Question

The reading of a spring balance when a block suspended from it in the air is $60\text{N}$. This reading changes to $40\text{N}$ when the block is fully submerged in water. Find the specific gravity of the block.
A) $3$
B) $2$
C) $6$
D) ${\displaystyle \frac{3}{2}}$

Answer 1

Hint:When the block is suspended from the spring balance and completely submerged in water, it experiences an upward thrust exerted by the water which causes the weight of the block to decrease. Specific gravity refers to the ratio of the weight of the block in the air to the change in the weight of the block when it is in water.

Formulas used:
-The change in the weight of a body when it is completely immersed in a liquid is given by, $\mathrm{\Delta }W=W_{air}-W_{liquid}$ where where $W_{air}$ is the weight of the body in the air and $W_{liquid}$ is the weight of the body in the liquid of the same volume.
-The specific gravity of a body is given by, $G={\displaystyle \frac{W_{air}}{\mathrm{\Delta }W}}$ where $W_{air}$ is the weight of the body in the air and $\mathrm{\Delta }W$ is the change in the weight of the body when it was fully submerged in some liquid of the same volume.

Complete step by step answer.
Step 1: Sketch a rough figure depicting the block suspended in air and in water and list the readings of the spring balance in the two mediums.
The figure given below describes the block suspended in air and in water.

When the block is suspended in air, the spring balance reads $W_{air}=60\text{N}$ .
However, when the same block was suspended in water the reading changed to $W_{water}=40\text{N}$ .
This is because of the upward thrust acting on the block as shown in the figure.
Step 2: Express the change in weight of the block.
Now the change in weight of the block in the two mediums will be $\mathrm{\Delta }W=W_{air}-W_{water}$ ------- (1)
Substituting for $W_{air}=60\text{N}$ and $W_{water}=40\text{N}$ in equation (1) we get, $\mathrm{\Delta }W=60-40=20\text{N}$
Thus the change in weight of the block is $\mathrm{\Delta }W=20\text{N}$ .
Step 3: Express the relation for the specific gravity of the block.
The specific gravity of the given block can be expressed as $G={\displaystyle \frac{W_{air}}{\mathrm{\Delta }W}}$ -------- (2)
Substituting for $W_{air}=60\text{N}$ and $\mathrm{\Delta }W=20\text{N}$ in equation (2) we get, $G={\displaystyle \frac{60}{20}}=3$ .
Thus we obtain the specific gravity of the block as $G=3$ .

So the correct option is A.

Note: Alternate method

We have the weight of the block in the air as $W=60\text{N}$ and that in water as $W^{\prime }=40\text{N}$ .
Archimedes principle gives the upward thrust $F_b$ experienced by the block to be equal to the weight of the water displaced.
Now the force due to gravity is the weight of the block in air and as seen from the figure it is $F_g=60\text{N}$ .
Then the weight of the block in water will be the sum of the forces acting on it i.e., $W^{\prime }=F_g-F_b$ .
$\Rightarrow F_b=F_g-W^{\prime }$ -------- (A)
The specific gravity $G$ of the block can also be defined as the ratio of the force of gravity to the upward thrust experienced by the block i.e., $G={\displaystyle \frac{W}{F_b}}$ ------- (B)
Substituting equation (A) in (B) we get, $G={\displaystyle \frac{W}{F_g-W^{\prime }}}$ ------ (C)
Substituting for $W=F_g=60\text{N}$ and $W^{\prime }=40\text{N}$ in equation (C) we get, $G={\displaystyle \frac{60}{60-40}}=3$ .
Thus we have the specific gravity of the block as $G=3$ and the correct option is A.