Search any question & find its solution
Question:
Answered & Verified by Expert
A \(0.5 \mathrm{~kg}\) block of brass (density \(=8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\) ) is suspended from a string. What is the tension in the string if the block is completely immersed in water? \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
Options:
Solution:
2909 Upvotes
Verified Answer
The correct answer is:
\(\frac{35}{8} \mathrm{~N}\)
Mass of block, \(m=0.5 \mathrm{~kg}\)
Density, \(\rho=8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)
Volume of block, \(V=\frac{m}{\rho}=\frac{0.5}{8 \times 10^3}=6.25 \times 10^{-5} \mathrm{~m}^3\)
When block is fully immersed in water, then upthrust force on the block
\(\begin{aligned}
F_1 & =m g=V \cdot \rho_w \times g \\
& =6.25 \times 10^{-5} \times 10^3 \times 10=0.625 \mathrm{~N}
\end{aligned}\)
\(\therefore\) Tension in the string
\(\begin{aligned}
T & =m g-F_1 \text { (upthrust force) } \\
& =0.5 \times 10-0.625=4.375 \mathrm{~N}=\frac{35}{8} \mathrm{~N}
\end{aligned}\)
Density, \(\rho=8 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\)
Volume of block, \(V=\frac{m}{\rho}=\frac{0.5}{8 \times 10^3}=6.25 \times 10^{-5} \mathrm{~m}^3\)
When block is fully immersed in water, then upthrust force on the block
\(\begin{aligned}
F_1 & =m g=V \cdot \rho_w \times g \\
& =6.25 \times 10^{-5} \times 10^3 \times 10=0.625 \mathrm{~N}
\end{aligned}\)
\(\therefore\) Tension in the string
\(\begin{aligned}
T & =m g-F_1 \text { (upthrust force) } \\
& =0.5 \times 10-0.625=4.375 \mathrm{~N}=\frac{35}{8} \mathrm{~N}
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.