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Question: An object of specific gravity \(\rho\) is hung from a massless string. The tension of a string is T....

An object of specific gravity ρ\rho is hung from a massless string. The tension of a string is T. The object is immersed in water so that one half of its volume is submerged. The new tension in the string is
A. (2ρ+12ρ)T(\dfrac {2\rho +1}{2\rho})T
B. (2ρ−12ρ)T(\dfrac {2\rho -1}{2\rho})T
C. (ρ−1ρ)T(\dfrac {\rho -1}{\rho})T
D. (ρ+1ρ)T(\dfrac {\rho +1}{\rho})T

Explanation

Solution

To solve this problem, first find the tension on the string when the object was in air. Then, find the tension on the string when the object is immersed in the water. Now, the tension will be the difference between the weight of the object and the buoyancy force. Weight of the object is equal to the tension on the string. So, substitute the initial tension on the string in this equation. And calculate the new tension on the string in terms of the initial tension on the string.
Complete solution step-by-step:

Complete answer:
Let the volume of the body be V
The initial tension in the string be T
The new tension in the string be T0{T}_{0}
When the object is in air, the tension on the string is given by,
T=ρVgT = \rho Vg …(1)
When the object is immersed in the water, the tension of the string becomes,
T0{T}_{0} = Weight of the object – Buoyant force
Weight of the object is equal to the tension on the string.
Substituting the values we get,
T0=ρVg−12Vg{T}_{0}= \rho Vg - \dfrac {1}{2}Vg
⇒T0=Vg(ρ−12)\Rightarrow {T}_{0} = Vg (\rho - \dfrac {1}{2})
⇒T0=Vg(2ρ–12)\Rightarrow {T}_{0}= Vg (\dfrac {2\rho – 1}{2})
Multiplying and dividing right-hand side by ρ\rho we get,
T0=ρVg(2ρ–12ρ){T}_{0} =\rho Vg (\dfrac {2\rho – 1}{2\rho})
Now, substituting the equation. (1) in above equation we get,
T0=T(2ρ–12ρ){T}_{0} =T (\dfrac {2\rho – 1}{2\rho})
Hence, the new tension in the string is T(2ρ–12ρ) T (\dfrac {2\rho – 1}{2\rho}).

So, the correct answer is “Option B”.

Note:
Students should remember that it is assumed that the mass is uniformly distributed over the length of the string. And it is assumed that the tension is also uniformly distributed over the length of the string When we place an object at one end of the string, downward force is exerted due to gravity and an upward force is exerted on the string which is called tension. The tension on the string is equal to the weight of the object.