KCET · Physics · Mechanical Properties of Fluids
Spheres of iron and lead having same mass are completely immersed in water. Density of lead is more than that of iron. Apparent loss of weight is \(w_{1}\) for iron sphere and \(w_{2}\) for lead sphere, Then, \(\frac{w_{1}}{w_{2}}\) is
- A \(>1\)
- B \(=1\)
- C become 0 and 1
- D \(=0\)
Answer & Solution
Correct Answer
(C) become 0 and 1
Step-by-step Solution
Detailed explanation
We have,
Relative density \(=\frac{\text { Actual weight }}{\text { Loss of weight }}\)
\((R D)_{\text {iron }}=\frac{w}{w_{1}}\) or \(\quad w_{1}=\frac{w}{(R D)_{\text {iron }}}\)
\((R D)_{\text {lead }}=\frac{w}{w_{2}}\) or \(\quad w_{2}=\frac{w}{(R D)_{\text {lead }}}\)
\(\frac{w_{1}}{w_{2}}=\frac{(R D)_{\text {iron }}}{(R D)_{\text {lead }}}>1\)
Relative density \(=\frac{\text { Actual weight }}{\text { Loss of weight }}\)
\((R D)_{\text {iron }}=\frac{w}{w_{1}}\) or \(\quad w_{1}=\frac{w}{(R D)_{\text {iron }}}\)
\((R D)_{\text {lead }}=\frac{w}{w_{2}}\) or \(\quad w_{2}=\frac{w}{(R D)_{\text {lead }}}\)
\(\frac{w_{1}}{w_{2}}=\frac{(R D)_{\text {iron }}}{(R D)_{\text {lead }}}>1\)
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