Solution

We know that the acceleration due to gravity \( g’ \) at a height \( h \) above the Earth’s surface is given by:

\[ g’ = g \left( \frac{R} {R + h} \right)^2 \]

To find \( h \) when \( g’ = \frac{g}{2} \), we solve:

\[ \frac{g}{2} = g \left( \frac{R} {R + h} \right)^2 \]

Simplifying, we find \( h = R \).

Solution

Using the kinematic equation for uniformly accelerated motion:

\[ s = ut + \frac{1}{2} a t^2 \]

Here, \( s = 2 \, \text{m} \), \( u = 0 \), \( a = 9.81 \, \text{m/s}^2 \), and \( t = 0.1 \, \text{s} \).

The distance \( h \) from the roof to the top of the window is given by:

\[ h = 0 + \frac{1}{2} \times 9.81 \times (0.1)^2 = 0.04905 \, \text{m} \]

Solution

Please refer to the graphs in your textbook or notes, as this format does not support graphical content. However, you can expect the following:

• Distance-time graph: A downward-opening parabola.
• Velocity-time graph: A straight line with a positive slope.
• Acceleration-time graph: A horizontal line representing constant acceleration.

Solution

The acceleration due to gravity \( g’ \) on the new planet is given by:

\[ g’ = g \left( \frac{2M} {(2R)^2} \right) = \frac{g}{2} \]

The maximum kinetic energy \( K \) is then:

\[ K = m \times g’ \times h = 10 \times \frac{9.81}{2} \times 10 = 490.5 \, \text{J} \]

Solution

The apparent weight in the liquid is 51 g, so the loss of weight in the liquid is \( 75 \, \text{g} – 51 \, \text{g} = 24 \, \text{g} \).

The apparent weight in water is 67 g, so the loss of weight in water is \( 75 \, \text{g} – 67 \, \text{g} = 8 \, \text{g} \).

The density \( \rho \) of the liquid is given by:

\[ \rho = \frac{\text{Loss in liquid}}{\text{Loss in water}} \times \text{Density of water} = \frac{24}{8} \times 1000 = 3000 \, \text{kg/m}^3 \]

Solution

If \( \frac{1}{4} \) of the volume is above water, \( \frac{3}{4} \) of the volume is submerged.

\[ \rho_{\text{body}} = \frac{3}{4} \times \rho_{\text{water}} = \frac{3}{4} \times 1000 \, \text{kg/m}^3 = 750 \, \text{kg/m}^3 \]

Solution

The block floats with \( \frac{1}{2} \) of its volume submerged in water and \( \frac{1}{4} \) submerged in the liquid. The density \( \rho_{\text{liquid}} \) is then:

\[ \rho_{\text{liquid}} = \frac{1}{2} \times \rho_{\text{water}} = \frac{1}{2} \times 1000 \, \text{kg/m}^3 = 500 \, \text{kg/m}^3 \]

Solution

a. If the masses are the same, the density \( S \) of the mixture is the average of the densities:

\[ S = \frac{S_1 + S_2}{2} \]

b. If the volumes are the same, the density \( S \) of the mixture is given by:

\[ S = \frac{2 \times S_1 \times S_2}{S_1 + S_2} \]

Solution

The volume of water displaced is \( V = \text{Length} \times \text{Breadth} \times \text{Height} \).

\[ V = 3 \, \text{m} \times 2 \, \text{m} \times 0.01 \, \text{m} = 0.06 \, \text{m}^3 \]

The mass \( m \) of the man is \( \rho \times V \times g \), where \( \rho = 1000 \, \text{kg/m}^3 \) and \( g = 9.81 \, \text{m/s}^2 \).

\[ m = 0.06 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 = 588.6 \, \text{kg} \]

Solution

If the masses are the same, the density \( S \) of the mixture is the average of the densities:

\[ S = \frac{1 \, \text{g/cm}^3 + 2 \, \text{g/cm}^3}{2} = 1.5 \, \text{g/cm}^3 \]

Solution

In equilibrium, the loss of weight in water for both bodies must be the same:

\[ \text{Loss of body 1} = \text{Loss of body 2} \] \[ (36 \, \text{g} – \rho_1 \times V_1) = (48 \, \text{g} – \rho_2 \times V_2) \]

We know \( \rho_1 = 9 \, \text{g/cm}^3 \) and \( V_1 = \frac{36 \, \text{g}}{9 \, \text{g/cm}^3} = 4 \, \text{cm}^3 \).

Solving, we find \( \rho_2 = 6 \, \text{g/cm}^3 \).

Clarification

When two bodies are in equilibrium when suspended in water, the buoyant force on each must be equal to its weight, and therefore the loss of weight of each body due to the buoyant force must also be equal.

Loss of Weight

The loss of weight of each body when submerged in water is equal to the weight of the water displaced by that body. Mathematically, the loss of weight \( \Delta W \) is:

\[ \Delta W = V \times \rho_{\text{water}} \times g \]

Applying to Both Bodies

Let’s consider the two bodies, A and B.

For body A:

• Mass \( m_A = 36 \, \text{g} \)
• Density \( \rho_A = 9 \, \text{g/cm}^3 \)

For body B:

• Mass \( m_B = 48 \, \text{g} \)
• Density \( \rho_B \) is unknown

Finding Volumes

Volume \( V_A = \frac{m_A}{\rho_A} = \frac{36}{9} = 4 \, \text{cm}^3 \)

Volume \( V_B = \frac{m_B}{\rho_B} \) (We want to find \( \rho_B \))

Equating Loss of Weight in Water

We know that the loss of weight in water for both bodies should be equal for them to be in equilibrium:

\[ \Delta W_A = \Delta W_B \] \[ V_A \times \rho_{\text{water}} \times g = V_B \times \rho_{\text{water}} \times g \] \[ V_A = V_B \] \[ 4 \, \text{cm}^3 = \frac{48 \, \text{g}}{\rho_B} \]

Solving this equation will give us \( \rho_B \), the density of the second body.