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Weight Quiz 2

physics

 A block completely immersed in a liquid is floating in the liquid and experiences a buoyant force of $F_{b} = 78.4\,N$. What is the mass of the block? Further, if the apparatus is taken to the moon, what will be the buoyant force acting on the block? $(g_{moon}=1.6\,ms^{-2})$

 Use Archimedes’ principle to obtain the weight of the block, following which you can find its mass.Then reapply the same principle to determine the buoyant force acting on the block on the moon.  

  Given $F_{b} = 78.4N$ According to Archimedes’ principle, when a block is immersed in a liquid, the buoyant force acting on it is equivalent to the weight of the block. $W = F_{b} \\ \Rightarrow mg = F_{b} \\ \Rightarrow m = \dfrac{F_{b}}{g} \\ $ Plugging in the values we get: $\\ m = \dfrac{78.4}{9.8} = 8kg \\$ Now, when the whole apparatus is taken to the moon, the buoyant force acting on it will be equal to its weight on the moon, i.e., $F^{\prime}_{b} = W = mg_{moon} \\ \Rightarrow F^{\prime}_{b} = 8 \times 1.6 \\ \therefore F^{\prime}_{b} = 12.8\,N \approx 0.16\,F_{b}$

 If the body of mass $m$ and density $\rho_b$ floats in a liquid of density $\rho_l$ then  determine the maximum mass that can be added to the body so that it does not sink.   

 The buoyant force will be equal to the weight of the block when it floats. Equate the two while including the additional mass in order to arrive at an expression for the same. 

  $\text{Let the volume of the body be }V_{b} = \dfrac{m}{\rho_{b}}, \\ \text{Let the maximum mass that can be added be }m^{\prime}.\\  \therefore,\text{ total mass }M = m +m^{\prime} \\ \text{A body floats in a liquid when the buoyant force acting on it is equal to its weight, i.e.,} \\ F_{b} = W  \\ \Rightarrow \rho_{l}V_b g = Mg \\  \Rightarrow \rho_l V_b = M \\ \Rightarrow \rho_l \dfrac{m}{\rho_b} = m+m^{\prime} \\ \Rightarrow m^{\prime} = \dfrac{m\rho_l}{\rho_b} – m \\ \therefore m^{\prime} = m\left(\dfrac{\rho_l}{\rho_b} -1\right)$

 A mass $m$ is suspended from a string that experiences a tension $T$ as shown in the figure.This setup is taken in an elevator that is going up with an acceleration $a$ and the tension in the string is now $T^{\prime}$. What is the apparent weight of the mass? What is the acceleration of the elevator?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b9a6cf08-25fb-440a-9174-63c3bdf27a873558148259981449535.png"/> 

 Use free body diagrams to arrive at equivalences between the string tension and weight or apparent weight in each case and rearrange them appropriately. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/0ca36456-1939-4f13-bae1-620eb5df38c87141224689991049692.png"/><br>$\text{In the first case we see that the weight of the mass is balanced by the tension in the string:} \\ W= T \Rightarrow mg = T \\ \text{In the second case, the weight of the mass is apparent due to a net acceleration of }(g+a): \\ W^{\prime} = m(g+a)= mg+ma = T+ma \\ \text{Now, we see that tension in the string is equivalent to the apparent weight of the mass} \\ T^{\prime} = W^{\prime} \\ \Rightarrow T^{\prime} =T+ma \\ \therefore a = \dfrac{T^{\prime}-T}{m}$

 An object of mass $1\,kg$ is suspended from a spring balance and another object placed on a physical balance measures a weight of $2\,kg$ at equilibrium. As such, both the spring balance and the physical balance are taken in an elevator which moves down with an acceleration of $2\,ms^{-2}$.  What will be the change in the reading of the spring balance? What will be the change in the equilibrium position of the physical balance? 

 Find the net acceleration for the spring balance and find its apparent weight to arrive at the change in its reading. For the physical balance, remember that the weight of the object it measures is relative to the standard weights.  

 $\text{We are given masses }m_1=1\,kg, m_2=2\,kg \\ \text{Acceleration of the elevator }a = 2\,ms^{-2}\text{ downwards.} \\ \text{Let us first look at the spring balance:} \\ \text{Initial reading: }W_{i} = mg = 1 \times 9.8 = 9.8\,N \\ \text{Elevator reading: }W_{f} = mg-ma = m(g-a) = 1(9.8-2) = 7.8\,N \\ \text{Change in reading:  }9.8-7.8 = 2\,N \\$Now, the physical balance works on the principle of comparison of weights. So, as far as the equilibrium position of the physical balance is concerned, though the apparent weight it measures may be lesser than the actual weight, the standard weights also assume the same apparent weight since they are subjected to the same relative net acceleration and hence the equilibrium position remains unchanged for the physical balance and will show the original reading.

 An exoplanetary atmosphere consists of 800 moles of carbon, 600 moles of hydrogen, 300 moles of oxygen and 100 moles of helium. What is the weight that this atmosphere exerts on the surface of the planet? How much weight would it exert over the surface of the Earth? $($Express in newtons $(N))$. $(g_{exo} = 5ms^{-2})$

 Convert the moles of gases into grams and then find the weight with all of the gas masses combined with respect to the acceleration due to gravity on the exoplanet and the earth respectively.

 $\text{Given: The exoplanetary atmosphere consists of masses of,} \\ \text{Carbon: }m_C = 800 moles = 800 \times 12 = 9600g = 9.6kg \\ \text{Hydrogen: }m_{H_2} = 600moles = 600 \times 2 = 1200g = 1.2kg \\ \text{Oxygen: }m_{O_2} = 300moles = 300 \times 32 =  9600g = 9.6kg \\ \text{Helium: }m_{He} = 100moles = 100 \times 4 = 400g = 0.4kg \\ \text{The weight exerted by this atmosphere on the exoplanet is given as:} \\ W = \left(m_C + m_{H_2} + m_{O_2}+m_{He}\right)g_{exo} \\ \Rightarrow W = (9.6+1.2+9.6+0.4)\times 5 = 104N \\ \text{The weight it would exert over the same area on earth would be:} \\ W = \left(m_C + m_{H_2} + m_{O_2}+m_{He}\right)g_{earth} \\ \therefore W = (9.6+1.2+9.6+0.4)\times 9.8 \approx 204N $