Question

A glass tank is 25cm long, 20cm wide, 30cm high and contains water. The surface of the water is 5 cm below the top of the tank.

After a solid metal spherical ball B1 has carefully been placed into the tank, the surface of the water is 3cm below the top of the tank.


(The formula for the volume of a sphere (ball) is $\frac{4}{3} \pi r^{3}$ )


10.1) Calculate the volume of the ball.

Hint: The volume of the ball is equal to the volume of water displaced by the ball.


10.2) Calculate the radius of the ball. Leave your answer in terms of $\pi$ and a surd if necessary.


10.3) Suppose the radius of a second solid metal ball, B2; is half the radius of ball B1: Suppose ball B2 was put into the tank of water instead of ball B1: Would the surface of the water be 4cm below the top of the tank? Explain your answer

Answer 1

Step-by-step explanation:

A glass tank is 25cm long, 20cm wide, 30cm high and contains water. The surface of the water is 5 cm below the top of the tank.

Water tank is in the shape of cuboid:

Volume of cuboid:

$l \times b \times h$

After a solid metal spherical ball B1 has carefully been placed into the tank, the surface of the water is 3cm below the top of the tank.

Change in height of water level after placing the spherical ball in it = 2 cm

Volume of water raised = Volume of spherical Ball

10.1) Calculate the volume of the ball

Answer:

$= 25 \times 20 \times 2 \\ \\ = 25 \times 40 \\ \\ volume \: of \: ball B1= 1000 \: {cm}^{3}$

10.2) Calculate the radius of the ball. Leave your answer in terms of $\pi$ and a surd if necessary.

Answer:

$\frac{4}{3} \pi {r}^{3} = 1000 \\ \\ {r}^{3} = \frac{1000 \times 3}{4\pi} \\ \\ r = 10 \Bigg({ \frac{3}{4\pi} }\Bigg)^{ \frac{1}{3} } \: cm$

10.3) Suppose the radius of a second solid metal ball, B2; is half the radius of ball B1: Suppose ball B2 was put into the tank of water instead of ball B1: Would the surface of the water be 4cm below the top of the tank? Explain your answer

Answer: B2 is half the radius of B1

$r_1 = 10 \Bigg({ \frac{3}{4\pi} }\Bigg)^{ \frac{1}{3} } \: cm \\ \\ r_2 = 5 \Bigg({ \frac{3}{4\pi} }\Bigg)^{ \frac{1}{3} } \: cm \\ \\ volume \: of \: B2 = \frac{4}{3} \pi\Bigg( {5 \Big({ \frac{3}{4\pi} }\Big)^{ \frac{1}{3} }}\Bigg)^{3} \\ \\ = \frac{4\pi}{3} \times ({5})^{3} \times \frac{3}{4\pi} \\ \\ = 125 \: {cm}^{3}$

On putting that ball let the change in height of water is h,so

$125 = 20 \times 25 \times h \\ \\ h = \frac{125}{20 \times 25} \\ \\ h = \frac{5}{20} \\ \\ h = \frac{1}{4} \: cm \\ \\ = 0.25 \: cm$

No,the surface of water is 4.75 cm below.

Hope it helps you.