A spring balance reacts $200gF$ when carrying a lump of lead in air. If the lead is now immersed with half of its volume in brine solution, what will be the new reading of the spring balance? Specific gravity of lead and brine $11.4$ and $1.1$ respectively.
A. $190.4gF$
B. $180.4gF$
C. $210gF$
D. $170.4gF$
Answer
594.9k+ views
Hint: The initial reading of the spring balance will be equal to the weight of the lump of lead. After the lead is half immersed in brine solution, the weight of the brine solution will decrease. Using mass equals density multiplied by volume, calculate the mass of lead immersed in solution and thus calculate the resultant mass.
Complete step by step answer:
The initial weight of the lump of lead will be:
${W_1} = mg$
Here, ${W_1}$ is the weight of the lump of lead in the initial given condition.
$m$ is the initial mass of the lump of lead
$g$ is the acceleration due to gravity;
As the initial reading on the spring is $200gF$ thus, the initial weight of the lump of lead is
${W_1} = 200gF$ --equation $1$
The total volume of the lump of lead will be:
$V = \dfrac{m}{{{\rho _l}}}$
Here, ${\rho _l}$ is the density of the lump.
When the lump of lead is half immersed in the brine solution, the solution will exert a force in the upward direction on the lump. Hence, the effective weight will decrease. The effective weight ${W_2}$ now will be:
${W_2} = {W_1} - \dfrac{{\rho gV}}{2}$
Here, $\rho $ is the density of the brine solution.
Since, only half of the volume is immersed.
Substituting the values from above equations, we get
$ \Rightarrow {W_2} = mg - \dfrac{{\rho gm}}{{2{\rho _l}}}$
$ \Rightarrow {W_2} = mg\left( {1 - \dfrac{\rho }{{2{\rho _l}}}} \right)$
$ \Rightarrow {W_2} = 200\left( {1 - \dfrac{{1.1}}{{2 \times 11.4}}} \right)$
$ \Rightarrow {W_2} = 190.35gF$
$ \Rightarrow {W_2} \simeq 190.4gF$
This will be the new reading in the spring.
So, the correct answer is “Option A”.
Note:
Only half of the lump of lead was immersed in the brine solution.
Volume, mass and density are related to each other and one can be expressed in terms of the other two.
The lump was immersed in the liquid hence, it will have a feeling of weightlessness like we feel when going down in the lift.
Complete step by step answer:
The initial weight of the lump of lead will be:
${W_1} = mg$
Here, ${W_1}$ is the weight of the lump of lead in the initial given condition.
$m$ is the initial mass of the lump of lead
$g$ is the acceleration due to gravity;
As the initial reading on the spring is $200gF$ thus, the initial weight of the lump of lead is
${W_1} = 200gF$ --equation $1$
The total volume of the lump of lead will be:
$V = \dfrac{m}{{{\rho _l}}}$
Here, ${\rho _l}$ is the density of the lump.
When the lump of lead is half immersed in the brine solution, the solution will exert a force in the upward direction on the lump. Hence, the effective weight will decrease. The effective weight ${W_2}$ now will be:
${W_2} = {W_1} - \dfrac{{\rho gV}}{2}$
Here, $\rho $ is the density of the brine solution.
Since, only half of the volume is immersed.
Substituting the values from above equations, we get
$ \Rightarrow {W_2} = mg - \dfrac{{\rho gm}}{{2{\rho _l}}}$
$ \Rightarrow {W_2} = mg\left( {1 - \dfrac{\rho }{{2{\rho _l}}}} \right)$
$ \Rightarrow {W_2} = 200\left( {1 - \dfrac{{1.1}}{{2 \times 11.4}}} \right)$
$ \Rightarrow {W_2} = 190.35gF$
$ \Rightarrow {W_2} \simeq 190.4gF$
This will be the new reading in the spring.
So, the correct answer is “Option A”.
Note:
Only half of the lump of lead was immersed in the brine solution.
Volume, mass and density are related to each other and one can be expressed in terms of the other two.
The lump was immersed in the liquid hence, it will have a feeling of weightlessness like we feel when going down in the lift.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

Which gas is abundant in air class 11 chemistry CBSE

10 examples of friction in our daily life

