David Sang and Graham Jones Solutions for Chapter: Matter and Materials, Exercise 11: EXAM-STYLE QUESTIONS

A boy stands on a platform of area $0.050{ \mathrm{m}}^2$ and a manometer measures the pressure created in a flexible plastic container by the weight $W$ of the boy, as shown.

The density of water is $1000 \mathrm{kg}{ \mathrm{m}}^{-3}$. Determine the weight $W$ of the boy.

The diagram shows water in a container filled to a depth of $0.50 \mathrm{m}$. The density of water is $1000 \mathrm{kg}{ \mathrm{m}}^{-3}$.

Calculate the pressure at X on the base of the container.

This diagram shows water in a container filled to a depth of $0.50 \mathrm{m}$. The density of water is $1000 \mathrm{kg}{ \mathrm{m}}^{-3}$.

 Explain why the pressure at X must be equal to the pressure at Y.

This diagram shows water in a container filled to a depth of $0.50 \mathrm{m}$. The density of water is $1000 \mathrm{kg}{ \mathrm{m}}^{-3}$.

(c) Explain why the force downwards on the base of the container is larger than the weight of the liquid in the container.

A light spring that obeys Hooke's law has an unstretched length of $0.250 \mathrm{m}$. When an object of mass $2.0 \mathrm{kg}$ is hung from the spring the length of the spring becomes $0.280 \mathrm{m}$. When the object is fully submerged in a liquid of density $1200 \mathrm{kg}{ \mathrm{m}}^{-3}$, the length of the spring becomes $0.260 \mathrm{m}$. Calculate the spring constant of the spring.

A light spring that obeys Hooke's law has an unstretched length of $0.250 \mathrm{m}$. When an object of mass $2.0 \mathrm{kg}$ is hung from the spring the length of the spring becomes $0.280 \mathrm{m}$. When the object is fully submerged in a liquid of density $1200 \mathrm{kg}{ \mathrm{m}}^{-3}$, the length of the spring becomes $0.260 \mathrm{m}$. Calculate the upthrust on the object.

A light spring that obeys Hooke's law has an unstretched length of $0.250 \mathrm{m}$. When an object of mass $2.0 \mathrm{kg}$ is hung from the spring the length of the spring becomes $0.280 \mathrm{m}$. When the object is fully submerged in a liquid of density $1200 \mathrm{kg}{ \mathrm{m}}^{-3}$, the length of the spring becomes $0.260 \mathrm{m}$. Calculate the volume of the object.

A light spring that obeys Hooke's law has an unstretched length of $0.250 \mathrm{m}$. When an object of mass $2.0 \mathrm{kg}$ is hung from the spring the length of the spring becomes $0.280 \mathrm{m}$. When the object is fully submerged in a liquid of density $1200 \mathrm{kg}{ \mathrm{m}}^{-3}$, the length of the spring becomes $0.260 \mathrm{m}$. Calculate the density of the material from which the object is made.