Maths 4 Marks Questions

$\therefore$ Radius of a semi-circular sheet, $\text{r}=\frac{28}{2}=14\text{cm}$

Since, a semi-cicular sheet of metal is bent to form an open cup.

Let the radius of a conical be R.

$\therefore$ Circumference of base of cone = Circumference of semi-circle $2\pi\text{R}=\pi\text{r}$

$\Rightarrow\ \ 2\pi\text{R}=\pi\times14\Rightarrow\text{R}=7\text{cm}$

Now, $\text{h}=\sqrt{\text{l}^2-\text{R}^2}=\sqrt{14^2-7^2}\ \ \ \ [\therefore\text{l}^2=\text{h}^2+\text{R}^2]$

$=\sqrt{196-49}=\sqrt{147}=12.1243\text{cm}$

Volume (capacity) of conical cup $=\frac{1}{3}\pi\text{R}^2\text{h}$

$=\frac{1}{3}\times\frac{22}{7}\times7\times7\times12.1243=622.38\text{cm}^3$

Hence, the capacity of an open conical cup is 622.38cm³.

Mass per unit volume is called the density.

Density (D) $=\frac{\text{Mass}}{\text{Volume}}\ \ \Rightarrow\ \ \text{Volume}=\frac{\text{Mass}}{\text{Density}}$

Here density is same because the spheres are of same matal.

$\therefore\ \ \frac{\text{V}_1}{\text{V}_1}=\frac{\frac{5920}{\text{D}}}{\frac{740}{\text{D}}}\Rightarrow\frac{\frac{4}{3}\pi\text{r}^3_1}{\frac{4}{3}\pi\text{r}^3_2}=\frac{5920}{740}$

$\Rightarrow\frac{\text{r}^3_1}{\text{r}^3_1}=\frac{5920}{740}\Rightarrow\frac{\text{r}^3_1}{\Big(\frac{5}{2}\Big)^3}=\frac{5920}{740}\ \ \ [\because\text{r}_2=\frac{1}{2}\times5]$

$\Rightarrow\text{r}^3_1=\frac{5920}{740}\times\frac{125}{8}=125$

$\Rightarrow\text{r}_1=(125)^{\frac{1}{3}}=5\text{cm}$

Hence, the radius of the larger sphere = 5cm.

$\therefore$ Internal radius of hemispherical tank = 14m ÷ 2 = 7m

Volume of hemispherical tank $=\frac{2}{3}\pi\text{r}^3=\frac{2}{7}\times\frac{22}{7}\times(7)^3$

$=\frac{44\times49}{3}=718.66\text{m}^3.$

The tank contains 50 kilolitres of water = 50,000 litres $=\frac{50,000}{1,000}\text{m}^3=50\text{m}^3$

Volume of water pumped into the tank = 718.66m³ - 50m³ = 668.66m³.

Its outer radius (R) = 16 ÷ 2 = 8cm

Thickness of iron sheet = 2cm

And its inner radius (r) = 8cm - 2cm = 6cm.

Height of cylinder = 100cm

Quantity of iron used in making the = Volume of hollow cylinder $=\pi(\text{R}^2-\text{r}^2)\text{h}$

$=\frac{22}{7}\times(8^2-6^2)\times100=\frac{22}{7}(64-36)\times100$

$=\frac{22}{7}\times28\times100=8800\text{cm}^2$

Thickness of one circular plate = 3cm.

As the plates are placed one above the other, so the height of the cylinder formed by placing 30 plates = 30 × 3 = 90cm

1. Total surface area of circular $=2\pi\text{rh}+2\pi\text{r}^2$

$=2\pi\text{r}(\text{r+h})=2\times\frac{22}{7}\times14(14+90)=44\times208=9152\text{cm}^2$

2. Volume of the cylinder $=\pi\text{r}^2\text{h}=\frac{22}{7}\times14\times14\times90=55440\text{cm}^3$

Volume of cube = (4)³ = 4 × 4 × 4 = 64cm³

As cube contains a sphere touching its sides, so the diameter of the sphere = 4cm.

Radius of sphere $=\frac{4}{3}\pi\text{r}^3=\frac{4}{3}\times\frac{22}{7}\times(2)^3$

$=\frac{88\times8}{21}=\frac{704}{21}=33.52\text{cm}^3$

Volume of gap in between them = 64cm³ - 33.52cm³ = 30.48cm³.