Question

A metal block of volume 500cm^3 and density 2 g cm^-3 is suspended from a spring balance and one fourth of its volume is immersed in water what will be the reading on the spring balance in newtons .(take g = 10 ms^-3)

Answer 1

It will be 8.750 N .

In Pearson it is given right . Try it yourself by the concept and also try to solve the by putting values in the question.

Answer 2

Answer:

The correct answer is $8.750N.$

Explanation:

Given:

A metal block of volume $500cm^3$ and density $2 g cm^-3$ is suspended from a spring balance and one-fourth of its volume is immersed in water what will be the reading on the spring balance in newtons . (take g $= 10 ms^-3$)

Step 1

Volume of metal block $(v)=500 \mathrm{~cm}^{3}$

Density (s) $=2 \mathrm{~g} / \mathrm{cm}^{3}$

$9=10 \mathrm{~m} / \mathrm{s}^{2}$

Step 2

We know that

Apparent weight = weight of metallic block - upthrust

$& V \lg \frac{-V}{4} \times d \omega \times g=V g\left(d-\frac{d \omega}{4}\right)$

$=& 500 \times 10^{-6} \times 10\left(2 \times 10^{3}-\frac{1 \times 10^{3}}{4}\right)$

$=& 500 \times 10^{-5}\left(\frac{8 \times 10^{3}-10^{3}}{4}\right)$

$=& 8.750 \mathrm{~N} \text$

Therefore, the correct answer is $8.750N.$

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