratio = π×11

$\sf{Q2. }$

The volume of a sphere is numerically equal to its surface area, then its

diameter is

$\sf{A2}$

$\mathsf {volume \: of \: sphere = \frac{4}{3} \pi \: {r}^{3} }$

$\mathsf {surface \: area \: of \: sphere \: = 4\pi {r}^{2} }$

According to the question :

$\sf{volume \: of \: sphere = surface \: of \: area}$

$\mathsf{ \frac {4}{3} \pi \: { {r}^{3} } = 4 \: \pi \: {r}^{2} }$

$r = 3$

We know that diameter is double the radius so .

$\sf \: d = 3 \times 2$

$\sf \: d = r \times 2$

$\sf \: d = 6$

Q3.

The total surface area of a cone of radius 2r and slant height l/2

A3.

$\sf{Total \: surface \: area = \pi \: r \: l \: + \pi \: {r}^{2} }$

Radius = 2r

l is given as 1/2

$T.S.A = \pi \: r \: (l + r)$

$T.S.A = \pi \: + 2r( \frac{1}{2} + 2r)$

$T.S.A = \pi \times 2r(1 + \frac{4r}{2} )$

$T.S.A = \pi \: r \: (l + 4r)$

$SURFACE \: AREA = \pi \: r \: (l + 4r)$

Q5. The total surface area of a cube is 96 ^

.Then radius of cube is

A5.

Let the sides be X , y , z

According to question :

(a) (b) (c) = (xy) (yz) (zx)

abc = (xyz)²

$xyz = \sqrt{abc}$

volume of cuboid = abc= √xyz

Q5.

The total surface area of a cube is 96 ^

.Then side of cube is

A5.

$T.S.A \: of \: cube = 6 \times {a}^{2}$

According to question :

$6 \times {a}^{2} = 96cm$

${a}^{2} = \frac{96}{6}$

${a}^{2} = \frac{16}{1}$

${a}^{2} = 16$

${a} = \sqrt{16}$

$a = 4$

Side = 4.

Q1.

If a spherical balloon grows to twice its radius when inflated, then the ratio of

the volume of the inflated balloon to the original balloon is.

A1.

let's assume the radius to be x

$\mathsf {volume \: of \: original \: sphere = \frac{4}{3} \pi \: {r}^{3} }$

$\mathsf{volume \: when \: inflated \: = \frac{4}{3} \pi \: {2r}^{3}}$

$Ratio = \dfrac{volume \: when \: inflated}{volume \: when \: original} = \dfrac{ \dfrac{4}{3} \: \pi \: 8 \: {r}^{3} }{ \dfrac{4}{3} \: \pi \: {r}^{3} } = \frac{8}{1}$

Ratio = 8:1

Q7. a square paper of side 25 cm is rolled to form a cylinder, then its curved

surface area is

A7.

$C.S.A \: of \: cylinder = 2\pi \: rh$

since , the square paper of side is rolled to form cylinder , it's circumference would be 25cm and height would also be 25cm.

$Circumference = 2\pi \: r$

$25 = 2 \times \frac{22}{7} \times \times r$

$\frac{1}{r} = \frac{44}{175}$

$r = 3.97$

We know formula of C.S.A

$C.S.A = 2\pi \: r \: h$

$C.S.A = \dfrac{2 \times 22 \times 3.97 \times 25}{7}$

$C.S.A = \dfrac{4367}{7}$

$C.S.A = 623.857$

Q9. The number of cubes whose edge measures 3 cm, that can be formed by

melting a cubical block of metal of edge 15 cm is

A9.

Number of cube that can be formed = volume of cube/ Volume that is being formed

$\implies \dfrac{ {15}^{3} }{ {3}^{3} }$

= 125 cubes.

There can be 125 cubes formed out of..

Q10. The number of cubes whose edge measures 3 cm, that can be formed by

melting a cubical block of metal of edge 15 cm is

A10. It would remain same.