Question

An object of volume 40 cm$^{3}$ and in the form of a cylinder is tied to the hook of a spring balance records the weight to be 125 gf. The object lowered in a liquid of density 0.8 GCM[tex]^{–3} [)tex], such that half of it is immersed in the liquid. What is the new reading on the spring balance ?​

Answer 1

$\underline{Correct \: Question:}$

An object of volume 40 cm$^{3}$ and in the form of a cylinder is tied to the hook of a spring balance records the weight to be 125 gf. The object lowered in a liquid of density 0.8 GCM$^{–3}$, such that half of it is immersed in the liquid. What is the new reading on the spring balance ?

$\underline{Answer:}$

New reading of the spring balance is 109 gf.

$\underline{Explanation:}$

Given:-

Initial reading on the spring balance = 125 gf

Volume of the object = 40 cm$^{3}$

So, volume of the object immersed in liquid = 40/2 = 20 cm$^{3}$

Volume of the liquid displaced (V) = 20 cm$^{3}$

Density of the liquid (d)

= 0.8 gm$^{–3}$

Mass of liquid displaced (m) = V × d = (20 cm$^{3}$) × (0.8 gm$^{–3}$)

So, weight of liquid displaced (m) = 16 g.

$\mathbb {Now \: by \: Archimedes' \: Principle}$

Upthrust = Weight of liquid displaced = 16 gf.

Hence, new reading of the spring balance

= (125 – 16) = 109 gf.